Showing that a given metric is Riemannian

[demonelite123]

So I have the hyperboloid in $\mathbb{R}^3$ given by $x^2 + y^2 - z^2 = -1$. I define the scalar product of 2 tangent vectors (a1, b1, c1) and (a2, b2, c2) at a point as a1a2 + b1b2 - c1c2.

I want to show that this defines a Riemannian metric on the hyperboloid. I know that a Riemannian metric must be linear, symmetric, and positive definite. Linearity and symmetry I have checked easily but I am confused on the positive definite portion. The problem tells me to show that it is a Riemannian metric but it seems like it does not satisfy the positive definite requirement. The matrix associated with $a_1^2 + b_1^2 - c_1^2$ is just the 3x3 identity matrix with the (3, 3) entry being -1 instead of 1. Thus, the eigenvalues of this matrix are 1, 1, and -1 so it seems to me that this is actually indefinite since the eigenvalues are not all the same sign.

Am I correct in saying this or is there something I am missing? thanks!

 

[mighty2000]

Thank you for your question. You are correct in noticing that the matrix associated with the scalar product you defined is not positive definite. However, this does not necessarily mean that it is not a Riemannian metric.

In order to show that a metric is positive definite, we need to check that for any tangent vector v at a point p, the scalar product of v with itself is greater than zero. In this case, we have a scalar product defined for any two tangent vectors at a point, not just for a tangent vector with itself. Therefore, we cannot use the eigenvalues of the matrix to determine whether it is positive definite or not.

To show that this scalar product defines a Riemannian metric, we need to check that it satisfies three conditions: linearity, symmetry, and positive definiteness. We have already established that it is linear and symmetric. Now, to show that it is positive definite, we need to show that for any tangent vector v at a point p, the scalar product of v with itself is greater than zero. This can be done by plugging in the components of v into the scalar product formula and simplifying the expression to show that it is greater than zero.

I hope this clarifies your confusion. Keep up the good work in your studies of Riemannian geometry!