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Let's have an interval expression for Friedmann Universe with 3-metric of a sphere, $$ ds^{2} = c^{2}dt^{2} - c^{2}\frac{ch^{2}(Ht)}{H^{2}}\left( d\rho^{2} + sin^{2}(\rho )(d\theta^{2} + sin^{2}(\theta )d\varphi ^{2})\right), $$ with metric tensor, which is given by $$ \hat {\mathbf g} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -ch^{2}(Ht) & 0 & 0 \\ 0 & 0 & -ch^{2}(Ht) \sin^2 (\rho ) & 0 \\ 0 & 0 & 0 & -ch^{2}(Ht) \sin^2 (\rho ) sin^{2} (\theta ) \\ \end{bmatrix}. $$ What physical sense has transformation $$ \hat {\mathbf T}, \quad |\hat {\mathbf T} = 1|, \quad T_{\alpha \beta } = T_{\beta \alpha}, $$ which is given by $$ \hat {\mathbf T}\hat {\mathbf g} \hat {\mathbf T} = \hat {\mathbf g}? $$ Maybe, the only space part of transformations determines the motion on the sphere, but what means the time part? It's something like Lorentz transformations, but for non-inertial systems.

The trivial solution for $\hat {\mathbf T}$ is $\hat {\mathbf T} = diag(1, 1, 1, 1)$. But maybe there exist other solutions, don't they?

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Well $T$ is in the symmetric part of $SL(4,\mathbb R)$, whatever thid might be. You can switch some signs of $diag(1,1,1,1)$, but I don't see another option just yet. In two dimensions you have sinh and cosh functions as entries, but they are no global isometriy of $g=((g_{11},0),(0,g_{22}))$ except for the simple case you mention. I have not much hope actually. – NikolajK Jun 26 '13 at 23:00
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Note that you can write $g$ as $g = D \eta D$, where :

$\eta$ = Diag$(1,-1,-1,-1)$,

$D$= Diag $(1,ch(Ht),ch(Ht) sin (\rho),ch (Ht)sin (\rho)sin (\theta))$

So you have : $T(D \eta D)T = D \eta D$, that is $(D^{-1}TD)\eta(DTD^{-1}) = \eta$

That is : $(DTD^{-1})^t\eta(DTD^{-1}) = \eta$, because $D$ and $T$ are symmetric.

So $DTD^{-1}$ is an element of $SO(1,3)$

So, the set $S$ of matrix $T$, is the symmetric part, of the set $\Sigma$ of matrix $D^{-1}XD$, where $X$ is an element of $SO(1,3)$.

An obvious subset of $S$ is the set $S'$ of symmetric matrix of $SO(1,3)$ which commute with $D$ (in this case $T =X$).

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