Oh right, this is supposed to be a math blog, isn’t it?

So, I was trying to understand the proof in do Carmo (which is a great book, incidentally) of Synge-Weinstein, and I tried to write an expository paper on it. Unfortunately I can’t get the formatting perfect in WordPress since some of the equations go out of the margins, but most if it should look ok. I thus made a PDF of this too. I dressed it up in formal article formatting just for the heck of it.

We give an exposition of the proof of a few results in global Riemannian geometry due to Synge and Weinstein using variations of the energy integral.

**1. Introduction **

One of the big refrains of modern Riemannian geometry is that curvature determines topology. Recall, for instance, the basic Cartan-Hadamard theorem that a complete, simply connected Riemannian manifold of nonpositive curvature is diffeomorphic to under the exponential map. We proved this basically by showing that is nonsingular under the hypothesis of nonnegative curvature (using Jacobi fields) and that it was thus a covering map (the latter part was relatively easy). More difficult, and relevant to the present topic, was the Bonnet-Myers theorem, which asserted the *compactness* of a complete Riemannian manifold with bounded-below, positive Ricci curvature. The proof there showed that a long enough geodesic could not minimize energy (by using the second variation formula—recall that the second variation formula is intimately connected with curvature), and therefore could not minimize length. Since the distance between two points in a complete Riemanninan manifold is the length of the shortest geodesic between them (Hopf-Rinow!), this implied a bound on the diameter.

Today, however, we’re going to assume at the outset that the manifold in question is already compact. One of the theorems will be that *a compact, even-dimensional orientable manifold of positive curvature is simply connected*. In particular, there is no metric of everywhere positive sectional curvature on the torus .

How will we do this? Well, first consider the universal cover . The covering transformations of are all smooth, and we can endow with a metric in a natural way such that these are isometries, and has positive curvature—hence, by comppleteness (a covering manifold of a complete manifold is also complete, easy exercise) and the Bonnet-Myers theorem, is compact. It is also orientable since we can pull back the -orientation. If is not simply connected, then we can find a nontrivial covering transformation .

But, we will show, using the that an *isometry of a compact, oriented, even-dimensional manifold* admits a fixed point. In particular, does, which means that it is the identity, contradiction.

**2. The statement **

We will now begin work on the more general fixed-point theorem.

So, we’re going to start with a compact oriented -dimensional Riemannian manifold of positive sectional curvature and an isometry .

Theorem 1 (Weinstein)Suppose is as above and preserves orientation if is even and reverses orientation if is odd. Then has a fixed point.

The hypothesis about the dimension seems a little odd, but it comes from linear algebra used in the proof.

**3. The strategy **

Here is the strategy. Let be the metric. By compactness, there is such that is minimal. Assuimng this minimum is nonzero, we consider the minimal geodesic from to and construct a variation of it joining points . By its construction and the second variation formula, we will show that for small, which contradicts minimality.

So, how are we going to whisk this variation out of thin air? We will construct a parallel vector field on , perpendicular to , and let

In order that connects to , we need (assuming is parametrized by ).

**4. Construction of the vector field **

Proposition 2There exists a parallel vector field on , perpendicular to , such that .

The first step, paradoxically enough, will be to prove that itself satisfies these conditions (except orthogonality), in other words that:

Lemma 3.

*Proof:* Now is a geodesic starting at , and if we show that the piecewise smooth broken geodesic (concatenation) is actually smooth, we will have established the first step.

Pick some point in the middle of . Then . But there is a path from to of the same length , namely traversed starting at to . For instance, we could take and then traverse the curve from to , for a total distance of . This means that is smooth, hence so is ; the only point in doubt was at . In particular the left and right-hand derivatives match, so .

*Proof:*

There was, in fact, method to this madness. We are now going to use this fact and linear algebra to construct the vector field . So, the goal is to find some vector such that the transformation obtained by first applying (and sending to ) and then parallel translating back along has an eigenvector perpendicular to —which we just proved is a fixed point. Then the parallel field extending can be taken as our , which proves the lemma. Now consider the subspace . Now is an isometry so fixes , and is of dimension one smaller. Also (and hence ) preserves (resp. reverses) orientation if is odd (resp. even). By now invoking the following result from linear algebra, such a vector falls into our lap.

Lemma 4 (Linear algebra)Let be an orthogonal linear transformation of a real vector space . Suppose fixes orientation if is odd and reverses it if is even. Then has a nontrivial fixed point.

This will be proved later (in the appendix). Anyway, we now can use Proposition 1.

**5. The second variation formula **

**5.1. The approach **

Recall that we have defined the variation ; by what has been discussed, for all . In particular, we have paths between and . Recall also the *energy* of a piecewise-smooth path ; we shall use this in the sequel because it is easier to work with than the length (which has annoying square roots). Now

because has a minimum at . Indeed, —since moves at constant speed, being a geodesic—and by Schwarz’s inequality. When we prove

it will follow that there is some small with but

contradiction.

**5.2. Proof of the variation formula **

First, let us recall a more general version of the second variation formula and a sketch of the proof. Let be a geodesic, a smooth variation of (*not necessarily fixing endpoints*) with variation vector field . Then

This becomes (where, by abuse of notation denotes differentiation w.r.t. )

i.e.

Differentiate with respect to again:

We shall now analyze each term separately. The first two terms become

The last term becomes

Since is a geodesic, evaluation at of this yields

which in total yields

This is the version of the second variation formula that we shall use.

**6. Computation of the variation **

Now let’s apply the formula to the constructed in the proof of Weinstein’s theorem. Fortunately, most of the mess clears up. By parallelism, , so all that we are left with is

by hypothesis on the sectional curvature and since are orthogonal. It now follows, as discussed previously, that is not minimal, which gives a contradiction.

**7. Consequences **

Theorem 5 (Synge)Let be a compact -dimensional Riemannian manifold of positive curvature.

- If is even and is oriented, then is simply connected.
- If is odd, then is orientable.

*Proof:* We have already discussed case a) in the introduction. In case b), if is not orientable, then there is an orientable *double cover* . The manifold is compact, has an induced Riemannian metric of positive curvature, and has an orientation-reversing covering transformation when considered as a convering space of . This transformation must thus have no fixed points, which contradicts Weinstein’s theorem.

**8. Appendix: Proof of the linear algebra lemma **

For convenience, we restate the lemma:

*Let be an orthogonal linear transformation of . Suppose fixes orientation if is odd and reverses it if is even. Then has a nontrivial fixed point.*

*Proof:* First, in either case, the nonreal eigenvalues of occur in conjugate pairs, so the product of nonreal eigenvalues is positive. All the real eigenvalues are since is orthogonal.

- is odd. Then and has an odd number of real eigenvalues; they thus cannot all be .
- is even. Then and has an even number of real eigenvalues; they thus cannot all be .

October 27, 2010 at 9:50 pm

Which Do Carmo by the way? Do Carmo’s diff. geometry book or Do Carmo’s Riemannian geometry book? There’re two!

October 28, 2010 at 3:35 pm

The one on Riemannian geometry.

November 18, 2011 at 4:35 pm

I think there is a typo in the intro. Cartan-Hadamard theorem applies to manifolds of non-positive rather than non-negative curvature.

November 18, 2011 at 8:08 pm

Fixed, thanks.

January 19, 2012 at 6:02 pm

Hi, I want make a remark and a question. First, the compactness in Synge can be replaced by completness by the Soul Theorem of Perelman. And the question: can you do the same in Weinstein Thm?

Thanks.

January 19, 2012 at 6:26 pm

I haven’t thought about this in years and don’t remember this at all. Sorry! You might try mathoverflow.