2
$\begingroup$

The generalized Gauss-Bonnet theorem says that:

Let $M$ be a closed oriented Riemannian manifold with an even dimension $n$, then $$ \int_{M}\Omega=\chi(M). $$

In that formula, $\chi$ is the euler characteristic and $$ \Omega =\frac{(2\pi)^{-\frac{n}{2}}}{2^{n}(\frac{n}{2})!}\sum_{\sigma,\tau\in S_{n}}g^{-1}\mathrm{sgn}(\sigma)\mathrm{sgn}(\tau)R_{\sigma(1)\sigma(2)\tau(1)\tau(2)}\cdots R_{\sigma(n-1)\sigma(n)\tau(n-1)\tau(n)}, $$ where $g$ is the determinant of the matrix of the metric and, if $R$ is the curvature tensor $$ R_{i,j,k,l}=\langle R(e_{i},e_{j})e_{l},e_{k}\rangle. $$

First of all, I would like to know if the formula that I have written for $\Omega$ is the correct one. Now, does this coincide with the Gauss-Bonnet Theorem in the two dimensional case? I mean, if $n=2$, then, taking into account the properties of the curvature tensor $$ \Omega=\frac{1}{2\pi}\cdot\frac{R_{1212}}{g}. $$ Is this $\frac{1}{2\pi}\mathcal{K}$ if $M$ is a surface in $\mathbb {R}^{3}$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.