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

03 Sep 2025

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

$$N=⟨a,c,bd⟩⊴F\{a,b\}\times F\{c,d\}$$

which is the kernel of the homomorphism

$$\begin{array}{rl}F\{a,b\}\times F\{c,d\}& \to \mathbb{Z}\\ a,c& \mapsto 0\\ b& \mapsto 1\\ d& \mapsto -1\end{array}$$

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

$⌜$ 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 $G=⟨x_1,\dots ,x_n\mid R_1,\dots ,R_m⟩$ 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 $G=⟨x_1,\dots ,x_n\mid R_1,\dots ,R_m⟩$ 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 $⌟$

So, to show that our $N=⟨a,c,bd⟩⊴F\{a,b\}\times F\{c,d\}$ 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

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

$$E_{pq}^2=H_p(Q;H_q(N))\Rightarrow H_{p+q}(G)$$

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

$$\begin{array}{ccccccccc}⋮& & ⋮& & ⋮& & ⋮& & ⋰\\ H_2(Q;H_0(N))& & H_2(Q;H_1(N))& & H_2(Q;H_2(N))& & H_2(Q;H_3(N))& & \cdots \\ H_1(Q;H_0(N))& & H_1(Q;H_1(N))& & H_1(Q;H_2(N))& & H_1(Q;H_3(N))& & \cdots \\ H_0(Q;H_0(N))& & H_0(Q;H_1(N))& & H_0(Q;H_2(N))& & H_0(Q;H_3(N))& & \cdots \end{array}$$

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 $\phi \in GL(V)$.

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

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

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

$$0\to V\stackrel{1-t}{\to }V\to 0$$

where $t$ acts by the automorphism $\phi$, giving the claim. $⌟$

In case $V$ is finite dimensional (as a $\mathbb{Q}$ vector space), then it’s easy to see that $\dim V^{\mathbb{Z}}=\dim \text{Ker}(1-\phi )$ and $\dim V_{\mathbb{Z}}=\dim (V/\text{Im}(1-\phi ))=\dim V-\dim \text{Im}(1-\phi )$ 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 $\phi$ 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 $V/(1-t)$ which is isomorphic to $\mathbb{Q}$. So $V^{\mathbb{Z}}$ is $0$-dimensional and $V_{\mathbb{Z}}$ is $1$-dimensional.

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

$$\begin{array}{ccccccccc}⋮& & ⋮& & ⋮& & ⋮& & ⋰\\ H_2(\mathbb{Z};H_0(N))& & H_2(\mathbb{Z};H_1(N))& & H_2(\mathbb{Z};H_2(N))& & H_2(\mathbb{Z};H_3(N))& & \cdots \\ H_1(\mathbb{Z};H_0(N))& & H_1(\mathbb{Z};H_1(N))& & H_1(\mathbb{Z};H_2(N))& & H_1(\mathbb{Z};H_3(N))& & \cdots \\ H_0(\mathbb{Z};H_0(N))& & H_0(\mathbb{Z};H_1(N))& & H_0(\mathbb{Z};H_2(N))& & H_0(\mathbb{Z};H_3(N))& & \cdots \end{array}$$

becomes

$$\begin{array}{ccccccccc}⋮& & ⋮& & ⋮& & ⋮& & ⋰\\ 0& & 0& & 0& & 0& & \cdots \\ H_0(N{)}^{\mathbb{Z}}& & H_1(N{)}^{\mathbb{Z}}& & H_2(N{)}^{\mathbb{Z}}& & H_3(N{)}^{\mathbb{Z}}& & \cdots \\ H_0(N{)}_{\mathbb{Z}}& & H_1(N{)}_{\mathbb{Z}}& & H_2(N{)}_{\mathbb{Z}}& & H_3(N{)}_{\mathbb{Z}}& & \cdots \end{array}$$

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 $N=⟨a,c,bd⟩$ 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

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

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

$$\begin{array}{ccccccccc}⋮& & ⋮& & ⋮& & ⋮& & ⋰\\ 0& & 0& & 0& & 0& & \cdots \\ \mathbb{Q}& & {\mathbb{Q}}^3& & H_2(N{)}^{\mathbb{Z}}& & H_3(N{)}^{\mathbb{Z}}& & \cdots \\ \mathbb{Q}& & {\mathbb{Q}}^3& & H_2(N{)}_{\mathbb{Z}}& & H_3(N{)}_{\mathbb{Z}}& & \cdots \end{array}$$

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 $E^{\mathrm{\infty }}$ page of our spectral sequence! General theory tells us that $H_n(G)=\underset{p+q=n}{⨁}{E}_{pq}^{\mathrm{\infty }}$, 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):

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

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 $N=⟨a,c,bd⟩\le F\{a,b\}\times F\{c,d\}$ 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 💖

---