Understanding the Definition of Isotropic Spaces in Riemannian Manifolds

[Wledig]

Why does the constraint:
$$R_{ijkl}=K(g_{ik} g_{jl} - g_{il}g_{jk})$$
Imply that the resulting space is maximally symmetric? The GR book I'm using takes this relation more or less as a definition, what is the idea behind here?

 

[Paul Colby]

Well, if $K$ is constant, so is $R_{ijkl}$ since $\nabla_\mu g_{ij} = 0$

 

[Wledig]

Not sure I see how this leads to the space being isotropic and homogeneous. And why should it be proportional to the curvature?

 

[zinq]

I find it frustrating to be faced with math questions that are not explained well at all.

What is assumed about the space? Is it 3-dimensional, as the title seems to imply? (A 3-dimensional manifold is sometimes called a 3-manifold, but only 3-dimensional Euclidean space is called "3-space" with no qualification.) Is it compact? What does the original poster mean by the use of the phrase "maximally symmetric"?

And it would help to make your notation and summation conventions explicit instead of implied.

Without clarity on these issues it can be difficult or impossible to answer questions like this.

(But: If a space is both isotropic and homogeneous, then it must have constant sectional curvature. This implies that, up to uniform scaling, its universal (i.e., simply connected) covering space is one of

a) the unit n-sphere,

b) n-dimensional Euclidean space, or

c) n-dimensional hyperbolic space

— with constant sectional curvatures equal to 1, 0, or -1, respectively.)

 

[Infrared]

But: If a space is both isotropic and homogeneous, then it must have constant sectional curvature.

I don't think this is true. Isn't $\mathbb{C}P^n$ for $n>1$ (with the Fubini-Study metric) a counterexample?

 

[Wledig]

Without clarity on these issues it can be difficult or impossible to answer questions like this.

Sorry, I assumed the concepts and notation were well known. Here's how Hobson puts it:

"A maximally symmetric space is specified by just one number - the curvature K, which is independent of the coordinates. Such constant curvature spaces must clearly be homogeneous and isotropic."

So it's like I've said, he takes this the constraint to be more or less a definition, which I'm trying to justify.

 

[zinq]

No. What I wrote is true: CPⁿ is homogenous but for n > 1 it's not isotropic. (Homogeneity ensures that all points look the same re curvature, and isotropy ensures that, at a given point, all sectional curvatures are equal.)

E.g., see https://en.wikipedia.org/wiki/Fubini%E2%80%93Study_metric#Curvature_properties.

 

[Infrared]

No. What I wrote is true: CPⁿ is homogenous but for n > 1 it's not isotropic. (Homogeneity ensures that all points look the same re curvature, and isotropy ensures that, at a given point, all sectional curvatures are equal.)

Are you sure? https://en.wikipedia.org/wiki/Isotropic_manifold gives $\mathbb{CP}^n$ as an example of an isotropic manifold.

The group $U(n+1)$ acts on $\mathbb{CP}^n$ by isometries, and the subgroup fixing a point $p$ is a copy of $U(n)$ acting in the usual way on $T_p\mathbb{CP}^n=\mathbb{C}^n.$ Since $U(n)$ acts transitively on the unit sphere in $\mathbb{C}^n$, we have isotropy.

To preserve sectional curvatures, you would want the isotropy group of a point to act transitively on the set of $2$-planes, not just the set of unit vectors.

 

[zinq]

Clearly I was assuming a stronger definition of "isotropic" than the one that others use for a Riemannian n-manifold: Each point x has a neighborhood U such that the group of isometries of U that fix x is the orthogonal group O(n).

 

[Infrared]

Clearly I was assuming a stronger definition of "isotropic" than the one that others use for a Riemannian n-manifold: Each point x has a neighborhood U such that the group of isometries of U that fix x is the orthogonal group O(n).

I don't know how it was "clear" that you were using your own invented definition of isotropic, especially since it's not the definition OP is using. The condition OP imposes on the metric implies that the manifold is (homogeneous and) isotropic in the usual sense, but not that it has constant sectional curvature as in your post 4.