Chris Grossack's Blog

Is a Random Perfect Group Nontrivial?

Fri, 20 Mar 2026 00:00:00 +0000

So it’s finals season, and earlier today some of the younger grad students were asking me for help studying for their topology finals. One of their practice problems was to build a cell complex with one 0-cell, two 1-cells, and two 2-cells which has nontrivial $\pi_1$ but trivial $H_1$. In principle, this isn’t very hard to do – I encourage you to think about it for a while to see what kind of group you’re looking for… Then see if you can look in the literature for a source of such a group.

Last chance to think about it yourself…

Ok, the fact that we have two 1-cells means our fundamental group will have two generators, say $a$ and $b$. Then the two 2-cells will give us two relations, say $R$ and $S$. Since we know that $H_1$ is the abelianization of $\pi_1$, this means we’re looking for a perfect group with a presentation by two generators and two relations.

If you try to build one by hand for a while (and I encourage you to try!), it’s actually somewhat difficult to do! I ended up looking up “small presentations of perfect groups” (or something like this) and quickly found Campbell, Kawamata, Miyamoto, Robertson, and Williams’s Deficiency Zero Presentations for Certain Perfect Groups which is full of examples¹. I also posted about this on mastodon, and Omar Antolín had a characteristically helpful response! He mentioned that the binary icosahedral group gets the job done², and these relations are the kind of thing you have a chance of finding by hand!

This brings up an interesting question, though – In some sense you would expect that two “random” relations should get the job done… Obviously things can’t be completely random, since we need the abelianized relations to be a matrix with determinant $\pm 1$ – otherwise our $H_1$ will have torsion… But what if we condition on this property. Does a “random” pair of relations work?

It’s been a minute since I’ve written a genuinely short blog post, and I was curious enough to write up some sage code anyways³, so I thought it could be fun to do it together!

As a quick exercise: What do I mean when I say that “we need the abelianized relations to be a matrix with determinant $\pm 1$ – otherwise our $H_1$ will have torsion”?

Ok, so here’s the plan:

We’ll take as input a number $N$. Then we’ll iterate through all pairs of relations $R,S$ which are words in the alphabet $\{a, a^{-1}, b, b^{-1} \}$ of length $N$ then we’ll see what fraction of those with vanishing $H_1$ also have nonvanishing $\pi_1$. If our conjecture is right then this fraction should get closer to $1$ as $N$ gets bigger.

Let’s do it!

This basically does exactly what you expect, haha. The main things to take note of are the @fork decorator when checking if a group is trivial. Since this is undecidable in general we bail on the computation if it lasts longer than thirty seconds… This creates some serious overhead on each loop, which is basically the same overhead involved in parallelising… So we might as well parallelise! This is what the @parallel decorator does. My laptop only has a measly 8 cores, so that’s how many I tell it to use. The reseed_rng flag is to make sure each process gets its own RNG, otherwise our 10,000 “random” runs will all be the same!

So what does the final scatter plot look like?

The trend is definitely upwards, which makes sense. Interestingly the code only starts seeing examples once the relations have length $7$. The smallest example I’m aware of is the binary icosahedral group that Omar told me about, which happens to have two relations of length $7$!

Even with relations of length $17$, though, only a tiny $0.4\%$ of the samples with trivial $H_1$ had an interesting $\pi_1$… I still think that this ratio should approach $1$ as the size of the relations gets large? But I was really hoping this data would be more suggestive, haha. That said, there’s a lot of problems with the data!

My laptop starts to struggle after $N=8$, which corresponds to relations of length $17$. I’ve uploaded the raw output if you’re interested in looking at it, but the main thing to note is how often we bail on computations. This is almost certainly throwing off our statistics, but I don’t really know how. After all, we’re now conditioning on both “trivial $H_1$” and “can be checked to be (non)trivial in at most 30 seconds on my ten year old laptop”. Here’s a table:

Relation Length	Nontrivial $\pi_1$	Total Runs	Killed Computations
1	0	10000	0
3	0	10000	0
5	0	10000	0
7	8	10000	0
9	10	10000	2
11	18	10000	19
13	27	10000	183
15	34	9998	664
17	47	9995	1726

So at this point the possible error we’re incurring by killing computations that take longer than $30$ seconds is dwarfing the number of groups we’re able to quickly prove are (non)trivial. Plus when we kill computations there seems to be a small chance that GAP throws some kind of exception that I’m not sure how to handle… This is the reason some later runs have slightly fewer than $10,000$ trials. I thought about increasing the timeout to a minute, or even five minutes, and running it again overnight to see if we can push things a bit further? But I decided I don’t really care, haha. I’m supposed to be writing a thesis, after all.

Further optimizing this would make a great project for somebody else to try, though! I think that it should be quite easy to get better data, and a lot of it! If you decide to look into this, definitely reach out and let me know what you find ^_^. In that vein, here’s a few take-home problems that you might want to play around with. They all look fun (at least to me) and if I had the time to spend I would probably think about them for a few weeks.

Can you further optimize this code to get more data? There’s a few obvious things to try:

writing in pure GAP, rather than a python package
somehow checking triviality without computing the cardinality of a nontrivial group
have a better computer than me, with lots of cores for parallel computation

You might have noticed that all our trials were on odd relation lengths… Experimentally it looks like even relation lengths never give perfect groups!

It’s pretty easy to prove that this really is the case. I’ll include a quick proof in a spoiler tag, but you might want to play around with it yourself for a minute.

solution

We know there's $2n$ many letters in each of the relations $R$ and $S$, coming from the alphabet $\{a,A,b,B\}$. We'll write $R_a$ for the number of times $a$ shows up in $R$, and similarly for $S_B$, etc. Then working mod $2$ we have $$R_a + R_A + R_b + R_B = 2n \equiv_2 0$$ so that (remembering that $+$ and $-$ are the same mod $2$) $$R_a - R_A \equiv_2 R_b - R_B$$ Recall that the abelianization of $G = \langle a,b \mid R, S \rangle$ vanishes if and only if the following matrix has determinant $\pm 1$: $$ \begin{pmatrix} R_a - R_A & S_a - S_A \\ R_b - R_B & S_b - S_B \end{pmatrix} $$ but working mod $2$ again and using the above observation say that $R_a - R_A \equiv_2 x \equiv_2 R_b - R_B$ and $S_a - S_A \equiv_2 y \equiv_2 S_b - S_B$. Then the mod-2 determinant of the above matrix is $xy - xy = 0$ so that this matrix can never have determinant $\pm 1$!

For a harder problem, which I don’t know how to solve, you might ask if the fraction of presentations with nontrivial $\pi_1$ approaches $1$ at all! Or even better, you might ask what the asymptotic behavior is as the number of relations gets large!

I asked about this on mathoverflow today, and I’m very excited to see what people have to say! I really don’t know much combinatorial group theory, so no matter what the conversation turns into I’m quite likely to learn something.

Alright! This is my first really short blog post in… quite a while, haha. I wrote the code pretty quickly, and once I figured out how to get the long computation to time out properly (and let it run all evening yesterday) the rest of the post came together in like two hours. I love problems like this, so it was easy to get nerdsniped by it.

Now it’s back to checking some details for my thesis. Next week is spring break, so I’m hoping to really sit down and get a lot done before teaching starts back up.

OH! And that reminds me! I mentioned this in a draft for a blog post on representation theory in analysis, but I haven’t mentioned it anywhere that’s live yet, so I should say it here! I got a postdoc!! 🎉🎉

This Fall I’ll be going to Montana State University to work with Sam Gunningham, David Ayala, and probably Ryan Grady too. I’ll be thinking about all sorts of fun things like factorization homology, “quantum” geometric langlands, topological field theories, and more! Everyone at the MSU campus has been so nice to me, and even though I’m an island girlie through and through I’m oddly excited to experience real winter for the first time in my life, haha. I’m tearing up a little bit writing this because I’m so happy to get to go there and work with them.

Alright, thanks for reading everyone! It’s back to the dissertation grind now, but this was really fun to think about for a day or two.

Stay safe, and we’ll talk soon 💖

As an aside, this search also pulled up Bray, Conder, Leedham-Green, and O’Brien’s Short presentations for alternating and symmetric groups which shows how to get presentations for alternating and symmetric groups with 2 generators and $\mathsf{PolyLog(n)}$ many relations (at least that’s my understanding – I haven’t read this paper closely at all).

This really scratches some kind of asymptotic computer science itch in my brain! ↩
He also mentioned he knows about this group because it’s the fundamental group of the Poincaré homology sphere! This makes it especially reasonable that he might have thought of it, since the original context for this problem is about a cell complex with $\pi_1$ but no $H_1$, and that’s exactly (one of) the defining properties of the homology sphere! ↩
Checking if a presentation gives the trivial group is famously impossible, but since we’re restricting ourselves to two generators and two relations I’m hopeful that sage can handle these cases for us! ↩

Talk -- Factorization Homology and Quantum Character Stacks

Fri, 20 Feb 2026 00:00:00 +0000

~~Today~~ Yesterday in the Representation Theory Seminar at UCR I gave a talk about Factorization Homology and how it lets us compute a “Quantum Character Stack”. This is all based on a great paper, Integrating Quantum Groups Over Surfaces by Ben-Zvi, Brochier, and Jordan, which I’ve been reading and rereading for the last few years. It’s been a while since I’ve written up my thoughts after a talk, so I figured I’d do that here to take a break from thesis writing.

I have an old post going through a talk I gave on factorization homology almost exactly 2 years ago back in March 2024, which might give a longer perspective on these ideas. I’m going to be fairly terse here because I want to get to the fun computation (which I’ll put in a sister post), and I also want to write this in just a few hours.

The talk was kind of a whirlwild, haha. Especially for my audience, I needed to explain some basics about stacks and the rough idea of factorization homology before I could even hope to get to the actual definition of the quantum character stack! That’s a big ask for an hour long talk, but I think I did alright. I asked my friend Shane how he thought it went, and he very graciously said that I did a good job telling a story and showing that some could, in the abstract, compute things like this… but I didn’t actually show the audience how they can compute with it. I think that’s a fair review, and is pretty consistent with my experience writing and giving the talk. Every professor that I talked to said that it was really good, though, which made me happy. Thankfully that’s also pretty consistent with my experience giving the talk, haha.

I’ve given other really dense talks before, and I remember coming off a bit… energetic, lol. I was pleased that I think I managed to fit a lot of material into this talk while still appearing somewhat collected at the white board. If nothing else, I didn’t end the talk out of breath, haha.

Anyways, enough about my thoughts, let’s get to the talk itself!

The beginning of the talk was meant to motivate stacks to the audience – particularly some younger grad students who have asked me about them before. I actually have a looooong post about stacks in the works, where I talk about how to think of them, how to compute with them, and why you might care. I’ve had to put it on the back burner while I work on my thesis, but hopefully some day I’ll finish it up, since I have a lot of Thoughts™.

Given a surface $\Sigma$ and a reductive group $G$, we would like to have a space whose points are representations of $\pi_1 \Sigma$ valued in $G$. To do this we can look at $\text{Hom}(\pi_1 \Sigma, G)$ and then quotient out by “change of basis” given by conjugation in $G$. This has the extra benefit of removing the reliance of $\pi_1 \Sigma$ on a choice of base point, since a change of base point leads to a conjugate representation.

There are a few things you could mean by the quotient $\text{Hom}(\pi_1 \Sigma, G) \big / G$. The first and most naive is to literally take the space and quotient out by the orbit equivalence relation. This gives a space that isn’t even Hausdorff (and it makes a nice exercise to see why!) so this isn’t great. The more subtle approaches are both based on the observation that a function on $X \big / G$ should be the same thing as a $G$-equivariant function on $X$. If you haven’t seen this before it’s worth taking a second to think about why this should be true!

If you’re a 20th century algebraic geometer you would define the Character Variety $\text{Ch}(\Sigma,G)$ as $\text{Spec} \big (\mathcal{O}(\text{Hom}(\pi_1 \Sigma, G))^G \big )$. This literally means “the space whose ring of functions is $G$-equivariant functions on $\text{Hom}(\pi_1 \Sigma,G)$”.

If you’re a 21st century geometer, you’re likely to de-emphasize $\mathbb{C}$-valued functions on $X$ (like $\mathcal{O}(X)$) for $\mathsf{Vect}_\mathbb{C}$-valued functions. These assign a vector space to every point in a way that “varies smoothly”, and the way to make this precise is via sheaves! So you find yourself interested in something like $\text{QCoh}(X)$. In this case, you might want to define the Character Stack $\underline{\text{Ch}}(\Sigma,G)$ to be “the space whose category of quasicoherent sheaves is $G$-equivariant sheaves on $\text{Hom}(\pi_1 \Sigma, G)$”.

It turns out that these two spaces are generally not the same! Let’s look at the simplest case where $\Sigma$ is just a disk. Then $\pi_1 \Sigma$ is the trivial group, so $\text{Hom}(\pi_1 \Sigma, G)$ is a point, with ring of functions given by $\mathbb{C}$. Then the $G$-action on this space (and this on the ring of functions) is trivial, so that the $G$-equivariant functions are still $\mathbb{C}$ and $\text{Ch}(\text{Disk},G) = \star$ is a point. In particular, its category of quasicoherent sheaves is just $\mathsf{Vect}$.

But what about the character stack $\underline{\text{Ch}}(\text{Disk},G)$? Well now we define its category of quasicoherent sheaves to be $G$-equivariant sheaves on $\text{Hom}(\pi_1 \Sigma, G) = \star$. So this is $\mathsf{Vect}^G$, which is not $\mathsf{Vect}$! Indeed, when we say that a vector space is $G$-equivariant, what do we mean? We mean that $g \cdot V$ should be “the same as $V$” for every $g \in G$, but the notion of sameness for vector spaces is isomorphism! So saying that $g \cdot V$ is “the same as $V$” is saying we have isomorphisms $\varphi_g : g \cdot V \cong V$. Of course, $G$ is still acting trivially on $\text{Hom}(\pi_1 \Sigma, G) = \star$, so $g \cdot V = V$ and so $\varphi_g$ is an isomorphism of $V$ with itself! These isomorphisms are supposed to be compatible, so we find that $\mathsf{Vect}^G$, the category of $G$-equivariant vector spaces, is actually the category $\text{Rep}(G)$ of vector spaces equipped with a $G$-action! The space whose category of sheaves in $\text{Rep}(G)$ is usually called $\mathsf{B}G$, and we’ll do this too¹.

The character stack is better behaved in certain ways. It’s smooth (in a stacky sense) while the character variety usually isn’t² (this is why it’s common to restrict to a “smooth locus” containing most representations). Moreover, the character stack for $\Sigma$ can be computed by gluing together the character stacks on an open cover for $\Sigma$, while the character variety has no such nice local-to-global property. The character variety also relies crucially on $G$ being reductive, while the character stack works for all groups $G$.

It’s a famous result of Goldman that the smooth locus of the character variety admits a symplectic structure, which quantizes to the ($G$-)skein algebra for $\Sigma$! It turns out the character stack also admits a symplectic structure (in a stacky sense) and it’s natural to want to quantize this too. It should have something to do with skein theory… But what?

Let’s change topic for a moment and talk about Factorization Homology. Again, we’ll be much more terse here than we probably should be.

The notion of an $E_n$-algebra in a monoidal $k$-category $\mathcal{C}$ interpolates between noncommutative algebras ($E_1$) and commutative algebras ($E_\infty$). When $n \gt k$ these stabilize so that $E_{k+1}$ algebras are already “fully commutative”. Since we spend a lot of time working in $1$-categories (like $\mathsf{Set}$ and $\mathsf{Vect}$) we only really see the distinction between $E_1$ (noncommutative algebras) and $E_2 = E_\infty$ (commutative algebras).

However, if we work in a familiar $2$-category like $\mathsf{Cat}$ then we can see a bit further! Now an $E_1$-algebra is a monoidal category, an $E_2$-algebra is a braided monoidal category, and an $E_3 = E_\infty$-algebra is a symmetric monoidal category.

At this point in the talk I said some words introducing braided monoidal categories and why they might be called that, but it’s starting to get late so I think I won’t say those words now. You can read all about this somewhere like here.

Precisely, let $\mathsf{Disk}^n$ be the ($\infty$-)category whose objects are disjoint unions of $n$-disks and whose morphisms are (spaces of) smooth embeddings. Then a functor from $\mathsf{Disk}^n \to \mathcal{C}$ which sends disjoint union to the tensor product in $\mathcal{C}$ is exactly an $E_n$-algebra in $\mathcal{C}$!

Since $\text{Disk}^n$ is a full subcategory of the category of all $n$-manifolds with smooth embeddings, we can try to extend a functor $\text{Disk}^n \to \mathcal{C}$ (read: an $E_n$-algebra $A$) to a functor $\text{Man}^n \to \mathcal{C}$. The free way to do this is via left Kan extension, and this is how we define factorization homology!

