Further, compact Lie groups are completely classified, so there's kind of a lot of conditions you can put on your underlying space... "It's on the list of compact Lie groups"
We are assuming the manifold to be smooth to begin with, right? So the proof would go by adapting the given smooth structure to the continuous group operation?
@AndrewThompson No, it means any topological manifold with a top. group structure admits a smooth structure under which that top. group structure becomes a Lie group structure.
@MikeMiller I think I read something like that when I skimmed one of the thousand Springer books I recently downloaded, but I don't recall whether $M$ was assumed to be orientable
The standard proof uses Euler classes. Those only make sense on oriented manifolds. I'm trying to extend the result now; it's probably true for everything.
@MikeMiller I don't think I am going to state the preimage theorem for Banach spaces. I rarely know anything about them (i.e., I don't want to embarrass myself!), and it's taking me far off from calculus. I think I'd just go back to bore myself with calculations :(
@BalarkaSen: Fine. Here is the statement of IFT I care about.
Given a map $U \to V$, open subsets of Euclidean spaces, such that $f(0)=0$, $df(0)$ is surjective, there is a chart around $0 \in U$ such that in this chart, $f$ is a projection.
That is to say, one of the canonical projections of Euclidean spaces $\Bbb R^{n+k} \to \Bbb R^n$, that kills off the last $k$ coordinates, say.
(All I'm really saying here is: you can find a chart in which it's linear.)
Similarly if $df(0)$ is injective you can find a chart in $V$ such that $f$ is locally the canonical inclusion of Euclidean spaces.
For Banach spaces the assumption in the first part is instead that the kernel of $df$ has a complement. For Hilbert spaces this is automatically true; for Banach spaces it needn't be.
eg $c_0 \subset \ell^\infty$
the first is the space of sequences that converge to 0, the latter bounded sequences
In the second part you want the image to have a complement.
(A complement to, uh, $X$ means a subspace $Y$ such that $X \cap Y = 0$, $X+Y = U$; that is, $X$ is a part of a direct sum decomposition)
Let $E$ be a (smooth) vector bundle over $M$, and $\sigma$ a section thereof. Hirsch proves that you can mildly perturb $\sigma$ to be transverse to the zero section. In particular, this is true of $TM$. (Counting the intersection number of this section with the zero section gives you the Euler characteristic; in general this idea gives you the Euler class. You need orientability here to be able to take a signed count. Otherwise you just get a Stiefel-Whitney class.)
Now suppose $M$ has a $G$-action, $G$ a finite group, and $E$ a vector bundle over $M$ (that this $G$-action lifts to). In particular, $E=TM$. (You may need $G$ to act freely; I don't remember. In any case, the following is true if it does.) Then given a $G$-equivariant section $\sigma$ you can perturb it (through $G$-equivariant sections) to one transverse to the 0 section.
In particular $G = \Bbb Z/2$, $M$ is the oriented double cover of a non-orientable manifold, $E=TM$.
Now what you get is a $\Bbb Z/2$-equivariant section of $TM$ on the oriented double cover that has a discrete set of zeros. ($\Bbb Z/2$-equivariant sections are the same thing as sections that descend to the non-orientable manifold.) Because $\Bbb Z/2$ acts freely, the zeroes pair off by the group action.
Pick half of the zeroes (so that together with their 'pairs' they form all the zeroes). Pick a chart in which all these zeroes are contained. You can do this. Now you have a ball $B$ in the interior of which all the zeroes are contained and, picking $B$ small enough, a ball $i(B)$ containing the other zeroes. Outside these balls you have a nonvanishing section of $TM$.
The point now is that the assumption that $TM$ has a nonvanishing section implies that you can modify the section over $B$ to make it nonvanishing. (It's something like Poincare-Hopf.) Then do the same modification to the other side. you now have a $\Bbb Z/2$-equivariant nonvanishing section, so it descends to a nonvanishing section to the manifold below, as desired.
If the non-orientable guy is $N$ I used here that $0=2\chi(N) = \chi(M)$.
@iwriteonbananas: I wouldn't bother digesting it. It can all be phrased better. Ask me again in a week and a half and I'll do my best to have a correct and well-phrased proof; you might post it on main so there's a canonical reference. There is little need to talk about $G$-equivariant transversality.
@iwriteonbananas: When I modify it over the ball, I need that the signed sum of zeroes is zero. I know that $\chi(M) = 0$, so when I take the signed sum over the whole thing, I get zero. But I have two balls worth of zeroes. Call the signed sum of the zeroes in the first ball $z(B)$. Then $z(i(B)) = (-1)^n z(B)$; if $n$ is even, that they sum to zero tells me $z(B) = 0$.
We have $2k$ integers greater than or equal to $j\geq0$
$a_1+a_2+\dots + a_k=n$ and $b_1+b_2+\dots + b_k=n$.
If for all $1\leq i\leq k$ we have $|n/k-a_i|\leq|n/k-b_i|$. Then $\sum\limits_{i=1}^n\binom{a_i}{j}\leq \sum\limits_{i=1}^n\binom{b_i}{j}$.
The same can be said if we replace binomial...
@MikeMiller I'm not sure if it's appropriate for me to ask on main. I was just skimming through various books, wildly jumping to results that sound cool and go far beyond my scope. If you feel like writing up an answer, I'll post a question though.
@iwriteonbananas: I am now no longer sure it applies for every non-orientable manifold. My money is that if it holds for 3-manifolds, it always holds, but I also put 50% odds that it fails there.
Anything else about topology you want to know before I go back to my hole? :P
So if you have a path $\gamma: x \to y$, it gives you an isomorphism $T_x \to T_y$ of inner product spaces. This isomorphism depends on the choice of path
I think it might be good to push forward and see how it's used and then think about the why.
@Stan: I have two finite dimensional inner product spaces. There is an isomorphism $U \to V$. But I would prefer to be given an isomorphism than to come up with one.
I kinda gave up on Spivak fairly quickly. I went at the pace of the university and I'm pretty sure I've now covered it all, @Hakim. I recall you sending an e-mail a while back and I'm sorry I haven't replied yet! I've become a master of procrastination in your absence :-b
@Kasper Out of curiosity, would you happen to know how to sum $$\sum_{k\text{ odd}}kq_1^{(k-1)/2} q_2^{(k-1)/2}p_1 + \sum_{k\text{ even}}kq_1^{k/2} q_2^{k/2-1}p_2$$ where $p_1, p_2 \in [0, 1]$ and $q_i = 1 - p_i$, $k \geq 1$ an integer?
We have $2k$ integers greater than or equal to $j\geq0$
$a_1+a_2+\dots + a_k=n$ and $b_1+b_2+\dots + b_k=n$.
If for all $1\leq i\leq k$ we have $|n/k-a_i|\leq|n/k-b_i|$. Then $\sum\limits_{i=1}^n\binom{a_i}{j}\leq \sum\limits_{i=1}^n\binom{b_i}{j}$.
The same can be said if we replace binomial...
