# A question related to the degree of vector field

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?

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@Chen: I tried to help you fix the TeX and display issues. It turns out you are editing at the same time, which caused massive breakage. Please see Revision 7 by me and the side box on "How to write math" to see how to make your formulae properly displayed. Also note that TeX notation is not available outside the "math environment" delimited by dollar signs, so for Poincaré you will have to actually type in the unicode character, or copy and paste from a utility like math.ntnu.no/~stacey/code/latexToUTF/utf.php – Willie Wong Mar 1 '11 at 10:57
@Wong: Thanks very much! – Chen Mar 1 '11 at 11:08
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 $\mathbb{R}^n$. – Benoît Kloeckner Mar 1 '11 at 18:27
Also, I guess $p$ is an isolated zero. – Pietro Majer Mar 1 '11 at 19:09
@Majer: Yes, $p$ is an isolated zero. @Kloechner: the map $\eta_{p}$ is to unitization the tangent vector on $\partial B_{p}(\varepsilon)$. – Chen Mar 2 '11 at 4:19