How to prove a manifold is simply connected?… using geometry

I was Looking at another questions title, and given the tag of DG, I thought it would read a little more like this one. Or at least that answers to this question would be answers to that question.

There are many different techniques one might use to show a manifold is simply conencted, I am only interested in a specific brand of them: those involving Differential Geometry. I borrowed a book from my friends physics professor, I think it was called Comparison theorems in Riemannian Geometry by Cheeger and Ebin, and the content was foreign to me. The goal of the book, if I recall properly, seemed to be proving theorems about $\pi_1(M)$ by examining geometric quantities like curvature.

This is amazing to me, and I am curious what the main idea could be. I currently only know of one relationship between geometric quantities and homotopy invariants, that is the fabled Chern-Weil theory. (fabled because I don't understand it... yet)

1. What are the precise statements of such types of results?

2. Which parts of the hypothesis do what? (for example what does compactness help you do here since it doesn't do a whole lot on the homotopy side(maybe that is wrong))

3. Other than other evidence/theorems, like Gauss-Bonnet, why would someone expect such results?

4. Can we get interesting results going the other way? How does $\pi_1$ affect things like curvature?

5. Can we say anything about higher homotopy groups?

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Have you looked at Dehn's Algorithm for hyperbolic groups? This is one of the earliest connections between curvature and the fundamental group and illustrates one of the main sources of the connection very well, IMO. I mean, that the fundamental groups of hyperbolic manifolds have such simple algorithms for things like the word problem, this is rather fundamental. –  Ryan Budney Dec 8 '10 at 10:34
A quick answer to question 2 -- compactness is very important on the homotopy side! It ensures that the fundamental group is finitely presented. –  Sam Nead Dec 9 '10 at 10:38

One also has the classical Bonnet-Myers and Synge theorems.

The Bonnet-Myers theorem states that if $M$ is complete with Ricci curvature bounded below $\delta > 0$, then $M$ is compact with finite fundamental group.

The idea of the proof is the opposite of the one in Matt E's answer. In positive (sectional) curvature, geodesics tend to get stuck together. One uses this to show that geodesics past a certain length (something like $\pi/\delta^2$) cannot continue to minimize.

It follows that if $B_r(0)\subseteq T_p M$ is any closed ball with radius $r > \pi/\delta^2$, then $exp(B_r(0)) = M$. In particular, you've written $M$ as the compact image of a continuous function so it's compact.

The theorem about the fundamental group is an immediate corollary: look at the universal cover $\tilde{M}$. One can "pull back" the metric so that the covering map is a local isometry. Hence, the same curvature estimates apply to $\tilde{M}$, so it, too, must be compact. But a compact manifold can only finitely cover something, so the fundamental group of $M$ is finite.

To push this to Ricci, one uses a nice trick: if $\text{Ric} > 0$, then the sectional curvature in some directions must be $> 0$, and one uses these directions only to show geodesics eventually stop minimizing.

The Synge theorems have a different style of proof. Here's one version of the theorem: suppose $M$ is compact with positive curvature, and $f:M\rightarrow M$ is an isometry. Suppose $\text{dim}(M)$ is even and $f$ is orientation preserving OR $\text{dim}(M)$ is odd and $f$ is orientation reversing. Then $f$ has a fixed point.

The proof is straightforward: suppose not. Choose $p\in M$ with $d(p,f(p))$ as small as possible. Choose a minimal geodesic from $p$ to $f(p)$. One then computes variations of the geodesic (this uses the parity of the dimension, if I recall correctly), and shows that some nearby geodesic is smaller, contradicting your choice of $\gamma$.

As a corollary, one learns that the fundamental group of an even dimension positively curved compact manifold is either trivial or $\mathbb{Z}/2\mathbb{Z}$. For, suppose $M$ satisfies all the hypothesis. Look at the deck group action on $\tilde{M}$. If it's orientation preserving, it has a fixed point, but the only deck transformation with a fixed point is the identity. If it's orientation reversing, then it's square is orientation preserving, and hence is the identity, so all elements are of order 2 in $\pi_1$. If there are 2 orientation reversing maps, then their product is orientation preserving and hence the identity, so there is at most one element of order 2.

Likewise, in odd dimensions, the corollary is that $M$ must be orientable, since the deck group must acts on $\tilde{M}$ by orientation preserving maps.

Finally, to just randomly answer one of your other questions, curvature can affect higher homotopy groups. As Matt E pointed out, in negative (really, nonpositive) curvature, the universal cover is $\mathbb{R}^n$, implying all higher homotopy groups vanish. In positive curvature, this is a very important and currently unsolved problem.

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Dear Jason, Thank you very much for this informative answer! –  Matt E Dec 9 '10 at 0:55

One example of how geometry affects $\pi_1$: a compact negatively curved manifold will have infinite fundamental group.

Proof: Negative curvature implies that the universal cover of such a manifold $M$ is diffeomorphic to $\mathbb R^n$. If $\Gamma = \pi_1(M)$, then $M = \mathbb R^n/\Gamma$, and so since $M$ is compact, the group $\Gamma$ must be infinite. QED

Speaking more vaguely, negative curvature makes geodesics diverge, and so tends to create a lot of $\pi_1$ (when the manifold is also compact, so that geodesics are forced to wrap around, and so have a chance to close up). On the other hand, positive curvature focuses geodesics, and so will tend to make $\pi_1$ smaller. A result somewhat related to this philosophy (that positive curvature implies simpler topology) is Gromov's theorem that the Betti numbers of a non-negatively curved manifold are uniformly bounded (the bound depends on the dimension, but nothing else). (This survey paper by Wilking gives a nice discussion of this result, and many related results about non-negative curvature and its implications for topology.)

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Thanks for the excellent answer! –  Sean Tilson Dec 10 '10 at 3:59

Brendle and Schoen use Ricci flow to prove a smooth version of the Sphere Theorem: An n-manifold with pinched positive curvature is diffeomorphic to a quotient of the $n$-sphere. (In particular, the fundamental group is finite.)

Very much in the other direction (ie regarding part four of your question) Shmuel Weinberger and co-authors have results of the form: Suppose that $M$ is a compact Riemannian manifold so that $\pi_1(M)$ has unsolvable word problem. Then the set of geodesics in $M$ is "wildly complicated". Some of this material is discussed in his book "Computers, Rigidity, and Moduli: The Large Scale Geometry of Riemannian Moduli Space".

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Of course, to get simply connected the much older version of the sphere theorem (which gives homeomorphism to the sphere) will suffice; and this is exactly the kind of result that is discussed in Cheeger and Ebin, if memory serves. –  Matt E Dec 8 '10 at 14:06
I thought simply connected was an assumption that goes into the problem. What if one applies everything to $\mathbb{R}P^n$ which is locally isometric to $S^n$? –  Jason DeVito Dec 8 '10 at 15:15
@Jason - Oops! Right you are. I'll make an edit. –  Sam Nead Dec 8 '10 at 15:34
@Matt - right - that is why I had "diffeomorphic" in italics. :) –  Sam Nead Dec 8 '10 at 15:36
Correction to my comment above: simply connected should be replaced by finite fundamental group. –  Matt E Dec 8 '10 at 16:32