Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I've just finished my first course in differential geometry, so forgive me if this is maybe a silly or well-known question, but given any, say, diffeomorphism of $n$-manifolds $\phi:M\rightarrow N$, I was wondering whether the map $f:M\rightarrow GL_n(\mathbb{R})$ defined by $f(p)=d\phi_p$ has any important or nice properties? One guess I had was that further sending $p$ to det$(d\phi_p)$ could say something about orientation? I had been thinking about the Gauss map and how it's a map from any surface $S$ to the sphere $\mathbb{S}^2$, and I was trying to come up with other canonical ways of creating maps of surfaces and/or manifolds to "special" ones.

share|improve this question
4  
You cannot define such a map unless you are able to pick fields on $M$ and on $N$ such that they are a basis at each point of $M$ and $N$ (because the matrix of a linear map depends on a choice of bases). You'd need both $M$ and $N$ to be [parallelizable]{mathworld.wolfram.com/Parallelizable.html). What you can do is construct a bundle $\hom(TM,TN)$ and define your $f$ to be a section of this bundle: this is a standard trick to avoid to pick bases. –  Mariano Suárez-Alvarez Dec 15 '09 at 3:54
5  
Comment to the last part of your question. An n-manifold M embedded into R^N defines canonically a map from M to Grassmannian Gr(n,N): the image of a point p in M is a tangent space at that point as a subspace of R^N. It is an analog of Gauss map for higher dimensional manifolds without orientation and metric. –  Evgeny Shinder Dec 15 '09 at 4:30
    
Pff, real men work with frechet manifolds. –  Harry Gindi Dec 15 '09 at 5:00
    
Thanks for your explanation, Mariano - makes sense! And thanks for another canonical map Evgeny :) Is there a standard name for a section of hom($TM$,$TN$)? –  Zev Chonoles Dec 15 '09 at 6:42
    
If you take M = N, so have a diffeomorphism f : M -> M, then you get a map from M to the "conjugation space" of GL_n(R): mod out by the relation of similarity. This isn't a nice manifold, though. –  Loop Space Dec 15 '09 at 8:56

1 Answer 1

up vote 9 down vote accepted

The question can be interpreted to ask, given a diffeomorphism $\phi:M \to N$, what topological information in its derivative, interpreted as a bundle map? If the bundle map is used carefully, the answer is that there is a lot of information. There is a principle that, in dimension $n \ne 4$, there is enough information to completely tell when a homeomorphism can be improved to a diffeomorphism. That is a hard result if you are comparing PL homeomorphisms to diffeomorphisms, and a very hard result if you are comparing topological homeomorphisms to diffeomorphisms. It is easier to sketch that in principle you get important obstructions to smoothing a homeomorphism. The obstructions exist in all dimensions, the only difference being that in dimension 4 they aren't everything.

Suppose that $M$ and $N$ are smooth $n$-manifolds, or topological $n$-manifolds, or PL $n$-manifolds, and $\phi$ is an equivalence. Then in all of these cases, $\phi$ has a formal "derivative" that is locally a germ of a (continuous, PL, or smooth) homeomorphism. The "derivative" is a map between "tangent" bundles (which are actually most natural as microbundles) to match the germ picture, rather than ordinary fiber bundles. If you're afraid of germs (who isn't), a germ is no more than an equivalence class of local maps: Two local maps are the same if they agree on a neighborhood. The germ derivative of a map is basically a fake, tautological derivative.

On the other hand, the group $\text{Diff}(n)$ of germs of diffeomorphisms that fix the origin is homotopy equivalent to the true derivative group $GL(n,\mathbb{R})$, which is homotopy equivalent to the orthogonal group $O(n)$. The other two groups of germs are called $PL(n)$ and $TOP(n)$.

Now suppose, to take the best-behaved case, that $\phi$ is a PL homeomorphism between two smooth manifolds $M$ and $N$ with smooth triangulations. It has a formal derivative $D\phi$ that is a map between tangent $PL(n)$ microbundles. You could ask, roughly speaking, whether you can homotop $D\phi$ to make it a true smooth derivative in the available $\text{GL}(n,\mathbb{R})$ bundle, or equivalently an $O(n)$ bundle. This is a necessary condition to be able to improve $\phi$ itself to a diffeomorphism. The more subtle result is that it is also a sufficient condition. In this sense, the relative homotopy theory $PL(n)/O(n)$ exactly measures the non-uniqueness of smooth structures on manifolds. (I learned about this stuff from Rob Kirby, but it looks like it is covered in some form by Jacob Lurie.)

There is an older situation in which the derivative of a map gives you an obstruction to something, and it turns out to be the only obstruction. Namely, if $M$ is a smooth $n$-manifold, you can ask whether it immerses in $\mathbb{R}^k$; or you could ask whether two immersions are equivalent through an isotopy of immersions. Clearly if $\phi:M \to \mathbb{R}^k$ is an immersion, the homotopy type of the derivative $d\phi$ (among maximum rank choices) is an invariant of $\phi$, and clearly there is no immersion if there is no available homotopy class. The theorem of Smale and Hirsch is that these are the only obstructions to existence and equivalence of immersions, unless $M$ is closed and the dimensions $n$ and $k$ are equal. This theorem is illustrated very nicely in the math movie Outside In.

Another way to say it is that you can formaly weaken both questions by decoupling maps from their derivatives, but still get the correct answer.

share|improve this answer
    
+1 for the germ-pun :-) –  Johannes Hahn Sep 2 '11 at 12:26

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.