I Ricci curvatures determine Riemann curvatures in 3-dimension

[tommyxu3]

Hello~

For usual Riemann curvature tensors defined: $R^i_{qkl},$ I read in the book of differential geometry that in 3-dimensional space, Ricci curvature tensors, $R_{ql}=R^i_{qil}$ can determine Riemann curvature tensors by the following relation:
$$R_{\alpha\beta\gamma\delta}=R_{\alpha\gamma}g_{\beta\delta}-R_{\alpha\delta}g_{\beta\gamma}+R_{\beta\delta}g_{\alpha\gamma}-R_{\beta\gamma}g_{\alpha\delta}+\frac{R}{2}(g_{\alpha\delta}g_{\beta\gamma}-g_{\alpha\gamma}g_{\beta\delta})$$
where $R$ is the scalar curvature defined by $g^{ql}R_{ql}.$

I come up with a way by expanding every Ricci tensor to the linear combinations of Riemann curvature to show the relation holds for $R_{1212}$ and $R_{1213},$ and other situations are up to permutations, which is obviously a very direct and without any beauty. My problem is if there exists any other way to see this fact, more intuitively, or more generally.

Thanks in advances~

 

[Ben Niehoff]

At the end of the day, it's going to be some sort of "permute the indices and add the terms together" type of procedure, and will probably take as much work as what you've already done.

However, here is a hint: A lot of dimension-dependent identities (maybe even all of them?) in $d$ dimensions can be derived by looking at the totally antisymmetric symbol in $d+1$ dimensions,

$$\varepsilon_{\mu_1 \ldots \mu_{d+1}} = \varepsilon_{[\mu_1 \ldots \mu_{d+1}]}$$
which of course must vanish identically, since it antisymmetrizes $d+1$ indices. So, anything you contract with it must automatically vanish. Stated another way, take any expression with enough indices, antisymmetrize $d+1$ of them, and you must get zero. So, just write out the antisymmetrization, and you get a multi-term identity.

The definition of the Ricci tensor is (using Latin indices now because they're faster to write)

$$R_{ab} = g^{ef} R_{eafb}$$
Clearly, in 3 dimensions, one can get an identity by doing the following:

$$g_{[cd} R_{ab]} = g_{ef} g_{[cd} R^e{}_a{}^f{}_{b]} = 0$$
This might give you what you want if you write out the terms, but I haven't checked. If it doesn't work, you can probably still think of something that does.

 

[tommyxu3]

Hello~ Thanks for your opinions.

One of my ideas is that for Ricci and Riemann both have 6 independent components and Ricci can be expressed as the linear combinations of Riemann, then I just have to check if the equation right when multiplying $g^{\alpha\gamma}$ and take sum for $\alpha$ and $\gamma.$

Does this work?

 

[Daniel Mahler]

The Riemann tensor being determined by the Ricci tensor is equivalent to the Weyl tensor vanishing, the proof of which is discussed in the thread https://www.physicsforums.com/threads/weyl-tensor-on-3-dimensional-manifold.128275/