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Could you help me with the following problem, please?

I have the next proposition of the book An Introduction to Riemannian Geometry with Applications to Mechanics and Relativity of Leonor Godinho and José Natário.

Proposition 1.8 If $M$ is isotropic at $p$ and $x:V\to \mathbb{R}^{n}$ is a coordinate system around $p$, then the coefficients of the Riemannian curvature tensor at $p$ are given by \begin{equation}R_{ijkl}(p)=-K_{p}(g_{ik}g_{jl}-g_{il}g_{jk}).\end{equation}

For the proof of this proposition, we define the 4-tensor $A$ on $T_{p}M$ \begin{equation}A=\sum_{i,j,k,l=1}^{n}-K_{p}(g_{ik}g_{jl}-g_{il}g_{jk})dx^{i}\otimes dx^{j}\otimes dx^{k}\otimes dx^{l}.\end{equation}

I have to verify that $A$ satisfies the same symmetry properties of the Riemannian curvature tensor $R$, given by:

  1. $R(X,Y,Z,W)+R(Y,Z,X,W)+R(Z,X,Y,W)=0;$
  2. $R(X,Y,Z,W)=-R(Y,X,Z,W);$
  3. $R(X,Y,Z,W)=-R(X,Y,W,Z);$
  4. $R(X,Y,Z,W)=R(Z,W,X,Y).$

I proved properties $2$, $3$ and $4$ with no problem, but I can't proved property $1$. I proved property $1$ when $n =2$, but for the case of any $n$, I have a problem with indices. I cannot write this case the correct way to proof this property. I tried to use induction but to no avail.

How could I proof this in a correct and formal way?

Cal22
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1 Answers1

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Hint Your tensor $A$ can be re-written this way: $$ A(X,Y,Z,W) = -K_p \left(g(X,Z)g(Y,W) - g(X,W)g(Y,Z) \right). $$ Can you go on from there?

Comment In a general Riemannian manifold, the tensor $\overline{R}$ defined by $$ \overline{R}(X,Y,Z,W) =g(X,Z)g(Y,W) - g(X,W)g(Y,Z) $$ is a curvature-like tensor that has the property that if $(M,g)$ has constant sectionnal curvature $\kappa$, then $R^g = \kappa \overline{R}$. Note that I used Besses's convention, that is $R(X,Y) = \nabla_{[X,Y]} - [\nabla_X,\nabla_Y]$ and $\sec(X,Y) = R(X,Y,X,Y)$ for two orthonormal vectors. If your convention is the opposite, then the sectionnal curvature is $R(X,Y,Y,X)$ and if $(M,g)$ has constant sectionnal curvature $\kappa$, then $R^g = -\kappa \overline{R}$.

Didier
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  • Yes, thanks! It never occurred to me to use the metric for this. – Cal22 May 05 '21 at 18:00
  • Could you say what your definition of $R(X,Y,X,Y)$ is? And your formula for the sectional curvature isn't quite right if you don't normalize the vectors $X$ and $Y$ somehow. – Deane May 05 '21 at 19:51
  • I have this definition: $R(X,Y,Z,W)=g(R(X,Y)Z,W)$ where $R(X,Y)Z=\nabla_{X}\nabla_{Y}Z-\nabla_{Y}\nabla_{X}Z-\nabla_{[X,Y]}Z$. On the other hand, the formula for sectional curvature is $\frac{R(X_{p},Y_{p},X_{p},Y_{p})}{g(X_{p},X_{p})g(Y_{p},Y_{p})-g(X_{p},Y_{p})^{2}}$. – Cal22 May 05 '21 at 20:51
  • @Deane I did not put all the details, this was just to say that this is not Lee's convention, for example. Of course you're right, $X$ and $Y$ have to be normalized. Also, $R(X,Y,Z,T) = g(R(X,Y)Z,T)$ for me. – Didier May 05 '21 at 21:12
  • According to my calculations, the sectional curvature of the plane spanned by two orthonormal vectors $X$ and $Y$ is $K(X,Y) = g(R(X,Y)Y,X)$, which is the opposite sign from your formula. The correct order of the vectors or indices has caused me a lot of grief over the years. These days, when I write indices, my convention is $R(\partial_k,\partial_l)\partial_j = R^i{}{jkl}\partial_i$, so therefore, $R{ijkl} = g_{ip}R^p{}{jkl} = g(\partial_i,R(\partial_k,\partial_l)\partial_j)$. With this convention, $R{ijij}$ is the sectional curvature if the basis is orthonormal. – Deane May 05 '21 at 23:39
  • @Deane Are you referring to my comment, or to Andrés Soledispa's one? – Didier May 06 '21 at 06:32
  • @Didier, it appears that my comment is directly relevant to Andrés Soledispa's comment and not yours. I didn't read your explanation of Besse's definition carefully. Regarding that, I suggest that you not use it, since it's quite unconventional. Besse clearly rigged it to make things "work", but the ordering of the terms is unnatural, since $[\nabla_X,\nabla_Y]$ is the crucial term and the other expresses the torsion free condition. What I suggest is either avoiding the notation $R(X,Y,Z,W)$ or let $R(X,Y,Z,W) = g(X,R(Z,W)Y)$. – Deane May 06 '21 at 14:38
  • I would also note that the notation I propose is consistent with the differential form version, where the curvature $2$-form is defined to be $$\Omega^i_j = d\omega^i_j + \omega^i_p\wedge\omega^p_j = \frac{1}{2}R^i{}_{jkl}\omega^k\wedge\omega^l$$ – Deane May 06 '21 at 14:40
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    @Deane Well, I always used Besse's convention (I guess it's a French cultural taste). It never has been a problem as I always check what convention is used when I read a paper and try to use author's convention when needed. Also, as I try to work as much as possible coordinate-freely (I think it's again a cultural taste), it is usually not a problem: what is important is that we all agree that the sphere has positive curvature, I guess. – Didier May 06 '21 at 14:44
  • @Didier, I just checked the book of Gallot-Hulin-Lafontaine and a survey by Besson on the Ricci flow. It appears that indeed the French use the opposite convention of what I'm used to seeing. For some reason I never noticed that before. – Deane May 06 '21 at 15:04