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

Let $M$ be a manifold equipped with a pair of surjective submersions $N_1 \stackrel{p_1}{\leftarrow} M \stackrel{p_2}{\rightarrow} N_2$ where $dim N_1 = dim N_2 = n$. Then we can find, for any point $m\in M$, a chart $U_1$ around $p_1(m)$ and a local section $s\colon U_1 \to M$ of $p_1$ such that $s(p_1(m)) = m$ and $s(U_1)$ is transverse to the fibres of $p_2$. In thinking about this we can clearly reduce to the case $N_1 = N_2 = \mathbb{R}^n$. We can even restrict attention to $M = \mathbb{R}^m$, and then it becomes a problem of linear algebra, namely finding a basis on $\mathbb{R}^m$ for which the submersions are both projections onto $n$ coordinates (we already know this separately).

I suspect that for infinite-dimensional vector spaces, and Frechet spaces in particular (really anything above Hilbert spaces in the usual hierarchy) this sort of result will not hold, and so for Frechet manifolds one cannot construct the analogous local section.

In more detail, I'm fairly sure that given a diagram of Frechet spaces

$$V_1 \stackrel{pr_1}{\leftarrow} V_1 \times F_1 \simeq W_1 \simeq V_2 \times F_2 \stackrel{pr_1}{\to} V_2$$

where $V_1$ is known to be isomorphic to $V_2$, one cannot in general find a (nonlinear?) map $F\colon V_1 \to V_2\times F_2$ (which is a section, and passes through the origin) such that $pr_1\circ F\colon V_1 \to V_2$ is injective an isomorphism. However, I'd like to see a counterexample (or, if I'm wrong, a proof that we can do it!). (EDIT: the map might need be be non-linear in this case because the definition of submersions doesn't use tangent spaces like in the finite-dimensional case. But linear would be good.)

VERSION 2: well it turns out that I have, in addition to the $V_i$s being isomorphic, I have $F_1$ isomorphic to $F_2$. Kudos to Andrew for guessing this would be the case for the application I have in mind.

share|improve this question
    
I seem to remember the tongue-in-cheek use, perhaps only for a while, of an "ask-johnson" tag. Perhaps we need an "ask-stacey" tag? ;-) –  Yemon Choi Jun 26 '12 at 3:37
    
:D tried that by email, with no response yet, so I thought I'd let everyone else have a go... –  David Roberts Jun 26 '12 at 3:39
    
Apparently, I do not understand the question: Otherwise, just take the isomorphism $J:V_1 \to V_2$ and set $F(x)=(J(x),0)$ so that $pr_1 \circ F =J$. –  Jochen Wengenroth Jun 26 '12 at 8:09
    
Whoops! I meant to say F is a section. That's the point of the whole thing. –  David Roberts Jun 26 '12 at 9:00

1 Answer 1

(I'm strongly tempted to leave this as a comment, as it's a rather trivial counter-example and it may result in the question being changed to avoid it, but I want to take advantage of the better formatting of answers.)

Here's a trivial counter-example. Let $V$ be a space which is isomorphic to one of its hyperplanes, in that we have an isomorphism $V \cong V \oplus \mathbb{R}$ (see shift space on the nLab). There are lots of these around and most of the "usual suspects" for locally convex topological vector spaces have this property (especially including spaces like $C^\infty(\mathbb{R},\mathbb{R}^n)$). We take $V_1 = V_2 = W_1 = V$, $F_1 = \lbrace 0\rbrace$, and $F_2 = \mathbb{R}$. The isomorphisms as $W_1 = V_1$ on the left, and $W_1 = V \cong \mathbb{R} \oplus V = \mathbb{R} \oplus V_2$ on the right. Then $\operatorname{pr}_1 \colon W_1 \to V_1$ is the identity and $\operatorname{pr}_1 \colon W_1 \to V_2$ is the composition $W_1 = V \cong \mathbb{R} \oplus V \to V$.

The only option for the section is the identity, since $W_1 \to V_1$ is an isomorphism, whence the induced map $V_1 \to V_2$ is the map $V_1 = V \cong \mathbb{R} \oplus V \to V$ which is not injective.

share|improve this answer
    
Hmm, that \emph{is} an easy counterexample. The problem is, it doesn't feel like a global counterexample, only a local counterexample (to quote Lakatos). That is to say, what I thought was the essence of my problem, and which suffices in the finite-dimensional case, isn't all the information I have. But at least I know that I do have to use more than just the submersions as I mentioned here. –  David Roberts Jun 26 '12 at 12:12
    
Yes, that's why I put my first comment in - I suspected that there would be extra information that would exclude this situation, such as $F_1$ and $F_2$ being isomorphic. –  Loop Space Jun 26 '12 at 13:11
    
+1 for the "whence" in that final sentence... –  Vidit Nanda Jun 26 '12 at 14:57
    
Hmm, yes, F_1 and F_2 do seem to be isomorphic in my case. Well spotted! –  David Roberts Jun 26 '12 at 21:49
1  
You'd better send this counterexample back to the fiery chasm from whence it came... ;-) –  David Roberts Jun 27 '12 at 1:00

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.