# Gauss Theorem - Riemannian Geometry

I want to prove Gauss' Theorem:

"Let p $\in$ M and x,y orthonormal vectors of $T_p M$. So k(x,y)-$\bar{k}$(x,y)= < B(x,x),B(y,y)> - |B(x,y)|$^2$.

From Riemannian Geometry, Manfredo do Carmo

We have:

$X,Y \in \Gamma (TU)$ where $U$ is a neighborhood in $M$

$\bar{X},\bar{Y}$ are extensions of $X,Y$ in $\bar{M}$

$\nabla$ is a Riemannian connection in $M$

$\bar{\nabla}$ is a Riemannian connection in $\bar{M}$

$H_\eta(x,y)= < B(x,y),\eta >, x,y\in T_p M$

$B(X,Y)=\bar{\nabla}_{\bar{X}} \bar{Y} - \nabla _{X} Y$

I'm stuck in this step:

$< \bar{\nabla} _ \bar{Y} \bar{\nabla}_\bar{X} \bar{X},Y> = \sum_i < H_i(X,X)\bar{\nabla}_\bar{Y} E_i + \bar{Y} H_i (X,X) E_i, Y> + <\bar{\nabla} _ \bar{Y} \nabla_X X,Y>$

And i have to prove that this is:

$< \bar{\nabla} _ \bar{Y} \bar{\nabla}_\bar{X} \bar{X},Y> = - \sum_i H_i(X,X)H_i (Y,Y) + <\bar{\nabla} _ \bar{Y} \nabla_X X,Y>$

Can someone help with a hint of how to go to this step?

I think i can do $<\bar{Y} H_i (X,X) E_i, Y>=0$ because $E_i$ and $Y$ are orthogonal.