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 $\{1,O\_1\}$?



Isomorphism between the Burnside ring $A(G)$ and the zeroth $G$-equivariant stable homotopy $\pi^{G}_0(S^0)$
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.)