The “factorization homology of $M$ with coefficients in $A$”, denoted by $\int_M A$, is defined to be the value of the left Kan extension $\text{Lan}(A)$ on $M$. This admits a “pointwise” formula to compute it, but it’s much much better to use excision! Like any good homology theory, factorization homology has a notion of Mayer-Vietoris for computation.

At this point I included a computation of $\int_{S^1} A$ for an algebra $A$ in $\mathsf{Vect}$ and showed that it recovers the Hochschild homology $HH_0(A)$. Even though I was starting to run out of time, I couldn’t help but mention one of my favorite facts about this too! If you view $A$ as an algebra in chain complexes which happens to be concentrated in degree $0$ then $\int_{S^1} A$ instead computes a derived enhancement, which happens to be $CHH_\bullet(A)$ – the entire complex of Hochschild chains! Since factorization homology is functorial and $S^1$ acts on itself, we get an induced $S^1$-action on $CHH_\bullet(A)$. An $S^1$-action on a chain complex is the data of a new differential on that complex, and it’s natural to ask what differential we get on Hochschild homology from this game! Well the HKR theorem says that $HH_\bullet(A)$ is the algebraic de Rham complex on $\text{Spec}(A)$ (when $A$ is commutative), and the differential coming from this $S^1$-action is exactly the de Rham differential!

Again, I think I want to finish this post up quickly, so I won’t include a copy of this computation here… I feel bad about it, though, so I’ll say that this is done in Hiro Tanaka’s fantastic series of talks starting here.

At this point we’re finally ready to bring the threads together!

One of the main ideas in the Integrating Quantum Groups Over Surfaces paper is that

\[\text{QCoh}(\underline{\text{Ch}}(\Sigma,G)) = \int_\Sigma \text{Rep}(G)\]

The computation I’m including in the sister post is a very explicit very special case of this computation where you can really get your hands on everything. Well… I at least sketch it, haha. You can see if you read that post. I originally planned to include this computation in the talk, but at this point I only had about 5 minutes left and I wanted to make sure I said something about the quantum character stacks in the title of the talk!

The point is that $\text{Rep}(G)$ is symmetric monoidal, so is an $E_\infty$ algebra, but we’re only integrating it over the measly $2$-manifold $\Sigma$! So we could get by with an $E_2$-algebra, which is less commutative! Working in $\mathsf{Cat}$ this means we want a braided monoidal category, and my favorite example is the category $\text{Rep}_q(G)$ of representations of a quantum group!

So now we see what to do:

Generalizing the above formula, we want to say that

\[\text{QCoh}(\underline{\text{Ch}}_q(\Sigma,G) = \int_\Sigma \text{Rep}_q(G)\]

But what does this really mean?

Remember earlier when I said that the 21st century approach to geometry is to focus on the (derived) category of sheaves? Well just like Grothendieck said that every (commutative) ring should count as functions on a space, we might bravely hope that every dg-category should be sheaves on a space! It turns out that one can push this idea very far, and this is one of the modern approaches to Noncommutative Geometry. See, for example, Kontsevich’s fantastic article Geometry in dg-Categories from the equally fantastic book New Spaces in Mathematics – every chapter is a banger.

This “noncommutative” perspective on geometry is what will let us make sense of the quantum character stack as a geometric object, even though we really only have access to what its category of sheaves should be.

At this point I basically had to stop the talk, but I rushed to say a few last minute things that I hoped would convince the audience that this is something that you can get your hands on.

Obviously I mentioned Juliet Cooke’s thesis, where she shows that there’s a concrete skein category defined in terms of tangles in the thickened $\Sigma \times I$ modulo local relations coming from the quantum group $G_q$. This should be compared to the classical skein algebra which is defined in terms of links in $\Sigma \times I$ modulo those same local relations. It turns out that this skein category presents the quantum character stack in the sense that the factorization homology $\int_\Sigma \text{Rep}_q(G)$ is the cocompletion of the skein category.

Also, the Barr-Beck yoga says that any category which looks like a category of algebras should be one, and indeed there’s an algebra object³ $A_\Sigma$ in $\text{Rep}_q(G)$ so that $\int_\Sigma \text{Rep}_q(G)$ is “just” a category of modules over $A_\Sigma$ (internal to $\text{Rep}_q(G)$, of course) and from a combinatorial presentation of $\Sigma$ Ben-Zvi, Brochier, and Jordan are able to compute explicit presentations of this internal algebra!

Alright, it’s a quick epilogue today. I would normally put the title, abstract, and slides here, but because it was an internal seminar and I gave a chalk talk I actually have none of those things, haha⁴.

Thanks for hanging out, everyone! It feels good to write about something that’s not my thesis, and I’m excited to go and write the sister post with this computation! That will have to wait a bit, though, since now it’s dinner time (I succeeded in writing this post in about three hours) and then I’m going climbing with some friends.

Stay safe, and we’ll chat soon ^_^.

You might be familiar with a different notion of $\mathsf{B}G$ from homotopy theory. In that world you take a contractible space with a free $G$-action and then quotient by it to get a “classifying space” where maps from $X$ to $\mathsf{B}G$ are principal $G$-bundles on $X$.

Up to homotopy a contractible space is a point, so this homotopy-theoretic $\mathsf{B}G$ is also “a point quotiented by $G$” just like our algebro-geometric example. Much of the same intuition goes into thinking about these two notions of $\mathsf{B}G$, but you have to remember that their implementations are different!

Since a lot of my readers are familiar with topos theory, I’ll say here that the category $G\text{-}\mathsf{Set}$ is a topos, and we often denote it by $\mathsf{B}G$ for this same reason. Indeed, the topos $\mathsf{B}G$ thinks its category of vector spaces is $\text{Rep}(G)$ so this is secretly the algebro-geometric example again. ↩
It’s a fun (but possibly tricky) exercise to compute the dimension of the tangent space at a generic point, then at the trivial representation. Remember that the tangent space to $\text{Ch}(\Sigma,G)$ at a representation $\rho : \pi_1 \Sigma \to G$ is given by the group cohomology
\[T_\rho \text{Ch}(\Sigma,G) = H^1(\pi_1 \Sigma, \mathfrak{g})\]
where $\mathfrak{g}$ is a $\pi_1 \Sigma$-module by composing $\rho$ with the adjoint action of $G$ on $\mathfrak{g}$.

For example let’s take $\Sigma$ to be a punctured torus, whose fundamental group is free on two generators $a$ and $b$, and let’s take $G$ to be $SL_2(\mathbb{C})$. Then a point in $\text{Ch}(\Sigma,G)$ is a representation $\rho$ up to conjugation, is a pair of matrices $A,B \in SL_2$ up to simultaneous conjugation (these are the images of $a$ and $b$ under $\rho$).

Now $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{C})$ is the space of $2 \times 2$ trace $0$ matrices, and the adjoint action of $G$ on $\mathfrak{g}$ is conjugation! So $\mathfrak{sl}_2$ becomes a $F_2$-module (read: a $\pi_1 \Sigma$-module) by $a \cdot M = A^{-1} M A$ and $b \cdot M = B^{-1} M B$. From here you can compute the group cohomology $H^1(F_2, \mathfrak{sl}_2)$ explicitly, and you’ll see that the generic dimension is not the dimension when $A = B = \text{Id}$. ↩
In fact this algebra comes as a kind of $\text{Rep}q(G)$-valued endomorphism object of the quantum structure sheaf… But I didn’t have time to say any of that at the end of the talk. See the _Integrating Quantum Groups Over Surfaces paper for more. ↩
Actually, writing this has reminded me that I never wrote a talk debrief for my JMM talk about Fukaya categories and my thesis work… Maybe I’ll write a very belated post about that, especially since it does have a title, an abstract, and slides!

We’ll see, though. I’ve been ridiculously busy lately. ↩

An Overly Explicit Computation with Character Stacks

Fri, 20 Feb 2026 00:00:00 +0000

In the main post I go over a talk I gave ~~today~~ yesterday explaining how factorization homology relates to (quantum) character stacks. In this post I want to do a sample computation which gives some justification that the main theorem from that talk (which is not mine) actually does work!

Recall from that blog post that we can compute the category of sheaves on the character stack $\underline{\text{Ch}}(\Sigma,G)$ by integrating $\mathsf{B}G$ over $\Sigma$ in the sense of factorization homology:

\[\text{QCoh}(\underline{\text{Ch}}(\Sigma,G)) \simeq \int_\Sigma \text{Rep}(G)\]

Let’s take a simple surface (like the annulus) and a simple group (like $GL_1 = \mathbb{C}^\times$) and just… check!

There’s a lot of problems with what I’m doing here… For one, I’m not actually sure if I’m working $\infty$-categorically or $1$-categorically! It’s possible that when I say $\text{QCoh}$ and $\text{Vect}$ and what not I really mean the derived category of quasicoherent sheaves and chain complexes, etc… that seems likely to me since I’m using a lot of formal gluing and duality statements, which tend to work better in the derived world… I really need to sit down and learn enough algebraic geometry to appreciate the differences between the derived and underived stories, but unfortunately that will have to wait for another day.

Another problem is that I’m only going to naively check that the objects of these categories seem to agree. It’s almost certainly possible to work with the arrows by hand as well and check that they agree, but I don’t have time to do that right now. Plus, in case we are working in a derived setting, I definitely don’t have time to check the infinitely many higher cells!

So with these caveats, I’ll follow in the footsteps of Jim Dolan and just press on doing computations to see if something meaningful pops out.

Let’s start with the left hand side.

When $\Sigma$ is the annulus and $G$ is $\mathbb{C}^\times$ we see that

\[\begin{aligned} \underline{\text{Ch}}(\Sigma,G) &\simeq \text{Hom}(\pi_1 \Sigma, G) \big / G \\ &\simeq \text{Hom}(\mathbb{Z}, \mathbb{C}^\times) \big / \mathbb{C}^\times \\ &\simeq \mathbb{C}^\times \big / \mathbb{C}^\times \\ &\simeq \mathbb{C}^\times \times \mathsf{B}\mathbb{C}^\times \end{aligned}\]

The last step is because here $\mathbb{C}^\times$ is acting on itself by conjugation – which is the trivial action since $\mathbb{C}^\times$ is abelian. We know a point with the trivial action gives rise to a $\mathsf{B}\mathbb{C}^\times$ and how we have $\mathbb{C}^\times$ many such points, so that we get a $\mathbb{C}^\times \times \mathsf{B}\mathbb{C}^\times$ in total.

So now we want to compute quasicoherent sheaves on this space, but products on the geometric side are coproducts (read: tensor products) on the algebraic side so we get

\[\begin{aligned} \text{QCoh}(\underline{\text{Ch}}(\Sigma,G)) &\simeq \text{QCoh}(\mathbb{C}^\times \times \mathsf{B}\mathbb{C}^\times) \\ &\simeq \text{QCoh}(\mathbb{C}^\times) \otimes \text{QCoh}(\mathsf{B}\mathbb{C}^\times) \\ &\simeq \text{QCoh}(\mathbb{C}^\times) \otimes \text{Rep}(\mathbb{C}^\times) \\ &\simeq \text{Rep}(\mathbb{Z}) \otimes \bigoplus_\mathbb{Z} \mathsf{Vect} \end{aligned}\]

In the last step we’ve used the fact that $\mathbb{C}^\times = \text{Spec}(\mathbb{C}[t^\pm])$ so that its category of sheaves is just modules over $\mathbb{C}[t^\pm]$, which we recognize as the group algebra of $\mathbb{Z}$. So such a module is exactly a vector space with an action of $\mathbb{Z}$. In fact this is some kind of Fourier duality, where sheaves on $\mathbb{C}^\times$ (some kind of circle) are interchanged with representations of $\mathbb{Z}$ (the dual group).

We’ve also used the fact that $\text{QCoh}(\mathsf{B}G)$ is $\text{Rep}(G)$, where for us $G = \mathbb{C}^\times$. Moreover, by Fourier duality again, $\text{Rep}(\mathbb{C}^\times)$ is the same as $\text{QCoh}(\mathbb{Z})$, but a sheaf on a discrete set like $\mathbb{Z}$ is just a choice of vector space above each point, so that we get the category of $\mathbb{Z}$-graded vector spaces.

Now of course we would expect tensor products to distribute over sums, and we know that $\mathsf{Vect}$ is the monoidal unit, so that at the end of the day we’ve computed

\[\text{QCoh}(\underline{\text{Ch}}(\Sigma,G)) \simeq \bigoplus_\mathbb{Z} \text{Rep}(\mathbb{Z})\]

the category of quasicoherent sheaves on the character stack is the category of $\mathbb{Z}$-graded $\mathbb{Z}$-modules! Which almost sounds like the kind of thing one can understand and compute with!

Next, the right hand side.

We compute factorization homology by excision – by cutting a surface into disks glued along collared embeddings and using the fact that the factorization homology of a disk with coefficients in $\mathcal{A}$ is just $\mathcal{A}$ again.

So we decompose the annulus into disks glued along their collared boundary

and then we can inductively compute the factorization homology by a Mayer-Vietoris style argument. Concretely, say you have an $E_2$-algebra $\mathcal{A}$ and you’re interested in computing factorization homology over the annulus. We want to do this in an oriented sense, and out of laziness I’m going to draw the annulus as though it’s a circle. Then we get

where as usual I’m writing $\mathcal{A}^\text{op}$ to mean $\mathcal{A}$ with the reversed algebra structure: $a \cdot^\text{op} b$ is defined to be $b \cdot a$. This makes the computation look more like the Hochschild homology computation it is, but could possibly be confusing since we’re about to work in the case where $\mathcal{A} = \text{Rep}(\mathbb{C}^\times)$ is a monoidal category! So just remember that in what follows $\text{Rep}(\mathbb{C}^\times)^\text{op}$ doesn’t mean the opposite category! It means the same category with the definition of $\otimes$ flipped – $M \otimes^\text{op} N$ is $N \otimes M$.

Then for us, we find ourselves wanting to compute

\[\text{Rep}(\mathbb{C}^\times)^\text{op} \boxtimes_{\text{Rep}(\mathbb{C}^\times)^\text{op} \boxtimes \text{Rep}(\mathbb{C}^\times)} \text{Rep}(\mathbb{C}^\times)\]

Here I’m switching over to $\boxtimes$ for the tensor product of categories in order to keep it straight with the monoidal product $\otimes$ inside a particular category later.

Now again Fourier duality makes this look less scary, since we know that $\text{Rep}(\mathbb{C}^\times) \simeq \bigoplus_\mathbb{Z} \mathsf{Vect}$! We need to know the monoidal structure now, and it will come as no surprise that we swap the “pointwise” monoidal structure on $\text{Rep}(\mathbb{C}^\times)$ with the “convolution” monoidal structure on $\bigoplus_\mathbb{Z} \mathsf{Vect}$!

To see why, consider the one dimensional $\mathbb{C}^\times$ representation $V_n$ where $z \in \mathbb{C}^\times$ acts by scalar multiplication by $z^n$. This is the generator for the grade $n$ copy of of $\mathsf{Vect}$. Now what is $V_n \otimes V_m$ supposed to be? Well $z$ acts diagonally so that $z \cdot (v_n \otimes v_m) = (z^n v_n) \otimes (z^m v_m) = z^{n+m} (v_n \otimes v_m)$ and $V_n \otimes V_m$ is $V_{n+m}$.

So cashing out the $\text{Rep}(\mathbb{C}^\times)$ for $\bigoplus_\mathbb{Z} \mathsf{Vect}$ everywhere we get

\[\left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right )^\text{op} \boxtimes_{ \left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right )^\text{op} \boxtimes \left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right ) } \left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right )\]

Where the action (on objects) of $\left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right )^\text{op} \boxtimes \left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right )$ on $\left ( \bigoplus_\mathbb{Z} \mathsf{Vect} \right )$ is by

\[(V_n \boxtimes V_m) \cdot V_i \simeq V_n \otimes V_i \otimes V_m\]

just like the bimodule action for the ordinary Hochschild homology! So, using the presentation for the relative tensor product of categories, we see that our factorization homology is generated by objects of the form

\[V_i \boxtimes V_j\]

where we add in isomorphisms

\[(V_n \otimes V_i \otimes V_m) \boxtimes V_j \cong V_i \boxtimes (V_m \otimes V_j \otimes V_n)\]

From this description it’s clear that, up to isomorphism, we can always make the $V_j$ into $V_0$ so that a normal form for objects of our category is

\[V_i \boxtimes V_0\]

We can do this by choosing $n = -(m+j)$, which uses up one of our two degrees of freedom. Of course, we can still freely choose $m$, which gives us more isomorphisms! When $j=0$ for simplicity (or, given the above discussion, without loss of generality), we get isomorphisms

\[(V_n \otimes V_i \otimes V_{-n}) \boxtimes V_0 \cong V_i \boxtimes (V_{-n} \otimes V_0 \otimes V_n) \cong V_i \boxtimes V_0\]

So we have integer many generating objects $V_i \boxtimes V_0$, each of which carries integer many automorphisms given by conjugating by $V_n$ for $n \in \mathbb{Z}$… Moreover, these automorphisms are compatible since $V_n \otimes V_m \cong V_{n+m}$ so that conjugating by $V_n$ and then $V_m$ is the same as conjugating by $V_{n+m}$.

