# I have a metric, I need to find the curvature. How do I continue?

I have the manifold $$M \times N$$, where $$M, N$$ are 2-dim manifolds, with coordinates $$(x_1, x_2)$$ and $$(x_3, x_4)$$ respecrively. On $$M\times N$$ i have the metric

$$g(\mathbb x, \mathbb x') = \begin{bmatrix} \phi(\mathbb x) & 0 \\ 0 & \psi(\mathbb x') \end{bmatrix}$$

So $$ds^2_{M\times N} = ds^2_M+ds^2_N$$

$$ds^2_{M\times N}(x_1, x_2, x_3, x_4) =$$

$$= \phi_{11} dx_1^2 + \phi_{22}dx_2^2+2\phi_{12}dx_1dx_2 +$$

$$+ \psi_{33}dx_3^2+\psi_{44}dx_4^2 + 2\psi_{34}dx_3dx_4$$

The curvature is $$R_{ijkl}\\$$

I know by symmetry that $$R_{11ks} = R_{22ks} = R_{33ks} = R{44ks} = 0$$

$$R_{ij11} = R_{ij22}=R_{ij33} = R_{ij44} = 0$$

Then

$$R_{ijks} = -R_{jiks}$$

$$R_{ijks} = -R_{ijsk}$$

So $$R_{jiks} = R_{ijsk}$$

which implies

$$R_{ijks} = R_{ksij} = R_{ksji}$$

Now, if $$i=k$$ and $$j=s$$,

$$R_{ijij} = -R_{ijji} = R_{jiji} = - R_{jiij}$$

So the first 6 independent symbols to calculate are

$$R_{1212} \quad R_{1313} \quad R_{1414}$$

$$R_{2323} \quad R_{2424} \quad R_{3434}$$

The other 14 independent symbols are $$R_{1213} \quad R_{1214} \quad R_{1223}$$

$$R_{1224} \quad R_{1234} \quad R_{1314}$$

$$R_{1323} \quad R_{1324} \quad R_{1334}$$

$$R_{1423} \quad R_{1434} \quad R_{2324}$$

$$R_{3423} \quad R_{3424}$$

I am not sure whether all those symbols i wrote are the correct independent ones. I know the formula for Christoffel symbols, then I need to calculate all those 20 symbols, but I know most of them will be zero. Am I doing good? Because i think i am overdoing and making mistakes.

EDIT ($$\textbf{about Christoffel's symbols and sectional curvature}$$):

Considering the symmetries $$\Gamma_{aij} = \Gamma_{aji}$$, we need to calculate the following Christoffel symbols:

$$\Gamma_{111} \quad \Gamma_{112} \quad \Gamma_{122} \quad \Gamma_{211} \quad \Gamma_{212} \quad \Gamma_{222}$$

$$\Gamma_{333} \quad \Gamma_{334} \quad \Gamma_{344} \quad \Gamma_{433} \quad \Gamma_{434} \quad \Gamma_{444}$$

$$\Gamma_{111} = \partial_1g_{11}/2 \quad \Gamma_{112} =\partial_2g_{11}/2\quad \Gamma_{122} = (2\partial_2g_{12} - \partial_{1}g_{22})/2$$

$$\Gamma_{211} = (2\partial_1g_{21} - \partial_2g_{11})/2 \quad \Gamma_{212} = \partial_1g_{22}/2 \quad \Gamma_{222} = \partial_2 g_{22} /2$$

$$\Gamma_{333} = \partial_3g_{33}/2 \quad \Gamma_{334} = \partial_4g_{33}/2 \quad \Gamma_{344} = (2\partial_4g_{34} - \partial_{3}g_{44})/2$$

$$\Gamma_{433} = (2\partial_3g_{43} - \partial_4g_{33})/2 \quad \Gamma_{434} = \partial_3g_{44}/2 \quad \Gamma_{444} = \partial_4 g_{44}/2$$

And what about the sectional curvature $$S$$? Being $$M, N$$ bidimensional, we only have one tangent plane, so one sectional curvature. In dimension 2, we know that the Ricci tensor coincides with the sectional curvature.

