Solving for metrics that are Einstein, i.e that satisfy $R_{\mu \nu} = \Lambda g_{\mu \nu}$ is highly non-trivial as soon as $g_{\alpha \beta}$ is allowed to have off-diagonal components. However, there have been some results relating to the existence of coordinate systems that diagonalise the metric (K. P. Tod, Class. Quantum Grav. 9 1992). In physics we expect certain components of the metric to obey certain kinds of symmetry relations (axisymmetry, existence of certain Killing vectors...) depending on the phenomenon that we are interested in. If one can show that these properties can be encoded in a diagonal metric under certain coordinate conditions then solving for the components in that particular basis becomes much simpler. The famous example here is the Newman-Janis application to the Schwarzschild solution to generate the Kerr metric via a complex coordinate transformation.

In classical matrix group theory one has the Jordan-Chevalley decomposition theorem which states (loosely) that any matrix can be written as the sum of a diagonalizable and nilpontent component: for any $M \in GL(n,\mathbb{C})$ we can write $M = D + N$. This result has extensions to semi-simple Lie algebras $\mathfrak{g}$.

I know that, at least for Banach algebras with unit, one can define a Riesz functional calculus analogous to the Cauchy integral formula to define functions of bounded, linear operators; let $a \in \mathfrak{A}$ with unit $I$ and $f$ holomorphic then:

$f(a) = \frac {1} {2 i \pi} \oint_{\Gamma} f(z) (zI - a)^{-1} dz$

With $\Gamma$ a rectifiable Jordan curve. One can think of the Einsteinian condition being a kind of differential equation with some differential operator $\mathfrak{D}$ lying in a Banach algebra $\mathfrak{A}$ since the Ricci tensor $R^{\mu}_{\nu}$ consists of derivatives of $g^{\mu}_{\nu}$.

Now for the question, can one decompose the metric tensor into a sum of diagonalizable (in the existence of coordinate sense) and another part? What results, if any, exist regarding such a decomposition? If the space-time does not permit a coordinate system $\{X^{\mu}\}$ under which the metric tensor is diagonal, can one trace this back to a physical property?

For simplicity we can take the dimension of the space-time to be 4.

One can formulate the theory of general relativity (and metric-affine gravity theories) in terms of gauge theory language, i.e, general relativity (in 4-dimensions) can be formulated as a gauge theory over $\mathfrak{su}(2,\mathbb{C})$. This is why I suspect these kinds of results to exist, but am unaware of any.

  • $\begingroup$ Local coordinates in which the metric tensor is diagonal are traditionally called orthogonal systems. Darboux wrote a book about triply orthogonal systems, i.e. such coordinates, just for Euclidean space of dimension 3. But he wasn't looking at general metrics or decompositions, so I guess that isn't really relevant. $\endgroup$ – Ben McKay May 2 '14 at 8:51
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    $\begingroup$ I don't know useful results for computing the Ricci tensor of the sum of two quadratic forms, so I can't help you there. What we do know is that in dimension $4$ the diagonalizable metrics depend on 8 functions of 4 variables locally while all metrics depend on 10 functions of 4 variables locally, so there have to be at least two independent identities that diagonalizable metrics satisfy that aren't satisfied by the general metric. They will be high order, though. In any case, not all Einstein metrics are diagonalizable: The Fubini-Study metric on $\mathbb{CP}^2$ is not locally diagonalizable. $\endgroup$ – Robert Bryant May 2 '14 at 11:06
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    $\begingroup$ DeTurck and I (projecteuclid.org/euclid.dmj/1077303804) proved that any 3-dimensional Riemannian metric is locally diagonalizable. We were told at the time that diagonal co-ordinates had already been studied and found useful in general relativity (maybe by Chandrasekhar?). But I've never understood why the Einstein equations might be easier to study in diagonal co-ordinates. $\endgroup$ – Deane Yang May 2 '14 at 13:05
  • $\begingroup$ Perhaps the Einstein equations from scratch aren't simpler in the diagonal 'gauge' because there may be highly non-trivial dependencies. Usuaully though one takes a solution already found and modifies it to allow for extra freedoms that should be permitted on physical grounds. Also, solving perturbation equations such as: Klein-Gordon (wave), Regge-Wheeler, Dirac... all contain many less cross terms in the diagonal system. Thank you for that reference Deane; much appreciated. This means using a 3+1 ADM split one can always find a triply orthogonal system for the spatial metric $\endgroup$ – Arthur Suvorov May 4 '14 at 22:32
  • $\begingroup$ Perhaps I have given this question an unfortunate title; I was asking more along the lines of the existence of a sum decomposition of the form $ g_{\mu \nu} = S_{\mu \nu} +N_{\mu \nu}$ where $S$ is diagonalizable and $N$ has some other properties. I am aware that some metrics do not admit such systems; thank you Robert for the Fubini example. I drew the comparison with the Jordan-Chevalley decomposition to see perhaps what Nμν should look like here. Since we have an analogous 'diagonalizable' construction, maybe Nilpotent means some derivative of $N$ vanishes eventually. $\endgroup$ – Arthur Suvorov May 4 '14 at 22:54

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