Thus we have $\mathbb{Z}$-many generating objects, each of which carries a $\mathbb{Z}$-action!

So, at least at the level of objects, it’s believable that when $\Sigma$ is the annulus that $\int_\Sigma \text{Rep}(\mathbb{C}^\times)$ is $\mathbb{Z}$-graded $\mathbb{Z}$-modules!

Of course, this is exactly what we were expecting, since if you recall from the previous section we computed the same thing for $\text{QCoh}(\underline{\text{Ch}}(\Sigma,\mathbb{C}^\times))$!

Lastly, let’s say what the “correct” way to do this is. Tom Gannon showed me how to do this when I talked to him about this yesterday. Obviously, any mistakes come from my incorrectly remembering some details, or incorrectly filling in others.

We compute the factorization homology side again… But this computation will work for any group $G$, and we’ll write it in that level of generality.

As before, we define $\underline{\text{Ch}}(\text{Annulus}, G)$ to be $\text{Hom}(\mathbb{Z}, G) \big / G$, which is $G \big / G$ – the quotient of $G$ by the conjugation action on itself. We want to compare the category of quasicoherent sheaves on this stack to the factorization homology $\int_{\text{Annulus}} \text{Rep}(G)$.

As before, excision tells us that we can compute $\int_{\text{Annulus}} \text{Rep}(G)$ by the relative tensor product

\[\text{Rep}(G)^\text{op} \boxtimes_{\text{Rep}(G)^\text{op} \boxtimes \text{Rep}(G)} \text{Rep}(G)\]

but recall that $\text{Rep}(G) \simeq \text{QCoh}(\mathsf{B}G)$, so we can rewrite this as

\[\begin{aligned} \text{QCoh}(\mathsf{B}G) \boxtimes_{ \text{QCoh}(\mathsf{B}G) \boxtimes \text{QCoh}(\mathsf{B}G)^\text{op}} \text{QCoh}(\mathsf{B}G)^\text{op} &\overset{(1)}{\simeq} \text{QCoh}(\mathsf{B}G \times_{\mathsf{B}G \times \mathsf{B}G^\text{op}} \mathsf{B}G^\text{op}) \\ &\overset{(2)}{\simeq} \text{QCoh}(G \big \backslash (G \times G^\text{op}) \big / G) \\ &\overset{(3)}{\simeq} \text{QCoh}(G \big / G) \end{aligned}\]

In step $1$ we use the fact that the (relative) tensor product of sheaves corresponds to sheaves on the pullback. In step $2$ we compute the pullback of classifying spaces. Indeed, if we wanted the (homotopy) pullback $\star \times_{\mathsf{B}(G \times G^\text{op})} \star$ (remember that $\mathsf{B}$ preserves products) this would just be the loopspace of $\mathsf{B}(G \times G^\text{op})$, which is $G \times G^\text{op}$. But we don’t have $\star$, we actually have $\mathsf{B}G = \star \big / G$. So we have one left $G$ action (from the $\mathsf{B}G$) and one right $G$ action (from the $\mathsf{B}G^\text{op}$) and we end up with the double quotient space $G \big \backslash (G \times G^\text{op}) \big / G$… Apparently if you work through the details both of these $G$ actions are diagonal, by left/right multiplication respectively, but I haven’t tried to understand why that is. I also don’t entirely know that I’m keeping track of the “op”s correctly, and while we’re at it I’m fuzzy on the details of this “act like we’re pulling back to a pair of points and then quotient by the $G$ actions after” stuff too… I’m sure it’s formal, and I just haven’t gotten familiar with the rules for symbol pushing yet, so let’t not worry about it.

Regardless, we find ourselves with $G \big \backslash (G \times G^\text{op}) \big / G$ where the action is by $h \cdot (g_1, g_2) \cdot k = (h g_1 k, k g_2 h)$. Then we can use one degree of symmetry by taking $k = h^{-1} g_2^{-1}$. This tells us that in the quotient we should identify $(g_1,g_2)$ with $(h g_1 g_2^{-1} h^{-1}, e)$. Note that the multiplication order is kind of funny because of all the “op”s floating around. Of course we have another degree of freedom to use by choosing $h$, and as before this gives us an action of $G$ on itself by conjugation! So altogether we see that $G \big \backslash (G \times G^\text{op} ) \big / G$ is equivalent to $G \big / G$, the quotient of $G$ by the conjugation action! But this is exactly what we expect $\underline{\text{Ch}}(\text{Annulus}, G)$ to be!

This is encouraging since this calculation looked superficially like the kind of thing we were doing by hand back in the objectwise calculation in the last section!

Ok, thanks for reading all! This was a very quick-and-dirty post, and I’m not sure I got the details right with the objectwise argument in the first half, or with the “op”s in the second half. I would normally spend a few days (or weeks) reading up on stacks, properties of $\text{QCoh}$, this Fourier/Cartier/Pontryagin/Whatever duality that I was using, and check that everything really does work the way I think it does… But I want to get this out at the same time as the main post. Plus I’ve already had to put two posts on the backburner which are probably mostly right but need some details checked (one is about the stack semantics for the internal logic of (1-)topoi, and one is an explicit computation of the Goldman bracket for character varieties), and I really don’t want a third post in that limbo space. This could easily become one of many technical computations stuck in my drafts, but maybe it’s nice to share some quick-and-dirty computations too, since I do a lot of those in the privacy of my own home, haha. A lot of math looks polished once it gets presented, but at least for me the early days of learning a subject often look like this.

Stay safe everyone, and stay warm!

$F_2 \times F_2$ is Incoherent -- A Polite Spectral Sequence Computation

Wed, 03 Sep 2025 00:00:00 +0000

Yesterday I watched my friend Jialin Wang defend her thesis, and as part of her background section she mentioned that the group $F_2 \times F_2$ is incoherent in the sense that it has a subgroup that’s finitely generated and not finitely presented. I was curious how one might prove something like this, and in the original paper (Stallings’s Coherence of 3-Manifolds Fundamental Groups) this fact is boiled down to an “exercise which can be performed with the help of [a] spectral sequence”. I’ve been slowly trying to make spectral sequences feel like friends, so this seemed like the perfect thing to work out quickly and turn into a blog post!

I’ve been doing a lot of writing lately, with two papers that I want out by the end of the summer and a new result (which will be my thesis) that I want out by the end of the year and an NSF proposal¹, and even more stuff that I’m not talking about yet… So everyone in my life has heard me do nothing but complain about writing for the last month, haha. Weirdly, though, I’ve been itching to write a blog post! Maybe because it’s so informal, or maybe because it’s something I know I can finish, or maybe it’s because I’m mainly sick of writing about the same thing all day. No matter what it is, I’m happy to be here, and happy to have the excuse to share something cool ^_^.

There’s no way I can give an introduction to spectral sequences that’s better than Vakil’s notes, so I won’t even try. I highly recommend everyone give those a read at least once in your mathematical life, especially if you’re planning to do anything that might require you to actually use spectral sequences “in the wild”. Going forwards in this post, I’ll assume that you know the basics of what spectral sequences are, and how (roughly) to compute with them, but if you’re feeling brave and know a bit about homology you can probably already understand a fair amount of the post.

First, though, a few words about our goal. We’re trying to show that a product of free groups, $G = F_2 \times F_2$, is not coherent. To do this, we need to find a subgroup of $G$ which is finitely generated but not finitely presented. Stalling’s original paper tells us that we should look at

\[N = \langle a, c, bd \rangle \trianglelefteq F\{a,b\} \times F\{c,d\}\]

which is the kernel of the homomorphism

\[\begin{align} F\{a,b\} \times F\{c,d\} &\to \mathbb{Z} \\ a,c &\mapsto 0 \\ b &\mapsto 1 \\ d &\mapsto -1 \end{align}\]

This subgroup is obviously finitely generated (since we defined it in terms of $3$ generators!) so we need to show that it isn’t finitely presented! The key insight will be that

Every finitely presented group has finitely generated $H_2$.

$\ulcorner$ Recall that the group homology $H_\bullet(G;M)$ is isomorphic to the “usual” homology² of its Eilenberg-MacLane Space $K(G,1)$ with coefficients in the local system associated to the $G$-module $M$. Now if $G = \langle x_1, \ldots, x_n \mid R_1, \ldots, R_m \rangle$ is finitely presented, we can explicitly build a $K(G,1)$ as follows:

First add a loop for every generator $x_i$. Then each relation $R_i$ is a word in the generators, thus is a loop in our space, and we glue in a disk with boundary given by $R_i$. Note that this makes the loop vanish in $\pi_1$ so that we’ve forced the fundamental group of this space to be $G$. Finally we inductively add in higher cells to kill the higher homotopy groups, since we want our $K(G,1)$ to be aspherical.

Now we compute $H_2(G;\mathbb{Z}) = H_2(K(G,1); \mathbb{Z})$ using this description of the cell structure of $K(G,1)$. Since $G = \langle x_1, \ldots, x_n \mid R_1, \ldots, R_m \rangle$ was finitely presented, we see that there’s $n$ many $1$-cells and $m$-many $2$-cells in $K(G,1)$. So the group of $2$-cycles is a subgroup of $\mathbb{Z}^m$, the free abelian group on our (finite) set of $2$-cells, and is itself finitely generated. Quotienting out the boundaries gives $H_2$, so we win since the quotient of a finitely generated group is still finitely generated.³ $\lrcorner$

So, to show that our $N = \langle a, c, bd \rangle \trianglelefteq F\{a,b\} \times F\{c,d\}$ isn’t finitely presented, we just have to show its $H_2$ isn’t finitely generated. We can simplify the discussion by computing $H_2(N; \mathbb{Q})$ instead, since fields make homological algebra much easier and the dimension of $H_2(N; \mathbb{Q})$ (as a $\mathbb{Q}$-vector space) is a lower bound on the number of generators for $H_2(N;\mathbb{Z})$ (do you see why?⁴). So with this in mind

It suffices to show that $H_2(N; \mathbb{Q})$ is not finite dimensional as a $\mathbb{Q}$-vector space!

Unless otherwise stated, all homology groups have coefficients in $\mathbb{Q}$ (with the trivial action) for the rest of this post.

If we could find some nice description of the isomorphism type of $N$ then we could maybe compute its $H_2$ directly… But why spend the effort looking? We already have a short exact sequence

\[1 \to N \to F\{a,b\} \times F\{c,d\} \to \mathbb{Z} \to 1\]

and the homologies of $F_2 \times F_2$ and $\mathbb{Z}$ should be easier to compute by hand. Experience shows there should be some way to relate the homologies of $N$, $F_2 \times F_2$, and $\mathbb{Z}$, and indeed we’re saved by the Hochschild-Serre Spectral Sequence!

This says that whenever we have a short exact sequence

\[1 \to N \to G \to Q \to 1\]

we get a spectral sequence

\[E^2_{pq} = H_p(Q; H_q(N)) \Rightarrow H_{p+q}(G)\]

relating the homology of $G$ to the homologies of $Q$ and $N$. See Ch. VII.6 in Brown’s classic textbook for more details.

Concretely this means that we can compute the homology of $G$ in terms of “nested” homology groups: $Q$ acts on $N$ by conjugation, and this induces an action of $Q$ on $H_q(N)$ – thus it makes sense to look at the homology of $Q$ with coefficients in $H_q(N)$! The spectral sequence gives a close relationship between $H_n(G)$ and the collection of “nested” homologies $H_p(Q; H_q(N))$ with $p+q = n$.

Precisely, the $E^2$-page of the spectral sequence is

\[\begin{gather} \vdots && \vdots && \vdots && \vdots && ⋰ \\ H_2(Q; H_0(N)) && H_2(Q; H_1(N)) && H_2(Q; H_2(N)) && H_2(Q; H_3(N)) && \cdots \\ H_1(Q; H_0(N)) && H_1(Q; H_1(N)) && H_1(Q; H_2(N)) && H_1(Q; H_3(N)) && \cdots \\ H_0(Q; H_0(N)) && H_0(Q; H_1(N)) && H_0(Q; H_2(N)) && H_0(Q; H_3(N)) && \cdots \\ \end{gather}\]

In our case, we know that $Q = \mathbb{Z}$ has particularly simple homology. Recall that a $\mathbb{Q}\mathbb{Z}$-module is just a $\mathbb{Q}$-vector space $V$ with a $\mathbb{Z}$ action. That is, it’s just a vector space $V$ with a choice of automorphism $\varphi \in GL(V)$.

For any $\mathbb{Q}\mathbb{Z}$-module $(V, \varphi)$, we compute

$H_\bullet(\mathbb{Z}; V) = \begin{cases} V_\mathbb{Z} = V \big / (1-\varphi) V & \bullet = 0 \\ V^\mathbb{Z} = \text{Ker}(1-\varphi) & \bullet = 1 \\ 0 & \text{otherwise} \end{cases}$

$\ulcorner$ Writing $\mathbb{Q}[t^\pm]$ for $\mathbb{Q}\mathbb{Z}$, we build a free resolution of $\mathbb{Q}$

\[0 \to \mathbb{Q}[t^\pm] \overset{1-t}{\to} \mathbb{Q}[t^\pm] \overset{1}{\to} \mathbb{Q} \to 0.\]

This tells us that $H_\bullet(\mathbb{Z}; V)$ is the homology of

\[0 \to V \overset{1-t}{\to} V \to 0\]

where $t$ acts by the automorphism $\varphi$, giving the claim. $\lrcorner$

In case $V$ is finite dimensional (as a $\mathbb{Q}$ vector space), then it’s easy to see that $\dim V^\mathbb{Z} = \dim \text{Ker} (1 - \varphi)$ and $\dim V_\mathbb{Z} = \dim \left ( V \big / \text{Im}(1 - \varphi) \right ) = \dim V - \dim \text{Im}(1 - \varphi)$ are equal, so that these two vector spaces are isomorphic.

In case $V$ is infinite dimensional, though, this can fail! Let $V = \mathbb{Q}[t^\pm]$, of countable dimension, and let $\varphi$ be the (invertible) “multiply by $t$” operator. The fixed points $V^\mathbb{Z}$ of this operator are the laurent polynomials $p$ so that $p = t \cdot p$ (read: so that $(1-t) \cdot p = 0$), and the only option is $p=0$. The co-fixed points $V_\mathbb{Z}$ are given by $V \big / (1-t)$ which is isomorphic to $\mathbb{Q}$. So $V^\mathbb{Z}$ is $0$-dimensional and $V_\mathbb{Z}$ is $1$-dimensional.

When $V$ is finite dimensional as a $\mathbb{Q}$-vector space we compute

\[H_0(\mathbb{Z};V) \cong V_\mathbb{Z} \cong V^\mathbb{Z} \cong H_1(\mathbb{Z};V)\]

as vector spaces. So if these are not isomorphic, then $V$ must be infinite dimensional!

This lets us start evaluating the terms of our spectral sequence:

\[\begin{gather} \vdots && \vdots && \vdots && \vdots && ⋰ \\ H_2(\mathbb{Z}; H_0(N)) && H_2(\mathbb{Z}; H_1(N)) && H_2(\mathbb{Z}; H_2(N)) && H_2(\mathbb{Z}; H_3(N)) && \cdots \\ H_1(\mathbb{Z}; H_0(N)) && H_1(\mathbb{Z}; H_1(N)) && H_1(\mathbb{Z}; H_2(N)) && H_1(\mathbb{Z}; H_3(N)) && \cdots \\ H_0(\mathbb{Z}; H_0(N)) && H_0(\mathbb{Z}; H_1(N)) && H_0(\mathbb{Z}; H_2(N)) && H_0(\mathbb{Z}; H_3(N)) && \cdots \\ \end{gather}\]

becomes

\[\begin{gather} \vdots && \vdots && \vdots && \vdots && ⋰ \\ 0 && 0 && 0 && 0 && \cdots \\ H_0(N)^\mathbb{Z} && H_1(N)^\mathbb{Z} && H_2(N)^\mathbb{Z} && H_3(N)^\mathbb{Z} && \cdots \\ H_0(N)_\mathbb{Z} && H_1(N)_\mathbb{Z} && H_2(N)_\mathbb{Z} && H_3(N)_\mathbb{Z} && \cdots \\ \end{gather}\]

Moreover, we know that our $H_0(N; \mathbb{Q}) = \mathbb{Q}$ and $H_1(N; \mathbb{Q}) = N_\text{ab} \otimes \mathbb{Q} = \mathbb{Q}^3$, since the abelianization of $N = \langle a, c, bd \rangle$ is isomorphic to $\mathbb{Z}^3$. While we’re here we can compute that the conjugation action of $Q = \mathbb{Z}$ on $N$ induces the trivial action on $H_0(N)$ and $H_1(N)$.

