6
$\begingroup$

Ok so I seem to be missing something here.

I know that the number of independent coefficients of the Riemann tensor is $\frac{1}{12} n^2 (n^2-1)$, which means in 2d it's 1 (i.e. Riemann tensor given by Ricci Scalar) and in 3d it's 6 (i.e. Riemann Tensor given by Ricci tensor).

But why does that constrain the Riemann tensor to only be a function of the metric? Why not a tensorial combination of derivatives of the metric?

What I mean is why is the Riemann tensor in 2D of the form \begin{align} R_{abcd} = \frac{R}{2}(g_{ac}g_{bd} - g_{a d}g_{b c}) \end{align}

and in 3D, \begin{align} R_{abcd} = f(R_{ac})g_{bd} - f(R_{ad})g_{bc} + f(R_{bd})g_{a c} - f(R_{bc})g_{ad} \end{align} where $f(R_{ab}) = R_{ab} - \frac{1}{4}R g_{ab}$?

Wikipedia says something about the Bianchi identities but I can't work it out. A hint I got (for the 2d case at least) was to consider the RHS (the terms in parenthesis) and show that it satisfies all the required properties of the Riemann tensor (sraightforward) and proceed from there - but, I have not been able to come up with any argument as to why there must be a unique tensor satisfying those properties.

Of course I could brute force it by computing $R_{abcd}$ from the Christoffel symbols etc., but surely there must be a more elegant method to prove the statements above.

Help, anyone? I haven't been able to find any proofs online - maybe my Googling skills suck.

$\endgroup$
  • 3
    $\begingroup$ What do you mean by derivatives of the metric? $\nabla g =0$ so you can't see anything like that. The only derivatives allowed come in the form of $R_{\mu\nu\rho\sigma}$ because it's the only tensor around that includes derivatives of the metric. So in 2d, the fact there's one component + the symmetries Riemann restricts Riemann to be proportional to $g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}$, the question is how to fix the overall normalization. To do that take the double trace of both sides, you'll see the only consistent overall normalization is $R/2$. $\endgroup$ – Andrew Sep 20 '13 at 11:59
  • 1
    $\begingroup$ Your formula for 3d is not correct. Look at this, and make n=2,3, and see why en.wikipedia.org/wiki/Ricci_decomposition $\endgroup$ – Cristi Stoica Sep 20 '13 at 12:02
  • $\begingroup$ Maybe I should mention that, in the same place en.wikipedia.org/wiki/Ricci_decomposition you will find the explanation you are looking for, in terms of decomposition in irreducible representations. You will see why for n<4 the formula becomes simpler, for n=2 even simpler than n=3. $\endgroup$ – Cristi Stoica Sep 20 '13 at 12:56
  • $\begingroup$ @CristiStoica Ah ok thanks. I just copied that equation (for the 3D case) from the solution of a problem set I found on-line, so maybe it's wrong... $\endgroup$ – nervxxx Sep 20 '13 at 13:02
  • 2
    $\begingroup$ ah lol after thinking about it for so long it finally struck me that the answer is indeed very simple - because there's only 1 independent component, that means that there's only 1 basis tensor for the space of all such tensors with the properties that the Riemann tensor possesses. Since $g g - g g $, which we pulled out of a hat, works, it's the only basis tensor and so $R$ must be proportional to $g g - g g $. thanks everyone! $\endgroup$ – nervxxx Sep 20 '13 at 13:13
5
$\begingroup$

The simplicity of geometry in lower dimensions is because the Riemann curvature tensor could be expressed in terms of simpler tensor object: scalar curvature and metric (in 2d) or Ricci tensor and metric (in 3d). That fact, of course, does not alter the possibility to write Riemann tensor (as well as Ricci tensor and scalar curvature) as a combination of metric derivatives. But each term in such a combination is not a tensor - only the whole object.

Now, let us recapture, why in lower dimensions we are able to reduce the Riemann tensor to a combination of lower rank tensor objects.

For 3d case, definition of Ricci tensor:

$$ R_{ab} = R_{abcd}g^{ac}$$

contains 6 independent components, exactly the number of independent components in Riemann tensor. So, this equation could be reversed, thus expressing $R_{abcd}$ in terms of $R_{ab}$ and $g_{ab}$.

In 2d case we could similarly start with definition of Ricci scalar:

$$ R = R_{ab} g^{ab} ,$$

and reverse it expressing $R_{ab}$ through $g_{ab}$ and $R$. The next step would be to express Riemann tensor with $g_{ab}$ and $R_{ab}$ (and thus through scalar $R$ only).

$\endgroup$
  • $\begingroup$ I think he means partial derivatives, not covariant derivatives of the metric. $\endgroup$ – Cristi Stoica Sep 20 '13 at 14:47
  • $\begingroup$ In higher dimensions, metric derivatives come into play, don't they? These are partial derivatives, not covariant derivatives, written in a way that the resulting expression is tensorial. After all, the Riemann tensor can be built up from Christoffel symbols and its derivatives, so a general Riemann tensor contains $g, \partial g, \partial^2 g$. $\endgroup$ – nervxxx Sep 20 '13 at 14:52
  • 1
    $\begingroup$ Yes, obviously, the Riemann tensor could be written as combination of derivatives of metric. What we try to explore here is the possibility of constructing Riemann tensor from some simpler (i.e. of lower rank) tensor objects. (Edited my answer to stress this point). $\endgroup$ – user23660 Sep 20 '13 at 15:01
  • $\begingroup$ @nervxxx: the Ricci tensor is ALREADY a combination of derivatives of the metric. $\endgroup$ – Jerry Schirmer Sep 20 '13 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.