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

Given two smooth manifolds $M$ and $N$, it is known that if $M$ is compact, then $C^\infty(M,N)$ is a Fréchet manifold whose tangent space at $f \in C^\infty(M,N)$ is the space $$T_f C^\infty(M,N) = \Gamma(M,f^*TN)$$ of all smooth vector fields $\xi(x) \in T_{f(x)}N$ along $f$.

Suppose we are given a foliation $\mathcal{F}$ of codimension $q$ on $N$ and let $C^\infty_{\mathcal{F}}(M,N)$ be the subset of somooth functions $f:M \to N$ tangent to the leaves. Is $C^\infty_{\mathcal{F}}(M,N)$ still a Fréchet manifold? If it is the case, what is its tangent space at a point?

If it is too general to answer, can we at least say something about the case where $\mathcal{F}$ is a foliation of codimension $1$?

share|improve this question
2  
What does it mean for a function to be tangent to the leaves? Is it that the image of the tangent map is always tangent to a leaf? Thanks - Michael –  Michael Murray Oct 6 '12 at 14:31
    
Yes. Thanks for asking. –  Hsueh-Yung Lin Oct 7 '12 at 10:23
1  
Isn't that equivalent to the image being contained in a single leaf? –  Igor Khavkine Oct 7 '12 at 15:25
    
Yes Igor, you're right. –  Hsueh-Yung Lin Oct 7 '12 at 20:16
add comment

1 Answer

I don't know about the Frechet structure, but can say a few words about the tangent space: if $f_t:M\to N$ is a time dependent family of maps contained in the leaves of $N$ for each $t$, then $\frac{d}{dt} f^*(\alpha)=0$ for any one-form $\alpha \in \Omega^1(N)$ vanishing on the leaves. So a necessary condition for a relative vector field $X$ along a map $f:M\to N$ to be a tangent vector in the space $C^\infty_{\mathcal{L}}(M,N)$ is $L_X(\alpha)=0$ for all one forms $\alpha$ vanishing on the foliation (here $L_X$ is the lie derivative along relative fields which may be defined by $d\circ i_x+i_x\circ d$).

I think (but am not sure) that this condition is also sufficient. But these tangent spaces might change dimension from one point to the other. Think of a moebius strip with the obvious foliations by circles. The middle circle cannot be deformed into any of the others without breaking it up, so in this case the tangent space are only deformations which stay in the same leave, while the other leaves allow small deformations.

share|improve this answer
    
Thanks for your answer. It seems that in your example of möbius strip, the moduli space has at least two connected component with different "dimension". Maybe it is more appropriate to ask whether each connected component of the moduli space is a Fréchet manifold and calculate their tangent space at each point. –  Hsueh-Yung Lin Oct 17 '12 at 9:11
    
You are right. I realize that I wasn't distinguishing clearly between the space of maps $S^1\to M$ tangent to leaves, and the moduli space of leaves of the foliation. Thinking it over I would guess that the latter is connected and one dimensional space, while the former has several connected components. Example: two component consisting of maps winding around the middle circle with winding number +1 or -1, which cannot be deformed outside of this leave. But maps winding around other leaves with winding number 1 can be deformed into a map winding twice around the circle. [cont.] –  Michael Bächtold Oct 17 '12 at 11:53
    
[cont.] But it still seems that each of these connected components might be a smoot infinite dimensional manifold. Originally I was expecting to get something singular, now I'm not sure and maybe indeed these spaces are always smooth frechet? –  Michael Bächtold Oct 17 '12 at 12:02
add comment

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.