Riemannian Geometry

Question

I come from a background of having done undergraduate and graduate courses in General Relativity and elementary course in riemannian geometry.

Jurgen Jost's book does give somewhat of an argument for the the statements below but I would like to know if there is a reference where the following two things are proven explicitly,

1. That the sectional curvature of a 2-dimensional subspace of a tangent space at a point on the Riemannian manifold is independent of the choice of basis. That is the definition of the sectional curvature depends only on the choice of the 2-dim subspace.

2. That the sectional curvature determines the Riemannian curvature fully.

Secondly can one give me a reference where I can see how in practice is sectional curvature computed. To a first timer to this subject it is not obvious how one does a calculation on "all" 2-dimensional subspaces of a high-dimensional space. Especially when people talk of manifolds with "constant sectional curvature". How are they realized?

I would like to see some explicit examples to understand this point.

Further some studies about homogeneous spaces (needed to understand some issues in Quantum Field Theory) got me to the following 4 very non-trivial ideas in Riemannian manifolds which I am stating in my own way here ,

1. That the isometry group of a Riemannian manifold is always a lie group.

2. The isotropy subgroup of any point on a Riemannian manifold under the smooth transitive action of its own isometry group on itself is a compact subgroup. (The context being what is called a "Riemannian Homogeneous Space")

3. {This point was earlier framed in a way which made the bi-implication false as pointed out by some people} The formulation should be as follows.

   A Riemannian Homogeneous Space is a riemannian manifold on which the isometry group acts transitively. Now the theorem is that such a space is compact IFF its isometry group is compact.

   Thats the statement whose intuition I am looking for.

   Apologies for the confusion caused.

4. { This question too was not framed properly. Basically I could not figure out how to write the nabla for the connection! It should be as Jose has pointed out.}

   A riemannian manifold is locally symmetric if and only if the Riemann curvature tensor is parallel with respect to the Levi-Civita connection.

Can one give me the intuition behind these or give me specific references where these are proven in explicit details?

Answer 1

To get a better feel of the Riemann curvature tensor and sectional curvature:

1. Work through one of the definitions of the Riemann curvature tensor and sectional curvature with a $2$-dimensional sphere of radius $r$.
2. Define the hyperbolic plane as the space-like "unit sphere" of $3$-dimensional Minkowski space, defined using an inner product with signature $(-,+,+)$. Work out the sectional and Riemann curvature of that
3. Repeat #1 and #2 for the $n$-dimensional sphere and hyperbolic space, as well as flat space

Sectional curvature determines Riemann curvature:

That the sectional curvature uniquely determines the Riemann curvature is a consequence of the following:

1. The Riemann curvature tensor is a quadratic form on the vector space of $\Lambda^2T_xM$
2. The sectional curvature function corresponds to evaluating the Riemann curvature tensor (as a quadratic form) on decomposable elements of $\Lambda^2T_xM$
3. There is a basis of $\Lambda^2T_xM$ consisting only of decomposable elements

Added in response to Anirbit's comment

Perhaps you shouldn't try to compute the curvature too soon. First, make sure you understand the Riemannian metric of the unit sphere and hyperbolic space inside out. There are many ways to do this. But the most concrete way I know is to use stereographic projection of the sphere onto a hyperplane orthogonal to the last co-ordinate axis. Either the hyperplane through the origin or the one through the south pole works fine. This gives you a very nice set of co-ordinates on the whole sphere minus one point. Work out the Riemannian metric and the Christoffel symbols. Also, work out formulas for an orthonormal frame of vector fields and the corresponding dual frame of 1-forms. Figure out the covariant derivatives of these vector fields and the corresponding dual connection 1-forms.

After you do this, do everything again with hyperbolic space, which is the hypersurface

$-x_0^2 + x_1^2 + \cdots + x_n^2 = -1$ with $x_0 > 0$

in Minkowski space with the Riemannian metric induced by the flat Minkowski metric. You can do stereographic projection just like for the sphere but onto the unit $n$-disk given by

$x_1^2 + \cdots + x_n^2 = 1$ and $x_0 = 0$,

where the formula for the hyperbolic metric looks just like the spherical metric in stereographic co-ordinates but with a sign change in appropriate places. This is the standard conformal model of hyperbolic space.

