Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One of my professors wrote the following open question on the blackboard:

If $M$ is a compact, connected smooth $4$-manifold such that $\pi_1(M) = 0$, $\pi_2(M) = 0$ (first two homotopy groups are trivial), does it follow that $M$ is diffeomorphic to the $4$-sphere?

and warned us, that if we managed to solve it, we would get an instant Ph.D. -- so, keen on getting a Ph.D. before my bachelor degree, I went to work immediately! ;-)

My first thought was the following: If one could endow $M$ with a Riemannian metric giving a Riemannian manifold with constant sectional curvature $1$, then by compactness $M$ would be a complete, connected, simply connected manifold of curvature $1$, which would imply the statement.

Now I obviously didn't get much further than this (my dreams were shattered!).

Anyways, this leads to the question:

"When is it possible to endow a smooth manifold with a metric which has some desired properties (i.e. constant curvature or bounded curvature)?"

Has there been much work on this? Are there any good books/papers I could take a look at (just to get some impression of how the experts approach this problem)?

I was also wondering whether the above is actually an approach to the problem taken by people working in the field? Or may it be completely hopeless to try and gain any control of the metric globally?

Well, as always I thank in advance for any comments, answers etc.

Best regards, S.L.

share|improve this question
2  
The index theorem has a lot of corollaries dealing with manifolds admitting certain types of metrics. A reference for some of the results is Lawson and Michelson's ``Spin Geometry." Some of the relevant stuff is on google books: books.google.com/…. –  Eric O. Korman Mar 3 '11 at 15:11

1 Answer 1

up vote 12 down vote accepted

I imagine you'd get more than a Ph. D!

I get the impression that it's generally hopeless to gain control of the metric globally. The problem is that if $g_1$ and $g_2$ are metrics, then so is $g_1 + g_2$, but the curvature of $g_1+g_2$ can be wildly different from $g_1$ and $g_2$.

Said another way, typically one uses a partition of unity argument to show that metrics exist on all smooth (paracompact) manifolds, but curvature is not well behaved under partitions of unity.

Another very relevant point is the Hopf conjecture: that $S^2\times S^2$ does not carry a metric of positive sectional curvature. If we were to adapt your method to this setting, we're just trying to prove positive curvature, which is a far cry from constant positive sectional curvature, but the efforts here have been fairly fruitless.

Also, Burkhard Wilking has shown $\mathbb{R}P^2\times\mathbb{R}P^3$ has a metric of "almost positive curvature" - it has an open dense set of points $U$ such that for any point in $U$, all sectional curvatures are positive. It follows from the classical Synge Theorem, that this example cannot be deformed to positive curvature everywhere. So, if you were to get your hands on a "sort of" nice metric on your hypothetical $S^4$, then there's still no reason to assume you can deform it to one even of positive curvature. (Though, to be clear, if $M$ is simply connected with a metric of almost positive curvature, it is still completely open whether or not this can be deformed to positive curvature.)

On the positive side, Jeff Cheeger showed the connect sum of any pair of $\mathbb{R}P^n$, $\mathbb{C}P^n$, $\mathbb{H}P^n$, or $\mathbb{O}P^2$ (of appropriate dimensions) carries a metric of nonnegative sectional curvature. To do this he looks at each of those manifolds minus a point, and shows that near the deleted point, one can deform the metric to be nonnegatively curved everywhere and locally isometric to the product $S^{k}\times \mathbb{R}$ near the removed point. This guarantees he can glue the metrics together on each piece and get a smooth nonnegatively curved metric.

share|improve this answer
    
Thanks a lot for this nice answer! You raised some very interesting points indeed. –  Sam Mar 3 '11 at 17:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.