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In $d$ dimensions, how many independent equations are contained in $R_{rsmn}=0$ in consideration of the Bianchi identity $\nabla_{[a}R_{bc]de}=0$?

This discussion reveals the independent equations contained in the Bianchi identity in consideration of the symmetries of the Riemann tensor, but I'm not quite sure how to use that number for this question.

I've given this a great deal of thought, but somehow, I don't seem to be making progress.

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  • $\begingroup$ “How many independent equations” is a complicated question when the constraints are of different type. Symmetries of the Riemann tensor limit the number of independent components at a point, while the differential Bianchi identity constrains Riemann tensor components at different points. Also, Bianchi identity can be repackaged as a wave-like operator for a spin 2 field (giving a smaller number of equations of a higher order). $\endgroup$
    – A.V.S.
    Commented Feb 6, 2023 at 19:06
  • $\begingroup$ Right, even so we can find the number of independent equations contained in the Bianchi identity by employing the symmetries of the Riemannian tensor (see link in the question). However, putting the latter to zero, and asking of how the B-identity effectively reduces the number of equations seems indeed difficult (just as in the case of how the Einstein equation being effectively reduced to $6$ equations from $10$ in four dimensions). $\endgroup$
    – John Doe
    Commented Feb 6, 2023 at 21:42
  • $\begingroup$ Though I'm convinced that folks on stackexchange have magic wands. $\endgroup$
    – John Doe
    Commented Feb 6, 2023 at 21:50

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First, note that in a $d$ dimensional spacetime, the Riemann tensor has $\frac{d^2(d^2-1)}{12}$ independent components, using the symmetries $R_{abcd}=-R_{bacd}=-R_{abdc}=R_{cdab}$ and Bianchi identity $R_{a[bcd]}=0$. (see eg Counting independent components of the Riemann curvature tensor).

Then I claim $\frac{d^2(d^2-1)}{12}$ is also the answer to your question. Here is my argument.

  • If you want to set $R_{abcd}=0$ at a point $x$, then you need to set $\frac{d^2(d^2-1)}{12}$ components to zero. The derivative of the Riemann tensor $\nabla_a R_{bcde}$ at the point $x$ is an independent rank-5 tensor.
  • If you want to set $R_{abcd}=0$ everywhere in the spacetime, then again using the freedom to specify that $\frac{d^2(d-1)^2}{12}$ components are zero everywhere is sufficient. The reason is that once the Riemann tensor vanishes everywhere, then you are dealing with a flat space, and the gradient of the Riemann tensor is automatically zero, so the Bianchi identify $\nabla_{[a} R_{bc]de}=0$ is automatically satisfied.
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    $\begingroup$ But it could be less than $d^2(d^2-1)/12$ (which is slightly different from the incorrect expression that you wrote), because the Bianchi identity relates the value of the Riemann tensor at different (although infinitesimally close) points on the manifold. It's true that the value of $\nabla R$ can be chosen independently of the value of $R$ at a single point, but not over the entire manifold. $\endgroup$
    – tparker
    Commented Feb 13, 2023 at 19:50
  • $\begingroup$ @tparker Thanks for the correction about $d^2(d^2-1)/12$ (wrote this too quickly). I could be wrong, but I would naively expect that at least for a $C^1$ differentiable curvature, you could have a manifold with $R_{abcd}=0$ except for an arbitrarily small region. Then the "amount by which the answer is less than $d^2(d^2-1)/12$" in the case where the whole manifold has zero curvature would be arbitrarily small. At least that's what I had in mind when I wrote the answer. I'd be happy to be shown wrong though :) I feel I'm cheating a bit since $R_{abcd}=0$ is a bit of a trivial case (...) $\endgroup$
    – Andrew
    Commented Feb 14, 2023 at 0:20
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    $\begingroup$ (...) On some level what this question is hinting at is something like, what data do you need to specify for the differential equation $\nabla_{[a} R_{bc]de}=0$ to have a unique solution, which I don't know how to answer in general. $\endgroup$
    – Andrew
    Commented Feb 14, 2023 at 0:21

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