Find all possible functions $f : \mathbb{Q} \rightarrow \mathbb{R}$ such that f is continous and has compact support.
Claim f := 0.
Suppose $f \neq 0$ then $f^{-1}(\{0\})$ is closed in $\mathbb{Q}$ thus we have that $f^{-1}(\{0\})$ is closed in Q. That is $\mathbb{Q} - f^{-1}(\{0\})$ is compact neighborhood of Q.
we know that all cpct subsets of Q has non-empty interior contradiction.
If $M$ is a manifold with universal covering $\widetilde{M} \to M$, then $\pi_1(M)$ acts on $\widetilde{M}$ and I can identify $M$ with $\Gamma \backslash \widetilde{M}$ where $\Gamma := \text{im}(\pi_1(M) \to \text{Homeo}(\widetilde{M}))$. Now if $U \subset \text{Homeo}(\widetilde{M})$ is a finite index subgroup, then $M$ is finitely covered by $(U \cap \Gamma) \backslash \widetilde{M}$, right?
@Silent I don't really see where $r$ is entering in the first place. You've got a unit circle, not a circle of radius $r$. (You could consider a circle of radius $r$ but that's not part of the theorem statement.)
If you did, though, then the substitution rule would give $z=re^{it}\implies dz=i re^{it}\,dt$
so you'd have an extra factor of $r$, canceling the one you're worried about
@TedShifrin kind of late to reply to what you said, but to show that $T_{(p,0)}TM\cong T_pM\oplus T_pM$ for the zero $v(p)=0$ of the vector field $v$, is the point to use the first isomorphism theorem on the pushforward of $\pi_{TM}\colon TM\to M$ at $p$?
The only canonical section a vector bundle has is the zero section.
There is no way to get a complement to the tangent space of the fiber at a general point $(p,v)$ without more structure (a connection on the vector bundle).
@TedShifrin isn't doing that like a brutal work? Can that really be done by pen and paper with ease? Or is it better to be done by some computational software? I consider there is some intuitive way to understand this. I think of two ways to understanding this: 1 every Lie group can be expressed in terms of an exponential function; 2. all group manifolds are parallelizable.
But I think the two points may be the result of that the Taylor series of any exponential function f(x) at every point $x_0$ converges to f(x) for all values of x, rather the reason of it. @TedShifrin
@CaptainBohemian: Software will never make a proof or understanding of why this is true.
The other things you're saying are off the deep end.
If $f(x) = P_n(x) + R_n(x)$, where $P_n$ is the $n$th Taylor polynomial, then the Taylor series is the limit of $P_n$ as $n\to\infty$. Thus, you must show that $R_n(x)\to 0$ as $n\to\infty$. No other way to make a rigorous argument.
If your approach to understanding the behavior of a single-variable function is to dive into Lie group theory and manifolds, something has probably gone wrong.
@anakhro: If you want to convince yourself of the complementation, it's fine in local coordinates on the vector bundle. One is the first $\Bbb R^n$ factor and the other is the other $\Bbb R^n$ factor. (By the way, there's nothing special here about the tangent bundle. It works for any vector bundle. Perhaps the general case would even be clearer.)
That's clean. I will try to do it generally like you said because maybe broadening the scope will make me less distracted by nuances of tangent bundles.
I once wrote a paper with one of my co-workers where we took pages to prove in total rigor (among other things) stuff that Arnold just asserted with a wave of his hand.
It was the beginning of a series of harder and harder papers, but it's still one of my favorites.
"Mathematics is a part of physics. Physics is an experimental science, a part of natural science. Mathematics is the part of physics where experiments are cheap."
I don't think Arnold is actually asserting mathematics is deeply experimental in the same sense as a scientist, as I think he of anyone would know that the experimental nature of physics is so that one can make an empirical claim, but in math there is no empirical claims.
"The Jacobi identity (which forces the heights of a triangle to cross at one point) is an experimental fact in the same way as that the Earth is round (that is, homeomorphic to a ball). But it can be discovered with less expense."
I mean, the question becomes what reason we have to believe that he intended things one way or another. He might've intended it as a metaphor, my guess is that he feels physics and geometry are extremely close and was gunning for something along those lines, he might actually be really bad at philosophy of math
If you read his mathematical methods for physicists, he conveys this.
Akin to Lockhart's "math is a beautiful simplification of our ugly world" kind of thing, Arnold is saying that math began as us modeling physical phenomenon (and shows no sign of stopping).
I dunno, when a hugely successful pure mathematician (and not hugely successful experimental physicist) starts making bold claims about the role of physics in math, you probably should feel that he's being blunt for a reason.
@MatheinBoulomenos do you think seeing application of mathematics outside of math is useful for all math students?
I disagree with you, @Mathein, because so much of mathematics comes from and leads to physics. I have a broader view of education than you do, I think.
And if you're going to teach students mathematics other than abstract algebra, the likelihood is that physics will be a motivation for some of what you teach.
That has more to do with employment and the fact that research even in pure math has become more computer-oriented. So, sure. But this is a totally different discussion.
@Daminark I mean that, [obviously] some people take offence to what Arnold said about the relation of math and physics, and definitely some would say it is deprecative of mathematics, the thing that Arnold is respected and known for. Why someone of Arnold's position would make strong statements like these when they meant to be taken in the deprecative way (to the detriment of their fame)---there would have to be some outlandish reason beyond being poor at epistemology.
That is to say, the bluntness Arnold displays is not that of someone who sucks at epistemology, but that of either someone who has some substantial life-altering reason for saying it (in the deprecative way), or someone who is simply not meaning it in that sense to begin with.
It definitely seems that Arnold is deprecating certain kinds of mathematics (axiomatic, algebraic...) because he prefers other kinds (physics-based, geometric)
I mean the question of whether Arnold believes verbatim the words he says is different from the question of the value judgment. In principle you could believe that math is a subset of physics and think that this doesn't say anything bad about math
@MatheinBoulomenos He's suggesting that only relying on axiomatic definitions for things while not appealing to the reason and intuition behind the axiomatic definition will be a detriment to the student.
This is true, though I imagine people still think the latter is still quite relevant since no reasonable definition of physics is likely to include, for instance, mathematical logic.
how this works vis. a vis quantum mechanics is probably confusing in its own right
e.g. quantum mechanics doesn't involve momentum and position in the same sense as in classical mechanics, but you still need to be able to talk about the results of experiments in those terms
