Equivariant Surgery problem

2010-07-27T16:37:41Z

I have a question about surgery.

Let $G= \mathbb{Z}_m \times \mathbb{Z}$ and $M$ be a oriented 3-manifold with G-action. i.e. There exists a map $f\colon M/G \to BG$, where $BG$ is classifying space.($BG=K(G,1)=K(\mathbb{Z}_m \times \mathbb{Z},1)).$ Therefore, we can regard $(M/G,f)\in \Omega^{SO}_3(K(G,1))$, where $\Omega^{SO}_3(X)$ is (3-dimensional oriented) bordism group of $X$. By using Leray-Serre spectral sequence, we can easily see that $\Omega^{SO}_3(K(G,1))=H_3(G)=\mathbb{Z}_m$ (By using Kunneth fomula) Therefore, $\exists$ r>0 such that $r(M/G,f)$ is nullbordant over $G$. (we can choose $r =m$) That is, there exists a 4-manifold $V$ with G-action such that $\partial V = M$ and $\partial (V/G) = M/G$ and induced map $g\colon V/G\to BG$ is an extension of $f\colon M/G \to BG$.

$\underline{Question}$ : Can I do a surgery on $V/G$ to make $\pi_1(V/G)=G$ without losing $G$ action on $V$?

Answer by Martin O for Equivariant Surgery problem

2010-07-27T19:50:55Z

Yes. You need to extend the map to $BG$ over the surgery cobordism, which is possible.

First let me add that all your $G$-actions seem to be free and that the boundary of $V$ would consist of $r$ copies of $M$.

Now for the surgeries: first take connected sum of V/G with 2 copies of $S^1\times S^2$'s (surgeries on embeddings $S^0\times D^3$) and extend the map to BG as $S^1\times S^2\to S^1$ and then to generators of $G=\pi_1(BG)$ to make $V'/G\to BG$ 1-connected. Then find $S^1$'s in $V'/G$ generating the kernel of $\pi_1$, thicken to embeddings of $S^1\times D^2$ (they have trivial normal bundle) and do surgery on them, extending the map to $BG$ by nullhomotopies of their images in $BG$.

Equivariant Surgery problem - MathOverflow

Equivariant Surgery problem

Answer by Martin O for Equivariant Surgery problem