After you understand this inside out, you can use these pictures to figure out why the $n$-sphere and its metric is given by $O(n+1)/O(n)$ and hyperbolic space by $O(n,1)/O(n)$ and why the metrics you've computed above correspond to the natural invariant metric on these homogeneous spaces. You can then check that the formulas for invariant metrics on homogeneous spaces give you the same answers as above.

Use references only for the general formulas for the metric, connection (including Christoffel symbols), and curvature. I recommend that you try to work out these examples by hand yourself instead of trying to follow someone else's calculations. If possible, however, do it with another student who is also trying to learn this at the same time.

If, however, you want to peek at a reference for hints, I recommend the book by Gallot, Hulin, and Lafontaine. I suspect that the book by Thurston is good too (I studied his notes when I was a student). For invariant Riemannian metrics on a homogeneous space, I recommend the book by Cheeger and Ebin (available cheap from AMS! When I was a student, I had to pay a hundred dollars for this little book but it was well worth it).

But mostly when I was learning this stuff, I did and redid the same calculations many times on my own. I was never able to learn much more than a bare outline of the ideas from either books or lectures. Just try to get a rough idea of what's going on from the books, but do the details yourself.

Answer 2

Here is one way to think about your first question which at least might provide a more geometric picture about what is going on.

I want to think about the curvature $R(X,Y)$ as parallel transport around the infinitesimal parallelogram $X \wedge Y$. If I drag a vector $Z$ around the parallelogram $X \wedge Y$, the result is $R(X,Y)Z$. Since the connection is metric, the map $Z \mapsto R(X,Y)Z$ is actually an infinitesimal rotation; this is the observation that $$\langle R(X,Y)Z, W\rangle = -\langle Z, R(X,Y)W\rangle$$

Now I want to define a new operator $S$ which acts bilinearly on pairs of 2-vectors. This will be $$S(X\wedge Y, Z \wedge W) = \langle R(X,Y)Z,W\rangle$$ where I have summed over some basis of 2-planes in $\bigwedge^2 T_pM$. Geometrically, $S$ reports how much the infinitesimal 2-plane $Z \wedge W$ rotates as it is dragged around the 2-plane $X\wedge Y$. To see that this is well-defined we need only to check $S(-,Z\wedge W) = -S(-, W \wedge Z)$. But this follows precisely because of the previous equation for $R$.

From here on, I'm going to use the metric to think of $S$ as $$S(X \wedge Y) = \sum_{2\text{-planes } Z\wedge W} \langle R(X,Y) Z,W \rangle ~Z\wedge W$$ The somewhat mysterious "pair swap" symmetry $\langle R(X,Y) Z, W\rangle = \langle R(Z,W)X,Y\rangle$ can now be interpreted as saying that the operator $S$ is symmetric. In particular, this means that we can take the spectral decomposition of $S$ to get a basis of orthogonal unit-area eigenplanes $X_i \wedge Y_i$, $$S(X_i \wedge Y_i) = \lambda_i \cdot X_i \wedge Y_i$$ The eigenvalues $\lambda_i$ are your sectional curvatures for this basis; any other sectional curvatures can be easily computed from these.

Note that knowledge of $S$ is now clearly sufficient to reconstruct the curvature tensor, since $$\langle R(X,Y)Z, W\rangle = \langle S(X\wedge Y), Z \wedge W \rangle$$ so in fact the sectional curvature tensor $S$ determines the usual curvature tensor $R$.

Answer 3

2) is very easy (assuming your manifold is connected; if not, it's false): you have an induced action of the isometries which fix the point x on the tangent space $T_xX$. This action preserves the metric, which is a positive definite inner product on this vector space. That is, this isometry subgroup is thus a closed subgroup of $SO(T_xX,g)$, which is a compact group. This actually also proves that it's a Lie group, since any closed subgroup of a Lie group is itself Lie.

This also makes it easy to prove the true direction of 3), since the isometry group acting on x gives a submersion with compact image and compact fibers, showing the group is compact.

Answer 4

do Carmo's "Riemannian Geometry"

Answer 5

The "only if" direction in 3 is incorrect: take standard $\mathbb R^2$ and introduce several bumps to make the isometry group trivial. The "if" direction is the Kobayashi-Nomidzu "Foundations of Differential Geometry", vol I, around Theorem 4.6 for references. There and in vol II you also find answers to both of your 1-2 questions, I think.

