Finite quotients of 3-manifold groups

Abstract

For $G$ and $H_{1},\dots , H_{n}$ finite groups, does there exist a 3-manifold group with $G$ as a quotient but no $H_{i}$ as a quotient? We answer all such questions in terms of the group cohomology of finite groups. We prove non-existence with topological results generalizing the theory of semicharacteristics. To prove existence of 3-manifolds with certain finite quotients but not others, we use a probabilistic method, by first proving a formula for the distribution of the (profinite completion of) the fundamental group of a random 3-manifold in the Dunfield-Thurston model of random Heegaard splittings as the genus goes to infinity. We believe this is the first construction of a new distribution of random groups from its moments.

Appendix:  Semicharacteristics

Let $M$ be a manifold of dimension $2n+1$ and $\kappa$ a field. The semicharacteristic of $M$ with coefficients in $\kappa$ is $\sum _{i=0}^{n} (-1)^{i} \dim H_{i}(M, \kappa )$.

Generalizing this, for $M$ a manifold with a surjection $\pi _{1}(M) \to G$ and associated covering $\tilde{M}\to M$ (equivalently, for $\tilde{M}$ a manifold with a free action of $G$), Lee [30, Definition 2.3] defined a semicharacteristic class $\sum _{i=0}^{n} (-1)^{i} [ H_{i}(\tilde{M},\kappa )]$ taken in the Grothendieck group of representations of $G$ over $\kappa$ modulo a certain subspace depending on the parity of $n$ and the characteristic of $\kappa$. The main result of [30] is that, modulo this subspace, the semicharacteristic is a bordism invariant [30, Theorem 2.7 and Theorem 3.8].

This result bears an obvious similarity to Lemma 2.11. The difference is that [30] considers the cohomology of the cover in $K$-theory, while we consider cohomology twisted by a representation. Cohomology twisted by a representation is the more powerful invariant: Recall that representations of a group are equivalent to modules of the group algebra and a module $P$ is projective if the functor $\operatorname{Hom}(P, -)$ is exact. We will check that the dimension of the $i$th cohomology twisted by each indecomposable projective module for the group algebra determines the class of the $i$th cohomology of the cover in $K$-theory. However, the $K$-theory does not determine the cohomology of non-projective modules.

Using this, we will show that our result implies the main result of [30] in the cases where they both apply. It would be interesting to find a suitable generalization of our result (equivalently, strengthening of Lee’s) to the higher-dimensional even characteristic case.

We begin by formally defining the subspace we quotient the Grothendieck group by. For $\kappa$ of characteristic $\neq 2$, and $n$ odd, the semicharacteristic is valued in the quotient of the Grothendieck group by the subspace generated by all symmetrically self-dual representations together with the regular representation, while for $\kappa$ of characteristic $\neq 2$, and $n$ even, the semicharacteristic is valued in the quotient of the Grothendieck group by the subspace generated by all symplectic representations together with the regular representation [30, Definitions 2.1 and 2.3]. For $\kappa$ of characteristic 2, the semicharacteristic is valued in the quotient of the Grothendieck group by the subspace generated by all even representations together with the regular representation [30, Theorem 3.8], where the even representations $V$ are those admitting a nondegenerate symmetric bilinear $G$-invariant form $\phi$ such that $\phi (x,tx)=0$ for all $x\in V$ and all $t\in G$ of order exactly 2 [30, p. 190].

We next recall that for $V$ an irreducible representation of a finite group $G$ over a field $\kappa$, there is a unique indecomposable projective module $\mathcal {P}(V)$ for $\kappa [G]$ that admits $V$ as a subrepresentation. This is also the unique indecomposable projective module for $\kappa [G]$ that admits $V$ as a quotient.

Lemma A.1

For $V$ a absolutely irreducible representation of $G$ over $\kappa$, the number of times $V$ appears in the Jordan-Hölder decomposition of $H_{i} (\tilde{M}, \kappa )$ is equal to $\dim H^{i} (M, \mathcal {P}(V))$.

Proof

Since projective modules are stable under duality, projective modules are also injective. Because $\mathcal {P}(V)$ is injective, $H^{i}(M, \mathcal {P}(V))= \operatorname{Hom}_{G} ( H_{i}(\tilde{M}, \kappa ), \mathcal {P}(V))$.

