# Ricci curvatures determine Riemann curvatures in 3-dimension

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• tommyxu3
This proof involves writing out the Riemann tensor in terms of the Ricci tensor, and using the fact that the Weyl tensor vanishes in 3 dimensions to simplify the expression. Therefore, the relation between the Riemann and Ricci tensors in 3 dimensions is not a coincidence, but rather a consequence of the special properties of 3-dimensional manifolds. In summary, in 3-dimensional space, the Ricci curvature tensor can be used to determine the Riemann curvature tensor through a specific relation derived from the Weyl tensor vanishing.
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.

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.

Spinnor
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?

## 1. What is the relationship between Ricci curvatures and Riemann curvatures in 3 dimensions?

The Ricci curvatures and Riemann curvatures are both measures of curvature in a 3-dimensional space. The Ricci curvature is a scalar value that describes the average curvature in a given direction, while the Riemann curvature is a tensor that describes the curvature in all directions at a given point. In 3 dimensions, the Ricci curvature determines the Riemann curvature, meaning that knowing the Ricci curvature at a point allows us to calculate the Riemann curvature at that point.

## 2. How are Ricci curvatures and Riemann curvatures calculated?

Ricci curvatures are typically calculated using the Ricci tensor, which is a symmetric 2-dimensional tensor that is derived from the Riemann curvature tensor. Riemann curvatures, on the other hand, are calculated using the Riemann curvature tensor, which is a 4-dimensional tensor that describes the curvature in all directions at a given point.

## 3. Can Ricci curvatures and Riemann curvatures be negative?

Yes, both Ricci curvatures and Riemann curvatures can be negative. Negative curvatures indicate that space is negatively curved, meaning that parallel lines will eventually converge. This is in contrast to positive curvatures, where parallel lines will diverge, and flat curvatures, where parallel lines will remain parallel.

## 4. How are Ricci curvatures and Riemann curvatures used in physics?

In physics, Ricci curvatures and Riemann curvatures are used to describe the curvature of spacetime in Einstein's theory of general relativity. In this theory, the curvature of spacetime is related to the distribution of matter and energy. By calculating the Ricci and Riemann curvatures, we can understand how matter and energy affect the curvature of spacetime and how this affects the motion of objects in the universe.

## 5. Are Ricci curvatures and Riemann curvatures the only measures of curvature in 3 dimensions?

No, there are other measures of curvature in 3 dimensions, such as the scalar curvature and the sectional curvature. However, Ricci curvatures and Riemann curvatures are two of the most commonly used measures, particularly in the context of general relativity and differential geometry.

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