Since $Q = \mathbb{Z}$ is generated by the image of $b$, the conjugation action on $N$ is literally conjugation by $b$. On the generators we compute

$a \mapsto b^{-1} a b = (bd)^{-1} a (bd)$
$c \mapsto b^{-1} c b = c$
$bd \mapsto b^{-1} (bd) b = bd$

In $H_0(N) = \mathbb{Q}$ the group $N$ doesn’t even make an appearance, so the induced action is trivial. On $H_1(N) = N_\text{ab} \otimes \mathbb{Q}$ we need to see what the conjugation action induces on the abelianization, but that becomes trivial since in $N_\text{ab}$ we have $(bd)^{-1} a (bd) = a$.

Since the $\mathbb{Z}$-action is trivial on $H_0(N) = \mathbb{Q}$ and $H_1(N) = \mathbb{Q}^3$ we learn that

$H_0(N)_\mathbb{Z} = H_0(N)^\mathbb{Z} = \mathbb{Q}$
$H_1(N)_\mathbb{Z} = H_1(N)^\mathbb{Z} = \mathbb{Q}^3$

So the $E^2$ page of our spectral sequence further reduces to

\[\begin{gather} \vdots && \vdots && \vdots && \vdots && ⋰ \\ 0 && 0 && 0 && 0 && \cdots \\ \mathbb{Q} && \mathbb{Q}^3 && H_2(N)^\mathbb{Z} && H_3(N)^\mathbb{Z} && \cdots \\ \mathbb{Q} && \mathbb{Q}^3 && H_2(N)_\mathbb{Z} && H_3(N)_\mathbb{Z} && \cdots \\ \end{gather}\]

The differential on the $E^2$ page points “up two, left one”, so we see that every differential is $0$. In fact it’s easy to see that all futher differentials vanish so that this is actually the $E^\infty$ page of our spectral sequence! General theory tells us that $H_n(G) = \bigoplus_{p+q = n} E^\infty_{pq}$, so we can compute $H_n(F_2 \times F_2)$ by summing over the $n$th diagonal in the above table⁵.

Of course, $H_n(F_2 \times F_2)$ is easy enough to compute by hand using the Künneth formula and the fact that $K(F_2, 1) = S^1 \vee S^1$ is a bouquet with two petals⁶. So we can compute it in two ways (directly via Künneth and “indirectly” via the spectral sequence) and compare to see what it tells us about $H_\bullet(N)$!

In particular, we learn (the left isomorphism comes from Künneth and the right isomorphism comes from the spectral sequence):

\[\begin{gather} \mathbb{Q} &\cong H_0(F_2 \times F_2) \cong& \mathbb{Q} \\ \mathbb{Q}^4 &\cong H_1(F_2 \times F_2) \cong& \mathbb{Q} \oplus \mathbb{Q}^3 \\ \mathbb{Q}^4 &\cong H_2(F_2 \times F_2) \cong& 0 \oplus \mathbb{Q}^3 \oplus H_2(N)_\mathbb{Z} \\ 0 &\cong H_3(F_2 \times F_2) \cong& 0 \oplus 0 \oplus H_2(N)^\mathbb{Z} \oplus H_3(N)_\mathbb{Z} \end{gather}\]

From the $H_2(F_2 \times F_2)$ computation, we learn that $H_2(N)_\mathbb{Z}$ must be $1$ dimensional. But from the $H_3(F_2 \times F_2)$ computation we learn that $H_2(N)^\mathbb{Z}$ must be $0$ dimensional!

Since the invariants and coinvariants have different diemnsions, our earlier discussion shows that $H_2(N)$ must be infinite dimensional! This means $N$ cannot have been finitely presented, as desired ^_^.

Let’s take a second to reflect on what just happened, since there were a decent number of moving parts.

We wanted to show that $N = \langle a, c, bd \rangle \leq F\{a,b\} \times F\{c,d\}$ is not finitely presented. First, we showed that every finitely presented group $G$ has finitely generated $H_2(G;\mathbb{Z})$ (using a concrete model of $K(G,1)$) so that it suffices to show $H_2(N; \mathbb{Z})$ is infinitely generated. Since the dimension of $H_2(N;\mathbb{Q})$ is a lower bound for the number of $\mathbb{Z}$-generators, we can work over a field and show that $H_2(N; \mathbb{Q})$ is infinite dimensional. Next, we showed that $H_2(N)$ comes with a natural $\mathbb{Z}$ action, and argued that $H_2(N)$ must be infinite dimensional if the invariants and coinvariants $H_2(N)^\mathbb{Z}$ and $H_2(N)_\mathbb{Z}$ have different dimensions. Finally, using the Hochschild-Serre spectral sequence, we were able to compute that $H_2(N)_\mathbb{Z}$ is one dimensional while $H_2(N)^\mathbb{Z}$ is zero dimensional. This shows that $H_2(N)$ must be infinite dimensional, and we win!

This is a clever trick, and a fairly subtle one! It’s something I’ll have to try to remember, since the obvious approach is to try and compute the (co)invariants explicitly, but I’m not even sure⁷ how to compute the $\mathbb{Z}$-action on $H_2(N)$! This lets you get your hands on the infinite-dimensionality indirectly, which feels very useful.

Thanks for hanging out, everyone! It’s wild to think that just a short week ago I was in Bozeman, Montana meeting a bunch of cool people and giving a talk about my thesis. Then all in a row over labor day weekend I had two little dinner parties and went to the beach to swim with leopard sharks! It wasn’t very productive, but it was extremely good for the soul, haha. Now I have a few short days to try and get more done before I fly to Chicago for the Fall School on Quantizations and Lagrangians. Take care all, and stay safe. We’ll talk soon 💖

On the off chance the NSF still exists next year ↩
Depending on which book you read, this is either a definition or a theorem. See, for instance, the Introduction or Chapter II.4 in Brown’s book on Group Cohomology. ↩
In fact, there’s a whole hierarchy of finiteness conditions on a group $G$. We say that a group $G$ is “of type $F_n$” if its $K(G,1)$ has a finite $n$-skeleton. That is, if there’s only finitely many $0$-cells, finitely many $1$-cells, …, and finitely many $n$-cells.

In the body we really showed that being finitely presented means being type $F_2$… Well, we showed half of this. Showing the converse (that $F_2$-groups are finitely presented) isn’t so hard either, and uses essentially the same idea.

Note that being type $F_2$ implies that $H_2$ is finitely generated, since (as we said in the main body) then $H_2$ is a quotient of a subgroup of a finitely generated abelian group. But even if $G$ isn’t of type $F_2$, then $H_2$ might “accidentally” be finitely generated, if we have a big generating set but then quotient out by a similarly big set of boundaries. In fact, this really does happen! Bestvina and Brady constructed a group whose $H_2$ is finitely generated (indeed, whose $H_n$ is finitely generated for all $n$) yet which is not finitely presented! See Morse Theory and Finiteness Properties of Groups ↩
The universal coefficient theorem promises $H_2(N;\mathbb{Q}) = H_2(N;\mathbb{Z}) \otimes \mathbb{Q}$, which kills any torsion subgroups (so we don’t see those generators) but keeps the free abelian part. ↩
This is one place where working over a field really simplifies things! In general we would only know that $H_n(G)$ is some iterated extension with the groups showing up along the diagonals as subquotients. Thankfully over a field the only extensions are direct sums, so we learn that $H_n(G)$ must be the direct sum along the diagonal! ↩
If this isn’t obvious, it’s a fantastic exercise in algebraic topology! Can you compute the homology groups $H_\bullet \Big ( (S^1 \vee S^1) \times (S^1 \vee S^1) ; \mathbb{Q} \Big )$?

Again, you’ll want the Künneth formula to handle the product, and then you’ll want something like Mayer-Vietoris to handle the wedge sums.

To relate this to group homology, note that $K(G \times H, 1) \cong K(G, 1) \times K(H, 1)$, so that the Künneth formula also applies to group homology! ↩
Though I gave up almost immediately, since I want this post finished so I can go back to writing more important things ↩

Free Things Are Complicated (Especially the Sphere Spectrum!)

Sun, 20 Jul 2025 00:00:00 +0000

I’ve spent the last week at CT2025, which has just come to a close. It was great getting to see so many old friends and meet so many new ones, and every time I go to a CT I’m reminded of just how much category theory there is in the world, as well as just how much I enjoy all of it! Right before this I was in Antwerp for some Noncommutative Geometry, where I learned a ton and met even more new friends! Then next week I go to Bonn for my third conference in a row. I’m trying to stay energetic, and thankfully I have a few days off between CT and QTMART to help me rest up!

I want to write up a lot of things I learned over the last month, since I have a lot of new thoughts on noncommutative geometry, mirror symmetry, and deformation theory, all coming from just my time in Antwerp! I’ve also learned a lot at CT and talked to a lot of interesting people about interesting things, and I’m sure I’ll have even more to say after my time in Bonn.

I think organizing all of those thoughts are going to take a while, though (if I end up writing them down at all), but today I have a quick observation inspired by a few lovely conversations I had with Clark Barwick at CT. One of the many questions I asked him was if there’s a conceptual reason the Sphere Spectrum (read: the homotopy groups of spheres) is so darn complicated. He gave me an answer that’s obvious in hindsight, but which totally rearranged the way I think about things:

The Sphere Spectrum is complicated because it's *free*

I think I internalized a while ago that “free” constructions are fairly concrete. After all, you look at the syntax of whatever object you’re interested in, quotient out by the relations you want to be true and you’re done! Plus, mapping out of a free thing is as simple as possible, since it’s a left adjoint! All you have to do is find a (usually simpler) map from your generating set to a structure of interest and let the magic of category theory build your (usually more complicated) map for you…

Of course, this view is heavily influenced by the kind of free structures I have experience with, and the kinds of questions I was asking about them. I was thinking about free groups and monoids, which you can study with word combinatorics, free (dg-)algebras on (graded) vector spaces, which look like polynomials, free categories on graphs, free $k$-linear or dg-categories on categories, and free cauchy completions of these, all of which come from just looking at paths, linear combinations, concentrating things in degree $0$, or working with twisted closures to add shifts and cones and whatnot¹. I was thinking about relatively free constructions like the universal enveloping algebra, with its PBW-basis, or the right angled artin group attached to a (reflexive, simple) graph… All of these constructions feel like friends to me, in part because I know how to compute with them. Why would the sphere spectrum – the free spectrum on a single point – be so different?

The point is that I’ve internalized these constructions as being tractable because I’m usually mapping out of them, in the direction the category theory encourages. I’m also usually relying on serious “normal form” theorems that make computing with these things tractable, or I’m doing fairly simple combinatorics with my generating set before arguing that these extend in some obvious way to things defined on the whole free object. All of these constructions become much less friendly when you start mapping into them, or asking more difficult questions about their internal structure. In hindsight, I’ve even personally struggled with tons of questions about free structures in my research!

Free groups are extremely interesting from basically any perspective, with deep questions about their first order theory (Tarski’s Problem), the combinatorics relating their generating sets (The Andrews-Curtis Conjecture), or the coarse geometry of their outer automorphisms.

Free cauchy completions are obviously complicated when you want to understand them on their own terms! If you take an algebra $A$ and view it as a one-object dg-category, then its cauchy completion¹ is its whole category of perfect complexes! An enormous chunk of representation theory is, in that lens, dedicated to nothing more than the study of a certain, complicated, free construction!

I’ve personally given up² on a problem about relatively free constructions in right angled artin groups! These interpolate between free and free-abelian groups, and geometric group theory is teeming with interesting open problems about raags. For instance, can you understand, at the level of the underlying graphs, when one raag will embed into another? I thought about this off and on for a year before I started working seriously with my current advisor, and I made almost no progress at all.

I also spent some time working with the adjunctions

\[\{ \text{finite limit categories} \} \leftrightarrows \{ \text{finite product categories} \} \leftrightarrows \{ \text{symmetric monoidal categories} \}\]

It’s interesting to try and construct these explicitly, and to understand the essential images of the left adjoints. This amounts to understanding which essentially algebraic theories are actually algebraic, and which algebraic theories are actually props. One of the big difficulties here is that we have a relatively free construction which adds relations rather than just operations. Adding new operations tends to be a fairly mild thing to do – consider the free algebra on an abelian group, which sends $A$ to its tensor algebra $\bigoplus_n A^{\otimes n}$ where it’s easy to recover the $A$ you started with. If instead we want to add new relations or axioms, for instance by freely sending a group $G$ to its abelianization $G \big / [G,G]$, then we lose lots of information in this construction.

After some conversations with John Baez and Todd Trimble I came quite close to characterizing the image of the left adjoint between finite product categories and symmetric monoidal categories by factoring it into a “lossy” construction adding new axioms forcing the monoidal unit to be terminal and a much simpler construction which freely adds new operations corresponding to the product projections. I’ve had to put that project on hold while I focus on my thesis work, but I really want to come back and finish it soon.

Of course as soon as you’re interested in logic, you have to accept that free things are complicated! The freest version of any theory lives in its classifying category, where truth and provability coincide. Then proving anything at all about the free model gives immediate understanding about all other models of that theory! This is already true for groups, whose classifying finite-product category is just the category of finitely generated free groups and homomorphisms. We don’t usually think about it because the kinds of statements you prove in equational logic aren’t very deep. But if you look instead at the classifying topos for groups and ask geometric questions suddenly you’re able to do a lot more, and the game becomes much harder!

Perhaps this is clearest in the semantics of programming languages, where the free model (often called the “term model” in this context³) is the programming language, and checking whether two terms are equal in this free model literally amounts to evaluating two programs and seeing if their respective values agree. Because of this, many important structural results about a programming language (such as canonicity) can be proven by building another model whose semantics you understand, and then producing a section of the unique map from the free model.

Also coming from logic are various lattices, whose free models can be quite intricate. Famously the free modular lattice on $3$ generators has $28$ elements, while the free modular lattice on $4$ generators is infinite! Indeed, this lattice has an undecidable word problem⁴, so that no program can tell whether two descriptions of its elements are the same or not! Heyting algebras are extremely important for semantics of intuitionistic logic, yet already the free heyting algebra on one generator is infinite, and the free heyting algebra on two generators is famously complicated. The study of free heyting algebras is still ongoing and seems quite difficult (at least as an outsider). See, for instance, Almeida’s recent preprint Colimits and Free Constructions of Heyting Algebras through Esakia Duality.

With all this in mind, it shouldn’t be surprising at all that the sphere spectrum is so complicated! It’s the free spectrum on a point, and as such the only “relations” it will have are those that hold in all spectra! But of course spectra should obviously be complicated – They control all possible (co)homology theories for all spaces! So in this sense one should expect the internal structure of the sphere spectrum to be quite complicated, since any simplification would persist to something true of all cohomology theories.

Of course, it’s easy to be complicated without being interesting, and I still think it’s a bit of a miracle that the sphere spectrum should have all this intricate structure inside it. As I understand it, much of Chromatic Homotopy Theory came from trying to explain patterns in the homotopy groups of spheres, and this subject is now as famously intimidating to outsiders⁵ as it is famously fascinating once you put in the work to become an insider⁶.

Thanks for reading all! This really was a quick one for once, since I already had a lot of these examples floating around in my head. I really had all of the tools to realize that free things are obviously complicated in general, especially their “internal structure” that doesn’t ride the coattails of the universal property, but for some reason I just didn’t put it together until my conversation with Clark.

It’s always dangerous to say what I’m thinking about writing about, but I at least have one more short post planned from my time in Antwerp, and maybe a longer one too if I have the energy. I’m doing a ton of writing right now, since I have two half-finished papers that I want to submit by the end of the summer. I think I’ll be able to get it done, but between writing these and going to conferences it’s been a tiring month. It’s tiring in a fun way, though, and I really feel like I’ve been productive in a way that I haven’t felt in a little while. I’m excited to start crossing a lot of these projects off my long-term-todo-list, especially since I already have three more projects I want to start!

Regardless, I hope you’re having a more restful summer than I am! Stay safe, all, and we’ll talk soon ^_^.

(photo by the amazing Emily Roff, taken at the top of the main tower in Brno)

Actually it’s not completely obvious to me that the cauchy completion of a dg-category should be its idempotent triangulated closure… The cauchy completion will certainly be idempotent complete and triangulated, but in the well-named paper Cauchy Completeness for DG-Categories, Nicolić, Street, and Tendas show that to be cauchy complete you also need to be closed under “cokernels of protosplit chain maps”… I think this is some kind of split idempotent condition? But I haven’t read the paper closely enough to know for sure.

If you want to be guaranteed to be correct, instead of “cauchy completion” you can say “the idempotent closure of the twisted closure”. That’s still a free construction and it still gives you the derived category in the special case your dg-category is a ring. ↩ ↩²
At least for now ↩
Pun intended ↩
Which is made more interesting by the fact that if you look at the class of lattices coming from lattices of subgroups of abelian groups, the corresponding free lattice on $4$ generators does have solvable word problem!

This is remarkable since (as I understand it) modular lattices are called that because they look like lattices of submodules (in particular, sublattices of a lattice of subgroups of an abelian group).

