# Gauss Theorem in Riemannian geometry

Theorem. Let $p \in M$ and let $x,y$ be orthonormal vectors in $T_pM$. Then \begin{align} \tag{1} K(x,y)-\overline K(x,y)=\langle B(x,x),B(x,y)\rangle - |B(x,y)|^2. \end{align}

Proof. Let $X,Y$ be local orthogonal extensions of $x,y$, respectively, which are tangent to $M$; we denote the local extensions to $\overline M$ of $X,Y$ by $\overline X,\overline Y$. Then \begin{align} K(x,y)-\overline K(x,y) &= \langle \nabla_Y \nabla_X X-\nabla_X \nabla_Y X -(\overline \nabla_{\overline Y} \overline \nabla_{\overline X}\overline X-\overline \nabla_{\overline X} \overline \nabla_{\overline Y}\overline X) \rangle(p) \\ &\qquad+ \langle \nabla_{[X,Y]}X-\overline \nabla_{[\overline X,\overline Y]}\overline X,Y \rangle(p). \end{align} Observe, first of all, that the last term is zero, because $$\langle \nabla_{[X,Y]}X-\overline \nabla_{[\overline X,\overline Y]}\overline X,Y \rangle(p) = -\langle (\overline \nabla_{[\overline X,\overline Y]}\overline X)^N,Y \rangle(p) = 0.$$ On the other hand, if we denote by $E_1,\ldots,E_m$, $m=\dim \overline M-\dim M$, local orthonormal fields which are normal to $M$, we have $$B(X,Y)=\sum_i H_i(X,Y)E_i, \quad H_i=H_{E_i}, \quad i=1,\ldots,m.$$ Therefore, at $p$, \begin{align} \overline \nabla_{\overline Y} \overline \nabla_{\overline X} \overline X &= \overline \nabla_{\overline Y} \left(\sum_i H_i(X,X)E_i + \nabla_X X \right) \\ &= \sum_i \left\{H_i(X,X) \overline \nabla_{\overline Y}E_i + \overline Y H_i(X,X)E_i \right\} + \overline \nabla_{\overline Y} \nabla_X X. \end{align} Hence, at $p$, \begin{align} \tag{2} \langle \overline \nabla_{\overline Y} \overline \nabla_{\overline X} \overline X,Y \rangle = -\sum_i H_i(X,X)H_i(Y,Y)+\langle \nabla_Y \nabla_X X,Y \rangle. \end{align} Similarly, \begin{align} \tag{3} \langle \overline \nabla_{\overline X} \overline \nabla_{\overline Y} \overline X,Y \rangle = -\sum_i H_i(X,Y)H_i(X,Y)+\langle \nabla_X \nabla_Y X,Y \rangle. \end{align} Using $(2)$ and $(3)$, we obtain $(1)$.

This is from Riemannian Geometry by Manfredo do Carmo, pp. 130--131.

How can we justify $(2)$?

I am asking exactly this unanswered question, but in my post I provided do Carmo's proof of the theorem for convenience. I hope this is substantial enough so that this post is not marked as a duplicate.

And that previous question suggested using the fact that $E_i$ and $Y$ are orthogonal; maybe that fact will play a role in the author's justification.

Update: I just figured out that \begin{align} \overline \nabla_{\overline Y} \nabla_X X &= (\overline \nabla_{\overline Y} \nabla_X X )^T+(\overline \nabla_{\overline Y} \nabla_X X )^N \\ &= \nabla_Y \nabla_X X + (\overline \nabla_{\overline Y} \nabla_X X )^N, \end{align} so that \begin{align} \langle \overline \nabla_{\overline Y} \nabla_X X ,Y\rangle &= \langle \nabla_Y \nabla_X X ,Y\rangle+\langle (\overline \nabla_{\overline Y} \nabla_X X )^N ,Y\rangle \\ &=\langle \nabla_Y \nabla_X X ,Y\rangle, \end{align} as $(\overline \nabla_{\overline Y} \nabla_X X )^N$ is a vector field normal to $M$ and $Y$ is a vector field tangent to $M$. This resolves justifying the second term of $(2)$.

So I am now asking only how to justify the first term of $(2)$: that is, asking how is it that $$\langle \overline \nabla_{\overline Y}E_i + \overline Y H_i(X,X)E_i ,Y\rangle = -H_i(X,X)H_i(Y,Y),$$ in order to completely justify the derivation of $(2)$?

• $(E_i,Y)=0$ and use $Y(E_i,Y)=0$ – HK Lee Mar 5 '17 at 9:20
• @HKLee I applied the identities you suggested and I was able to derive the desired result. :) Now, I know that $(E_i,Y)=0$ because $E_i$ is normal to $M$ whereas $Y$ is tangent to $M$. But it is still not clear to me that $Y(E_i,Y)=0$; if I can ask, why is this true? – New day rising Apr 19 '17 at 6:25
• $(E_i,Y)=0$ on some open set in $M$ So it is constant along curve $c$ in $M$ s.t. $c'(0)=Y$. – HK Lee Apr 19 '17 at 7:06