Let $M$ be a closed, connected and oriented surface that is immersed in the sphere $\mathbb{S}^3$. Let $K$ denote the Gauss curvature of $M$ (the product of the principal curvatures) and denote by $K_{sec}$ the sectional curvature of $\mathbb{S}^3$ restricted to the tangent planes of $M$. I want to apply Gauss-Bonnet's Theorem to $M$. Should I integrate $K_{sec}$? That is, the following formula is true?
$$ \int_M K_{sec} \, \operatorname{d} \operatorname{vol}_M = 2 \pi \chi(M)$$
I suspect this is the case, because Gauss-Bonnet is intrinsic, whereas $K$ is extrinsic.
Best Answer
No, you need the Gaussian curvature of $M$ with the induced metric. (This is only going to be the product of the principal curvatures when $M$ is embedded in a flat space. Consider, for example, a great $2$-spheres in $S^3$, whose second fundamental form is identically $0$.) $K_{\text{sec}}=1$ for all $2$-planes in $T_pS^3$.
EDIT: There are classical formulas for the (intrinsic) curvature of a Riemannian surface in terms of its first fundamental form. This is also very nicely calculated using orthonormal moving frames to get the connection $1$-form $\omega_{12}$ and the intrinsic Gaussian curvature $K$ is given by the equation $-K\,dA = d\omega_{12}$. Indeed, using this equation, it's not hard to give a proof of the Gauss-Bonnet theorem using Stokes's Theorem.
In general, if $M\subset N$ is a submanifold of the Riemannian manifold $N$, then the Gauss equations relate the curvature of $M$ to the second fundamental form and the curvature tensor of $N$. In particular, when $N=S^3$ has constant curvature $1$, the formula you want is $$ K = k_1k_2 + 1.$$ Note that this checks with the basic examples we were discussing in comments.