So it suffices to prove that for a representation $W$, the number of times $V$ appears in the Jordan-Hölder decomposition of $W$ is equal to $\dim \operatorname{Hom}_{G}( W, \mathcal {P}(V))$.

Again because $\mathcal {P}(V)$ is injective, both sides are additive in exact sequences, so we may reduce to the case when $W$ is irreducible, and the statement is that for $W$, $V$ irreducible, $\operatorname{Hom}_{G}(W, \mathcal {P}(V)) \cong \kappa$ if $W \cong V$ and 0 if $W \ncong V$, which is standard. □

Lemma A.2

For $V$ a absolutely irreducible representation of $G$ over a field $\kappa$ of characteristic not two, $V$ is symmetrically self-dual if and only if $\mathcal {P}(V)$ is, and $V$ is symplectic if and only if $\mathcal {P}(V)$ is.

For $V$ a absolutely irreducible representation of $G$ over a field $\kappa$ of characteristic two, $\mathcal {P}(V)$ is symplectic if and only if $V$ is self-dual but not an even representation.

Proof

Fix $V$ an absolutely irreducible representation of $G$.

Let $f$ be a homomorphism $\mathcal {P}(V) \to \kappa [G]$ of left $\kappa [G]$-modules. Then $f$ defines an embedding $V \to \mathcal {P}(V) \to \kappa [G]$. Any such embedding must have the form $x \in V \mapsto \sum _{g\in G} a_{f}( g^{-1}\cdot x) [g]$ for some linear form $a_{f} \in V^{\vee}$. Furthermore, composition with $f$ defines a linear map $V \cong \operatorname{Hom}(\kappa [G], V) \to \operatorname{Hom}( \mathcal {P}(V), V) \cong \operatorname{Hom}(V, V) = \kappa$, and thus a linear form $b_{f}$ on $V$.

We claim that $b_{f} = \lambda a_{f}$ for some $\lambda \in \kappa ^{\times}$. To see this, note that both $f \mapsto a_{f}$ and $f \mapsto b_{f}$ are nontrivial homomorphisms $\operatorname{Hom}( \mathcal {P}(V),\kappa [G]) \to V^{\vee}$ that are equivariant for the right $G$ action of $\kappa [G]$. As a right $G$-module, $\operatorname{Hom}( \mathcal {P}(V),\kappa [G]) \cong \mathcal {P}(V)^{ \vee }\cong \mathcal {P}(V^{\vee})$ has a unique quotient isomorphic to $V^{\vee}$, so any two such homomorphisms differ by a $G$-invariant automorphism of $V^{\vee}$, i.e. by scalar multiplication.

Now fix one such $f$ that is a split injection. The pullback of any bilinear form on $\kappa [G]$ along $f$ gives a bilinear form on $\mathcal {P}(V)$. (All bilinear forms we consider will be $G$-invariant.) The pullback of a symmetric bilinear form is symmetric, and the pullback of a symplectic bilinear form is symplectic. Furthermore, because $f$ is split, every symmetric bilinear form on $\mathcal {P}(V)$ arises by pullback along $f$ from a symmetric bilinear form on $\kappa [G]$, and similarly with symplectic forms. Thus, to test when $V$ is symmetrically self-dual, we will calculate all symmetric bilinear forms on $\kappa [G]$ and check when the pullback of one along $f$ is symmetric, and similarly in the symplectic case.

We first describe the bilinear forms on $\kappa [G]$. These are parameterized by tuples $\mathbf {d}= (d_{g})_{g\in G}$ of coefficients in $\kappa$ associated to $g\in G$, and are given by the formula

$$ \langle \sum _{g\in G} a_{g}[g], \sum _{g\in G} b_{g} [g] \rangle _{ \mathbf {d}} = \sum _{g \in G} \sum _{h\in G} a_{g} b_{h} d_{ h^{-1} g } . $$

Then

$$ \langle \sum _{g\in G} a_{g}[g], \sum _{g\in G} b_{g} [g] \rangle _{ \mathbf {d}} = \langle \sum _{g\in G} b_{g} [g] , \sum _{g\in G} a_{g}[g] \rangle _{\overline{\mathbf {d}}} $$

where

$$ \overline{d}_{g} = d_{g^{-1}} $$

so $\langle , \rangle _{\mathbf {d}}$ is symmetric if $d_{g} =d_{g^{-1}}$ for all $g$ and symplectic if $d_{g} =- d_{g^{-1}}$ for all $g$ with, in characteristic 2, the additional condition $d_{e} =0$ where $e$ is the identity.