This is apparently proven in Herrmann’s On the Equational Theory of Submodule Lattices, and I learned all this from Ralph Freese the comments of this n-Category Cafe post. ↩
Such as myself. ↩
Which it seems like I might start doing soon, for a project I’m not ready to talk about yet. ↩

An Explicit Computation in Derived Algebraic Geometry

Sat, 21 Jun 2025 00:00:00 +0000

Earlier this week my friend Shane and I took a day and just did a bunch of computations. In the morning we did some differential geometry, where he told me some things about what he’s doing with symplectic lie algebroids. We went to get lunch, and then in the afternoon we did some computations in derived algebraic geometry. I already wrote a blog post on the differential geometry, and now I want to write one on the derived stuff too!

I’m faaaaar from an expert in this stuff, and I’m sure there’s lots of connections I could make to other subjects, or interesting general theorem statements which have these computations as special cases… Unfortunately, I don’t know enough to do that, so I’ll have to come back some day and write more blog posts once I know more!

I’ve been interested in derived geometry for a long time now, and I’ve been sloooowly chipping away at the prerequisites – $\infty$-categories and model categories, especially via dg-things, “classical” algebraic geometry (via schemes), and of course commutative and homological algebra. I’m lucky that a lot of these topics have also been useful in my thesis work on fukaya categories and TQFTs, which has made the time spent on them easy to justify!

I’ve just started reading a book which I hope will bring all these ideas together – Towards the Mathematics of Quantum Field Theory by Frédéric Paugam. It seems intense, but at least on paper I know a lot of the things he’s planning to talk about, and I’m hoping it makes good on its promise to apply its techniques to “numerous examples”. If it does, I’m sure it’ll help me understand things better so I can share them here ^_^.

In this post we’ll do two simple computations. In both cases we have a family of curves where something weird happens at a point, and in the “clasical” case this weirdness manifests as a discontinuity in some invariant. But by working with a derived version of the invariant we’ll see that at most points the classical story and the derived story agree, but at the weird point the derived story contains ~bonus information~ that renders the invariant continuous after all!

Ok, let’s actually see this in action!

First let’s look at what happens when we intersect two lines through the origin. This is the example given in Paugam’s book that made me start thinking about this stuff again.

Let’s intersect the $x$-axis (the line $y=0$) with the line $y=mx$ as we vary $m$. This amounts to looking at the schemes $\text{Spec}(k[x,y] \big / y)$ and $\text{Spec}(k[x,y] \big / y-mx)$. Their intersection is the pullback

and so since everything in sight is affine, we can compute this pullback in $\text{Aff} = \text{CRing}^\text{op}$ as a pushout in $\text{CRing}$:

Pushouts in $\text{CRing}$ are given by (relative) tensor products, and so we compute

\[k[x,y] \big / (y) \ \otimes_{k[x,y]} \ k[x,y] \big / (y-mx) \cong k[x,y] \big / (y, y-mx) \cong k[x] \big / (-mx)\]

which is $k$ when $m \neq 0$ and is $k[x]$ when $m=0$, so we’ve learned that:

When $m \neq 0$ the intersection of $\{y=0\}$ and $\{y=mx\}$ is $\text{Spec}(k)$ – a single point¹.
When $m = 0$ the intersection is $\text{Spec}(k[x])$ – the whole $x$-axis.

This is, of course, not surprising at all! We didn’t really need any commutative algebra for this, since we can just look at it!

The fact that the dimension of the intersection jumps suddenly is related to the lack of flatness in the family of intersections $k[x,y,m] \big /(y, y-mx) \to k[m]$. Indeed, this doesn’t look like a very flat family!

We can also see it isn’t flat algebraically since tensoring with $k[x,y,m] \big / (y, y-mx)$ doesn’t preserve the exact sequence²

\[0 \to k[m] \overset{- \cdot m}{\longrightarrow} k[m] \longrightarrow k \to 0\]

In the derived world, though, things are better. It’s my impression that here flatness is a condition guaranteeing the “naive” underived computation agrees with the “correct” derived computation. That is, flat modules $M$ are those for which $M \otimes^\mathbb{L} -$ and $M \otimes -$ agree for all modules $X$! I think that one of the benefits of the derived world is that we can pretend like “all families are flat”. I would love if someone who knows more about this could chime in, though, since I’m not confident enough to really stand by that.

In our particular example, though, this is definitely true! To see this we need to compute the derived tensor product of $k[x,y] \big / (y)$ and $k[x,y] \big / (y-mx)$ as $k[x,y]$-algebras. To do this we need to know the right notion of “projective resolution” (it’s probably better to say cofibrant replacement), and we can build these from (retracts of) semifree commutative dg algebras in much the same way we build projective resolutions from free things!

Here “semifree” means that our algebra is a free commutative graded algebra if we forget about the differential. Of course, “commutative” here is in the graded sense that $xy = (-1)^{\text{deg}(x) \text{deg}(y)}yx$.

For example, if we work over the base field $k$, then the free commutative graded algebra on $x_0$ (by which I mean an element $x$ living in degree $0$) is just the polynomial algebra $k[x]$ all concentrated in degree $0$. Formally, we have elements $1, \ x_0, \ x_0 \otimes x_0, \ x_0 \otimes x_0 \otimes x_0, \ldots$, and the degree of a tensor is the sum of the degrees of the things we’re tensoring, so that for $x_0$ the whole algebra ends up concentrated in degree $0$.

If we look at the free graded $k$-algebra on $x_1$, we again get an algebra generated by $x_1, \ x_1 \otimes x_1, \ x_1 \otimes x_1 \otimes x_1, \ldots$ except that now we have the anticommutativity relation $x_1 \otimes x_1 = (-1)^{1 \cdot 1} x_1 \otimes x_1$ so that $x_1 \otimes x_1 = 0$. This means the free graded $k$-algebra on $x_1$ is just the algebra with $k$ in degree $0$, the vector space generated by $x$ in degree $1$, and the stipulation that $x^2 = 0$.

In general, elements in even degrees contribute symmetric algebras and elements in odd degrees contribute exterior algebras to the cga we’re freely generating.

What does this mean for our example? We want to compute the derived tensor product of $k[x,y] \big / y$ and $k[x,y] \big / y-mx$. As is typical in homological algebra, all we need to do is “resolve” one of our algebras and then take the usual tensor product of chain complexes. Here a resolution means we want a semifree cdga which is quasi-equivalent to the algebra we started with, and it’s easy to find one!

Consider the cdga $k[x,y,e]$ where $x,y$ live in degree $0$ and $e$ lives in degree $1$. The differential sends $de = y$, and must send everything else to $0$ by degree considerations (there’s nothing in degree $-1$). This cdga is semifree as a $k[x,y]$-algebra, since if you forget the differential it’s just the free graded $k[x,y]$ algebra on a degree 1 generator $e$!

So this corresponds to the chain complex

\[\cdots 0 \to k[x,y]e \overset{d}{\to} k[x,y] \to 0\]

where $de = y$ is $k[x,y]$ linear so that more generally $d(pe) = p(de) = py$ for any polynomial $p \in k[x,y]$.

If we tensor this (over $k[x,y]$) with $k[x,y] \big / y-mx$ (concentrated in degree $0$) we get a new complex

\[\cdots 0 \to (k[x,y] \big / y-mx)e \to (k[x,y] \big / y-mx) \to 0\]

where the interesting differential sends $pe \mapsto ey$ for any polynomial $p \in k[x,y] \big / y-mx$. Some simplification gives the complex

\[0 \to k[x] \overset{- \cdot mx}{\longrightarrow} k[x] \to 0\]

whose homology is particularly easy to compute!

$H_0 = k[x] \big / mx$
$H_1 = \text{Ker}(mx)$

We note that $H_0$ recovers our previous computation, where when $m \neq 0$ we have $H_0 = k$ is the coordinate ring of the origin³ and when $m=0$ we have $H_0 = k[x]$ is the coordinate ring of the $x$-axis.

However, now there’s more information stored in $H_1$! In the generic case where $m \neq 0$, the differential $mx$ is injective so that $H_1$ vanishes, and our old “classical” computation saw everything there is to see. It’s not until we get to the singular case where $m=0$ that we see $H_1 = \text{Ker}(mx)$ becomes the kernel of the $0$-map, which is all of $k[x]$!

The version of “dimension” for chain complexes which is invariant under quasi-isomorphism is the euler characteristic, and we see that now the euler characteristic is constantly $0$ for the whole family!

Next let’s look at some kind of “hidden smoothness” by examining the singular curve $y^2 =x^3$. Just for fun, let’s look at another family of (affine) curves $y^2 = x^3 + tx$, which are smooth whenever $t \neq 0$. We’ll again show that in the derived world the singular fibre looks more like the smooth fibres.

Smoothness away from the $t=0$ fibre is an easy computation, since we compute the jacobian of the defining equation $y^2 - x^3 - tx$ to be $\langle -3x^2 - t, 2y \rangle$, and for $t \neq 0$ this is never $\langle 0, 0 \rangle$ for any point on our curve⁴ (We’ll work in characteristic 0 for safety). Of course, when $t=0$ $\langle -3x^2, 2y \rangle$ vanishes at origin, so that it has a singular point there.

To see the singularity, let’s compute the tangent space at $(0,0)$ for every curve in this family. We’ll do that by computing the space of maps from the “walking tangent vector” $\text{Spec}(k[\epsilon] \big / \epsilon^2)$ to our curve which deform the map from $\text{Spec}(k)$ to our curve representing our point of interest $(0,0)$.

Since everything is affine, we turn the arrows around and see we want to compute the space of algebra homs

\[k[x,y] \big / (y^2 - x^3 - tx) \to k[\epsilon] \big / \epsilon^2\]

so that the composition with the map $k[\epsilon] \big / \epsilon^2 \to k$ sending $\epsilon \mapsto 0$ becomes the map $k[x,y] \big / (y^2 - x^3 - tx) \to k$ sending $x$ and $y$ to $0$.

Since $k[x,y] \big / (y^2 - x^3 - tx)$ is a quotient of a free algebra, this is easy to do! We just consult the universal property, and we find a hom $k[x,y] \big / (y^2 - x^3 - tx) \to k[\epsilon] \big / \epsilon^2$ is just a choice of image $a+b\epsilon$ for $x$ and $c+d\epsilon$ for $y$, so that the equation $y^2 - x^3 - tx$ is “preserved” in the sense that $(c+d\epsilon)^2 - (a+b\epsilon)^3 - t(a+b\epsilon)$ is $0$ in $k[\epsilon] \big / \epsilon^2$.

Then the “deforming the origin” condition says that moreover when we set $\epsilon = 0$ our composite has to send $x$ and $y$ to $0$. Concretely that means we must choose $a=c=0$ in the above expression, so that finally:

The tangent space at the origin of $k[x,y] \big / (y^2 - x^3 - tx)$ is the space of pairs $(b,d)$ so that $(d \epsilon)^2 - (b \epsilon)^3 - t(b \epsilon) = 0$ in $k[\epsilon] \big / \epsilon^2$. Of course, this condition holds if and only if $tb=0$, so that:

When $t \neq 0$ the tangent space is the space of pairs $(b,d)$ with $b=0$, which is one dimensional.
When $t = 0$ the tangent space is the space of pairs $(b,d)$ with no further conditions, which is two dimensional!

Since we’re looking at curve, we expect the tangent space to be $1$-dimensional, and this is why we say there’s a singularity at the origin for the curve $y^2 = x^3$….. But what happens in the derived world?

Now we want to compute the derived homspace. As before, a cofibrant replacement of our algebra is easy to find, it’s just $k[x,y,e]$ where $x$ and $y$ have degree $0$, $e$ has degree $1$ and and $de = y^2 - x^3 - tx$. Note that in our last example we were looking at quasifree $k[x,y]$-algebras, but now we just want $k$-algebras! So now this is the free graded $k$-algebra on 3 generators $x,y,e$, and our chain complex is:

\[0 \to k[e] \overset{e \mapsto y^2 - x^3 - tx}{\longrightarrow} k[x,y] \to 0\]

We want to compute the derived $\text{Hom}^\bullet(-,-)$ from this algebra to $k[\epsilon] \big / \epsilon^2$, concentrated in degree $0$.

The degree $0$ maps are given by diagrams that don’t need to commute⁵!

Of course, such maps are given by pairs $(a + b \epsilon, c + d \epsilon)$, which are the images of $x$ and $y$. As before, since we want the tangent space at $(0,0)$ we need to restrict to those pairs with $a=c=0$ so that $\text{Hom}^0(k[x,y] \big / y^2 - x^3 - tx, \ k[\epsilon] \big / \epsilon^2) = k^2$, generated by the pairs $(b,d)$.

Next we look at degree $-1$ maps, which are given by diagrams

which are given by a pair $r + s\epsilon$, the image of $e$. Again, these need to restrict to the $0$ map when we set $\epsilon=0$, so that $r=0$ and we compute $\text{Hom}^{-1}(k[x,y] \big / y^2 - x^3 - tx, \ k[\epsilon] \big / \epsilon^2) = k$, generated by $s$.

So our hom complex is

\[0 \to \{(b\epsilon, d\epsilon)\} \to \{s \epsilon\} \to 0\]

where the interesting differential sends degree $0$ to degree $-1$ and is given by $df = d_{k[\epsilon] \big / \epsilon^2} \circ f - f \circ d_{k[x,y] \big / y^2-x^3-tx}$.

So if $f$ is the function sending $x \mapsto b \epsilon$ and $y \mapsto d \epsilon$ then we compute

\[(df)(e) = 0 \circ f - f(de) = -f(y^2-x^3-tx) = -tb \epsilon\]

So phrased purely in terms of vector spaces we see our hom complex is (living in degrees $0$ and $-1$):

\[0 \to \langle b, d \rangle \overset{(-t \ 0)}{\longrightarrow} \langle s \rangle \to 0\]

So we compute

$H^0 = \text{Ker} ((-t \ 0))$
$H^{-1} = \langle s \rangle \big / \text{Im}((-t \ 0))$

When $t \neq 0$, our map is full rank so that $H^0$ are the pairs $(b,d)$ with $b=0$ – agreeing with the classical computation. Then $H^{-1} = 0$, so again we learn nothing new in the smooth fibres.

When $t=0$, however, our map is the $0$ map so that $H^0$ is the space of all pairs $(b,d)$ is two dimensional – again, agreeing with the classical computation! But now we see the $H^{-1}$ term, which is $1$ dimensional, spanned by $s$.

Again, in the derived world, we see the euler characteristic is constantly $1$ along the whole family!

There’s something a bit confusing here, since there seem to be two definitions of “homotopical smoothness”… On the one hand, in the noncommutative geometry literature, we say that a dga $A$ is “smooth” if it’s perfect as a bimodule over itself. On the other hand, though, I’ve also heard another notion of “homotopically smooth” where we say the cotangent complex is perfect. I guess it’s possible (likely?) that these should be closely related by some kind of HKR Theorem, but I don’t know the details.

Anyways, I’m confused because we just computed that the curve $y^2 = x^3$ has a perfect tangent complex, which naively would make me think its cotangent complex is also perfect. But this shouldn’t be the case, since I also remember reading that a classical scheme is smooth in the sense of noncommutative geometry if and only if it’s smooth classically, which $y^2 = x^3$ obviously isn’t!

Now that I’ve written these paragarphs and thought harder about things, I think I was too quick to move between perfectness of the tangent complex and perfectness of the cotangent complex, but I should probably compute the cotangent complex and the bimodule resolution to be sure…

Unfortunately, that will have to wait for another day! I’ve spent probably too many hours over the last few days writing this and my other posts on lie algebroids. I have some kind of annoying hall algebra computations that are calling my name, and I have an idea about a new family of model categories which might be of interest… But checking that something is a model category is usually hard, so I’ve been dragging my feet a little bit.

Plus, I need to start packing soon! I’m going to europe for a bunch of conferences in a row! First a noncommutative geometry summer school hosted by the institute formerly known as MSRI, then CT of course, and lastly a cute representation theory conference in Bonn.

I’m sure I’ll learn a bunch of stuff I’ll want to talk about, so we’ll chat soon ^_^. Take care, all!

In fact we know more! This $k$ is really $k[x,y] \big / (x=0,y=0)$, so we know the intersection point is $(0,0)$. ↩
Indeed after tensoring we get
\[k[x,y,m] \big / (y, y-mx) \overset{- \cdot m}{\longrightarrow} k[x,y,m] \big / (y, y-mx) \longrightarrow k[x,y,m] \big / (y, y-mx, m) \to 0\]
since here $k \cong k[m] \big / (m)$. But then we can simplify these to
\[k[x,m] \big / (mx) \overset{- \cdot m}{\longrightarrow} k[x,m] \big / (mx) \longrightarrow k[x] \to 0\]
and indeed the leftmost map (multiplication by $m$) is not injective! The kernel is generated by $x$. ↩
Again, if you’re more careful with where this ring comes from, rather than just its isomorphism class, it’s $k[x,y] \big / (x,y)$, the quotient by the maximal ideal $(x,y)$ which represents the origin. ↩
The only place it could possibly be $\langle 0, 0 \rangle$ is when $y=0$, but the points on our curve with this property satisfy $x^3-tx=y^2=0$ so that when $t \neq 0$ the solutions are $(x,y) = (0,0), \ (\sqrt{t}, 0), \ (-\sqrt{t}, 0)$. But at all three of these points $\langle -3x^2 - t, 2y \rangle \neq \langle 0, 0 \rangle$. ↩
This is a misconception that I used to have, and which basically everyone I’ve talked to had at one point. Remember that dg-maps are all graded maps! Not just those which commute with the differential!

