Pseudo-Riemannian tensor and Morse index

  • #1
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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]?
 

Answers and Replies

  • #2
I like Serena
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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.
 
  • #3
Matterwave
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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.
 
  • #4
quasar987
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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]?
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.
 

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