Pseudo-Riemannian tensor and Morse index

Jimmy Snyder

Let [itex]g_{ij}[/itex] be a tensor, where [itex]0 \leq i,j \leq n[/itex]. The Morse index [itex]\mu[/itex] is the number of negative eigenvalues of g. On page 469 of Eberhard Zeidler's QFT III: Gauge Theory, it says that g is Riemannian if [itex]\mu = 0[/itex] and pseudo-Riemannian if [itex]0 < \mu < n[/itex]. Is this correct? If so, what kind of tensor is it when [itex]\mu = n[/itex]?

I like Serena

A Riemannian manifold requires the metric tensor to be positive-definite.
Positive-definite is equivalent to all eigenvalues being positive.
If one or more is negative, the tensor is indefinite.
It can still be used for a pseudo-Riemannian manifold.

The reason I can think of for a distinction for [itex]\mu = n[/itex], is that the tensor is negative-definite.

Matterwave

I'd imagine that if you had all negative eigenvalues, you could always just pick the negative of the metric and you'd have a Riemannian manifold again. I'm not sure such a sign would be physical from a practical point of view.

From a mathematical point of view, I'm not sure what it would mean.

quasar987

Are you certain you have this straight? What you call the Morse index is usually just called the signature of g. The signature is a well-defined concept for any symmetric bilinear form on a vector space. The Morse index on the other hand, is a number associated with a critical point p of a Morse function f on a manifold M. It is defined as the signature of the Hessian of f at p. The Hessian of f at p is the bilinear form whose matrix wrt some coordinate chart is the matrix of second partial derivatives. It is a symmetric bilinear form obviously so we may speak of its signature. (check) The Morse condition on f just means that this bilinear form is nondegenerate so it can have any signature btw 0 and n.

It just seems very strange to me why anyone would call the signature of a bilinear form the Morse index as there is nothing at all "Morsy" about it afaik.