Good morning/day/evening/night,
I was presented to the generalized Gauss-Bonnet-Chern theorem for hypersurfaces in Euclidean space;
For a closed, even dimensional manifold $M$ with dimension $n$ embedded in $\mathbb{R}^{n+1}$ we have
$$\int_M K\mathbb dV = \text{Volume }\mathbb{S}^n\cdot\frac{\chi(M)}{2}.$$
I wonder,
- How does this theorem look for compact manifolds with boundary?
- Can this theorem be generalized to hold for closed odd dimensional manifolds? (I assume there will be a problem here since the Euler characteristic vanishes.)
Both questions are for hypersurfaces in Euclidean space.
Any reference will be greatly appreciated.
Best Answer
Let $\nu: M \to \mathbb S^n$ be a normal unit vector field along $M$, then the derivative $d\nu$ of $\nu$ maps $T_pM$ to $T_{\nu(p)}S = \nu(p)^\perp = T_pM$ and the Gaussian curvature is given by $$K_p = \det(d\nu(p): T_pM \to T_pM)$$
Now the volume form on $M$ is given by $\mathrm{d}vol_M = \iota(\nu) \mathrm{d}vol_{\mathbb R^{n+1}}$, i.e. for tangent vectors $\xi_1,\dots, \xi_n \in T_pM$ we have $$\mathrm{d}vol_M(p)(\xi_1, \dots, \xi_n) = \mathrm{d}vol_{\mathbb R^{n+1}}(p)(\nu(p), \xi_1, \dots, \xi_n) = \det(\nu(p), \xi_1, \dots, \xi_n)$$
Now consider the pullback $\nu^\ast \mathrm{d}vol_{S^n}$ of $\mathrm{d}vol_{S^n}$ to $M$:
\begin{eqnarray*} \nu^\ast \mathrm{d}vol_{S^n}(p)(\xi_1, \dots, \xi_n) &=& \mathrm{d}vol_{S^n}\left(\nu(p)\right)\left(d\nu(p) \xi_1, \dots, d\nu(p)\xi_n\right) \\ &=& \det\left(\nu(p), d\nu(p) \xi_1, \dots, d\nu(p)\xi_n\right) \\ &=& K_p\cdot \det\left(\nu(p), \xi_1, \dots, \xi_n\right) \\ &=& K_p \; \mathrm{d}vol_M \end{eqnarray*}
Therefore $$\int_M K \; \mathrm{d}vol_M = \int_M \nu^\ast \mathrm{d}vol_{S^n} = \deg(\nu) \int_{S^n} \, \mathrm{d}vol_{S^n} = \deg(\nu) \cdot \text{Volume }S^n$$
For even $n$, we have $\deg(\nu) = \frac{\chi(M)}2$, so I guess one might consider this to be a generalization to odd dimensional manifolds.