Now a bilinear form $\mathcal {P}(V) \times \mathcal {P}(V) \to \kappa$ is nondegenerate if and only if the induced map $\mathcal {P}(V) \to \mathcal {P}(V)^{\vee}$ is injective, i.e. if its kernel is zero, which happens if and only if the induced map $V \to \mathcal {P}(V) \to \mathcal {P}(V)^{\vee}$ is injective. Thus, the pullback of $\langle , \rangle _{\mathbf {d}}$ is nondegenerate if and only if there exists $x \in V$ and $y\in \mathcal {P}(V)$ such that $\langle f(x), f(y) \rangle _{\mathbf {d}} \neq 0$.

Now the map $L_{f} \colon \kappa [G] \to V^{\vee }$ that sends $\alpha \in \kappa [G]$ to the linear form $x \mapsto \langle f(x), \alpha \rangle _{\mathbf {d}}$ is $G$-equivariant since $f$ and $\langle ,\rangle _{\mathbf {d}}$ are. Thus it defines an element of $V^{\vee}$. We calculate this element by evaluating at $\alpha =[e] \in \kappa [G]$.

For $x\in V$, $f(x) = \sum _{g\in G} a_{f}( g^{-1} x) [g]$ by definition. Thus

$$ L_{f}([e]) (x) = \langle f(x), [e] \rangle _{\mathbf {d}} = \sum _{g \in G} a_{f} (g^{-1}\cdot x) d_{ g }= a_{f} \Bigl( \sum _{g} d_{g} g^{-1} \cdot x \Bigr) . $$

The pullback of the bilinear form $\langle ,\rangle _{\mathbf {d}}$ is nondegenerate if and only if the composition of $L_{f}$ with $f \colon \mathcal {P}(V) \to \kappa [G]$ is nonzero.

If $V$ is not self-dual, then the homomorphism $\mathcal {P}(V) \to \kappa [G] \to V^{\vee}$ automatically vanishes and thus there is no such nondegenerate bilinear form. So suppose that $V$ is self-dual, so in particular there is a map $\gamma : V^{\vee} \to V$. The isomorphism $\operatorname{Hom}(\kappa [G], V) \cong V$ sends a linear map $L$ to $L([e])$ so it sends $\gamma \circ L_{f}$ to

$$ \gamma (L_{f}([e]) ) = \gamma ( x \mapsto a_{f} ( \sum _{g} d_{g} g^{-1} \cdot x ) )= \sum _{g\in G} d_{g} g\cdot \gamma (a_{f}) \in V. $$

So by the definition of $b_{f}$, the composition $\gamma \circ L_{f} \circ f \colon \mathcal {P}(V) \to \kappa [G] \to V^{ \vee}\to V$ is nonzero if and only if

$$ b_{f} \Bigl( \sum _{g \in G} d_{g} g\cdot \gamma (a_{f}) \Bigr) \neq 0 $$

and a nondegenerate symmetric (or symplectic) bilinear form exists if and only if $b_{f} ( \sum _{g \in G} d_{g} g\cdot \gamma (a_{f}) ) \neq 0$ for some $\mathbf {d}$ satisfying the conditions to be symmetric (or symplectic). To simplify this, note that $b_{f} ( x) = \langle \gamma (b_{f}), x \rangle _{V}$ for $\langle , \rangle _{V}$ the bilinear form on $V$, and recall $b_{f} = \lambda a_{f}$ so we can express the nonvanishing condition more simply as

$$ \sum _{g\in G }d_{g} \langle \gamma (a_{f}), g\cdot \gamma (a_{f}) \rangle _{V} \neq 0 . $$