The key point is that the differential on $\text{Hom}^\bullet(A,B)$ sends a degree $n$ map $f$ to
\[df = d_B \circ f - (-1)^n f \circ d_A\]
so that $df = 0$ if and only if $d_B f = (-1)^n f d_A$ if and only if $f$ commutes with the differential (in the appropriate graded sense).

This means that, for instance, $H^0$ of the hom complex recovers from all graded maps exactly $\text{Ker}(d) \big / \text{Im}(d)$, which are the maps commuting with $d$ modulo chain homotopy! ↩

Explicitly Computing The Action Lie Algebroid for $SL_2(\mathbb{R}) \curvearrowright \mathbb{R}^2$

Fri, 20 Jun 2025 00:00:00 +0000

This is going to be a very classic post, where we’ll chat about a computation my friend Shane did earlier today. His research is largely about symplectic lie algebroids, and recently we’ve been trying to understand the rich connections between poisson geometry, lie algebroids, lie groupoids, and eventually maybe fukaya categories of lie groupoids (following some ideas of Pascaleff). Shane knows much more about this stuff than I do, so earlier today he helped me compute a super concrete example. We got stuck at some interesting places along the way, and I think it’ll be fun to write this up, since I haven’t seen these kinds of examples written down in many places.

Let’s get started!

First, let’s recall what an action groupoid is. This is one of the main examples I have in my head for lie groupoids, which is why Shane and I started here.

If $G$ is a group acting on a set $X$, then we get a groupoid:

\[G \times X \rightrightarrows X\]

here we think of $X$ as the set of objects and $G \times X$ as the set of arrows, where

the source of $(g,x)$ is just $x$
the target of $(g,x)$ is $g \cdot x$, using the action of $G$ on $X$
the identity arrow at $x$ is $(1,x)$
if $(g,x) : x \to y$ and $(h,y) : y \to z$, then the composite is $(hg,x) : x \to z$
if $(g,x) : x \to y$ then its inverse is $(g^{-1},y) : y \to x$.

Action groupoids are interesting and important because they allow us to work with “stacky” nonhausdorff quotient spaces like orbifolds in a very fluent way. See, for instance, Moerdijk’s Orbifolds as Groupoids: An Introduction, which shows how you can easily define covering spaces, vector bundles, principal bundles, etc. on orbifolds using the framework of groupoids.

The point is that a groupoid $E \rightrightarrows X$ is a “proof relevant equivalence relation” in the sense that $E$ keeps track of all the “proofs” or “witnesses” that two points in $X$ are identified, rather than just the statement that two points are identified. Indeed, we think of $e \in E$ as a witness identifying $s(e)$ and $t(e)$ in $X$. Then reflexivity comes from the identity arrow, symmetry comes from the inverse, and transitivty comes from composition. The “proof relevance” is just the statement that there might be multiple elements $e_1$ and $e_2$ which both identify $x$ and $y$ (that is, $s(e_1) = s(e_2) = x$ and $t(e_1) = t(e_2) = y$). By keeping track of this extra information that “$x$ and $y$ are related in multiple ways” we’re able to work with a smooth object (in the sense that both $E$ and $X$ are smooth) instead of the nonsmooth, or even nonhausdorff quotient $X/E$.

Next, we know that a central part of the study of any lie group $G$ is its lie algebra $\mathfrak{g}$. This is a “linearization” of $G$ in the literal sense that it’s a vector space instead of a manifold, which makes it much easier to study. But through the lie bracket $[-,-]$ it remembers enough of the structure of $G$ to make it an indispensable tool for understanding $G$. See, for instance, Bump’s excellent book Lie Groups or Stillwell’s excellent Naive Lie Theory. With this in mind, just playing linguistic games we might ask if there’s a similarly central notion of “lie algebroid” attached to any “lie groupoid”. The answer turns out to be yes, but unlike the classical case, not every lie algebroid comes from a lie groupoid! We say that not every lie algebroid is integrable.

This, finally, brings us to the computation that Shane and I did together:

As explicitly as possible, let’s compute the lie algebroid coming from the action groupoid of $SL_2(\mathbb{R}) \curvearrowright \mathbb{R}^2$.

Let’s start with a few definitions coming from Crainic, Fernandes, and Mărcuț’s Lectures on Poisson Geometry.

A Lie Algebroid on a manifold $M$ is a vector bundle $A \to M$ whose space of global sections $\Gamma(A)$ has a lie bracket $[-,-]_A$, equipped with an anchor map $\rho : A \to TM$ to the tangent bundle of $M$ that’s compatible with the lie bracket in the sense that for $\alpha, \beta \in \Gamma(A)$ and $f \in C^\infty(M)$ we have

$[\alpha, f \cdot \beta]_A = f \cdot [\alpha,\beta]_A + (\rho(\alpha) f) \cdot \beta$

Then, given a lie groupoid $E \rightrightarrows M$ with source and target maps $s,t : E \to M$ and identity map $r : M \to E$, its lie algebroid is explicitly given by letting

$A = r^* \text{Ker}(dt)$, which is a vector bundle over $M$
$[-,-]_A$ coming from the usual bracket on $TE$ (since $\text{Ker}(dt)$ is a subbundle of $TE$)
$\rho : A \to TM$ given by $ds$ (which is a map $TE \to TM$, thus restricts to a map on $\text{Ker}(dt)$, and so gives a map on the pullback $A = r^* \text{Ker}(dt)$)

If this doesn’t make perfect sense, that’s totally fine! It didn’t make sense to me either, which is why I wanted to work through an example slowly with Shane. For us the action groupoid is

\[SL_2(\mathbb{R}) \times \mathbb{R}^2 \rightrightarrows \mathbb{R}^2\]

where

\[s (M,v) = Mv\] \[t (M,v) = v\] \[r (v) = (\text{Id}, v)\] \[(M_2, M_1 v) \circ (M_1, v) = (M_2 M_1, v)\]

First let’s make sense of $A = r^* \text{Ker}(dt)$.

We know that $dt : T(SL_2(\mathbb{R}) \times \mathbb{R}^2) \to T\mathbb{R}^2$, that is, $dt : TSL_2(\mathbb{R}) \times T\mathbb{R}^2 \to \mathbb{R}2$, and is the derivative of the map $t : (M,v) \mapsto v$. If we perturb a particular point $(M_0, v_0)$ to first order, say $(M_0 + \delta M, v_0 + \delta v)$, then projecting gives $v_0 + \delta v$, which gives $\delta v$ to first order. So the kernel of this map are all the pairs $(\delta M, \delta v)$ with $\delta v = 0$, and we learn

\[\text{Ker}(dt) = TSL_2(\mathbb{R}) \times \mathbb{R}^2\]

where we’re identifying the zero section of $T\mathbb{R}^2$ with $\mathbb{R}^2$ itself.

Now to get $A$ we’re supposed to apply $r^*$ to this.

By definition, this means the fibre of $r^* \text{Ker}(dt)$ above the point $v$ is supposed to be the fibre of $\text{Ker}(dt)$ over $r(v) = (\text{id},v)$. But this fibre is $T_{\text{id}}SL_2(\mathbb{R}) \times \{v\}$, so that the pullback is a trivial bundle with fibre $\mathfrak{sl}_2(\mathbb{R})$:

\[A = \mathfrak{sl}_2(\mathbb{R}) \times \mathbb{R}^2\]

viewed as a trivial fibre bundle over $\mathbb{R}^2$. (Recall that the lie algebra $\mathfrak{sl}_2(\mathbb{R})$ is defined to be the tangent space of $SL_2(\mathbb{R})$ at the identity).

Next we want to compute the bracket $[-,-]_A$. Thankfully this is easy, since it’s the restriction of the bracket on $TSL_2(\mathbb{R}) \times T\mathbb{R}^2$. Of course, an element of $A$ comes from $T_\text{id}SL_2(\mathbb{R}) = \mathfrak{sl}_2(\mathbb{R})$ and the zero section of $T\mathbb{R}^2$, so we get the usual lie bracket on $\mathfrak{sl}_2(\mathbb{R})$ in the first component and the restriction of the bracket on $T\mathbb{R}^2$ to $\{0\}$ in the second slot. This zero section piece isn’t doing anything interesting, and so after identifying this bundle with the trivial bundle $\mathfrak{sl}_2(\mathbb{R}) \times \mathbb{R}^2$ we see the bracket is just the usual bracket on $\mathfrak{sl}_2(\mathbb{R})$ taken fibrewise.

Lastly, we want to compute the anchor map. This caused me and Shane a bit of trouble, since we need to compute $ds$ at the identity, and we realized neither of us knew a general approach for computing a chart for $SL_2(\mathbb{R})$ near $\text{id}$!

In hindsight for a $k$ dimensional submanifold of $\mathbb{R}^n$ the idea is obvious: Just project onto some well-chosen $k$-subset of the usual coordinate directions! I especially wish it were easier to find some examples of this by googling, so I’m breaking it off into a sister blog post which will hopefully show up earlier in search results than this one will.

The punchline for us was that $SL_2(\mathbb{R})$ is defined to be

\[\left \{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathbb{R}^4 \ \middle | \ ad-bc = 1 \right \}\]

So a chart near a point $(a_0, b_0, c_0, d_0)$ can be computed by looking at the jacobian of $f : \mathbb{R}^4 \to \mathbb{R}$ with $f(a,b,c,d) = ad-bc$ evaluated at the matrix of interest $(a_0, b_0, c_0, d_0)$. Since $1$ is a regular value for $f$, the jacobian will have at least one nonzero entry, and by locally inverting that coordinate we’ll get our desired chart!

Since we want a chart near the identity, we compute the jacobian of $ad-bc$ at the identity to be

\[\begin{aligned} \left . J f \right |_{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}} &= \left . \langle d, -c, -b, a \rangle \right |_{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}} \\ &= \langle 1, 0, 0, 1 \rangle \end{aligned}\]

We see that $\frac{\partial f}{\partial a} \neq 0$, so that locally near $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ the manifold looks like

\[\left \{ \begin{pmatrix} a & b \\ c & \frac{1 + bc}{a} \end{pmatrix} \right \} \cong_\text{diffeo} \{ (a,b,c) \mid a \neq 0 \} \subseteq \mathbb{R}^3\]

and this is our desired local chart!

Now… Why were we doing this? We wanted to compute the anchor map from $A = r^* \text{Ker}(dt)$ to the tangent bundle $T\mathbb{R}^2$. This is supposed to be $ds$ (restricted to this subbundle).

So how can we compute this? In the main body, I’ll make some identifications that make the presentation cleaner, and still show what’s going on. If you want a very very explicit version of this computation, take a look at this footnote¹.

Well $s : SL_2(\mathbb{R}) \times \mathbb{R}^2 \to \mathbb{R}^2$ is the map sending $(M,v) \mapsto Mv$.

Explicitly, if we fix an $(x,y)$, this is the map

\[\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto (ax+by, cx+dy)\]

From the previous discussion, we can write this as a map in local charts $\{(a,b,c) \in \mathbb{R}^3 \mid a \neq 0 \} \to \mathbb{R}^2$ given by

\[s : (a,b,c) \mapsto \left ( ax+by, cx+ \frac{1+bc}{a}y \right )\]

and now it’s very easy to compute $ds$. It’s just the jacobian

\[\begin{pmatrix} x & y & 0 \\ \frac{-(1+bc)}{a^2}y & \frac{c}{a} y & x + \frac{b}{a} y \end{pmatrix}\]

but we only care about the value at the identity, since $A$ comes from pulling back this bundle along $r : v \mapsto (\text{id},v)$. So evaluating at $(a,b,c) = (1,0,0)$ we find our anchor map is

\[\begin{pmatrix} x & y & 0 \\ -y & 0 & x \end{pmatrix}\]

Moreover, by differentiating $\begin{pmatrix} a & b \\ c & \frac{1+bc}{a} \end{pmatrix}$ in the $a$, $b$, and $c$ directions and evaluating at $(1,0,0)$ we see that the basis $(\partial_a, \partial_b, \partial_c)$ for the tangent space to $(1,0,0)$ in our chart gets carried to the following basis for the tangent space at the identity matrix in $SL_2(\mathbb{R})$:

\[\partial_a \mapsto \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \quad \quad \partial_b \mapsto \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \quad \quad \partial_c \mapsto \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\]

which we recognize as $H$, $E$, and $F$ respectively.

But this is great! Now we know that the lie algebroid of the action groupoid of $SL_2(\mathbb{R}) \curvearrowright \mathbb{R}^2$ is given by the trivial bundle $\mathfrak{sl}_2(\mathbb{R}) \times \mathbb{R}^2 \to \mathbb{R}^2$ with the usual bracket on $\mathfrak{sl}_2$ taken fibrewise, and the anchor map $\rho : \mathfrak{sl}_2 \times \mathbb{R}^2 \to T\mathbb{R}^2$ sending (in the fibre over $(x,y)$)

$\rho_{(x,y)}(H) = (x, -y)$
$\rho_{(x,y)}(E) = (y, 0)$
$\rho_{(x,y)}(F) = (0, x)$

(which is the standard representation of $\mathfrak{sl}_2(\mathbb{R})$ on $\mathbb{R}^2$, viewed in a kind of bundle-y way.)

Thanks for hanging out, all! It was fun to go back to my roots and write a post that’s “just” a computation. This felt tricky while Shane and I were doing it together, but writing it up now it’s starting to feel a lot simpler. There’s still some details I don’t totally understand, which I think will be cleared up by just doing more computations like this, haha.

Also, sorry for not spending a lot of time motivating lie algebroids or actually doing something with the result of this computation… I actually don’t totally know what we can do with lie algebroids either! This was just a fun computation I did with a friend, trusting that he has good reasons to care. I’ve been meaning to pester him into guest-writing a blog post (or very confidently holding my hand while I write the blog post) about lie algebroids and why you should care. As I understand it, they give you tools for studying PDEs on manifolds which have certain mild singularities. This is super interesting, and obviously useful, and so I’d love to spend the time to better understand what’s going on.

Stay safe, and if you’re anywhere like Riverside try to stay cool!

If you want to be super duper explicit, our function $s$ sends $SL_2(\mathbb{R}) \times \mathbb{R}^2 \to \mathbb{R}^2$, which in a chart around the identity looks like the function
\[\{(a,b,c) \in \mathbb{R}^3 \mid a \neq 0 \} \times \mathbb{R}^2 \to \mathbb{R}^2\]
given by
\[(a,b,c,x,y) \mapsto \left ( ax+by, cx+ \frac{1+bc}{a}y \right )\]
now we differentiate to get $ds : TSL_2(\mathbb{R}) \times T\mathbb{R}^2 \to T\mathbb{R}^2$ sending $(a,b,c,\delta a,\delta b,\delta c, x, y, \delta x, \delta y)$ to $(ax+by, cx+\frac{1+bc}{a}y, ?, ?)$

Where the two $?$s are the output of matrix multiplication against the jacobian of $s$:
\[\begin{pmatrix} x & y & 0 & a & b \\ \frac{-(1+bc)}{a^2}y & \frac{c}{a} y & x + \frac{b}{a} y & c & \frac{1+bc}{a} \end{pmatrix} \begin{pmatrix} \delta a \\ \delta b \\ \delta c \\ \delta x \\ \delta y \end{pmatrix}\]
Then we’re supposed to restrict this to $\text{Ker}(dt)$, which are the points $(a,b,c, \delta a, \delta b, \delta c, x, y, 0, 0)$. Since $\delta x = \delta y = 0$, we don’t even bother writing those entries of the matrix, and that’s how we get $ds$ as written in the main body.

Now, as in the main body, we pull this bundle back along $r : v \mapsto (\text{id},v)$, which in our chart is $(x,y) \mapsto (1,0,0,x,y)$ so that our bundle $A$ (with its structure map to $\mathbb{R}^2$) is
\[A = \{(1,0,0, \delta a, \delta b, \delta c, x, y, 0, 0)\} \to \{(x,y)\} = \mathbb{R}^2\]
which, in the main text, we identify with $\mathfrak{sl}_2 \times \mathbb{R}^2 = \{(\delta a, \delta b, \delta c, x, y)\} \to \{(x,y)\} = \mathbb{R}^2$

so we learn that our anchor map $ds$ is the restriction of the above map $ds$ to this pulled back subbundle, and sends
\[(\delta a, \delta b, \delta c, x, y) \mapsto (x,y,?,?)\]
where, again the $?$s are the result of the matrix multiplication
\[\begin{pmatrix} x & y & 0 \\ -y & 0 & x \end{pmatrix} \begin{pmatrix} \delta a \\ \delta b \\ \delta c \end{pmatrix}\]
which brings us back to the result of the main body.

