What is the relationship between eigenvalues and the metric in linear spaces?

[lethe]

the signature of a metric is often defined to be the number of positive eigenvalues minus negative eigenvalues of the metric.

this definition has always seemed a little suspicious to me. eigenvalues are defined for endomorphisms of a linear space, whereas the metric is a bilinear functional on the vector space. it is not clear to me how one would write down an eigenvalue equation for the metric.

insofar as the metric can always be written as a matrix in local coordinates, i guess you can just shutup and pretend that it is a linear transformation, and calculate its eigenvalues, but it seems to me to be a suspicious procedure.

any thoughts?

 

[arcnets]

This will be nothing new to you, but if you have any metric on a vector space V, it can be written as
|v| = <v|M|v>
where M is a linear operator on V, so it sure must have eigenvalues...

 

[Hurkyl]

I think lethe is referring to the fact that, to be perfectly strict, when applying the metric to a vector, the result is a dual vector.


In any case, the definition I've read for the signature is to diagonalize the metric and look at the diagonal entries. For matrices, those are precisely the eigenvalues, but the definition works for any rank 2 tensor.

 

[arcnets]

Yes. But the problem may be that the 'number of eigenvalues' is not finite. For example, in quantum physics, the vector v may be a wavefunction. The metric M will then be some integral operator, and I can understand why lethe finds it suspicious to talk about a 'number of ... eigenvalues'.

 

[lethe]

Originally posted by arcnets
This will be nothing new to you, but if you have any metric on a vector space V, it can be written as
|v| = <v|M|v>
where M is a linear operator on V, so it sure must have eigenvalues...

umm.. well i ve seen the statement that any bilinear form can be written at v^tMw, but for now i don t want to talk about that, it seems very coordinate dependent.

as for your equation, the way i am used to it, in bra-ket notation (which is a coordinate independent notation), <v| already has incorporated the metric: it is the linear functional that takes |w> to g(v,w). in this case, |v|²=<v|M|v> if and only if M=1. this is another way of saying that <v|w> is just another notation for the inner product. in other words, your M matrix is extraneous.

Originally posted by arcnets
Yes. But the problem may be that the 'number of eigenvalues' is not finite. For example, in quantum physics, the vector v may be a wavefunction. The metric M will then be some integral operator, and I can understand why lethe finds it suspicious to talk about a 'number of ... eigenvalues'.

assume a finite dimensional vector space.

Originally posted by Hurkyl

In any case, the definition I've read for the signature is to diagonalize the metric and look at the diagonal entries. For matrices, those are precisely the eigenvalues, but the definition works for any rank 2 tensor.

yeah? actually, perhaps that is the definition that i know too. thing is, i see the word "diagonalize", and i think eigenvalue equation. i have no idea how to diagonalize any kind of second rank tensor, unless it is a linear transformation.

how would you diagonalize a second rank tensor in general?

 

[lethe]

oh duh. gram-schmidt.

 

[jeff]

Given a metric g, &exist; a basis {u_i} of T_p(M) at each point p of a connected manifold M with g(u_i,u_j) = &plusmn;&delta;_ij. Expressing one such basis in terms of another shows that the integer &sum;_i g(u_i,u_i) is basis-independent. So if g is continuous on M, &sum;_i g(u_i,u_i) is constant.

 

[lethe]

Originally posted by jeff
Given a metric g, &exist; a basis {u_i} of T_p(M) at each point p of a connected manifold M with g(u_i,u_j) = &plusmn;&delta;_ij. Expressing one such basis in terms of another shows that the integer &sum;_i g(u_i,u_i) is basis-independent. So if g is continuous on M, &sum;_i g(u_i,u_i) is constant.

i see. thanks.

and am i correct in saying that gram-schmidt is the way to show that there exists such a basis?

 

[jeff]

Originally posted by lethe
am i correct in saying that gram-schmidt is the way to show that there exists such a basis?

Yes, in that gram-schmidt is needed to prove that finite-dimensional inner product spaces have orthonormal bases.