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Let $V$ be a smooth manifold, $E \rightarrow V$ a vector bundle over $V$ and $\Gamma$ be a finite group acting nontrivially on $V$ and $E$. Let $s \in C^\infty(E)$ be a generic $\Gamma$-equivariant section.

Genericity of $s$ implies that the zero locus $Z := s^{-1}(0)$ of $s$ is a singular manifold.

The order of vanishing of $s$ determines a stratification of $Z$ into bits $Z_k$, where $Z_k$ is given by all $z \in Z$, such that $s$ vanishes of order $k$ at $z$, that is $$Z_k = \lbrace z \in Z: \nabla s(z) =0, \dots, \nabla^{k}s(z) =0, \nabla^{k+1}(z) \neq 0 \rbrace$$

My Question is: are the $Z_k$'s nonisngular manifolds? To be more precise, I am interested in the $Z_k$-bit with $k$ largest so that $Z_k \neq \emptyset$.

My feeling is that $Z_k$ does not necesseraly have to be a (nonsingular) manifold, because it could consist of many complicated different bits, but I couldn't come up with a proof.

Any References would also be appreciated.

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What do you mean by "genericity implies that the zero locus is singular"? Usually genericity implies regluarity... – diverietti May 1 '12 at 17:36
Well yes, but you still have the group action. Therefore the zero locus is a (singular) manifold. – Spinorbundle May 1 '12 at 19:18
Not really... In your setting one could very well take the trivial action. – diverietti May 2 '12 at 7:06
Sorry for that! I forgot to mention that the group acts nontrival on V and E... – Spinorbundle May 2 '12 at 13:06
@Spinorbundle: In your terminology, is a smooth manifold a particular type of singular manifold? – Mark Grant May 2 '12 at 13:42

I also do not think that your notion of genericity is correct (since it does not give the right answers if $\Gamma$ is trivial). However, I am also pretty sure that your way to stratify $s^{-1}(0)$ is wrong. Indeed, it is natural to set $Z_0:= Z\setminus Z_1$ (otherwise you do not even get a stratification of $Z$). Then it could easily happen that $Z_1=\emptyset$ but $Z_0$ is singular. For instance, take the base equal to ${\mathbb R}^2$, the action of $\Gamma={\mathbb Z}_2$ on ${\mathbb R}^2$ generated by reflection in a line $L$, trivial line bundle over ${\mathbb R}^2$ with trivial action of $\Gamma$ on the fiber. The point is that $\nabla s$ will vanish in the direction normal to $L$ and will be (generically) nonzero along $L$.

On the other hand, Y-G. Oh defines in normally polynomial sections, where the notion of genericity is the standard one (since spaces of sections of bounded degrees are now finite-dimensional). Then the correct stratification of $s^{-1}(0)$ is by the subgroups of $\Gamma$ (stratum of a point $x$ in the base is given by its stabilizer in $\Gamma$). Then in Proposition 35.13 Oh proves that (provided that the degrees of polynomial sections are high enough) each stratum of $s^{-1}(0)$ for a generic polynomial section $s$ is a smooth submanifold of the expected dimension.

I do not think that the answer given in Oh's paper is optimal since ideally, one would like the set of "generic sections" to be not only dense in the space of all smooth sections (which Oh proves) but also open, which, of course, will not be the case in his setting. However, if one is interested in having an abundant supply of sections whose zero sets are stratified by manifolds of "expected" dimensions, then his construction suffices. This is useful, for instance, if one were to develop an obstruction theory for orbifold bundles. One should be able use the notion of equivariant transversality developed by Bierstone in to get the better result.

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Thanks for the link. I don't understand what you mean with "wrong". I am not claiming that such a stratification (perhaps the word is misleading) exists, or that the V_k's have to be nonempty, but assume we have a V_k with k maximal, is it then true that V_k is a nonsingular manifold? – Spinorbundle May 2 '12 at 13:13
It is wrong in the sense that it does not yield a stratification by smooth manifolds, while there is another stratification that does work. In my example Z is singular, one can easily modify it to get example where $Z_1$ is singular. – Misha May 2 '12 at 13:46

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