2
$\begingroup$

One way to understand the eigenvalues of a matrix (rank 2 tensor) in a space with a Euclidean metric is that they are the rotation invariant aspects of the matrix. That is, they're the rank 2 generalization of the rank 1 vector's length. Perhaps most importantly, given a matrix $M$ you can show that $$ M_{ij} = [V^{-1}]_{ik} \Lambda_{kl} V_{lj}, $$ where $V$ is the matrix of eigenvectors, and $\Lambda$ is the diagonal matrix of the eigenvalues.

When you switch over to a Minkowski space, you run into the issue that each rank 2 tensor has four versions, depending on contra/co-variant status of the two indices: e.g. $T_{\mu\nu},\ T^{\mu}_{\hphantom{\mu}\nu},\ T^{\hphantom{\mu}\nu}_{\mu}, \text{ and }T^{\mu\nu}$. The defining equation for eigenvalues is: $$ M v = \lambda v,$$ with $\lambda$ the scalar eigenvalue. Adding tensor indices to the equation can yield \begin{align} M^{\mu}_{\hphantom{\mu}\nu} v^\nu &= \lambda v^\mu \text{ or} \\ M_{\mu}^{\hphantom{\mu}\nu} v_\nu &= \lambda v_\mu, \end{align} because those are the only combinations that have consistent indices.

This is consistent with the fact that it is known that the two Lorentz invariant combinations of the electromagnetic field are $\epsilon_0\mathbf{E}^2 - \mathbf{B}^2 / \mu_0$ and $\mathbf{E}\cdot\mathbf{B}$, and the eigenvalues of $F_{\mu\nu}\eta^{\nu\alpha}$ can be expressed in terms of these, whilst the eigenvalues of $F_{\mu\nu}$, calculated naïvely, are manifestly not Lorentz invariant.

All that said, how do we understand the eigenvalues of the metric tensor, $g_{\mu\nu}$? Unlike other tensors, we can't just raise or lower the indices of $g_{\mu\nu}$ in a meaningful way, since $g_{\mu\nu}g^{\nu\alpha} = \delta_\mu^{\hphantom{\mu}\alpha}$. I understand that $\sqrt{-\mathrm{det}(g)}\mathrm{d}^4x$ is Lorentz invariant, and that's the square root of the product of the eigenvalues of $g_{\mu\nu}$ times the coordinate volume element, to it should be Lorentz invariant.

$\endgroup$
3
  • $\begingroup$ Let me restrict Riemannian metrics first. I'm not sure if this is the kind of answer you are looking for, but I remember thinking about this question and (if I remember correctly), if you fix a point $p$ on your coordinate system, the eigenvectors of $[g_{ij}]$ at $p$ are precisely the vectors that are orthogonal with respect to $g_{ij}$ and some new coordinate-dependent metric $\delta_{ij}$ (basically $\delta_{ij}$ treats the coordinate basis at $p$ as orthogonal even though according to $g$ it might not be). If you normalize these [...] $\endgroup$ Aug 16, 2023 at 3:39
  • $\begingroup$ [...] vectors with respect to $\delta_{ij}$, I think the the $g_{ij}$-lengths of them will be the eigenvalues. I'll see if this works for pseudo-Riemannian metrics. If this isn't what you're looking for, or if I'm not correct, I apologize. $\endgroup$ Aug 16, 2023 at 3:39
  • $\begingroup$ @MaximalIdeal That's exactly the sort of thing I'm looking for, but I'm specifically interested in manifolds that have indefinite metrics (esp. locally Minkowski). I'm also primarily interested in the eigenvalues, though the eigenvectors are useful to know, too. $\endgroup$ Aug 16, 2023 at 3:44

1 Answer 1

2
$\begingroup$

Briefly speaking, OP's issue is already at the level of linear algebra rather than differential geometry.

  1. A real symmetric $(0,2)$ or $(2,0)$ tensor does not have an invariant notion of eigenvalues, or even determinant or trace.

    Example: A dilation transformation of the basis vectors scales the eigenvalues.

    But the tensor does have an invariant notion of signature, cf. Sylvester's law of inertia.

  2. In contrast, a $(1,1)$ tensor has an invariant notion of determinant and trace.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.