Suppose we have an $n$-dim Riemannian space $V_n$ endowed with a metric. More precisely, it is defined a $(0,2)$-type tensor field in $V_n$: $$g_{\mu\nu} = g_{\mu\nu}(x),\quad x\in V_n$$ Questions.

  1. Fix point $x_0\in V_n$. Then how do we define an eigenvector of $g_{\mu\nu}$? On the one hand, one can write $$g_{\mu\nu}x^{\nu} = \lambda x^\mu,$$ which is kind of nonsense because this is not a tensor equation. On the other hand, it could be written as $$g_{\mu\nu}x^\nu = \lambda x_\mu$$ but this is true for every vector and one can conclude that all vectors are eigenvectors of metric with eigenvalue $\lambda = 1$.
  2. Let us fix $n=4$ and continue with pseudo Rimannian space of index 1. I.e. pass to the case of General Relativity. Suppose also we have Gaussian coordinates in $V_4$ at a point $M$ so $$g_{\mu\nu} = \eta_{\mu\nu} = \mathrm{diag}(-1, 1, 1, 1)$$ at that point.

    a) Isn't the fact that there are 4 eigenvalues, namely -1, 1, 1, 1, of this metric?

    b) How many eigenvectors has that metric? I have an opinion that infinitely many because I feel that Lorentz transformations change vector but leave the property to be an eigenvector.

    c) What is physical sense of eigenvectors and -values of a metric?

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    $\begingroup$ A priori, a Riemannian metric is a smooth family of symmetric bilinear forms $g_p: T_pM\otimes T_pM\to \Bbb R$ on the tangent spaces $T_pM$ (i.e. $g_p \in \text{Sym}^2(T^*_p M)$), not a smooth family of linear transformations $T_pM\to T_p M$. As such, there is no notion of eigenvector. To identify a bilinear form with a linear transformation, you will need to define an isomorphism $T^*_p M\cong T_p M$: The obvious choice for such an isomorphism is given by the metric itself---[comment cont.'d below] $\endgroup$ – Danu Jan 26 '18 at 12:33
  • $\begingroup$ This yields a composition $$ T^*_pM \otimes T_p M\to T_p M\otimes T_p M\to \Bbb R$$ where the first step is given by the inverse metric, namely by realizing that any covector $\alpha$ in $T_p^* M$ is of the form $g_p(v,-)$ for some $v\in T_pM$ and mapping $\alpha\mapsto v$. The second step is given by the metric, viewed as a bilinear form. This composition realizes $g_p$ as an element of $T_pM\otimes T_p^*M$ and hence as a linear transformation. In components, this linear map sends $(\alpha_\mu,v^\rho)\mapsto (g^{\nu\mu}\alpha_\nu,v^\rho)\mapsto \alpha_\mu v^\mu$ [comment cont'd below] $\endgroup$ – Danu Jan 26 '18 at 12:46
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    $\begingroup$ From here, it is not hard to work out explicitly that this means that the metric corresponds precisely to the identity map (for instance, by plugging in $v^\mu=(1,0,\dots,0)^T$ and varying the place of the $1$, and similarly for $\alpha_\mu$). In components, this translates to $g^{\mu\nu}g_{\nu\rho}=\delta^\mu_\rho$. $\endgroup$ – Danu Jan 26 '18 at 12:56
  • $\begingroup$ Thank you for your comments. I will try to understand its content. But. Why do not to post these comments as an answer? $\endgroup$ – LRDPRDX Jan 26 '18 at 13:11
  • $\begingroup$ Because I feel like I didn't really understand what you were asking. I just dumped some thoughts (admittedly, they turned out a bit longer than planned). $\endgroup$ – Danu Jan 26 '18 at 13:21

You almost answered your own question when you noted that the eigenvector equation is nonsense. It is not meaningful to speak of eigenvectors of the metric tensor. Eigenvectors are meaningful for mixed tensors, but the metric is a symmetric bilinear form that must be expressed by a purely co- or contra-variant tensor. In matrix notation, mixed tensors would transform as $\mathbf{M}'=\mathbf{PM}{{\mathbf{P}}^{-1}}$, but symmetric bilinear forms as $\mathbf{S}'=\mathbf{PS}{{\mathbf{P}}^{T}}$. Under the former, eigenvalues are invariant; but under the latter, only the numbers of positive, negative, and null eigenvalues are invariant.


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