Of course, most working differential geometers wouldn’t write out this much detail to do this computation! I think it might be helpful to some newcomers to the field, and I certainly found it clarifying to write down exactly what happened, even if Shane and I weren’t nearly this careful when we were doing this together at a whiteboard. ↩

How to Explicitly Compute Charts for a Levelset Submanifold

Fri, 20 Jun 2025 00:00:00 +0000

While doing a computation with my friend Shane the other day, we realized we needed to explicitly compute a local chart near the identity of $SL_2(\mathbb{R})$. It took us longer than I’d like to admit to figure out how to do this (especially since it’s so geometrically obvious in hindsight), and so I want to write down the process for future grad students looking to just do a computation! If you want to see what Shane and I were actually interested in, you can check out the main post here.

Ok, let’s hop right in! Say you have a $2$-manifold in $\mathbb{R}^3$ to start¹:

If we think of the purple manifold $M$ as being an open disk, representing a small neighborhood of some possibly larger 2-manifold, then we can see the projection onto the $xy$-plane is a diffeomorphism onto its image (an open disk in $\mathbb{R}^2$) while the projections onto the $yz$ and $xz$ planes are not open in $\mathbb{R}^2$! This is because the normal to $M$ is parallel to the $z$ axis inside $M$, (indeed, at the “top of the hill”) so the tangent plane at that point degenerates and projects to a line whenever we project onto a coordinate plane containing the $z$-direction.

For a more computational example, let’s try the hyperboloid $x^2 + y^2 - z^2 = 1$, and let’s see what happens near a few points.

The tangent plane at a point is controlled by the jacobian of the defining equation, which for us is $\langle 2x, 2y, -2z \rangle$.

This gives us three (disconnected) charts: $\{2x \neq 0\}$, $\{2y \neq 0\}$, and $\{-2z \neq 0\}$, which we can see visually here (and we also drop the unnecessary scalars):

These turn into 6 honest-to-goodness charts where we turn the disconnected condition $\{x \neq 0\}$ into the pair of connected conditions $\{x \gt 0\}$ and $\{x \lt 0\}$. Indeed it’s easy to see that the 6 connected components in the above pictures are all diffeomorphic to an open subset of $\mathbb{R}^2$, and we can see this algebraically by projecting onto the plane avoiding the nonzero coordinate.

On $\{x \gt 0 \}$, for example, we have an open set of the $yz$-plane, shown here in orange:

Algebraically, we compute this chart by noting on $\{x \gt 0 \}$, we can solve for $x$ and (using the positive square root, since $x \gt 0$) write our surface locally as

\[\{ (+\sqrt{1+z^2-y^2}, y, z) \mid z^2 - y^2 \gt -1 \}\]

which is diffeomorphic in the obvious way to its projection onto the $yz$-plane

\[\{ (y,z) \mid z^2 - y^2 \gt -1 \}\]

so this is one of our charts!

Similarly, we can look at $\{z \gt 0\}$, solve for $z$ and locally write our surface as

\[\{ (x,y,+\sqrt{x^2+y^2-1}) \mid x^2 + y^2 \gt 1 \}\]

which is diffeomorphic to $\{ (x,y) \mid x^2 + y^2 \gt 1 \}$ – another chart.

On the intersection of these charts, $\{x, z \gt 0 \}$, it’s now easy to write down our transition maps (if one is so inclined):

Here our charts are the diffeomorphisms

\[\{(+\sqrt{1+z^2-y^2},y,z) \mid z^2-y^2 \gt -1, \ z \gt 0 \} \to \{(y,z) \mid z^2-y^2 \gt -1, \ z \gt 0 \}\] \[\{(x,y,+\sqrt{x^2+y^2-1}) \mid x^2+y^2 \gt 1, \ x \gt 0 \} \to \{(x,y) \mid x^2+y^2 \gt 1, \ x \gt 0 \}\]

so it’s easy to compose them to see our transition maps between these charts are

\[(y,z) \mapsto (+\sqrt{1+z^2-y^2},y) : \{(y,z) \mid z^2-y^2 \gt -1, \ z \gt 0 \} \to \{(x,y) \mid x^2+y^2 \gt 1, \ x \gt 0 \}\] \[(x,y) \mapsto (y,+\sqrt{x^2+y^2-1}) : \{(x,y) \mid x^2+y^2 \gt 1, \ x \gt 0 \} \to \{(y,z) \mid z^2-y^2 \gt -1, \ z \gt 0 \}\]

As a (fun?) exercise, compute the $\{y \gt 0 \}$ chart, and the other two transition maps.

For another example, let’s take a look at $SL_2(\mathbb{R})$, which is defined to be $\left \{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mid ad-bc = 1 \right \} \subseteq \mathbb{R}^4$.

Then the jacobian of our defining map is $\langle d, -c, -b, a \rangle$, and we get charts corresponding to $\{d \neq 0 \}$, $\{-c \neq 0 \}$, $\{ -b \neq 0 \}$, and $\{ a \neq 0 \}$.

In the $\{d \neq 0\}$ chart, for instance, our defining equation looks like $a = \frac{1+bc}{d}$, so that $SL_2(\mathbb{R})$ looks locally like

\[\left \{ \begin{pmatrix} \frac{1+bc}{d} & b \\ c & d \end{pmatrix} \ \middle | \ d \neq 0 \right \} \cong_\text{diffeo} \{ (b,c,d) \in \mathbb{R}^3 \mid d \neq 0 \}\]

In the main post you can see how my friend Shane and I used this to compute the anchor map for a certain lie algebroid.

Again, it makes a nice exercise to explicitly write out the various charts and transition maps

What about a codimension 2 example?

Let’s go back to our happy little hyperboloid, and intersect it with the surface $xyz = 1$. That is, we want to consider the manifold

\[\left \{ (x,y,z) \ \middle | \ \begin{array}{c} x^2 + y^2 - z^2 = 1 \\ xyz = 1 \end{array} \right \}\]

This is the levelset of the map $\mathbb{R}^3 \to \mathbb{R}^2$ sending $(x,y,z) \mapsto (x^2 + y^2 - z^2, \ xyz)$ taking value $(1,1)$. So we compute the jacobian

\[\begin{pmatrix} 2x & 2y & -2z \\ yz & xz & xy \end{pmatrix}\]

and our charts are all the ways this matrix can have full rank. These conditions are:

$2x \neq 0$ and $xz \neq 0$
$2x \neq 0$ and $xy \neq 0$
$2y \neq 0$ and $yz \neq 0$
$2y \neq 0$ and $xy \neq 0$
$-2z \neq 0$ and $yz \neq 0$
$-2z \neq 0$ and $xz \neq 0$

If we look at the $\{2x \neq 0, \ xz \neq 0\}$ chart, we can ask sage to solve for $x$ and $z$ as functions of $y$:

So as in the previous hyperboloid example, we need to break this into four charts, depending on whether $x$ and $z$ are positive or negative.

Following the sage computation, in the $\{x \gt 0, z \gt 0\}$ chart, we can write our curve as

\[\left ( \sqrt{\frac{1}{2}} \sqrt{-\frac{y^{3} - y - \sqrt{y^{6} - 2 \, y^{4} + y^{2} - 4}}{y}}, \ y, \ \frac{2 \, \sqrt{\frac{1}{2}}}{y \sqrt{-\frac{y^{3} - y - \sqrt{y^{6} - 2 \, y^{4} + y^{2} - 4}}{y}}} \right )\]

which, by projecting out the $y$ coordinate, is diffeomorphic to the open subset of $\mathbb{R}$ where all these square roots are defined.

Ok, thanks for reading, all! This was extremely instructive for me, and hopefully it’s helpful to some of you as well! Sometimes it’s nice to just do some computations. Talk soon!

That’s right. I finally bought an ipad. This has already dramatically improved my paper-reading experience (I love you Zotero) and my remote teaching experience, and I’m so glad that the example-drawing experience is shaping up to be everything I wanted it to be as well! ↩

A Proof that there's No Constructive Proof of the Intermediate Value Theorem

Tue, 10 Jun 2025 00:00:00 +0000

The other day my friend Lucas Salim was asking me some questions about categorical logic and constructive math, and he mentioned he’d never seen a proof that there’s no constructive proof of the intermediate value theorem before. I showed him the usual counterexample, and since my recent blog post about choice was so quick to write I decided to quickly write up a post about this too, since I remember being confused by it back when I was first learning it.

The key fact is Soundness and Completeness of the topos semantics of constructive logic. This says that there is a way of interpreting the usual syntax of mathematics into a topos in such a way that

(Soundness) If you can constructively prove a statement, then its interpretation in every topos is true
(Completeness) If a statement is interpreted as true in every (elementary¹) topos, then there must exist a constructive proof

For a proof, see Chapter II of Lambek and Scott’s Introduction to Higher Order Categorical Logic or Section D4.3 of The Elephant.

This means that as long as we’re careful to avoid choice and excluded middle, anything we prove will be true when interpreted in any topos we like! Then there’s a mechanical procedure that lets us convert this interpretation into a corresponding statement in the “real world²”, and this gives us lots of “theorems for free” for each individual constructive theorem! A nice case study is given by the Weierstrass Approximation Theorem, which I gave a talk on years ago³. Since this theorem is constructively provable⁴…

by interpreting it in the effective topos we learn there’s a computer program $\mathtt{Approx}$ which takes as input a function $f : [0,1] \to \mathbb{R}$⁵ and an $\epsilon \gt 0$ and outputs the coefficients of a polynomial approximating $f$.
by interpreting it in a sheaf topos $\text{Sh}(\Theta)$ we learn that for any continuous family of functions $f_\theta(x) : \Theta \times [0,1] \to \mathbb{R}$, there’s locally⁶ a polynomial $p_\theta(x)$ whose coefficients are continuous functions of $\theta$ which approximates each $f_\theta(x)$
etc.

If you’re interested in learning how to externalize a statement in a topos to a statement about the real world, I highly recommend Ingo Blechschmidt’s excellent paper Exploring mathematical objects from custom-tailored mathematical universes which gives a high level overview of the topic, while still giving enough details to let you externalize a few statements of your own!

But where were we? The usual proofs of the intermediate value theorem aren’t constructive. See, for instance, Bauer’s Five Stages of Accepting Constructive Mathematics or Section 1 of Taylor’s A Lambda Calculus for Real Analysis for a discussion of how some common proofs fail (as well as great lists of constructively provable alternatives). Since the usual proofs seem to fail, we might guess that IVT is not provable constructively… but how could we prove this?

Say, towards a contradiction, that there were a constructive proof of the intermediate value theorem. Then it would be true in every topos, and thus its various externalizations would all be true in the real world. So to show that there isn’t a constructive proof, all we have to do is find a topos which doesn’t think it’s true!

Following many who came before me⁷, we’re going to use the topos of sheaves on $(-1,1)$. It’s been a minute since we’ve externalized a statement together, so let’s do it now!

In full symbolic glory⁸, the IVT says

\[\forall f : \mathbb{R} \to \mathbb{R} . \forall a,b \in \mathbb{R} . (a \lt b \land f(a) \lt 0 \land f(b) \gt 0) \to (\exists x \in \mathbb{R} . a \lt x \lt b \land f(x) = 0)\]

So now using the forcing language for $\text{Sh}((-1,1))$ (check out Theorem $1$ in Chapter VI.7 of Mac Lane and Moerdijk’s Sheaves in Geometry and Logic if you’re not sure what this means), we compute:

\[1 \Vdash \forall f : \mathbb{R} \to \mathbb{R} . \forall a,b \in \mathbb{R} . (a \lt b \land f(a) \lt 0 \land f(b) \gt 0) \to (\exists x \in \mathbb{R} . a \lt x \lt b \land f(x) = 0)\]

We cash out the universal quantifiers to get “for every open $U \subseteq (-1,1)$, and for every continuous $f : U \times \mathbb{R} \to \mathbb{R}$, $a : U \to \mathbb{R}$, $b : U \to \mathbb{R}$, we have…”

\[U \Vdash (a \lt b \land f(a) \lt 0 \land f(b) \gt 0) \to (\exists x \in \mathbb{R} . a \lt x \lt b \land f(x) = 0)\]

Next, cashing out the implication gives “for every open $U \subseteq (-1,1)$, and for every continuous $f : U \times \mathbb{R} \to \mathbb{R}$, $a : U \to \mathbb{R}$, $b : U \to \mathbb{R}$, so that for all $t \in U$ we know $a(t) \lt b(t)$, $f(t,a(t)) \lt 0$, and $f(t,b(t)) \gt 0$, we have…”

\[U \Vdash \exists x \in \mathbb{R} . a \lt x \lt b \land f(x) = 0\]

Finally, cashing out the existential quantifer and the stuff inside it we get the external statement:

The IVT is true inside $\text{Sh}((-1,1))$ if and only if the following is true externally:

For every open $U \subseteq (-1,1)$
for every continuous $f : U \times \mathbb{R} \to \mathbb{R}$, $a : U \to \mathbb{R}$, $b : U \to \mathbb{R}$
so that for all $t \in U$ we know $a(t) \lt b(t)$, $f(t,a(t)) \lt 0$, and $f(t,b(t)) \gt 0$
there is an open cover $\{U_\alpha\}$ covering $U$ with continuous functions $x_\alpha : U_\alpha \to \mathbb{R}$ so that
for all $t \in U_\alpha$, $a(t) \lt x_\alpha(t) \lt b(t)$ and $f(t,x_\alpha(t)) = 0$.

What a mouthful!

Of course, we’re trying to prove this fails, so all we have to do is find an open set $U$ and functions $f$, $a$, and $b$ satisfying the assumptions so that the conclusion fails. We’ll choose $U = (-1,1)$ to be the whole set, $a(t) = -2$ and $b(t) = 2$ to be constant functions, and $f(t,x) : (-1,1) \times \mathbb{R} \to \mathbb{R}$ to be

\[f(t,x) = \max(\min(t+x+1, t), t+x-1)\]

Then we see that, indeed, $f(t,a) \lt 0$ and $f(t,b) \gt 0$ for all $t \in (-1,1)$, so to prove the IVT fails in this topos we just need to show there’s no open cover on which $x(t)$ with $f(t,x(t)) = 0$ can be chosen continuously.

The idea is that no matter how hard we try, $x(t)$ cannot be continuous in a neighborhood of $t=0$. Indeed, here’s an animation showing how $x(t)$ changes as we change $t$:

You can see that when $t=0$ the root $x(t)$ jumps between $\pm 1$!

Indeed, if we plot $x(t)$ we get:

and it’s obvious that this is not the graph of a continuous function in any neighborhood of $t=0$.

As long as we’re showing pretty graphics, you can also visualize this whole function $f$ as a surface over the strip $(-1,1) \times \mathbb{R}$. Then choosing a $t$ amounts to choosing a “slice” of the surface, and we can see that where that slice intersects the $(t,x)$-plane jumps suddenly as we cross $t=0$. In this example the axes are labeled $x$ and $y$ rather than $x$ and $t$:

So where did we start, and where did we end? If the IVT were constructively provable, it would be true inside $\mathsf{Sh}((-1,1))$ and thus for our $f(t,x)$ we could find an open cover on which the zero $x(t)$ varies continuously in $t$. But this can’t possibly happen in a neighborhood of $t=0$, so we learn there is no constructive proof!

Buuuuut, all is not lost! Usually classical theorems do have constructive analogues, either by adding new assumptions, weakening the conclusion, or by finding a different statement of the theorem that’s more positive. Andrej Bauer’s paper Five Stages of Accepting Constructive Mathematics lists many possibilities.

For instance, one way to weaken the conclusion is to prove that for any $\epsilon$ you like, there’s an $x$ with $|f(x)| \lt \epsilon$. In our example, if we plot those $x$ so that $|f(t,x)| \lt \epsilon$ we get

and it’s easy to fit the graph of a continuous selection function $x(t)$ inside this thickened region.

Another approach is to recognize that the problem comes from $f$ “hovering” at $0$ when $t=0$. If we forbid this hovering, for instance by assuming $f$ is strictly monotone, then we can constructively prove the IVT (See Bauer’s Five Stages paper again).

There’s yet another version, coming from Abstract Stone Duality, where we say that whenever $f(a) \lt 0 \lt f(b)$, the compact subspace $Z_f = \{ x \in [a,b] \mid f(x) = 0 \}$ is occupied (Cor 13.11 in A Lambda Calculus for Real Analysis). This is a condition that’s weaker than inhabited but stronger than nonempty, which you can read about in Section 8 of the same paper. I don’t understand this condition very well, because I haven’t spent as much time thinking about ASD as I would like. Hopefully sometime soon I’ll find some time to work through some examples!

Edit (July 7, 2025):

Jim Kingdon recently started a thread in the metamath github talking about this, and told me about it over mastodon. In this thread, Mario Carneiro gave a slick proof that the IVT implies (analytic) LLPO, which should feel familiar. Recall that Analytic LLPO is the statement that $\forall t \in \mathbb{R} . t \geq 0 \lor t \leq 0$.

