Calculate the Gaussian curvature of hyperbolic space I would like to find a simple proof of the fact that the Gaussian curvature of hyperbolic space is -1 (I know in general it is just constant negative, but this special case would be enough.).
So far I have tried to use the first and second fundamental form of a parametrisation of the hyperboloid, and also this formule: $$K = \frac{F_{xx}F_{yy} - F_{xy}^2}{ (1+F_x^2+F_y^2)^2}$$ for $z = F(x,y)=x^2+y^2+1$ but in both cases I didn't get the right answer. I'm not sure if I can't do this or if I just made a calculation mistake.
Also I have read that the curvature of the two sheeted hypberboloid is positive and I'm very confused about that. We can use this as a model for hyperbolic space and therefore it should have negative curvature right?
 A: The hyperboloid does indeed have positive curvature if you endow it with the induced metric $dx^2+dy^2+dz^2$ of Euclidean 3-space it is embedded in.
The hyperboloid becomes a model of negatively curved hyperbolic space with a different metric, namely the metric $dx^2+dy^2-dz^2$. This is not a proper Riemannian metric on all of $\mathbb R^3$, since it is not positive definite, but it does become positive definite if you restrict it to the tangent planes of the two-sheeted hyperboloid.
The isometries of the hyperboloid model are now induced by linear automorphisms of $\mathbb R^3$, exactly those that preserve a sheet of the hyperboloid (those all preserve the metric too).
If you repeat the computations you performed before, but using the "inner product" $\langle v,w\rangle = v^1w^1 + v^2w^2 - v^3w^3$ instead of the usual one (and use Gram-Schmidt instead of a cross product to determine the normal), you'll get a different normal vector, you'll get different first and second fundamental forms, and you should end up with the curvature $K=-1$ you're expecting.