We now specialize to particular cases. In characteristic not two,

$$\begin{aligned} \sum _{g\in G }d_{g} \langle \gamma (a_{f}), g\cdot \gamma (a_{f}) \rangle _{V} ={}& \sum _{g\in G }d_{g} \langle g^{-1} \cdot \gamma (a_{f}), \gamma (a_{f}) \rangle _{V} \\ ={}& \sum _{g\in G }d_{g} \langle \gamma (a_{f}), g^{-1} \cdot \gamma (a_{f}) \rangle _{V} \cdot \textstyle\begin{cases} 1 & V \textrm{ symmetric} \\ -1 & V \textrm{ symplectic} \end{cases}\displaystyle \\ ={}& \sum _{g\in G }d_{g} \langle \gamma (a_{f}), g \cdot \gamma (a_{f}) \rangle _{V} \\ &{}\cdot \textstyle\begin{cases} 1 & V \textrm{ symmetric} \\ -1 & V \textrm{ symplectic} \end{cases}\displaystyle \cdot \textstyle\begin{cases} 1 & \langle , \rangle _{\mathbf {d}} \textrm{ symmetric} \\ -1 & \langle , \rangle _{\mathbf {d}} \textrm{ symplectic}. \end{cases}\displaystyle \end{aligned}$$

If $V$ is symmetric and $\langle , \rangle _{\mathbf {d}}$ is symplectic, or vice versa, then the signs don’t match and so $\sum _{g\in G }d_{g} \langle \gamma (a_{f}), g\cdot \gamma (a_{f}) \rangle _{V}$ is equal to its own negation and thus vanishes. Since a self-dual absolutely irreducible representation is either symmetric or symplectic, we see there is no nondegenerate symmetric form on $\mathcal {P}(V)$ unless $V$ is symmetric and no nondegenerate symplectic form on $\mathcal {P}(V)$ unless $V$ is symplectic. Conversely, for any nonzero $\gamma (a_{f})$, there is always some $h$ such that $\langle \gamma (a_{f}), h\cdot \gamma (a_{f}) \rangle _{V} \neq 0$ by irreducibility of $V$. In this case, we can take $d_{h} = 1$, and $d_{h^{-1}}=1$ if $V$ is symmetric or $d_{h^{-1}}=-1$ if $V$ is symplectic, and $d_{g}=0$ for $g \neq h, h^{-1}$, and this ensures $\sum _{g\in G }d_{g} \langle \gamma (a_{f}), g\cdot \gamma (a_{f}) \rangle _{V}\neq 0$.

In characteristic 2, the unique bilinear form on $V$ is necessarily symmetric. Thus if $\langle , \rangle _{\mathbf {d}}$ is symplectic, the contributions of $g$ and $g^{-1}$ to the sum $\sum _{g\in G }d_{g} \langle \gamma (a_{f}), g\cdot \gamma (a_{f}) \rangle _{V}$ are equal. Thus these contributions cancel each other unless $g = g^{-1}$, i.e. if $g$ has order dividing 2. Since $d_{e}=0$, we need only consider the contribution from $g$ of exact order 2. If $V$ is even, then the contribution vanishes by definition and so there are no nondegenerate symplectic forms. Conversely, if $V$ is not even then for some $g$ and $x$ we have $\langle x, g\cdot x \rangle _{V}$ is nonzero. Since $g$ has order 2, $\langle x, g\cdot x \rangle _{V}$ defines a Frobenius-semilinear form, so it vanishes for all $x$ outside a proper subspace. Choose $h$ such that $h\cdot \gamma (a_{f})$ is not in that subspace, and observe that $\langle \gamma (a_{f}), h^{-1}gh \cdot \gamma (a_{f})\rangle _{V} = \langle h \gamma (a_{f}), gh\cdot \gamma (a_{f})\rangle _{V} \neq 0$, so choosing $\mathbf {d}$ supported on $h^{-1} gh$ we construct a nondegenerate symplectic form. □

We are now ready to describe how the semicharacteristic studied by Lee is determined by the cohomology groups controlled in Lemma 2.11, and thus to deduce Lee’s theorem (except in the even characteristic $n>1$ case) from Lemma 2.11.

Lemma A.3

Let $G$ be a finite group and $\kappa$ a finite splitting field for $G$. Let $n$ be a natural number. We will always take $M$ to be a $2n+1$-dimensional oriented manifold with a homomorphism $\pi _{1}(M)\to G$.

