# Homogeneous and Isotropic But not Maximally Symmetric Space

Is this statement correct: "In a homogeneous and Isotropic space the sectional curvature is constant, while in a maximally symmetric space the Riemann Curvature Tensor is covariantly constant in all directions everywhere. A maximally symmetric space is always homogeneous and isotropic, but the reverse is necessarily not true." ?

If it is correct, can you provide examples of some metrics which are not maximally symmetric despite being homogeneous and isotropic?

You are correct, a spacetime need not be maximally symmetric to be homogenous and isotropic. Isotropy and homogeneity are restrictions on the spatial structure of the universe, which lead to spatial maximal symmetry. A fully maximally symmetric spacetime (such as the de Sitter spacetime) would also include the temporal component. As an example, our own universe is (to a good approximation) spatially maximally symmetric but not maximally symmetric as a whole.

A universe (thus, a Lorentzian manifold) is spatially maximally symmetric if it can be foliated by a one-parameter family of spacelike hypersurfaces $$\Sigma_t$$ that are maximally symmetric.

Isotropy and homogeneity constraints lead to a spatially maximally symmetric Universe with an FRW metric:

$$$$\mathrm{d}s^2 = -\mathrm{d}t^2 + a^2(t) g_{ij}\mathrm{d}x^i \mathrm{d}x^j.$$$$

This is of the form $$\mathbb{R} \times \Sigma$$ where $$\mathbb{R}$$ represents time and $$\Sigma$$ is a homogeneous and isotropic three-manifold. See this document for a more in-depth explanation with Killing vectors.

OP's question is closely related to their preceding one. I have actually provided a near-complete answer to the present question there, although it appears that OP's question predates the clarifying edit I have made to my answer there.

Nonetheless it might be worth it to provide a more in-depth answer.

The problem is that OP assumes that$$(\text{maximally symmetric})\Leftrightarrow \nabla_m R_{ijkl}=0.$$

This is not the case. First of all for maximally symmetric and symmetric spaces there exist both local and global versions of the definitions. In this answer we will only be considering local conditions and as an abuse of terminology, I will write "symmetric space" in place of "locally symmetric space" and "maximally symmetric space" in place of "locally maximally symmetric space".

A maximally symmetric space of dimension $$m$$ is characterized by the fact that it has $$m(m+1)/2$$ linearly independent Killing vector fields, which is the maximum possible number.

Killing vector fields are those which satisfy $$\nabla_i X_j + \nabla_j X_i=0,$$i.e. their covariant derivative is antisymmetric. By differentiating this condition, and using the cyclic identity for the curvature tensor, we get $$\nabla_{i}\nabla_{j}X_{k}=R_{\ ijk}^{l}X_{l}.$$ By introducing the antisymmetric variables $$Y_{ij}=Y_{[ij]}$$, one can write this as the system $$\nabla_i X_j=Y_{ij},$$ $$\nabla_i Y_{jk}=R^l_{\ ijk}X_l.$$

The initial conditions for this system consists of the values $$X_i(p), Y_{ij}(p)$$ at some point $$p$$ of which there are $$m(m+1)/2$$, so this is the maximum number of independent solutions of this system. But since the system is overdetermined, it might not admit the maximum number of solutions. It does so if and only if it is completely integrable.

According to the Frobenius theorem, the conditions for the complete integrability can be obtained by differentiating the equations, using the equations to eliminate the derivatives of the dynamical variables, and then the second derivatives must be symmetric, which in this case is equivalent to imposing that the second covariant derivatives satisfy the Ricci identities. The integrability conditions for the first equation is just the cyclic identity for the curvature tensor, so it gives no nontrivial conditions. For the second equation, we get $$\left(\nabla_{i}R_{\ jkl}^{m}-\nabla_{j}R_{\ ikl}^{m}\right)X_{m}=\left(\delta_{k}^{n}R_{\ lij}^{m}-\delta_{l}^{n}R_{\ kij}^{m}+\delta_{i}^{n}R_{\ jkl}^{m}-\delta_{j}^{n}R_{\ ikl}^{m}\right)Y_{mn}.$$

If we want complete integrability, then this equation should be satisfied for any possible value $$X_i$$ and $$Y_{ij}$$ take at one point, so the integrability conditions separate into two independent groups: $$\nabla_{i}R_{\ jkl}^{m}=\nabla_{j}R_{\ ikl}^{m}\qquad(\ast)$$and$$\delta_{k}^{n}R_{\ lij}^{m}-\delta_{l}^{n}R_{\ kij}^{m}+\delta_{i}^{n}R_{\ jkl}^{m}-\delta_{j}^{n}R_{\ ikl}^{m} = \delta_{k}^{m}R_{\ lij}^{n}-\delta_{l}^{m}R_{\ kij}^{n}+\delta_{i}^{m}R_{\ jkl}^{n}-\delta_{j}^{m}R_{\ ikl}^{n}.\qquad(\ast\ast)$$

