4
$\begingroup$

Let $G$ be a compact Lie group. I know that the Burnside ring $A(G)$ is isomorphic to the zeroth $G$-equivariant stable homotopy $\pi^{G}_0(S^0)$. What is the isomorphism between $A(G)$ and $\pi^{G}_0(S^0)$ and if I consider an element of $\pi^{G}_0(S^0)$ then how can I determine the corresponding one in $A(G)$? For example consider the one dimensional orthogonal group $O_1$, then $A(O_1)$ (as a group) is a free abelian group with $\{1,O_1\}$ as a basis. Let $\{+1,-1\}$ be the set of degrees that determine a homotopy class in $\pi^{O_1}_0(S^0)$ then how can I determine the corresponding element to $\{+1,-1\}$ in $A(O_1)$ in terms of the elements of the basis $\{1,O_1\}$?

$\endgroup$

2 Answers 2

10
$\begingroup$

Expanding on the answer by user43326, and covering the compact Lie case: The map $A(G) \to \pi^G_0(S^0)$ indeed comes from the Pontrjagin-Thom construction. Each generator of $A(G)$ is a $G$-orbit $G/H$ (with $G/NH$ finite). Take an embedding $G/H\to V$ in some $G$-representation $V$, with normal bundle $\nu$. This induces a map $f\colon S^V \to T\nu \to S^V$, where the first map is the Pontrjagin-Thom map and the second comes from the inclusion of $\nu$ in the trivial bundle together with the projection $G/H\to *$. (This second map is equivariantly null-homotopic if $G/NH$ is not finite, which helps explain the restriction on the orbits that appear as generators.) The class of $f$ is the corresponding element in $\pi^G_0(S^0)$.

As an example, consider the case asked about, where $G=\mathbb{Z}/2$, and consider the characterization of an element $f\in \pi^G_0(S^0)$ in terms of the nonequivariant degrees of $f$ and $f^G$. The trivial orbit $G/G$ gives a stable homotopy element with degrees both $+1$. For the nontrivial orbit $G/e$, embed in the line $L$ with nontrivial $G$-action, say as the unit sphere $S(L)$. The resulting map $f\colon S^L\to S^L$ wraps around twice, so has nonequivariant degree 2, while the degree of $f^G$ is 0.

The map $\pi^G_0(S^0) \to A(G)$ is a good deal trickier to describe rigorously, because it wants to involve transversality, and equivariant transversality is not very well-behaved. Suffice it to say that it is possible to take a representative map $f\colon S^V\to S^V$ and make it "transverse" to the origin, in the sense that $f^{-1}(0)$ is a disjoint union of orbits of $G$ and $f$ itself can then be described as a (signed) sum of maps as above. (We know this has to be true because of the isomorphism of $A(G)$ and $\pi^G_0(S^0)$, but trying to prove the isomorphism using this argument is what's tricky.)

$\endgroup$
4
$\begingroup$

This is described in Chapter 5 of Lewis, May, Steinberger "Equivariant Stable homotopy theory" (with contribution by McClure). Basically the idea is that a generator of $A(G)$ is a $G$-set, so 0-dimensional $G$-manifold, thus by applying Pontrjagin-Thom construction, one gets an element of stable cohomotopy, although in their book they describe the map in terms of Spanier-Whitehead dual.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.