1
$\begingroup$

Let $M$ is a $2n$-dim smooth Riemannian manifold, $\eta$ is a vector field on $M$, $p\in M$ is a isolated zero of $\eta$. Then we can define a map from $\partial B_{p}(\varepsilon)$ to $S^{2n-1}(1)$ by $\eta$: $$\eta_{p}=\frac{\eta}{|\eta|}: \partial B_{p}(\varepsilon)\rightarrow S^{2n-1}(1)$$

Here $B_{p}(\varepsilon)$ is a ball about of $p$ with radius $\varepsilon$.

Let $f_1,\cdots,f_{2n-1}$ be an orthonormal basis of $T(\partial B_{p}(\varepsilon))$, let $f^{*}_1,\cdots,f^{*}_{2n-1}$ be the metric dual basis.

We can find that

$$\eta^{*}\wedge(\nabla^{TM}_{f_1}\eta)^{*}\wedge(\nabla^{TM}_{f_{2n-1}}\eta)^{*}$$

$$= (d{\rm vol}_{g^{TM}})\int^{B}\eta^{*}\wedge (\nabla^{TM}_{f_1}\eta)^{*}\wedge\cdots\wedge (\nabla^{TM}_{f_{2n-1}}\eta)^{*}$$

then

$$f^{\ast}_{1}\wedge\cdots\wedge f^{\ast}_{2n-1}\int^{B}\eta^{\ast}\wedge (\nabla^{TM}_{f_1}\eta)^{\ast}\wedge\cdots\wedge (\nabla^{TM}_{f_{2n-1}}\eta)^{\ast}=\eta^{\ast}_{p}\omega ,(1) $$

here $\omega$ be the volume form on $S^{2n-1}(1)$.

I find this formula (1) in the article “$\eta$-invariants and the Poincaré-Hopf index formula”. But I don't know how to get it. Is there some way to compute it?

$\endgroup$
4
  • 1
    $\begingroup$ You should explain your notations, because the map you first define does not make sense to me (you should explain how you identify the various tangent spaces with $\mathbb{R}^n$. $\endgroup$ Mar 1, 2011 at 18:27
  • $\begingroup$ Also, I guess $p$ is an isolated zero. $\endgroup$ Mar 1, 2011 at 19:09
  • $\begingroup$ @Majer: Yes, $p$ is an isolated zero. Kloechner: the map $\eta_{p}$ is to unitization the tangent vector on $\partial B_{p}(\varepsilon)$. $\endgroup$
    – Chen
    Mar 2, 2011 at 4:19
  • $\begingroup$ @Kloechner: Could you tell me why the map is no sense? Why i need to identify the various tangent spaces with $\mathbb{R}^{n}$? Thanks $\endgroup$
    – Chen
    Mar 2, 2011 at 13:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.