For the second condition $$(\ast\ast)$$, take the trace on $$n,k$$, which gives $$(m-1)R_{klij}=g_{ik}R_{jl}-g_{jk}R_{il}.\qquad(\sharp)$$ The left hand side is antisymmetric in $$k,l$$, therefore the right hand side must be so as well, from which we get $$g_{ik}R_{jl}-g_{jk}R_{il}=-g_{il}R_{jk}+g_{jl}R_{ik}.$$ Taking the trace on $$i,k$$ here gives $$R_{ij}=\frac{R}{m}g_{ij}.$$

If we substitute this back into $$(\sharp)$$, we get $$R_{ijkl}=\frac{R}{m(m-1)}\left(g_{ik}g_{jl}-g_{il}g_{jk}\right).\qquad(\sharp\sharp)$$

Now, one can deduce a truckload of info from these relations:

Maximal symmetry is the same as constant curvature:

Take also the first integrability condition $$(\ast)$$, contract over $$i,m$$, then again over $$j,l$$. We get $$0=(m-1)\nabla_k R,$$ so $$R$$ is constant. If this is substituted back in $$(\sharp\sharp)$$, we get that the space has constant curvature.

Conversely, if the space is constant curvature, then $$\nabla_m R_{ijkl}=0$$, so $$(\ast)$$ is identically satisfied, whereas it is straightforward, but laborous to verify that $$(\ast\ast)$$ is also satisfied, thus the Killing equation is completely integrable, and thus there is a maximal set of independent Killing vector fields.

Homogeneity and isotropy: The space is homogeneous if for each pair of infinitesimally nearby points, there is an infinitesimal isometry mapping one of the points into the other. This is of course the "local" notion of homogeneity, but as stated in the beginning of this answer, I will only consider local stuff. Translated to math, this means that given any point $$p$$ and any tangent vector $$v^i$$ at $$p$$, there is a Killing vector field $$X^i$$ such that $$X^i(p)=v^i$$ and $$\nabla_j X^i(p)=0$$ (this latter condition essentially means that the flow of $$X$$ does not rotate the vectors, but act only as translation, but admittedly, this is rather handwavy, modern differential geometers do this entire topic rather differently).

So then the condition for homogeneity is precisely $$(\ast)$$.

Likewise, we say that the space is isotropic about a point $$p$$ if there are infinitesimal isometries fixing $$p$$ but specifying an arbitrary infinitesimal rotation at $$p$$. This then means that it is possible to find a Killing vector field $$X^i$$ such that $$X^i(p)=0$$ and $$\nabla_i X_j(p)$$ is arbitrary.

The condition for isotropy at $$p$$ is that $$(\ast\ast)$$ should hold at $$p$$, so from this we can deduce that if the space is isotropic about $$p$$, we have $$R_{ijkl}(p)=\frac{R(p)}{m(m-1)}(g_{ik}(p)g_{jl}(p)-g_{il}(p)g_{jk}(p)),$$ but only at $$p$$.

But we then see that if the space is homogeneous and also isotropic about all points, then this precisely amounts to $$(\ast)$$ and $$(\ast\ast)$$, so that the space is then maximally symmetric and thus also constant curvature.

Schur's theorem:

Suppose that the space is isotropic about all points. Then $$(\sharp\sharp)$$ is satisfied, but $$(\ast)$$ not necessarily. Nonetheless, we can use the twice-contracted Bianchi identity on $$(\sharp\sharp)$$ to get $$0=\left(\frac{1}{m}-\frac{1}{2}\right)\nabla_i R,$$ which means that if $$m>2$$, isotropy about all points also implies homogeneity.

Symmetric spaces are not maximally symmetric:

A characterization of symmetric spaces is that $$\nabla_m R_{ijkl}=0$$. If this is inserted into the integrability conditions, we find that $$(\ast)$$ is satisfied, so in particular, the space is (infinitesimally) homogeneous. But there is absolutely no reason why this condition should also make $$(\ast\ast)$$ true.

An example of a symmetric space that isn't maximally symmetric (i.e. constant curvature) is $$\mathbb {CP}^n$$ for $$n>1$$ with the Fubini-Study metric.