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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.

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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.

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  • $\begingroup$ $K$ is defined to be the product of the principal curvatures...You fix a normal vector field $\eta$, consider the shape operator and take its determinant.. $\endgroup$ Aug 28 '18 at 1:45
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    $\begingroup$ That only gives you the correct result, as I said, when you're in a flat space. This is is how it's presented in $\Bbb R^3$, but it's wrong in general. I gave you an explicit counterexample in my answer. :) The general Gauss Bonnet (for surfaces and for higher even dimensions) is for the intrinsic Gaussian curvature. $\endgroup$ Aug 28 '18 at 1:49
  • $\begingroup$ If $M$ is a great $2$-sphere, then it is totally geodesic, i.e., the shape operator vanishes. Thus, the principal curvatures are zero. Where is the flaw? $\endgroup$ Aug 28 '18 at 1:52
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    $\begingroup$ The flaw is that by your reasoning the Euler characteristic of the sphere would be $0$!!! $\endgroup$ Aug 28 '18 at 1:54
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    $\begingroup$ You’re just wrong. Try any torus. Or prove the theorem ... $\endgroup$ Aug 28 '18 at 2:01
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Ted Shifrin already gave the explanation. You can deduce it from definition. Choose a local O.N. frame $\{e_1,e_2\}$ on $M$, with a normal vector $N$, it gives a O.N. frame for $S^3$. By definition \begin{eqnarray*} 1=K_{sec}=\tilde{K}(e_1,e_2)&=&<\tilde{R}(e_1,e_2)e_1,e_2>\\ &=& <R(e_1,e_2)e_1,e_2>+h_{12}^2-h_{11}h_{22}\\ &=& K_M-k_1k_2 \end{eqnarray*} Where we choose $e_i$ to be the principal directions. So $K_M=k_1k_2+1$ is the Guassian curvature of $M$ with the induced metric. Only in flat space you can say the Guassian curvature is the product of principal curvatures, but in our case, the ambient manifold is constantly curved space.

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