• (1) If $n$ is odd and $\kappa$ has characteristic $\neq 2$, the class of $\sum _{i=0}^{n} (-1)^{i} [ H_{i}(M,\kappa )]$ in the Grothendieck group of representations of $G$ over $\kappa$, modulo the classes of symmetrically self-dual representations, is determined by $\sum _{i=0}^{n} (-1)^{i} \dim H^{i}(M, V) \ \ \mathrm{mod}\ 2$ for even-dimensional symplectic representations $V$ of $G$ over $\kappa$ that are projective. In particular, it is an invariant of the class of $M$ in the oriented bordism group of $BG$.

• (2) If $n$ is even and $\kappa$ has characteristic $\neq 2$, the class of $\sum _{i=0}^{n} (-1)^{i} [ H_{i}(M,\kappa )]$ in the Grothendieck group of representations of $G$ over $\kappa$, modulo the classes of symplectic representations and the regular representation, is determined by $\sum _{i=0}^{n} (-1)^{i} \dim H^{i}(M, V) \ \ \mathrm{mod}\ 2$ for even-dimensional symmetrically self-dual representations $V$ of $G$ over $\kappa$ that are projective. In particular, it is an invariant of the class of $M$ in the oriented bordism group of $BG$.

• (3) If $n$ is odd and $\kappa$ has characteristic $\neq 2$, the class of $\sum _{i=0}^{n} (-1)^{i} [ H_{i}(M,\kappa )]$ in the Grothendieck group of representations of $G$ over $\kappa$, modulo the classes of even representations, is determined by $\sum _{i=0}^{n} (-1)^{i} \dim H^{i}(M, V) \ \ \mathrm{mod}\ 2$ for even-dimensional symplectic representations $V$ of $G$ over $\kappa$ that are projective and lift to $\operatorname{ASp}_{\kappa}(V)$. In particular, if $n=1$ then it is an invariant of the class of $M$ in the oriented bordism group of $BG$.

In the first and third cases, it is not necessary to mod out by the regular representation as the regular representation is symmetrically self-dual and, in characteristic 2, even.

Proof

We handle part (1) first, then describe how the arguments in the remaining cases differ.

A class in the Grothendieck group can be represented as $\sum _{V} m_{V}[V]$, the sum taken over irreducible representations $V$ of $G$ over $\kappa$, for some integers $m_{V}$.

Note that $V \oplus V^{\vee}$ is always symmetrically self-dual. Thus, two classes arising from two tuples of integers $m_{V}$, $m_{V}'$ are equivalent modulo the symmetrically self-dual representations if $m_{V} - m_{V^{\vee}} = m_{V}' - m_{V^{\vee}}'$ for all irreducible representations $V$ and $m_{V} - m_{V'}$ is even for all $V$ self-dual but not symmetrically self-dual. Indeed, in this case, the difference between the classes is a sum of irreducible symmetrically self-dual representations, sums of a non-self-dual representation and its dual, and even multiples of a symplectic representation, which are sums of a representation and its dual (itself).

When representing the class of $\sum _{i=0}^{n} (-1)^{i} [ H_{i}(M,\kappa )]$ in the Grothendieck group this way, $m_{V}(M) = \sum _{i=0}^{n} (-1)^{i} \operatorname{mult}_{V} H_{i}(M, \kappa )$. So to show this class, modulo symmetrically self-dual representations is determined by $\sum _{i=0}^{n} (-1)^{i} \dim H^{i}(M, V) \ \ \mathrm{mod}\ 2$ for even-dimensional symplectic representations of $G$ over $\kappa$ that are projective, it suffices to show that $m_{V}(M) - m_{V^{\vee}}(M)$ is determined, as is $m_{V}$ mod 2 for irreducible representations that are self-dual but not symmetrically self-dual.

For the first part, the fact that $m_{V} (M)= m_{V^{\vee}} (M)$ was already proven by Lee, using Euler characteristic and Poincaré duality arguments. For the second part, if $V$ is irreducible and self-dual but not symmetric, then it must be symplectic so by Lemma A.2, $\mathcal {P}(V)$ is symplectic Because $\mathcal {P}(V)$ is symplectic, it is even-dimensional. Thus by Lemma A.1,

$$ m_{V}(M) = \sum _{i=0}^{n} (-1)^{i} \operatorname{mult}_{V} H_{i}(M, \kappa ) = \sum _{i=0}^{n} (-1)^{i} H^{i}(M ,\mathcal {P}(V)) \quad \mathrm{mod}\ 2 $$

so $m_{V}(M)\ \ \mathrm{mod}\ 2$ is determined by $\sum _{i=0}^{n} (-1)^{i} H^{i}(M ,\mathcal {P}(V) )\ \ \mathrm{mod}\ 2$, and $\mathcal {P}(V)$ satisfies all of the assumed properties.

