Gaussian Curvature for Poincare upper half plane embedded in Minkowski

Question

I started with the mapping of Poincaré half plane in Minkowski space $\mathbb{R}^{(1,2)}$ by the mapping $f(x,y)\to(X_0,X_1,X_2)$ via $$ X_0=\frac{x^2+y^2+1}{2 y}\notag\\ X_1=\frac{x}{y}\notag\\ X_2=\frac{x^2+y^2-1}{2 y}.\notag\\ $$ The metric on $\mathbb{R}^{(1,2)}$ is the Minkowski metric given by $$ ds^2=-dX_0^2+dX_1^2+dX_2^2. $$ The induced metric on the Poincaré half plane model is $$ ds^2=\frac{dx^2+dy^2}{y^2}. $$ The orthonormal frame choosen for $\mathbb{R}^{(1,2)}$ is given by $$ e_1=\{\frac{X_1}{(X_0-X_2)},1,\frac{X_1}{(X_0-X_2)}\}\notag\\ e_2=\{\frac{(1-X_1^2-(X_0-X_2)^2)}{2(X_0-X_2)},-X_1,\frac{(1-X_1^2+(X_0-X_2)^2)}{2(X_0-X_2)}\}\notag\\ n=\{\frac{(1+X_1^2+(X_0-X_2)^2)}{2(X_0-X_2)},X_1,\frac{(1+X_1^2-(X_0-X_2)^2)}{2(X_0-X_2)}\}\notag\\ $$ where $e_1,e_2$ are tangent to the disk and are spacelike while the normal $n$ is chosen as timelike ($\langle n,n\rangle=-1$). The shape operator is given by $S(e_a)=-(\nabla_{e_a}n)^T$. For this case $\nabla=\partial$ the operator is given by $$ S(e_1)=-e_1,~~~S(e_2)=-e_2. $$
Thus the shape operator is proportional to the opposite identity matrix $S=-I$. The determinant of the shape operator is the Gaussian Curvature. For this case it is coming as $K=1$. But according to the literature it should be $K=-1$. I haven't found this discussion of embedding in space with signature elsewhere. Hence I am posting it here. As I am not very much familiar with this area any comment will be helpful.

Answer 1

The correct "definition" of the Gaussian curvature for surfaces embedded in Minkowski space (if you don't want to assume knowledge about the Riemann curvature tensor; in which case you can deduce the formula to come) is $$K = \varepsilon \det(S),$$where $\varepsilon$ is the "sign" of the normal direction to the surface, i.e., $1$ if the surface is timelike (so the normal direction is spacelike) and $-1$ if the surface is spacelike (so the normal direction is timelike).

You're right that in this case, $S$ equals minus the identity, whose determinant is $1$. But the unit normal field to $\mathbb{H}^2$ is just the position vector field, which is timelike. So you gain an extra minus sign, and $K=-1$. This adjustment is needed because the Minkowski metric is not Riemannian, but instead Lorentzian (even though it restricts to a Riemannian metric on $\mathbb{H}^2$).

Some books (such as Weinstein's "Introduction to Lorentz Surfaces") do not pay attention to this (so her notion of "Gaussian curvature" differs from the sectional curvature computed from the Riemann tensor by a sign on spacelike surfaces). This also reminds me that there's a book (I think it's Eisenhart's "Non-Riemannian Geometry", although I'm not sure) which doesn't care about different sign conventions for the curvature tensor, so he ends up having spheres with negative Ricci curvature, go figure.

Your question is more relevant than what it might sound, there's something here much deeper going on here. When I was learning this, I took a good two months to accept $\varepsilon$ in the definition precisely because of this example.