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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 $.

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1 Answer 1

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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.

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  • $\begingroup$ 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 $\{3, 4\}$? $\endgroup$
    – oxedex
    Nov 6, 2021 at 14:51
  • $\begingroup$ 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 $\{3,4\}$). $\endgroup$ Nov 6, 2021 at 15:47
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    $\begingroup$ I edited the post calculating explicitly the Christoffels and I added a section about sectional curvature. Hope it's correct. $\endgroup$
    – oxedex
    Nov 6, 2021 at 17:22
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    $\begingroup$ 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/… $\endgroup$ Nov 6, 2021 at 19:32
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    $\begingroup$ 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} $ $\endgroup$ Nov 6, 2021 at 19:34

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