Finally, by Lemma 2.11, $\sum _{i=0}^{n} (-1)^{i} H^{i}(M ,\mathcal {P}(V)) \ \ \mathrm{mod}\ 2$ is determined by the bordism class of $M$.

For part (2), the argument is similar, except for the following: First, we use the fact that $V \oplus V^{\vee}$ is always symplectic. Second, we prove that $\mathcal {P}(V)$ is symmetrically self-dual, and so we can no longer use the fact that $\mathcal {P}(V)$ is symplectic to guarantee it is even-dimensional. Instead we use the fact that we need only determine $\sum _{i=0}^{n} (-1)^{i} [ H_{i}(M,\kappa )]$ in the Grothendieck group modulo both the symmetrically self-dual representations and the regular representation.

Adding a copy of the regular representation does not affect $m_{V} - m_{V^{\vee}}$, but it swaps the parity of $m_{V}$ if $V$ is self-dual of odd multiplicity in $\kappa [G]$. Since the multiplicity of an irreducible representation $V$ in $\kappa [G]$ is equal to $\dim \mathcal {P}(V)$, adding a copy of the regular representation swaps the parity of $m_{V}$ for all irreducible representations $V$ with $\dim \mathcal {P}(V)$ odd. Thus, to determine the class modulo symplectic representations and the regular representation, it suffices to know $m_{V} -m_{V^{\vee}}$ for all irreducible representations $V$, $m_{V}\ \ \mathrm{mod}\ 2$ for all symmetrically self-dual irreducible representations $V$ with $\dim \mathcal {P}(V)$ even, and $m_{V} +m_{W} \ \ \mathrm{mod}\ 2$ for all pairs $V$, $W$ of symmetrically self-dual irreducible representations with $\dim \mathcal {P}(V)$, $\dim \mathcal {P}(W)$ odd.

Thus, in the second part, it suffices to know $\sum _{i=0}^{n} H^{i} (M, \mathcal {P}(V) )\ \ \mathrm{mod}\ 2$ where $\mathcal {P}(V)$ is projective, symmetrically self-dual, and even-dimensional, and in the third part, it suffices to know $\sum _{i=0}^{n} H^{i}(M, \mathcal {P}(V) \oplus \mathcal {P}(W))\ \ \mathrm{mod}\ 2$ where $\mathcal {P}(V) \oplus \mathcal {P}(W)$ is projective, symmetrically self-dual, and even-dimensional. So we still need consider only representations that satisfy all the assumed properties. Then we use Lemma 2.11 the same way.

For part (3) it is again similar to part (1). We now use the fact that $V \oplus V^{\vee}$ is even, which may be less familiar – the form $\langle (x_{1},y_{1}) , (x_{2},y_{2}) \rangle = x_{1} \cdot y_{2} + x_{2} \cdot y_{1}$ is symmetric, and for $g$ of order 2,

$$\begin{aligned} \langle (x,y) , g\cdot (x,y) \rangle &= \langle (x,y) , (g x,g y) \rangle = x \cdot g y + g x \cdot y \\ &= x \cdot g y + x \cdot g^{-1} y = x \cdot gy + x \cdot gy = 0 \end{aligned}$$

where we use $g= g^{-1}$ and the fact that the characteristic is two, so this form is even.

We can again use the argument that symplectic representations must be even-dimensional, but we now face the difficulty that Lemma A.2 ensures that $\mathcal {P}(V)$ is symplectic but we want the action of $G$ to lift to $\operatorname{ASp}_{\kappa}(V)$. However, the obstruction to such a lift is contained in $H^{2}(G, \mathcal {P}(V))$ which vanishes since $\mathcal {P}(V)$ is projective, so a lift always exists. Finally, here Lemma 2.11 is restricted to the $n=1$ case only. □

By combining Lemma 2.10 and Lemma A.3, we can check that the semicharacteristic vanishes in the odd characteristic $n=1$ case. Again, this requires only the projective case of Lemma 2.10, and the general case may be significantly stronger.