EDIT II ($$\textbf{forgot to add metric symbols}$$):

For $$M$$:

$$S(e_1, e_2) = \frac{Ric(e_1, e_1) + Ric(e_2, e_2)}{2}$$

$$Ric(e_1, e_1) = g(R(e_1, e_1)e_1,e_1) + g(R(e_1, e_2)e_2, e_1) =$$

$$= R^1_{111}g_{11} + R^2_{111}g_{21} + R^1_{122}g_{11} + R_{122}^2g_{21}$$

$$Ric(e_2, e_2) =$$

$$= g(R(e_2, e_1)e_1,e_2) + g(R(e_2, e_2)e_2, e_2)$$

$$= R^1_{211}g_{12} + R^2_{211}g_{22} + R^1_{222}g_{12} + R^2_{222}g_{22} =$$

This implies $$S(M) =$$

$$= \frac{g_{11}}{2}(R^1_{111} + R^1_{122}) + \frac{g_{22}}{2}(R^2_{211} + R^2_{222})$$

$$+\frac{g_{12}}{2}(R^2_{111} + R^2_{122} + R^1_{222} + R^1_{211})$$

Similarly for $$N$$.

Let $$R^{M}$$ be the Riemann curvature of $$M$$, and $$R^{N}$$ be the Riemann curvature of $$N$$. Since you have just a product manifold where the $$(x_1,x_2)$$ and $$(x_3,x_4)$$ coordinates are independent, the resulting Riemann curvature has the property:

$$R_{abcd} (x,x')=\\= \begin{cases} R^{M}_{abcd}(x)\quad \text{if} \quad \{a,b,c,d\}\subset\{1,2\} \\ R^{N}_{abcd}(x')\quad \text{if} \quad \{a,b,c,d\}\subset\{3,4\} \\ 0 \quad \text{in any other case (that is,} \\ \quad\; \text{ when indices come from both subspaces.)} \end{cases}$$

You still need to compute the Riemann curvature of the $$2$$-dimensional $$M$$ and $$N$$ submanifolds, but it's easier than working in the full $$4$$-dimensional manifold.

An informal justification for the above would be that parallel transport of a vector from the $$M$$ component of the tangent space along a curve formed in the $$N$$ component leaves it unchanged. Any change in the $$M$$-components of a vector must come from movement along the $$M$$-subspace.

• So the only two terms i have to calculate are $R_{1212}$ and $R_{3434}$. I still have troubles with the Christoffel's symbols. Now I have the formula for the $\Gamma_{ijk}$: $\Gamma_{ijk} = \frac{ \partial_k(g_{ij}) + \partial_j(g_{ik}) - \partial_i(g_{jk}) }{2}$ But, still, there are 8 Christoffels for $\{1, 2\}$ and 8 for $\{3, 4\}$? Nov 6, 2021 at 14:51
• Yes. There are $8$ index components for the $2D$ Christoffel symbol, tough with symmetry you see there are $6$ independent ones. The symmetry is specifically $\Gamma_{ijk} = \Gamma_{ikj}$, giving $\Gamma_{a12} = \Gamma_{a21}$ for $a\in\{1,2\}$. So you still need to compute $2*6 = 12$ components ($6$ for$\{1,2\}\;$ and $6$ for $\{3,4\}$). Nov 6, 2021 at 15:47
• I edited the post calculating explicitly the Christoffels and I added a section about sectional curvature. Hope it's correct. Nov 6, 2021 at 17:22
• If you're using the $(3,1)$ Riemann tensor $R^{a}_{bcd}$ (some prefer to use the $(4,0)$ one $R_{abcd}$ because it has more symmetries ), then you already have $1$ covariant and $1$ contra-variant index. You don't need to add the metric when contracting: $Ric(e_1, e_1) = R^{1}_{111}+ R^{2}_{121} = R^{2}_{121}$. Here is a formula for the $(4,0)$ Riemann tensor: wikimedia.org/api/rest_v1/media/math/render/svg/… Here is a formula for the $(3,1)$ one: wikimedia.org/api/rest_v1/media/math/render/svg/… Nov 6, 2021 at 19:32
• They are related by $R^{a}_{bcd} = g^{ax}R_{xbcd}$. So for example $R^{1}_{212} = g^{11} R_{1212} + g^{12} R_{2212} = g^{11} R_{1212}$ Nov 6, 2021 at 19:34