8
$\begingroup$

Let $V \rightarrow M$ be an oriented rank $k$ vector bundle over a compact orientd manifold $M$. Let $X \subset M$ be a compact topological subspace of $M$ that is a smooth oriented submanifold of dimension $k$, except possibly at a a set of points that have ``dimension'' less than or equal to $k-2$. More precisely, the set of singular points is contained inside a submanifold of dimension $k-2$ or less.

Let $s :X \rightarrow V$ be the restriction of a smooth section from $M$ to $V$. Assume that when restricted to $X$, the section vanishes only on the smooth points of $X$, and it vanishes transversally. Is it true that the number of zeros of $s$ inside $X$, counted with a sign is the Euler class of $V$ evaluated on the fundamental class of $X$ , ie $$ +-|s^{-1}(0)| = \int_{[X]} e(V) $$

We need $X$ to have singularities of dimension $k-2$ or less, to ensure that it defines a homology class $[X]$ (ie to make sure that the integration actually makes sense). Note that $[X]$ is an element of $H_k(M, \mathbb{Z})$ and $e(V) \in H^k(M, \mathbb{Z})$. So the expression makes perfect sense, even though $X$ is a singular space.

I believe this statement is true, but is there a reference for this fact?

$\endgroup$

2 Answers 2

3
$\begingroup$

You make several assumptions, one being that $X$ is a stratified space, carrying an orientation class. The answer to you question is yes, under more restrictive assumptions. Here they are.

  • The section $s$ vanishes transversally along $M$.
  • The zero set $s^{-1}(0)$ intersects the stratified space $X$ transversally.

The complete rigorous proof is a bit more involved, and the clearest argument I know is sheaf theoretic, and it involves a sheaf theoretic version of the Poincare duality, also known as Verdier duality. All the information you need you can find in B. Iversen's book Cohomology of Sheaves, especially chapters IX and X.

Addendum I realize you do not need these stringent conditions. Denote by $V_X$ the restriction of $V$ to $X$ and by $\tau_X$ its Thom class viewed as a class in the local cohomology of $V_X$ along $X$, $\tau_X\in H^k_X(V_X)$ (integer coefficients). We can then arrange that $\tau_X$ has support in a tiny neighborhood of $X$ in $V_X$. Then $e(V_X)=s^*\tau_X\in H^k(X)$ is supported in a tiny open neighborhood $N$ of $s^{-1}(0)\cap X$ in $X$, i.e., $e(V_X)$ is in the image of $H^k(X, X\setminus N)$ in $H^k(X)$. Now use the technology in Iversen to conclude that

$$ \langle e(V_X), [X]\rangle =\sum_{s(x)=0} \epsilon(x), $$

where $\epsilon(x)\in\{\pm 1\}$ is the local Euler number of $S$ at $x$, and $\langle-,-\rangle $ denotes the pairing between cohomology and homology. (This is a bit long.)

$\endgroup$
1
  • 1
    $\begingroup$ I was going to write something similar to your addendum. The detailed argument seems quite messy, but it might help to note the following: If $p\in X$ then the Thom class restricts to a generator in the local cohomology $H^k(V_p,V_p-0)$ of the fiber at zero, and the fundamental class $[X]$ restricts to a generator in the local homology $H_k(X,X-p)$ if $p$ is nonsingular (both of these groups being isomorphic to the integers). The sign of a zero $p\in s^{-1}(0)$ is then the product of these generators. $\endgroup$
    – Mark Grant
    Jan 15, 2013 at 14:08
0
$\begingroup$

This statement is true for "radial" vector fields with respect to $X$, see Theorem 5.11 here. Note that radial vector fields are allowed to vanish at singular points of $X$. If you go through the proof, maybe you can modify it so that it holds for all vector fields satisfying your conditions. The result itself might be contained in papers of M.Schwarz cited in Brasselet's notes linked above, take a look.

$\endgroup$
1
  • $\begingroup$ I looked a bit at that section. The problem is, I am asking my question for sections of a vector bundle, not vector fields (which are sections of the Tangent bundle). Is it conceivable that I can modify these arguments to hold for sections of vector bundles? $\endgroup$
    – Ritwik
    Jan 15, 2013 at 6:07

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.