Justifying "determinant" of second fundamental form using the definition of sectional curvature

Question

Question: if $M\subseteq \widetilde{M}$ is a non-degenerate hypersurface in a pseudo-Riemannian manifold $M$, then is it true that $$\langle X, \widetilde{\nabla}_XN\rangle =\langle Y, \widetilde{\nabla}_YN\rangle $$for all $X,Y$ tangent to $M$, where $N$ is a unit normal to $M$?

Context: Today I saw in class the definition of sectional curvature and I thought that I might check as a self-posed exercise that the Gaussian curvature of a non-degenerate surface in Minkowski space is really given by the "determinant" $$K = \frac{\langle {\rm II}(X,X), {\rm II}(Y,Y) \rangle - \langle {\rm II}(X,Y), {\rm II}(Y,X)\rangle }{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2},$$as in page $118$ of Kühnel's Curves, Surfaces, Manifolds book, where ${\rm II}$ is the vector-valued second fundamental form and $X$ and $Y$ form a basis of the tangent plane to the surface. This always bothered me, so let's get on to it.

Assume that $M$ is a non-degenerate submanifold of a pseudo-Riemannian manifold $\widetilde{M}$ (with induced metric), with Levi-Civita connections $\nabla$ and $\widetilde{\nabla}$. Then $$\nabla_XY = \widetilde{\nabla}_XY - {\rm II}(X,Y),$$where ${\rm II}$ is the second fundamental form, normal to $M$. Terms in blue below are normal, so they die in the process. Great, then $$\begin{align} K^\nabla_{{\rm span}(X,Y)} &= \frac{\langle R^\nabla(X,Y)Y,X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2} \\ &= \frac{\langle \nabla_X\nabla_YY - \nabla_Y\nabla_XY - \nabla_{[X,Y]}Y,X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2} \\ &= \frac{\langle \widetilde{\nabla}_X\nabla_YY - \color{blue}{{\rm II}(X,\nabla_YY)} - \widetilde{\nabla}_Y\nabla_XY + \color{blue}{{\rm II}(Y,\nabla_XY)} - \widetilde{\nabla}_{[X,Y]}Y+\color{blue}{{\rm II}([X,Y],Y)},X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2} \\ &= \frac{\langle \widetilde{\nabla}_X\nabla_YY - \widetilde{\nabla}_Y\nabla_XY - \widetilde{\nabla}_{[X,Y]}Y,X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2} \\ &= \frac{\langle \widetilde{\nabla}_X\widetilde{\nabla}_YY - \widetilde{\nabla}_X({\rm II}(Y,Y)) - \widetilde{\nabla}_Y\widetilde{\nabla}_XY +\widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_{[X,Y]}Y,X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2} \\ &\color{darkred}{\stackrel{!!!}{=}\frac{\langle R^\widetilde{\nabla}(X,Y)Y+ \widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_X({\rm II}(X,X)), X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2}}\\ &= K^\widetilde{\nabla}_{{\rm span}(X,Y)}+ \frac{\langle \widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_X({\rm II}(X,X)),X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2}\end{align}$$This seems nice in general, so now assume $\widetilde{M} = \Bbb R^3_1$ (hence $K^\widetilde{\nabla}_{{\rm span}(X,Y)}=0$) and that $M$ is a surface. If $\epsilon = \langle N,N\rangle$, where $N$ is a unit (hence non-lightlike) normal field to $M$, we have that $${\rm II}(X,Y) = -\epsilon \langle Y, \widetilde{\nabla}_XN\rangle N.$$

Discarding normal components, we have: $$\begin{align} \langle \widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_X({\rm II}(X,X)),X\rangle &= \langle \widetilde{\nabla}_Y(-\epsilon \langle Y, \widetilde{\nabla}_XN\rangle N) - \widetilde{\nabla}_X(-\epsilon \langle X, \widetilde{\nabla}_XN\rangle N),X\rangle \\ &= \epsilon \left( \langle X, \widetilde{\nabla}_XN\rangle \color{red}{\langle X, \widetilde{\nabla}_XN\rangle} - \langle Y, \widetilde{\nabla}_XN\rangle\langle X, \widetilde{\nabla}_YN\rangle\right). \end{align}$$On the other hand: $$\begin{align} \langle {\rm II}(X,X), {\rm II}(Y,Y) \rangle - \langle {\rm II}(X,Y), {\rm II}(Y,X)\rangle &= \langle \langle X, \widetilde{\nabla}_XN\rangle N, \langle Y, \widetilde{\nabla}_YN\rangle N \rangle - \langle \langle Y, \widetilde{\nabla}_XN\rangle N, \langle X, \widetilde{\nabla}_YN\rangle N\rangle \\ &= \epsilon \left( \langle X, \widetilde{\nabla}_XN\rangle \color{red}{\langle Y, \widetilde{\nabla}_YN\rangle} - \langle Y, \widetilde{\nabla}_XN\rangle\langle X, \widetilde{\nabla}_YN\rangle \right)\end{align}$$

Did I screw up somewhere in the end, or do we have $\langle X, \widetilde{\nabla}_XN\rangle =\langle Y, \widetilde{\nabla}_YN\rangle $?

Answer 1

As Anthony pointed in the comments, I changed $\widetilde{\nabla}_X({\rm II}(Y,Y))$ to $\widetilde{\nabla}_X({\rm II}(X,X))$ without realizing. I'll indicate the mistake in the original question in $\color{darkred}{\mbox{red/brown}}$. Here goes the right one:

$$\begin{align*} K^\nabla_{{\rm span}(X,Y)} &= \frac{\langle R^\widetilde{\nabla}(X,Y)Y+ \widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_X({\rm II}(Y,Y)), X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2} \\ &= K^\widetilde{\nabla}_{{\rm span}(X,Y)}+ \frac{\langle \widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_X({\rm II}(Y,Y)),X\rangle}{\langle X,X\rangle\langle Y,Y \rangle - \langle X,Y \rangle^2}. \end{align*}$$Now we use the expression ${\rm II}(X,Y) = -\epsilon \langle Y, \widetilde{\nabla}_XN\rangle N$ to get $$\begin{align}\langle \widetilde{\nabla}_Y({\rm II}(X,Y)) - \widetilde{\nabla}_X({\rm II}(Y,Y)),X\rangle &= \langle \widetilde{\nabla}_Y(-\epsilon\langle Y, \widetilde{\nabla}_XN\rangle N) - \widetilde{\nabla}_X(-\epsilon\langle Y, \widetilde{\nabla}_YN\rangle N),X\rangle \\ &= \epsilon \left( \langle Y, \widetilde{\nabla}_YN\rangle\langle \widetilde{\nabla}_XN,X\rangle - \langle Y, \widetilde{\nabla}_XN\rangle\langle \widetilde{\nabla}_YN,X\rangle\right),\end{align}$$as wanted. The computation actually holds for hypersurfaces in pseudo-Euclidean spaces $\Bbb R^n_\nu$ of arbitrary signature.