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Eigenvalues of the metric?

  1. Jun 28, 2003 #1
    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?
  2. jcsd
  3. Jun 28, 2003 #2
    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...
  4. Jun 28, 2003 #3


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    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.
    Last edited: Jun 28, 2003
  5. Jun 28, 2003 #4
    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'.
  6. Jun 28, 2003 #5
    umm.. well i ve seen the statement that any bilinear form can be written at vtMw, 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.

    assume a finite dimensional vector space.

    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?
  7. Aug 7, 2003 #6
    oh duh. gram-schmidt.
  8. Aug 10, 2003 #7


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    Given a metric g, &exist; a basis {ui} of Tp(M) at each point p of a connected manifold M with g(ui,uj) = &plusmn;&delta;ij. Expressing one such basis in terms of another shows that the integer &sum;i g(ui,ui) is basis-independent. So if g is continuous on M, &sum;i g(ui,ui) is constant.
    Last edited: Aug 11, 2003
  9. Aug 11, 2003 #8
    i see. thanks.

    and am i correct in saying that gram-schmidt is the way to show that there exists such a basis?
  10. Aug 11, 2003 #9


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