The degree of Gauss map

501 Views Asked by At

If $M$ is an $2m$-dimensional closed orientable hypersurface in $\mathbb R^{2m+1}$, then we have a Gauss map $G:M\rightarrow S^{2m}$.

I have known from my differential geometry book that deg$G=\frac{1}{2}\chi(M)$ where $\chi(M)$ is the Euler characteristics of $M$. So my question is: is there an easy way to prove this conclusion if we assume the Poincaré–Hopf theorem and why is this argument only applied to the even dimension?

1

There are 1 best solutions below

0
On

The Euler characteristic is zero for any odd-dimensional compact orientable manifold (this follows from Poincaré duality).