The Smith homomorphisms are a family of homomorphisms between equivariant bordism groups in different dimensions.

One example that is known to be an isomorphism is the map $$ \tilde\Omega_{d+1}^{\rm Spin}(B\mathbb{Z}/2)\to\Omega_d^{\rm TPin-}, $$ where TPin- denotes a tangential Pin- structure (equivalently a normal Pin+ structure) and tilde denotes reduced bordism. This is obviously a very useful map.

Analogously, we have a map $$ \tilde\Omega_{d+1}^{SO}(B\mathbb{Z}/2) \to \Omega_d^O. $$ This map also seems like an isomorphism checking a few cases, but I don't have a proof or reference.

General Definition

Let me give the general way of defining these maps. I'll stick with Spin bordism but you can replace Spin with any geometric structure. Start in $\Omega_d^{\rm Spin}(BG)$ with a Spun $d$-manifold $X$ with a map to $BG$. Pick a (real) representation $\rho$ of $G$, there will be a Smith homomorphism for each one. Associate to the map to $BG$ the $\rho$ bundle over $X$ and let $Y$ be the zero locus of a generic smooth section. If $Y$ is the wrong dimension, then the map sends $X$ to zero, otherwise we continue...

The normal bundle of $Y$ is the restriction of $\rho$, so $Y$ represents a class in $\Omega_{d-dim \rho}^{\rm Spin}(BG,\rho)$, denoting twisted spin bordism (a spin structure on $TY$ plus the pullback of $\rho$ by the map to $BG$). After checking some things (which I'm not so sure of in $\dim \rho$>1), this gives a well defined map $$ \Omega_d^{\rm Spin}(BG) \to \Omega_{d-\dim \rho}^{\rm Spin}(BG,\rho) $$ which reduces to the first map when $G=\mathbb{Z}/2$, $\rho$ is the sign representation, and we restrict $X$ to be trivial in ordinary spin bordism.

What is the kernel and cokernel of this map in terms of other bordism groups?

Cokernel of the Smith homomorphism

In http://projecteuclid.org/euclid.pjm/1102699433 they consider the first map for Spinc, and it looks like the argument on page 13 can be used to attack the general problem.

It goes like this. Let's stick with $\dim\rho = 1$ for now. Start with $Y$ representing a class in $\Omega^{\rm Spin}_d(BG,\rho)$. I will use $\rho$ also to denote the associated bundle on $Y$. Consider the unit sphere bundle $X$ of this plus a trivial bundle $\rho\oplus \underline{\mathbb{R}}$. If $\rho$ is 1-dimensional this is a circle bundle over $Y$. $\rho \oplus \mathbb{R}$ has tangent bundle $TY\oplus \rho \oplus \mathbb{R}$ expressed as a bundle over $Y$. This has a spin structure from $Y$ and $X$ gets the boundary spin structure when we think of $\rho\oplus \mathbb{R}$ as a disc bundle.

$X$ can also be thought of as the pullback of a circle bundle over $BG$ and inherits a map to $BG$ this way that restricts to the given map $Y \to BG$. Thus $Y \mapsto X$ gives a map $$ \Omega_d^{\rm Spin}(BG,\rho) \to \tilde\Omega_{d+1}^{\rm Spin}(BG). $$

Let's compose this with the Smith homomorphism. The bundle over $X$ associated to $\rho$ can be thought of as pulled back from $Y$. We can produce many sections $Y \to X$ and use these to make sections of $\rho$ over $X$ which have this copy of $Y$ as their zeroes. These seem reasonably generic, so it looks like for 1 dimension the Smith homomorphism is always surjective. Is this true?

Kernel of the Smith homomorphism

I'm not sure how to compute the kernel. For instance, if $H$ is the kernel of the representation, then any class whose $G$ bundle reduces to an $H$ bundle will be in the kernel of the homomorphism. Is this the whole kernel?

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    $\begingroup$ Is spun : spin :: oriented : orientable? I like it. $\endgroup$ – Qiaochu Yuan May 27 '14 at 21:20
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    $\begingroup$ yep, and you can guess what it is for pin :P $\endgroup$ – Ryan Thorngren May 27 '14 at 21:36
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    $\begingroup$ strung : string ? $\endgroup$ – Sam Gunningham May 27 '14 at 23:35
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    $\begingroup$ But can you guess ___ : fivebrane :: pinned : pin? $\endgroup$ – Chris Gerig May 28 '14 at 0:11
  • $\begingroup$ @ChrisGerig you got 'pinned' wrong ;-). But, perhaps 'fivebrung'? (incidentally: 'brung' was a colloquial past tense of 'bring', where I grew up) $\endgroup$ – David Roberts May 29 '14 at 0:39

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