# Zero locus of a generic smooth section

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.

-
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 http://www.math.wisc.edu/~oh/normallypolynomial.pdf 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.
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