In the following, I will refer by O'Neil to Barrett O'Neil's book Semi-Riemannian Geometry With Applications to Relativiy.
Consider the $M=\{ x \in \mathbb{R}^n_s | f(x)=r^2\}$ with $f(x)=g_s(x,x)$ and $g_s(x,y)=-\sum_{i=1}^{s}x_iy_i+\sum_{i=s+1}^{n}x_iy_i$. It is a hypersurface of the pseudo-Euclidean space $\mathbb{R}^n_s=(\mathbb{R}^n,g_s)$.
I would like to compute its shape operator (or Weingarten map). Here are my steps:
- The gradient of $f$ is grad$f(x)=\frac{2}{r}(-x_1,...,-x_s,x_{s+1},...,x_n)^t$
- By O'Neil (Proposition 4.17, p.106) $n(x)=\frac{1}{r}(-x_1,...,-x_s,x_{s+1},...,x_n)^t$ defines a unit normal vector field of $M$.
- The shape operator is then given by $-\widetilde{\nabla}_vn(x)$ with $\widetilde{\nabla}$ being the Levi-Civita connection of $\mathbb{R}^n_s$.
- Using the formula $\widetilde{\nabla}_VZ=\sum_iV[Z^i]\partial_i$ (with a local frame), I obtain the following for the shape operator: $S_x(v)=\frac{1}{r}\sum_{i=1}^sv_i\partial_i-\frac{1}{r}\sum_{i=s+1}^nv_i\partial_i$
However, O'Neil states $S=-I/r$. (Lemma 4.27, p.111)
What am I missing?
Addendum
I believe I made a silly mistake when computing the covariant derivative:
$\widetilde{\nabla}_vn\neq\frac{1}{r}(-\sum_{i=1}^sv_i\partial_i+\sum_{i=s+1}^nv_i\partial_i)$
Instead it should read
$\widetilde{\nabla}_vn=\frac{1}{r}(\sum_{i=1}^sv_i\partial_i+\sum_{i=s+1}^nv_i\partial_i)$.
Derivating $-x_i$ flips the sign again.