So, am I understanding correctly that you claim that for the section $s\colon SU\rightarrow SO(TM)$, $s^{\ast}\tilde{\omega}_1^2,...,s^{\ast}\tilde{\omega}_1^n$ form a coframe on each fiber?
@Thor. Yes. Again, go back to just a plain manifold to convince yourself. I started the whole thing days ago saying that if $e_1,\dots,e_n$ is an orthonormal frame on $M$, with dual coframe $\omega_i$, then $\omega_1\wedge\dots\wedge\omega_n$ is the volume form.
@Lelouch: When you say $n$-manifold you mean with boundary, explicitly. If the boundary is empty, this is clearly nonsense.
Oh, I guess when you say "depends only on the boundary" and the boundary is empty, the integral must be $0$?
@TedShifrin I'm really sorry if I'm not able to express myself clearly: Let $M$ be a manifold, and suppose you have a $n$-form $\omega$ on it. Fix any $n-1$ dimensional manifold $N'$. For any $n$-dimensional submanifold $N$ with boundary $N'$, we have the integral of the pullback of $\omega$ on $N$ to be dependant on only $N'$, and not $N$.
The point is that the point $x$ on the manifold is being replaced by the vector $e_1$, and so you need the connection to define differentiation of the section $e_1$ of the bundle.
@Thor: If you read about affine connections, you find that you keep track of the point as well as the tangent vectors. The so-called solder form then is the part of the connection that gives the dual coframe.
OK, so you have to assume the form is closed in the first place, @Lelouch. Then you're asking: If I have a closed form $\omega$ and its integral over all chains with the same boundary is the same, must $\omega$ be exact? So I can glue two such chains together and get a cycle (manifold without boundary, perhaps some corners along the boundary) and the integral over that cycle is $0$. If the integral of a closed form over all cycles is $0$, yes, it is exact.
That's because cohomology is the dual of homology.
But you cannot fix your $N'$. You need to know this universallly.
@TedShifrin Yes, I meant $N'$ was arbitrary (but that the value of the integral don't depend on $N'$). Thanks, but why do we need to assume it's closed in the first place ?
At any rate, @Thor, I'm saying that the two equations $"dx" = \sum\omega_j\otimes e_j$ and $\nabla e_1 = \sum\omega_1^j\otimes e_j$ are totally parallel. Pun intended.
Can you give a counterexample when it's not closed ?
Because for $1$-form (I think) closedness is not needed to assume a priori
assume it's connected, fix a point $p$, and define $f(q) = \int_{\gamma} \omega$ where $\gamma$ is joning $p$ and $q$. The condition implies this is well defined, and it's not hard to see $df = \omega$
Of course a non-closed form cannot be exact, so it is a necessary condition. I agree that you can deduce from Stokes's Theorem that path-independence implies closedness.
The same argument works in general. If $d\omega\ne 0$ around some point, I can create a little sphere around that point over which the integral of $\omega$ will be nonzero, and then splitting the sphere in two pieces gives different results.
At any rate, it's a consequence of the deRham theorem that if the integral over all $n$-cycles is $0$, then the form is exact. That's not easy to show directly.
If the closed form is not exact, then it represents a nonzero cohomology class, and hence there must be some cycle over which it has nonzero integral.
I guess this is the claim that $H^\ast_{dR}(M) \to (H_\ast(M;\Bbb R))^\vee$ is injective, and you're saying "at that point you may as well show they're isomorphic".
I certainly agree the proof strategy isn't as obvious for $* > 1$.
Deleted message: Yeah I'd like to do this all at the level of forms, no MV. By the time you've introduced Mayer-Vietoris you may as well prove the de Rham theorem.
Interesting how much of the argument is sorta buried in things like the five lemma.
What I always liked about sheaf arguments for the Dolbeault isomorphism, etc., is that you could chase things through the serpent and construct actual representatives.
@TedShifrin that makes sense. Maybe such a notion isn't used, but I'm curious about diffeomorphisms that permute angles. So the sets of angles are the same (membership)
His multivariable chapters are so sloppy. He teaches them to maximize/minimize by substituting (eg if you are on a surface $x^2 - y^2 - z^2 = 1$, then minimized distance-squared to 0 by minimizing $1+2y^2+2z^2$ over the plane instead). This fails in even the simplest examples, eg finding the extrema of $y^2$ over the unit circle. You miss two if you substitute!
Of course there's actual math hidden in why this substitution trick doesn't work, but since it doesn't in the simplest cases, why is it in the damn book?
I would encourage people to use Lagrange in general, but yes, parameterizing also works fine. I agree the issue is hidden boundary points.
There's theoretical errors too but it's irritating when even his computational examples lead people astray and I have to take time to explain why X bit of the book is misleading and/or wrong.
@TedShifrin It is nt given any information for $\alpha$. I thought that we could do: $$\left ( (n+1)^{\alpha}-n^{\alpha}\right )=n^{\alpha}\left ( \left (\frac{n+1}{n}\right )^{\alpha}-1\right )=n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )$$ But I realized that this doesn't help us, right?
@MaryStar: start with $\alpha$ a positive integer. Then what can you do?
@FakeMod: What do you know about two analytic functions that agree on a set of points with a limit point?
@MikeM: No, there are only the two points corresponding to the origin. The situation I was thinking about was with boundary issues.
@MaryStar: I think they're interested in the case $0<\alpha<1$. Try $\alpha=1/2$. This should be familiar.
user434058
7:50 PM
@TedShifrin That was just a curious question from a naïve beginner. I don't exactly understand the things you're referring to, and thus I think I am getting ahead of myself. I shall return again once I know the necessary prerequisites. :-)
do we use in that case the expansion of $ (n+1)^{\alpha}$ ?
