PseudoRiemannian tensor and Morse indexby Jimmy Snyder Tags: index, morse, pseudoriemannian, tensor 

#1
Nov1511, 02:50 PM

P: 2,163

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 pseudoRiemannian if [itex]0 < \mu < n[/itex]. Is this correct? If so, what kind of tensor is it when [itex]\mu = n[/itex]?




#2
Nov1611, 03:00 AM

HW Helper
P: 6,189

A Riemannian manifold requires the metric tensor to be positivedefinite.
Positivedefinite is equivalent to all eigenvalues being positive. If one or more is negative, the tensor is indefinite. It can still be used for a pseudoRiemannian manifold. The reason I can think of for a distinction for [itex]\mu = n[/itex], is that the tensor is negativedefinite. 



#3
Nov1611, 03:04 AM

P: 2,060

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
Dec211, 01:57 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

PseudoRiemannian tensor and Morse indexIt 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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