$\ulcorner$ Fix $t \in \mathbb{R}$, and again consider the function $f_t(x) : \mathbb{R} \to \mathbb{R}$ by

\[f_t(x) = \max(\min(t+x+t,t), t+x-1)\]

Then by the IVT, $f_t(x)$ has a zero $f_t(z) = 0$, and constructively we know that $z \lt \frac{1}{2} \lor z \gt \frac{-1}{2}$. But now if $z \lt \frac{1}{2}$ then $t \leq 0$ and if $z \gt \frac{-1}{2}$ then $t \geq 0$, giving the claim. $\lrcorner$

Mario also mentions that LLPO and Countable Choice is enough to prove IVT, so that in any settings where CC holds LLPO and IVT are equivalent⁹. Indeed

$\ulcorner$ Say $f(a) \lt 0$ and $f(b) \gt 0$.

For every $r \in \mathbb{Q}$, decide if $f(a + r(b-a))$ is $\geq 0$ or $\leq 0$. For each $r$ this is an application of LLPO, but there might be two options (if $f$ outputs a value that is both $\geq 0$ and $\leq 0$), so we have to to choose one of these for each of our countably many $r \in \mathbb{Q}$. So we’ve used both LLPO and CC to do this.

Now we can do binary search in $[a,b]$ to find a zero. The midpoints $m$ we check will always be of the form $a + r(b-a)$ for some $r \in \mathbb{Q}$, so to split our interval in half we can use our pre-made choices – recurring into the upper half of the interval if $f(m) \leq 0$ or the lower half if $f(m) \geq 0$.

Then the sequence of midpoints for our intervals is a cauchy sequence whose modulus of convergence we can compute, since after the first $n$ bisections all the future midpoints lie in an interval of width $2^{-n}(b-a)$. Since the dedekind reals are complete for these kinds of explicit cauchy sequences (see here, for instance), these converge to a real number $z$. As usual, continuity of $f$ implies that $f(z) = 0$. $\lrcorner$

Ok, thanks for reading, all! It’s nice to get a few quick posts up while I’m working on some longer stuff. I’m still thinking a lot about a cool circle of ideas involving Fukaya Categories, Skein Theory and T(Q)FTs, and Hall Algebras, and I’m slowly making progress on writing posts about all these fun things.

Now, though, I have to go run a review session for a calculus class, haha. I’ll try to resist telling them about the fascinating subtleties that show up when you try to do everything constructively.

Stay safe, and we’ll talk soon 💖

Usually on this blog when I talk about topoi I mean grothendieck topoi, but for this completeness result we really do need to allow more general elementary topoi with NNO. Indeed there are statements true in all grothendieck topoi that are not constructively provable (since they fail in some elementary topos). See here for a partial list.

I would actually love to know if there’s a reference for what one has to add to IHOL to get something sound and complete for grothendieck topoi… I spent some time looking, but the only thing I found was Topological Completeness for Higher-Order Logic by Awodey and Butz, but it seems like they use $1+1$ in place of $\Omega$, so this isn’t the usual interpretation of logic in a grothendieck topos (which is also where the classical completeness comes from). ↩
Maybe “base topos” would be less philosophically charged ↩
I was really fumbling around with topos theory back then, haha. I’m much more confident now, and in the last few years I’ve just worked through a lot more examples and done more computations and read more papers and generally just learned a lot. Rereading that post was surprisingly… nostalgic isn’t the right word… but it’s fun to see how much I’ve grown! ↩
The usual proof with bernstein polynomials works if we’re careful to check some constructively relevant details. I’ll copy it here in case the wikipedia article changes someday:

Fix $\epsilon > 0$.

We write $b_{k,n}(x) = \binom{n}{k} x^k (1-x)^{n-k}$, and note that:
1. $\sum_{k=0}^n b_{k,n} = 1$
2. $\sum_{k=0}^n \left ( x - \frac{k}{n} \right )^2 b_{k,n} = \frac{x(1-x)}{n}$
These are all provable by just expanding the left hand side, which is constructive.

We also fix a $\delta$ so that whenever $|x-y| \lt \delta$ we have $|f(x) - f(y)| \lt \epsilon$. This is because, constructively, every continuous function on a compact sublocale of $\mathbb{R}$ is uniformly continuous. Note that here we crucially need to be working with locales! (see, eg, Thm 10.7 in Taylor’s A Lambda Calculus for Real Analysis)

Lastly, we fix $M$ an upper bound for $\lvert f \rvert$. This is possible since $[0,1]$ is compact, overt, and inhabited (see Rmk 10.4 in Bauer and Taylor’s The Dedekind Reals in Abstract Stone Duality) thus the continuous $\lvert f \rvert$ admits a maximum (Thm 12.9 in A Lambda Calculus for Real Analysis).

Now let $B_n f = \sum_{k=0}^n f(k/n) b_{k,n}$. We compute:
\[\begin{aligned} \lvert f(x) - (B_n f)(x) \rvert &\overset{(a)}{=} \lvert \sum_{k=0}^n (f(x) - f(k/n)) b_{k,n}(x) \rvert \\ &\leq \sum_{k=0}^n \lvert f(x) - f(k/n) \rvert b_{k,n}(x) \\ &\overset{(b)}{\leq} \sum_{k \text{ s.t. } |x-k/n| \lt \delta} \lvert f(x) - f(k/n) \rvert b_{k,n}(x) + \sum_{k \text{ s.t. } |x-k/n| \gt \frac{1}{2}\delta} \lvert f(x) - f(k/n) \rvert b_{k,n}(x) \end{aligned}\]
In step (a) we use (1), and in step (b) we use the fact that $\forall x \in \mathbb{R} . x \lt \delta \lor x \gt \frac{\delta}{2}$ is constructively true (since the intervals overlap we don’t need excluded middle here!)

But we can bound the first sum by noticing in this region $|x-k/n| \lt \epsilon$ so that (using (1) again)
\[\begin{aligned} \sum_{k \text{ s.t. } |x-k/n| \lt \delta} \lvert f(x) - f(k/n) \rvert b_{k,n}(x) &\leq \sum_{k \text{ s.t. } |x-k/n| \lt \delta} \epsilon b_{k,n}(x) \\ &\leq \sum_{k=0}^n \epsilon b_{k,n}(x) \\ &= \epsilon \end{aligned}\]
And since $\lvert f(x) - f(k/n)\rvert \leq \lvert f(x) \rvert + \lvert f(k/n) \rvert \leq 2M$, we compute:
\[\begin{aligned} \sum_{k \text{ s.t. } |x-k/n| \gt \frac{1}{2}\delta} \lvert f(x) - f(k/n) \rvert b_{k,n}(x) &\leq \sum_{k \text{ s.t. } |x-k/n| \gt \frac{1}{2}\delta} 2M b_{k,n}(x) \\ &\overset{(c)}{\leq} 2M \sum_{k \text{ s.t. } |x-k/n| \gt \frac{1}{2}\delta} \left ( \frac{\delta}{2} \right )^{-2} \left ( x - \frac{k}{k} \right )^2 b_{k,n}(x) \\ &\leq \frac{8M}{\delta^2} \sum_{k=0}^n \left ( x - \frac{k}{k} \right )^2 b_{k,n}(x) \\ &\overset{(d)}{=} \frac{8M}{\delta^2} \frac{x(1-x)}{n} \\ &\leq \frac{8M}{\delta^2} \frac{1}{4n} \end{aligned}\]
In step (c) we use $|x - k/n| \gt \frac{\delta}{2}$ to say that $\left ( \frac{\delta}{2} \right )^{-2} \left ( x - \frac{k}{k} \right )^2 \geq 1$, so that we can multiply through by it and make our sum bigger. In step (d) we use (2), and at the end we use the fact that $x(1-x) \leq \frac{1}{4}$ on $[0,1]$.

Now since $\mathbb{R}$ is constructively archimedian (Defn. 1.1 in The Dedekind Reals in Abstract Stone Duality) we see $\exists n \in \mathbb{N} . \frac{2M}{\delta^2n} \lt \epsilon$.

Since these bounds were uniform in $x$, we learn that
\[\forall \epsilon \gt 0 . \exists n . \lVert f - B_n f \rVert_\infty \lt \epsilon\]
as desired. ↩
A real number $x$ in this topos is a program that eats a natural number $n$ and outputs a rational number $x(n)$. We think about this as a sequence of rational approximations $x(n)$ converging to some real number $x$. So a function $[0,1] \to \mathbb{R}$ in this topos is a program that takes as input a program $x$ (outputting rational approximations between $0$ and $1$) and outputs a new program $f(x)$ which outputs rational approximations. ↩
Because our statement of Weierstrass $\forall \epsilon . \forall f . \exists p . \lVert f - p \rVert_\infty \lt \epsilon$ includes an existential quantifier there’s no way to get around the fact that the $p$ we build is only defined locally on an open cover.

If you were more careful and gave a type theoretic proof that $\prod_\epsilon \prod_f \sum_p \lVert f - p \rVert_\infty \lt \epsilon$ then you could take $p$ to be a single polynomial defined on the whole of $\Theta$… I haven’t thought very hard about how possible this is (mainly because I haven’t spent much time thinking about what theorems about locales are provable in type theory), but I’m sure a talented undergraduate could figure it out. ↩
The earliest version of this example which I’ve seen comes from Stout’s Topological Properties of the Real Numbers Object in a Topos back in 1976. I think basically every other paper I’ve cited in this post gives some version of this example too, so it’s very well trodden ground! ↩
Well, not FULL glory I guess. Over on mastodon, Antoine Chambert-Loir pointed out that I (rather embarrassingly) forgot to mention the assumption that $f : \mathbb{R} \to \mathbb{R}$ is continuous!

Thankfully in my externalization I mention that we want $f : U \times \mathbb{R} \to \mathbb{R}$ to be continuous, which is what that would have externalized to, if I’d remembered to mention it! ↩
Recall that in the presence of countable choice, the dedekind and cauchy reals agree so that LLPO and analytic LLPO agree. ↩

An Empty Product of Nonempty Sets

Wed, 04 Jun 2025 00:00:00 +0000

A few days ago I saw a cute question on mse asking about a particularly non-intuitive failing of the axiom of choice. I remember when I was an undergrad talking to a friend of mine about various statements equivalent to choice, and being particularly hung up on the same statement that OP asks about – The product of nonempty sets is nonempty. I understood that there were models where the axiom of choice fails, and so in those models we must have some family of nonempty sets whose product is, somehow, empty! Now that I’m older and I’ve spent much more time thinking about these things, this is less surprising to me, but reading that question reminded me how badly I once wanted a concrete example, and so I’ll share one here!

This should be a pretty quick post, since I’ll basically just be fleshing out my answer to that mse question. But I think it’ll also be nice to have here, since these things can be hard to find when you’re first getting into logic and topos theory! Let’s get to it!

First, let’s remember that for any group $G$ the category $G\text{-}\mathsf{Set}$ of sets equipped with a $G$-action is a topos. Indeed you can see it as a presheaf topos, since $G\text{-}\mathsf{Set}$ is equivalent to the category of functors from $G \to \mathsf{Set}$ (viewing $G$ as a one-object category). We’ve talked about this topos before, and it’s wild to think how far I’ve come since writing that post!

The basic idea of $G\text{-}\mathsf{Set}$ as a topos is that any set theoretic construction we do to some $G$-sets again gives us $G$-sets! For instance, any (co)limits of $G$-sets will have a natural $G$-action. If $X$ is a $G$-set then its powerset $\mathcal{P}(X)$ has a $G$-action where if $A \in \mathcal{P}(X)$ we define $g \cdot A = \{g \cdot a \mid a \in A \}$, which is another element of $\mathcal{P}(X)$. If $X$ and $Y$ are $G$-sets then the set of functions $X \to Y$ is again a $G$-set where we say $(g \cdot_{X \to Y} f)(x) = g \cdot_Y f(g^{-1} \cdot_X x)$.

In particular, we can recover the “$G$-equivariant” constructions as the global elements! So even though $\mathcal{P}(X)$ contains all subsets of $X$ (not just the $G$-invariant subsets), if we look at the global elements (that is the maps $1 \to \mathcal{P}(X)$) we do get exactly the $G$-invariant subsets. Similarly while the set of functions $X \to Y$ sees all functions, the global elements of this set will pick out exactly the $G$-equivariant functions.

But $G\text{-}\mathsf{Set}$ has a coreflective subcategory given by those $G$-sets all of whose orbits are finite. The coreflector takes a $G$-set and just deletes all the infinite orbits, so we have an adjunction

\[(\iota : G\text{-}\mathsf{Set}_\text{finite orbits} \to G\text{-}\mathsf{Set}) \dashv (R : G\text{-}\mathsf{Set} \to G\text{-}\mathsf{Set}_\text{finite orbits})\]

which gives us a comonad $\iota R$ on $G\text{-}\mathsf{Set}$. This comonad is idempotent, and its category of coalgebras is equivalent to $G\text{-}\mathsf{Set}_\text{finite orbits}$. Then since $\iota$ is left exact (and so is $R$, since it’s a right adjoint), we see that $G\text{-}\mathsf{Set}_\text{finite orbits}$ is the category of coalgebras for a lex comonad on a topos, thus is itself a topos!

As a cute exercise, check that $\iota$ really is left exact!

Now doing computations in this topos is pretty easy! Finite limits and arbitrary colimits are computed as in $G\text{-}\mathsf{Set}$ since $\iota$ preserves these. Arbitrary limits and exponentials $Y^X$ come from coreflecting – that is $Y^X$ as computed in $G\text{-}\mathsf{Set}_\text{finite orbits}$ is just what we get by removing the infinite orbits from $Y^X$ as computed in $G\text{-}\mathsf{Set}$, and similarly for limits. The subobject classifier is just the usual set of truth values $\{ \top, \bot \}$ with the trivial $G$-action¹. In particular this topos is boolean, so set theory inside it is particularly close to the usual ZF set theory.

Now with this in mind, we can prove the main claim of this post:

In $\mathbb{Z}\text{-}\mathsf{Set}_\text{finite orbits}$, let $C_n$ be $\mathbb{Z}/n$ with its obvious $\mathbb{Z}$-action. Then each $C_n$ is inhabited² in the sense that $\exists x . x \in C_n$, and yet $\prod_n C_n = \emptyset$!

So this topos shows explicitly how, in the absence of choice, you can have a family of nonempty sets³ whose product is somehow empty!

The computation is actually quite friendly! To compute $\prod_n C_n$ in this topos, we first compute the product in the category of all $\mathbb{Z}$-sets, then throw away any infinite orbits.

But it’s easy to see that every orbit is infinite! Any element of the product will contain an element from every $C_n$, so that in any finite number of steps some large entry in this tuple won’t be back where it started.

Ok, this one was actually quite quick, which I’m happy about! My parents are visiting soon, and I’m excited to take a few days to see them ^_^. I have another shorter post planned which another grad student asked me to write, and I’ve finally actually started the process of turning my posts on the topological topos into a paper. I’m starting to understand TQFTs better, and it’s been exciting to learn a bit more physics. Hopefully I’ll find time to talk about all that soon too, once I take some time to really organize my thoughts about it.

Thanks for hanging out, all! Stay safe, and we’ll talk soon.

In case $G = \mathbb{Z}$ then it’s a kind of cute fact that this topos is equivalent to the topos of continuous (discrete) $\widehat{\mathbb{Z}}$-sets, where $\widehat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$. See, for instance, Example A2.1.7 on page 72 of the elephant. This gives another computationally effective way to work with this topos!

I’m pretty sure I convinced myself that more generally the category of $G$-sets all of whose orbits are finite should be equivalent to the category of continuous discrete $\widehat{G}$-sets, but I haven’t thought hard enough about it to say for sure in a blog post. ↩
Of course, there’s no global points for $n \neq 1$, since maps $1 \to C_n$ correspond to fixed points. But existential quantification is local, so that the topos models $\exists x \in C_n . \top$ if there’s some surjection $V \twoheadrightarrow 1$ and a map $V \to C_n$. ~~We can take $V = \mathbb{Z}$ with its left multiplication action on itself, and there is a map from $\mathbb{Z} \to C_n$.~~ Edit, June 5: Thanks to Kevin Carlson for pointing out that $\mathbb{Z}$ with its left multiplication action isn’t in this topos, since its orbit is infinite! We instead need, for each $n$, to look at the surjection from $C_n \twoheadrightarrow 1$ and then you can use the identity map from $C_n \to C_n$ to witness inhabited-ness.

If you’re more used to type theory, we don’t have $\Sigma_{x : C_n} \top$, since that would imply a global element. But despite this, we do have the propositional truncation $\lVert \Sigma_{x : C_n} \top \rVert$, so that an element of $C_n$ merely exists. ↩
Since this topos is boolean, nonempty and inhabited are actually synonyms here. Moreover, this “nonempty” is closer to how a lot of working mathematicians speak, so it felt right to use this wording here. ↩