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.

For two manifolds $M^m$ and $N^n$ with $m<n$ the Smale-Hirsch theorem says that the differential map $d:\operatorname{Imm}(M,N)\to\operatorname{Mon}(TM,TN)$ is a weak homotopy equivalence, where $\operatorname{Imm}(M,N)$ is the space of immersion from $M$ to $N$ with the compact-open $C^{\infty}$ topology and $\operatorname{Mon}(TM,TN)$ is the space of vector bundle maps which are monomorphisms on their fibers from the tangent bundle $TM$ of $M$ to $TN$. How exactly does this now imply that there is a regular homotopy from $\operatorname{id}:S^2\to\mathbb{R}^3$ to $-\operatorname{id}:S^2\to\mathbb{R}^3$? The original paper by Smale mentions the use of the Stiefel manifold $V_{3,2}$ through $\pi_2(V_{3,2})$ but I can't figure out how to get this into play with the modern formulation of the Smale-Hirsch theorem. Any help is very much appreciated.

Anyone mildly curious can also check out this movie of such an explicit eversion: http://www.youtube.com/watch?v=BVVfs4zKrgk

EDIT: Or could someone simply explain to me why $\operatorname{Mon}(TS^2,T\mathbb{R}^3)$ is path-connected? This must hold true since Smale claims that all immersions $S^2\to\mathbb{R}^3$ are regularly homotopic.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The space of maps $\operatorname{Mon}(TS^2,T\mathbb{R}^3)$ projects to the space of maps $\operatorname{Map}(S^2,\mathbb{R}^3)$ by forgetting the fiber maps (i.e. restricting to the zero sections). The fiber over a map $u$ is then the monomorphisms from $TS^2$ to $u^* T \mathbb{R}^3$. This is a locally trivial fibration with contractible base (by contracting to a constant map at $0$), hence a product. So it is homotopy equivalent to the fiber. To compute the fiber, take $u={\rm const}$. Get maps from $T S^2$ to trivial $\mathbb{R}^3$ bundle over $S^2$, which is the same as maps $T S^2$ to $\mathbb{R}^3$ monomorphic on the fibers.

This is the space of sections of a locally trivial fibration over $S^2$ with fiber the non-compact Stiefel manifold of all 2-frames in $\mathbb{R}^3$ (pick a local trivialization on $T S^2$ and see that on that trivializing chart you just need a 2-frame at over every point). This is associated fibration to $T S^2$, so is given by clutching function and since the clutching is twice the generator of $\pi_1 (V_{3,2})$, and so is trivial, the fibration is trivial. This means the space of sections is just maps $S^2$ to the fiber, which by Gram-Schmidt retracts to $V_{2,3}$. Hence $\pi_0$ is $\pi_2(V_{2,3})=0$, as wanted.

Maybe another way to replace/rephrase that last paragraph, is to note that $T S^2$ sits in the trival $\mathbb{R}^3$ bundle over $S^2$, as in $T S^2 \oplus \nu = {\rm trivial}$, and given orientations of everything (which we are implicitly assuming, actually), a monomorphism extends uniquely to a map of the $\mathbb{R}^3$-fiber to $\mathbb{R}^3$, where the vector in $\nu$ is the unit normal to the image of the original monomorphism. The space of such maps is a subset of 3 by 3 matrices, and a partial Gram-Schmidt says that the space of all invertible matrices retracts to it. The same argument in the $S^2$ family says the fiber is homotopy equivalent to the space of maps from $S^2$ to $GL(3)$, which is again by Gram-Schmidt homotopy equivalent to maps from $S^2$ to $SO(3)$. This gives same conclusion...

Also, the eversion is done in Eliashberg-Mashchev book on h-principle, in section 4.2.

share|improve this answer
    
What a fantastic answer, thank you very much! –  Martin Worsek Jul 5 '11 at 13:47

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.