1
$\begingroup$

Are there any interesting interpretations or meanings of the spectrum of the metric tensor (i.e. the eigenvalues or eigenvectors of its quadratic form)? Tensors of other types (e.g. the stress tensor) clearly have such concepts.

I saw this question (on Quora) that states that the concept itself is ill-defined. I was wondering if others had some thoughts.

However, as the question itself notes, there is the metric signature, which involves the signs of the eigenvalues of the metric is mentioned as one example where it makes sense. I suppose this is because it is invariant to changes in the basis; are there no other quantities like this?

I suppose also that the freedom given in choosing coordinate systems means that the interpretation will probably be with respect to the given system (or require a specific sort of coordinate system to be chosen), but correct me if I'm wrong.

  • 1
    Over a point you are asking if the eigenvalues of the matrix representing an inner product means anything. For one thing, they are not well defined for the depend on the basis. On the other hand, by changing bases you can assume they are all 1 – 2017-02-19
  • 0
    @MarianoSuárez-Álvarez Indeed, that is essentially what I meant in my last paragraph (and is the argument for ill-definedness by the linked question). But would that force the coordinate system to take a certain structure? And how does the metric signature then make sense? – 2017-02-19
  • 2
    The signature makes sense because of a theorem of Sylvester. See https://en.m.wikipedia.org/wiki/Sylvester's_law_of_inertia – 2017-02-19
  • 1
    Sylvester's Law of Inertia also tells us that signature is indeed the only "quantity like this", in a sense that it makes precise. – 2017-02-19
  • 3
    One can choose always choose, around an arbitrary point, a local frame $(E_a)$ such that the matrix representation of the metric with respect to the frame is a diagonal matrix with entries $\{1, \ldots, 1, -1, \ldots -1\}$, so that the metric components are $g_{ab} = \pm \delta_{ab}$. These are the *orthonormal local frames*, but in general they *cannot* be coordinate frames: If one can choose around every point orthonormal local frames that are also coordinate frames, then $g$ is locally flat. – 2017-02-19
  • 0
    @Travis thanks for your comments. I'm not quite sure I understand why you could not use the orthonormal frames you have described as moving coordinate frames, nor why Mariano above says one can assume all the eigenvalues are 1 (perhaps he meant $\pm 1$?). – 2017-02-19
  • 0
    For the first issue, see http://math.stackexchange.com/questions/2148532/lie-bracket-of-local-orthonormal-basis-of-vector-fields/2148694#2148694 . As for the second, I believe in his first comment that Mariano was referring to the Riemannian case, in which in claim is true. (Often by "inner product" one means a symmetric bilinear form that is not only nondegenerate but also positive definite.) – 2017-02-19
  • 0
    the key objects are: change of basis, change of components, diagonalization, change of coordinates, change of charts, change on tangent basis, etc – 2017-02-23
  • 0
    change in components of tangent vectors, rank two tensor change of basis and change of components – 2017-02-28

0 Answers 0