(This is a cross-post of this unanswered math.stackexchange question)

In Edmond's 1982 paper Surface Symmetry II, at the bottom of page 145, he writes:

"Corollary - If $G$ is a split nonabelian metacyclic group acting freely on a surface $M$, then the natural map $\textbf{B} : \mathcal{FG}(G,M)^*\rightarrow H_2(G,\mathbb{Z})$ is surjective.

Proof: Since $H_2(G)\cong\Omega_2(BG)$, each element of $H_2(G)$ is represented by a free action on some surface, which can be chosen to be connected..."

Here, $M$ is a closed oriented surface, and $\mathcal{FG}(B,M)^*$ is the set of orientation-preserving equivariant homeomorphism classes of free actions of $G$ on $M$. The quotient $M/G$ does not depend on the particular chosen action, so let $N := M/G$. Then the set $\mathcal{FG}(B,M)^*$ seems to be the same as the set of surjective homomorphisms $\pi_1(N)\twoheadrightarrow G$ modulo $\text{Aut}(\pi_1(N))$.

The map $\textbf{B} : \mathcal{FG}(G,M)^*\rightarrow H_2(G,\mathbb{Z})$ is the one given by Eric Wofsey's answer to this question. Namely, every element $\eta\in\mathcal{FG}(G,M)^*$ gives $M$ the structure of a $G$-torsor over $N$, hence we get a map $N\rightarrow BG$, and the image of $\eta$ in $H_2(G,\mathbb{Z}) = H_2(BG,\mathbb{Z})$ is defined to be the image of the fundamental class of $N$ under the induced map on homology $H_2(N)\rightarrow H_2(BG)$.

My question is - Could someone help to elucidate the final quoted sentence? The author never explicitly defines what $\Omega_2(BG)$ is, though it seems that it is some kind of "(co?)bordism group", but there seem to be many variants on (co?)bordism groups, and without understanding exactly what he means, it's difficult to understand why the isomorphism $H_2(G)\cong\Omega_2(BG)$ would result in the map $\textbf{B}$ being surjective.

From another angle - is it possible to argue that $\textbf{B}$ is surjective without appealing to the isomorphism with the (co?)bordism group?

Is it possible to describe the map $\textbf{B}$ group-theoretically? Ie, given an $\text{Aut}(\pi_1(N))$-orbit of a surjection $\pi_1(N)\twoheadrightarrow G$, how do we produce an element of $H_2(G)$? Here, if $G$ has the presentation $$1\rightarrow R\rightarrow F_n\rightarrow G\rightarrow 1$$ where $F_n$ is a free group of rank $n$, then we know that $H_2(G) = (R\cap[F,F])/[F,R]$.


1 Answer 1


The group $\Omega_d(X)$ being used is the following oriented bordism group. The objects are equivalence classes of closed oriented d-dimensional manifolds $N$ with a map $N \to X$. Two elements $N_1 \to X$ and $N_2 \to X$ are equivalent if there exists a compact oriented (d+1)-manifold $W$ with a map $W \to X$ such the restriction $\partial W \to X$ is homeomorphic to the disjoint union of the maps $N_1 \to X$ and $\overline N_2 \to X$, which is the same manifold with the same map but the opposite orientation.

There is always a natural map $\Omega_d(X) \to H_d(X)$ that sends $f: N \to X$ to the image $f_*([N])$ of the fundamental class in $H_d(X)$. The question of whether this map is surjective is Steenrod's problem, and it is always true when $d \leq 6$.

To finish off the proof of surjectivity, it suffices to observe the following. Given an element $\alpha \in H_2(BG)$, lift it to an element $\alpha \in \Omega_2(BG)$, represented by a map $N \to BG$ from a closed 2-manifold. Then you can construct a covering map $M' = N \times_BG EG \to N$ which is a principal $G$-bundle, giving you a construction of some manifold $M'$ with free $G$-action that gets turned back into your element in $H_2(G)$.

This only covers the sentence "Proof: Since $H_2(G)\cong\Omega_2(BG)$, each element of of $H_2(G)$ is represented by a free action on some surface" that you stated after "Proof", and does not show that you can take your manifold to be $M$ or that you can take it to be connected. For $M'$ to be connected you would need the map $N \to BG$ to give a surjection on $\pi_1(N) \to G$. You can do this by taking elements in $\pi_1(BG)$, representing them as loops, and attaching handles to $N$ that also wind around those loops.

So far as producing the map group-theoretically, one could try the following. Given a surjection $\pi_1(N) \to G$, take a presentation for the surface group $$ \pi_1(N) \cong \langle x_1, x_2, \dots, x_g, y_g \mid \Pi [x_i, y_i] = e\rangle $$ and lift the elements $x_i$ and $y_i$ into $F_n$. The image of the product of commutators $\Pi [x_i,y_i]$ in $R \cap [F,F] / [R,F]$ will be your desired element in $H_2(G)$. I don't know if there's a purely group-theoretic proof of surjectivity because I'm not sure if it relies on the particular type of group $G$ you're considering.

(These groups $\Omega$ were classically called cobordism groups, but because they are more closely related to homology than cohomology it has become common to refer to them as bordism groups. The terms are roughly interchangeable. There are analogues of cohomology theory built out of manifolds, but these are often called "bordism cohomology" because using either "cobordism" or "co-cobordism" would be a bad idea.)

  • $\begingroup$ Dear Tyler - Thank you so much for the answer! It definitely clears up a lot of my confusion. Is it easy to see that $\Omega_d(X)\rightarrow H_d(X)$ should be injective? (At least in the situation $\Omega_2(BG)\rightarrow H_2(BG)$ when $G$ is finite?) $\endgroup$ Dec 13, 2016 at 22:02
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    $\begingroup$ Dear Amy, just like with surjectivity the property of injectivity is not so easy to see and it is special to low dimensions (it is only true in dimensions strictly smaller than 4). If this helps, in both cases there is an Atiyah-Hirzebruch spectral sequence $H_p(X; \Omega_q(pt)) \Rightarrow \Omega_{p+q}(X)$ and so these results are consequences of being able to calculate the first few oriented bordism groups: $\Omega_0(pt) = \Bbb Z$ and $\Omega_q(pt)=0$ for $0 < q < 4.$ (Said differently, every closed orientable $d$-manifold is the boundary of an orientable $(d+1)$-manifold in these degrees.) $\endgroup$ Dec 14, 2016 at 5:32

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