Let $\pi\colon E\to B$ be a $G$-vector bundle and $s\colon B\to E$ be an equivariant smooth section such that $s^{-1}(0)$ is compact, where $G$ is a compact Lie group. From classical differential topology we know that for a smooth vector bundle $\pi\colon E\to B$ with a smooth section $s$, by a slight perturbation $p$ we can get the transversality of the perturbed section $s+p$. Moreover, we get a homology class $[(s+p)^{-1}(0)]\subset H_{*}(B;Z)$ which is independent of the choice of $p$. In the equivariant case is there similar result?

## 1 Answer

There are various approaches to defining equivariant Euler classes in the literature (see these course notes, or the works on equivariant orientations by authors such as Costenoble, May and Waner). Since I am no expert, I will not try to summarize these, but instead offer the following trivial example to illustrate that the "section transverse to zero section" approach to defining Euler classes may be doomed from the start.

Let $G=\mathbb{Z}/2$, and consider the two-dimensional real $G$-representation $V$ where the non-trivial element acts by $(x,y)\mapsto(-x,-y)$. This is a rotation, so preserves orientation. Regarded as a $G$-vector bundle $V\to \mathrm{pt}$, it seems that this should satisfy any reasonable definition of orientability of $G$-vector bundles. Thus it should have an equivariant Euler class $e(V)\in H^2_G(\mathrm{pt};\mathbb{Z})=H^2(BG;\mathbb{Z})$.

Now suppose we want to define the Euler class through transversality. We run into an immediate problem, in that there are **no** equivariant sections $\mathrm{pt}\to V$ other than the zero section itself (since $0$ is the only fixed point), and this is clearly not transverse to the zero section.

As Dylan remarked, the real problem is that transversality simply does not work in the same way for $G$-manifolds. There are lots of ways to try to circumvent this problem, though. Again, I'm afraid I'm not so knowledgeable, so I will just list some names: Wasserman, Costenoble--Waner, Field, Bierstone, tom Dieck, Illman and more.

**Edit:** It now occurs to me that there *is* an approach to the equvariant Euler class using transversality (I must have blanked it from my memory, as it appeared in my thesis!).

Say that a $G$-vector bundle $\xi\to X$ is *orientable* if the associated vector bundle $EG\times_G\xi\to EG\times_G X$ is orientable (here we are applying the functor given by the Borel construction, and $EG$ is some universal principal $G$-bundle). In this case $\xi$ has an equivariant Thom class $t^G_\xi\in H^n(EG\times_G\xi,EG\times_G\xi';\mathbb{Z})$ (where $\xi'$ denotes the subset of $\xi$ consisting of nonzero vectors).

Now we would like to say that $t^G_\xi$ is represented by the zero-section $i: EG\times_G X \hookrightarrow EG\times_G \xi$, and that the equivariant Euler class is represented by the inverse image under some transverse approximation $j$. This does not quite work, as $EG$ is not a smooth manifold in general (it is infinite dimensional). However, under very mild assumptions ($G$ compact Lie, perhaps?) we can find an $EG$ which is a *direct limit* of smooth free $G$-manifolds $EG^{(r)}$ of the indicated dimension.

As far as equivariant cohomology is concerned, for most purposes it usually suffices to work with these (large but) finite dimensional approximations. In particular, you can define the equivariant Euler class to be the class represnted by $j^{-1}(0)$ in $H^n(EG^{(r)}\times_G X;\mathbb{Z})$, where $j: EG^{(r)}\times_G X \to EG^{(r)}\times \xi$ is a smooth section transverse to the zero-section in the non-equivariant sense, and $r$ is sufficiently large compared to $n$.