Section 5.1 of Wald's *General Relativity*

[Alpha2021]

From the section[5.1] of 'Homogeneity and Isotropy' from General Relativity by Robert M. Wald (pages 91-92, edition 1984) whatever I have understood is that -

$\Sigma_t$ is a spacelike hypersurface for some fixed time $t$. The hypersurface is homogeneous.

The metric of whole space is $g$ and the form of the metric on hypersurface $\Sigma_t$ is $h$. Thus if $g$ is the metric of dimension $4$, then $\Sigma_t$ has dimension $3$. Next, Riemann curvature tensor $R_{ab}{}^{cd}$ is defined from the metric $h$ (Or, from the metric $g$, I am not sure about this). Now, if there is an antisymmetric tensor $A_{ij}$ defined on $\Sigma_t$, this tensor is transformed by $R_{ab}{}^{cd}$ as $A'_{ij} = R_{ij}{}^{cd} A_{cd}$. $R_{ij}{}^{cd}$ itself is antisymmetric with respect to its two indices $i$ and $j$. This transformation can be viewed as linear self-adjoint transformation. If we name this linear transformation as $L$ and vector space as $W$, then $L: W \to W$. This vector space $W$ can be spanned by eigenvector $L$. The corresponding eigenvalues must be equal because of isotropy. Thus $L$ can be expressed as multiple of the identity operator i.e. $L=K I$.

Then he suddenly claimed that

$$
R_{ab}{}^{cd} = K \delta^c{}_a \delta^d{}_b
$$

I have not understood how to claim this equation.

 

[PeterDonis]

Hi @Alpha2021 and welcome to PF!

For equations here we use LaTeX, and there is a LaTeX Guide link at the bottom left of the window you use to make posts. Please use it; it helps a lot to make your posts more readable when they have equations in them. I have edited your OP in this thread to use LaTeX.

 

[PeterDonis]

Riemann curvature tensor is defined from the metric $h$ (Or, from the metric $g$, I am not sure about this).

In this particular case, Wald is using the Riemann tensor on the surface of homogeneity $\Sigma_t$, meaning it is derived from the metric $h$ on that surface. He uses the notation ${}^{(3)}R$ for this reason.

I have not understood how to claim this equation.

You left out an essential part of that equation, the square brackets around the lower indexes on the RHS. The equation as Wald wrote it is (note that I am also using the ${}^{(3)}R$ notation described above):

$$
{}^{(3)} R_{ab}{}^{cd} = K \delta^c{}_{[ a} \delta^d{}_{b ]}
$$

The square brackets denote antisymmetrization, meaning that the above equation expands to:

$$
{}^{(3)} R_{ab}{}^{cd} = K \left( \delta^c{}_a \delta^d{}_b - \delta^c{}_b \delta^d{}_a \right)
$$

The expression in parentheses on the RHS is simply the identity operator in the vector space $W$ of 2-forms, i.e., the operator that takes any 2-form into itself. The antisymmetrization is necessary because we are dealing with 2-forms, which are antisymmetric (0, 2) tensors.

 

[Alpha2021]

@PeterDonis
Thanks a lot. I missed the latex option. I am sorry.

 

[Alpha2021]

Hi @PeterDonis,
Thanks a lot for your kind reply.
The first answer is clear to me.

About the second question - I am sorry for my mistake. But, I still have the confusion. I know Riemann tensor as
$R^{\eta }_{\; \mu \nu \alpha} = \frac{\partial \Gamma^{\eta}_{\mu \alpha}}{\partial x^{\nu}}+ \frac{\partial \Gamma^{\eta}_{\mu \nu}}{\partial x^{\alpha}} + \Gamma^{\eta}_{\kappa \nu }\Gamma^{\kappa }_{\mu \alpha} -\Gamma^{\eta}_{\kappa \alpha}\Gamma^{\kappa }_{\mu \nu}$

Then how has it transformed to $K\delta^c{}_{[a} \delta^d{}_{b]}$? What is its relation with equal eigen values?

 

[PeterDonis]

how has it transformed to $K\delta^c{}_{[a} \delta^d{}_{b]}$?

Because the constraints of homogeneity and isotropy mean that almost all of the components of the Riemann tensor of $\Sigma_t$ vanish. The only ones left are the ones in that equation.

What is its relation with equal eigen values?

Equal eigenvalues is one of the consequences of homogeneity and isotropy--specifically isotropy, which means that there can't be any direction in space that is different from the others. That requires all eigenvalues of ${}^{(3)} R$, considered as a linear transformation on the space $W$ of 2-forms, to be equal; if they weren't, the different eigenvalues would mean different directions in space had different properties.

 

[Alpha2021]

@PeterDonis
Thanks a lot for your kind reply.

 

[PeterDonis]

@PeterDonis
Thanks a lot for your kind reply.

You're welcome!