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In a gravity theory in spacetime, the metric has signature $− + +· · ·+$. Concretely this means that the metric tensor $g_{μν}$ may be diagonalized by an orthogonal transformation, i.e. $$(O^{-1})_{μ}^{\;a} = O^a_{\;μ}$$and $$g_{μν} = O^a_{\;μ}D_{ab}O^b_{\;ν}$$ with positive eigenvalues $λ^a$ in $D_{ab} = \textrm{diag}(−λ_0, λ_1, . . . , λ_{D−1})$.

The construction above, which involved only matrix linear algebra, allows us to define an important auxiliary quantity in a theory of gravity, namely $$e^a_μ(x) ≡\sqrt{λ^a(x)}O^a_μ(x).$$ Using this tetrad we can write $g_{μν}(x) = e^a_ μ(x)η_{ab}e^b_ν (x)$ , In the bold above:

  • Why would this mean that the metric tensor may be diagonalize by an orthonormal transformation?

  • What is meant by diagonalization here (mathematically)?

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    $\begingroup$ I don't understand your question. Your second equation says that $O$ diagonalizes $g$. Then you define $e$ as a scaled version of $O$. And then you wonder why it is said that $e$ comes from diagonalizing $g$, when your last displayed equation $g = e^{-1}\eta e$ is the same as the equation $\eta = e g e^{-1}$, i.e. $e$ diagonalizes $g$ (since $\eta$ is diagonal)? $\endgroup$
    – ACuriousMind
    Commented Jan 17, 2016 at 16:00
  • $\begingroup$ @ACuriousMind I said what does it mean for it to be diagonalized by an "orthonormal" transformation. What is orthonormal transformation supposed to mean here? I just needed to understand the text with some more details into mathematics. $\endgroup$ Commented Jan 17, 2016 at 18:02
  • $\begingroup$ @ACuriousMind Also note that $g$ is already a diagonal matrix, so why diagonalize it when it is already diagonal? $\endgroup$ Commented Jan 17, 2016 at 18:03
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    $\begingroup$ The concept of an orthogonal matrix and of diagonalization are standard linear algebra concepts. Relativity texts usually expect you to be familiar with linear algebra. $\endgroup$
    – ACuriousMind
    Commented Jan 17, 2016 at 19:38

2 Answers 2

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Let's go step by step as it seems you're missing some fundamentals.

We know from (linear) algebra, that a symmetric bilinear form can be transformed to a diagonal matrix with elements $e$ on the main diagonal $e\in \{0,1,-1\}$. The tripel counting the amount of times each number appears is called signature. If you didn't know that, check this.

Now, a metric tensor is a symmetric bilinear form, so we know it has a transform, so that we get its signature. By the way, from Sylvester's law of inertia follows, that the transform is an orthogonal transform, if the matrix is invertible.

I hope this answers the first question. I didn't completely get what your second question was... Diagonalisation is always the same thing.

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  • $\begingroup$ tThank you. I have to think about your answer some more and will get back within few hours. As for my second question it was how to obtain $e^a_μ(x)$ by diagonalization -- or at the very least, why would they say we can obtain it by diagonalization? How did they know that? (It has been sometime since I last dealt with linear algebra) $\endgroup$ Commented Jan 17, 2016 at 16:51
  • $\begingroup$ No problem :) When diagonalizing a matrix A you get a diagonal matrix $D_A$ with the eigenvalues on the main diagonal. But, and thats often very important, you get a matrix $S$, so that $D_A=S^{-1}AS$. Sometimes, as here, $S$ is orthogonal, this means $S^{-1}=S^T$. $\endgroup$
    – famfop
    Commented Jan 17, 2016 at 17:07
  • $\begingroup$ After a quick read of your answer, I don't think this addresses my question completely. My question was, why if the signature is $-++..+$ then the metric can be diagonalized? I can't relate between your answer and this and would appreciate that you elaborate. $\endgroup$ Commented Jan 17, 2016 at 17:08
  • $\begingroup$ When saying a matrix/bilinear form has a signature you're implicitly saying it can be diagonalized. Because, if it has a signature, it means the diagonal matrix (lets call it) $D_A =$ diag(signature) Example: minkowski: signature (+,-,-,-) -> diagonal matrix: diag(1,-1,-1,-1). $\endgroup$
    – famfop
    Commented Jan 17, 2016 at 17:13
  • $\begingroup$ Yes, true! In our case, this means $g_{\mu\nu}$ is a diagonal matrix, right? If so, what does the other mean when he says "may be diagonalized by an orthogonal transformation", I mean it is already a diagonal matrix, why you want to diagonalize it using some "orthogonal" transformation, which I have no clue what it means. $\endgroup$ Commented Jan 17, 2016 at 17:24
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In a gravity theory in spacetime, the metric has signature $− + +· · ·+$.

That's a convention. Other conventions are that it has signature $+ - - -$.

Concretely this means that the metric tensor $g_{μν}$ may be diagonalized

The signature doesn't tell you that it is diagonalizable. The fact that $g_{\mu\nu}=g_{\nu\mu}$ tells you that it is diagonalizable. Normal operators are diagonalizable, those are ones that commute with their adjoint. Since this one equals its adjoint, so it commutes with its adjoint, so it's normal, so it's diagonalizable.

$g_{μν}$ may be diagonalized by an orthogonal transformation, i.e. $$(O^{-1})_{μ}^{\;a} = O^a_{\;μ}$$and $$g_{μν} = O^a_{\;μ}D_{ab}O^b_{\;ν}$$ with positive eigenvalues $λ^a$ in $D_{ab} = \textrm{diag}(−λ_0, λ_1, . . . , λ_{D−1})$.

The signature is what told you that after you diagonalize it, the values on the diagonal will be one one negative and three positive values (with your convention on the signature).

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  • $\begingroup$ Are you trying to say that if we have a certain manifold, we can then impose on it any metric of signature that we can choose? $\endgroup$ Commented Jan 18, 2016 at 19:16
  • $\begingroup$ @Beyond-formulas In General Relativity it is part of the theory that the metric tensor is symmetric and that it has a signature of + - - - or - + + +. Being symmetric means it is diagonalizable. If it wasn't symmetric then you wouldn't have GR. And the signature also needs to be one of those two types or else you don't have GR. And once you pick one, that will tell you how many of the terms along the diagonal of the diagobalize version have positive versus negative numbers. Did you really think I meant anything else? $\endgroup$
    – Timaeus
    Commented Jan 18, 2016 at 19:22
  • $\begingroup$ I was asking about your first line, when you were talking about conventions. Like I can choose one convention and you can choose another and we can then make up a diagonal matrix out of that. Thus, I repeat my first question. If we are on a manifold, we choose the conventions that suit our theory and work toward diagonalizing them? In short, do we choose whatever metric we want on a a manifold? Is a a metric some additional imposed constraint? $\endgroup$ Commented Jan 18, 2016 at 19:31
  • $\begingroup$ @Beyond-formulas You need to speak more carefully. We usually don't call something a metric unless it is a non degenerate symmetric rank two bilinear form. But there are metrics with signatures like + + - - and those are not part of GR and so when you talk about using "whatever metric we want" then the answer must be no. But if you are asking about whether you use + - - - or whether you use - + + + then you are free to use either. And there are other sign conventions too, ones about the Riemann Tensor and others about the Einstein Tensor. $\endgroup$
    – Timaeus
    Commented Jan 18, 2016 at 19:35

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