# Pseudo-Riemannian tensor and Morse index

Let $g_{ij}$ be a tensor, where $0 \leq i,j \leq n$. The Morse index $\mu$ 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 $\mu = 0$ and pseudo-Riemannian if $0 < \mu < n$. Is this correct? If so, what kind of tensor is it when $\mu = n$?

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I like Serena
Homework Helper
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 $\mu = n$, is that the tensor is negative-definite.

Matterwave
Gold Member
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
Let $g_{ij}$ be a tensor, where $0 \leq i,j \leq n$. The Morse index $\mu$ 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 $\mu = 0$ and pseudo-Riemannian if $0 < \mu < n$. Is this correct? If so, what kind of tensor is it when $\mu = n$?