To see how sectional curvature is computed you need to go through a lot of examples.

The "if" direction in 4 is incorrect: there are manifolds of constant scalar curvature that are not locally symmetric.

Answer 6

Just to add some things to Igor Belegradek's post:

"1.That the isometry group of a Riemannian manifold is always a lie group."
This is the famous Myers-Steenrod theorem, proven in 1939 (Myers, S.B. and N.E. Steenrod: The group of isometries of a Riemannian manifold. The Annals of Mathematics, Vol 40, No. 2, April 1939, p. 400-416.)
It is in fact highly non-trivial, and I think you need that the manifold is connected

Your point "3.That isometry group of a Riemannian manifold is compact IFF the Riemannian manifold is compact." is as Igor pointed out false, the only thing which is right is the following

3.If the (connected) Riemannian manifold is compact then the isometry group is compact.
This is also a part of Myers-Steenrod theorem, and can be found in the reference above.

The "idea" of the proof is the following: (Let $(M,g)$ be a Riemannian manifold)

• Show that $(G=Iso(M,g), CO, op)$ is a locally compact topological transformationgroup.Here $CO$ is the compact-open topology, and $op: G \times M \rightarrow M$ the group action. Moreover $(M,g)$ compact implies $(G, CO, op)$ compact.
• Show that any tangential subgroup $H$ of $Diff(M)$ inherits a differentiable structure $[b]$ such that $(H,[b],op)$ ($op$ being the natural operation on $M$) is a Lie-Transformation group which is first-countable. The underlying topology $\tau$ is finer than $CO$-topology.
  (If $(M,g)$ has countable many connected components, $G$ is a tangential subgroup of $Diff(M)$)
• Show that the topology $\tau$ cannot be strictly finer than the $CO$-topology. (needs frame-bundles, etc.)

Answer 7

Two answers (by Deane Yang and Matt Noonan) have addressed the question about sectional curvatures determining the full curvature tensor, but they seem incomplete to me. The proofs I know use the additional symmetry $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$.

Of course there's a basis for $\Lambda^2 T_xM$ consisting of decomposable two-vectors, as Deane says, but knowing a quadratic form on a basis doesn't determine it – you need to know the bilinear form on pairs of basis elements. And in Matt's argument, why should the eigenvectors of $S$ be decomposable two-vectors?

Answer 8

I have two favourite books on differential geometry where you can find answers to your questions:

1. do Carmo's Riemannian Geometry (as suggested by David Lehavi)
2. Besse's Einstein manifolds

Let me just point out that your 4th point is not quite correct. The statement is that

A riemannian manifold is locally symmetric if and only if the Riemann curvature tensor (and not just the scalar curvature) is parallel with respect to the Levi-Civita connection; i.e., $\nabla R = 0$

This presupposes that by "locally symmetric" you understand that the geodesic symmetry (i.e., changing the sign of the parameter of the geodesic) is an isometry at every point, otherwise it is a definition of locally symmetric.

---

Edit (in response to Anirbit's comment)

This is indeed a result of Élie Cartan and in fact, as far as I understand the history, Cartan started his research on symmetric spaces by studying the question of which riemannian manifolds have parallel curvature. He then classified the irreducibles and found the well-known relationship to the classification of simple Lie algebras. I'm not sure when the characterisation in terms of the geodesic symmetry was introduced. The proof is not complicated. It is basically that the curvature tensor is invariant under the map which interchanges opposite points along a geodesic. In other words, if you fix a point $p $ in your manifold and look at a geodesic $\gamma$ through $p$ in the direction $X$, then if you follow the geodesic a 'time' $s$ you get to some point $p(s)$. But there is also a geodesic through the same point with direction $-X$ and if you follow that geodesic for a time $s$ you end up at a point $p(-s)$. The map which sends $p(s)$ to $p(-s)$ for any (small, say) $s$ leaves the curvature invariant. The covariant derivative of the curvature along $X$ at $p$ can be understood as the difference between the curvature parallel transported to $p(s)$ and that transported to $p(-s)$ divided by $s$ in the limit as $s\to 0$, but even before you take the limit, the difference vanishes.

Since you said your background is in relativity, I wonder whether you are not also interested in the case of locally symmetric spaces in lorentzian (or other indefinite) signature. In general signature this is still an open problem, but for lorentzian it was solved by Cahen and Wallach in this paper.