What does diagonalization mean here?

Question

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)?

Answer 1

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.

Answer 2

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).