# Doubt on Do Carmo proof of Gauss Formula

I was checking the proof of Gauss theorem in do Carmo's book on Riemannian Geometry (p.131) and got stuck in one of the steps. The theorem of Gauss as stated on Do Carmo goes as:

Let $p\in M$ and let $x,y$ be orthonormal vectors in $T_pM$. Then** $$K(x,y)-\bar{K}(x,y)=\langle B(x,x),B(y,y)\rangle - \vert B(x,y)\vert^2.$$

Proof:
Let $X,Y$ be local orthogonal extensions of $x,y$, respectively, which are tangent to $M$; we denote the local extensions to $\bar{M}$ of $X,Y$ by $\bar{X},\bar{Y}$. Then

$$K(x,y)-\bar{K}(x,y)=\langle \nabla_Y\nabla_X X -\nabla_X\nabla_Y X-(\bar{\nabla}_\bar{Y}\bar{\nabla}_\bar{X}\bar{X}-\bar{\nabla}_\bar{X}\bar{\nabla}_\bar{Y}\bar{X}),Y\rangle(p)+\langle\nabla_{[X,Y]}X-\bar{\nabla}_{[\bar{X},\bar{Y}]}\bar{X},Y\rangle(p)$$

The last term being zero

On the other hand, if we denote by $E_1,\cdots,E_m$,$m=\dim\bar{M}-\dim M$ local orthonormal fields which are normal to $M$, we have

$$B(X,Y)=\sum_i H_i(X,Y)E_i,\ \ H_i=H_{E_i}, \ \ i=1,\cdots, m.$$

Therefore, at $p$,

$$\bar{\nabla}_{\bar{Y}}\bar{\nabla}_{\bar{X}}\bar{X}=\bar{\nabla}_{\bar{Y}}(\sum_i H_i(X,X)€_i+\nabla_X X)\\ =\sum_i(H_i(X,X)\bar{\nabla}_{\bar{Y}}E_i+\bar{Y}H_i(X,X)E_i)+\bar{\nabla}_{\bar{Y}}\nabla_X X .$$ Hence, at $p$, $$\langle\bar{\nabla}_{\bar{Y}}\bar{\nabla}_{\bar{X}}\bar{X},Y\rangle=-\sum_iH_i(X,X)H_i(Y,Y)+\langle\nabla_Y\nabla_X X,Y\rangle.$$

My question: I do not understand how he gets to this expression from the last one. Where does the $H_i(Y,Y)$ (and the minus sign) come from? I think the second term in the sum vanishes as $E_i$ is normal to $Y$, but the first term got me stuck.

## 1 Answer

I believe do Carmo is using an unstated fact here. Not only can our vector fields $X$ and $Y$ be extended to $\bar{X}$ and $\bar{Y}$ in the ambient manifold $\bar{M}$, but they can be extended in such a way that their orthogonal components $\langle\bar{X},E_i\rangle$ and $\langle\bar{Y},E_i\rangle$ are constant. This can always be done, at least locally, and recall that $B(x,x)$ is independent of the choice of extension.

Then from the metric preserving property of the Levi-Civita connection, we have $$0=\bar{X}\langle E_i,\bar{Y}\rangle=\langle\bar{\nabla}_{\bar{X}}E_i,\bar{Y}\rangle+\langle E_i,\bar{\nabla}_{\bar{X}}\bar{Y}\rangle$$ and $$0=\bar{Y}\langle E_i,\bar{Y}\rangle=\langle\bar{\nabla}_{\bar{Y}}E_i,\bar{Y}\rangle+\langle E_i,\bar{\nabla}_{\bar{Y}}\bar{Y}\rangle.$$

So we have $\langle\bar{\nabla}_{\bar{X}}E_i,\bar{Y}\rangle=-\langle E_i,\bar{\nabla}_{\bar{X}}\bar{Y}\rangle=-H_i(X,Y)$ and $\langle\bar{\nabla}_{\bar{Y}}E_i,\bar{Y}\rangle=-\langle E_i,\bar{\nabla}_{\bar{Y}}\bar{Y}\rangle=-H_i(Y,Y).$

So starting with the equation

$$\bar{\nabla}_{\bar{X}}\bar{\nabla}_{\bar{Y}}\bar{X}=\left(\sum_i H_i(Y,X)\bar{\nabla}_{\bar{X}}E_i+\bar{X}H_i(Y,X)E_i\right)+\bar{\nabla}_{\bar{X}}\nabla_{Y}X$$ and taking the inner product with $Y$ gives us $$\langle\bar{\nabla}_{\bar{X}}\bar{\nabla}_{\bar{Y}}\bar{X},Y\rangle=\left(\sum_i H_i(Y,X)\langle\bar{\nabla}_{\bar{X}} E_i,Y\rangle+\bar{X}H_i(Y,X)\langle E_i,Y\rangle\right)+\langle \bar{\nabla}_{\bar{X}}\nabla_{Y}X,Y\rangle.$$

Then since $E_i$ is in the orthogonal complement, we have $\langle E_i,Y\rangle=0.$ Since $Y$ is in the tangent space to $M$, only the parallel component of $\bar{\nabla}_{\bar{X}}\nabla_{Y}X$ contributes and we have $\langle \bar{\nabla}_{\bar{X}}\nabla_{Y}X,Y\rangle=\langle \nabla_{X}\nabla_{Y}X,Y\rangle.$

Putting it all together, our equation becomes

$$\langle\bar{\nabla}_{\bar{X}}\bar{\nabla}_{\bar{Y}}\bar{X},Y\rangle=-\sum_i H_i(Y,X)H_i(X,Y)+\bar{X}H_i(Y,X)\langle E_i,Y\rangle+\langle \nabla_{X}\nabla_{Y}X,Y\rangle.$$

Similarly, we have

$$\langle\bar{\nabla}_{\bar{Y}}\bar{\nabla}_{\bar{X}}\bar{X},Y\rangle=-\sum_i H_i(X,X)H_i(Y,Y)+\bar{X}H_i(Y,X)\langle E_i,Y\rangle+\langle \nabla_{Y}\nabla_{X}X,Y\rangle,$$

as required.