In the case of $\alpha=1/2$ we get $\sqrt{n+1}-\sqrt{n}=\frac{\left (\sqrt{n+1}-\sqrt{n}\right )\left (\sqrt{n+1}+\sqrt{n}\right )}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$ which goes to 0
I still don't see why the pullbacks of $\tilde{\omega}_1^2,...,\tilde{\omega}_1^n$ form a coframe for the vertical subbundle. It would suffice to show that all the other connection forms pull back to $0$, which sounds plausible, but I'm not seeing it anyhow.
@MaryStar: Not quite right. There is a similar conjugate trick. But you shouldn't need to do that explicitly. I'm suggesting that you think of this as a function of $\alpha$ (and $n$).
@Thor: No, that's false (again, think about the base case), but you already have a spanning linearly independent set.
@TedShifrin I was listing off an example that's OK, as something he would use it for. He's careful to not give examples where things fail. But not to indicate why they don't fail.
The circle example is one where it fails. By projecting to the y-axis, he adds new boundary points that weren't present on the circle; but by not discussing this nobody thinks to check for them. :)
@TedShifrin Oh, in that case substitute nothing, as it's already a function of 1 variable. You could substitute it to being $1-x^2$ instead, same issue arises.
I mainly find it very irritating to have a book attached to the class that I occasionally have to tell students not to read. I suspect some portion of them are only reading it and not listening to me.
how is it analogous if the analogous statement in the base case is false? when all the connection forms pull back to $0$ (which can happen on the base), they won't form a basis of anything except the trivial space
@MikeM: I understand completely. It was humorous when one of our algebraists taught our algebra course out of my book and hated it. He bashed it all the time, and the students said so repeatedly in evaluations. Of course, most other instructors loved it. :P
@Thor: They do NOT in general pull back to $0$. Read my little section on moving frames on surfaces in my notes. All the geometry is in those forms on the base.
@Thor: For example, if you have a generic surface in $\Bbb R^3$ and choose $e_1,e_2$ as the principal directions, then $\omega_1^3 = k_1\omega_1$ and $\omega_2^3 = k_2\omega_2$, where $k_i$ are the principal curvatures.
I'm saying that all you care about is that the $\omega_i$ give a coframing, i.e., a basis for the cotangent space. In the sphere bundle case, you have the $\omega_i$ (in the base) and the $\omega_1^j$ (for the fiber) and that's a complete coframing.
@MaryStar: All I care about is that it's bounded, say by $2^\alpha$. Now what happens when you look at $n^\alpha(\dots)$?
Oh, you're right, @MaryStar. I'm too busy doing too many things here. You do need to see that $$n\left((1+1/n)^\alpha-1\right)$$ goes to $0$ as $n\to\infty$. Is that right?
And so are the $\omega_1^j$ for $S^{n-1}$ when $e_1$ is the point on the sphere, @Thor. It's identical. $x=e_1$.
Instead of $dx = \sum\omega_je_j$ we have $de_1=\sum\omega_1^je_j$.
This question is an old question from mathstackexchange.
Let $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $
And let
$ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $
It appears that
$$\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $$
Why is that so ?
Notice
$$\int_0^{2 \pi} \ln(\sin(x) + \f...
Ok, then we have that $n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\rightarrow n^{\alpha}\left ( 1-1\right )=n^{\alpha}\cdot 0$. If $\alpha>0$ then we have an undefinite form and if $\alpha<0$ then we have $0$. Is that correct? @TedShifrin
@MikeMiller $M$ oriented, Riemannian manifold, $SO(TM)$ oriented, orthonormal frame bundle, $SM$ sphere bundle, $\tilde{\pi}\colon SO(TM)\rightarrow SM$ projection on first coordinate. Let $\sigma$ be the fiber volume form on $SM$. I'm trying to understand why $\tilde{\pi}^{\ast}\sigma=\tilde{\omega}_1^2\wedge...\wedge\tilde{\omega}_1^n$ (the $\tilde{\omega}$ are the connection forms).
ls it possible to prove there is no bijection from the power set of the reals to the reals without the axiom of power set , the diagonal argument or induction from the fact there are more reals than integers ??
That's never an axiom or definition. It's proved by what you call the diagonal argument. You're going to have a lot of trouble proving it without ultimately just disguising the diagonal argument
@Ted so you're saying that the, when we identify the fiber with the regular sphere, the restriction of the pullbacks from the connection forms on the frame bundle agrees with the connection forms on the sphere coming from the frame that we get from our section?
Given any sequence of real numbers x1, x2, x3, ... and any interval [a, b], there is a number in [a, b] that is not contained in the given sequence.[B]
To find a number in [a, b] that is not contained in the given sequence, construct two sequences of real numbers as follows: Find the first two numbers of the given sequence that are in the open interval (a, b).
Denote the smaller of these two numbers by a1 and the larger by b1. Similarly, find the first two numbers of the given sequence that are in (a1, b1). Denote the smaller by a2 and the larger by b2. Continuing this procedure generates a sequence of intervals (a1, b1), (a2, b2), (a3, b3), ... such that each interval in the sequence contains all succeeding intervals — that is, it generates a sequence of nested intervals. This implies that the sequence a1, a2, a3, ... is increasing and the sequence b1, b2, b3, ... is decreasing.[10]
@TedShifrin So I think I figure it out! From Bernoulli's inequality we get $\left (1+\frac{1}{n}\right )^{\alpha}\geq 1+\frac{a}{n}$. Let the original sequence be $a_n$. Then we have that $$a_n\geq n^{\alpha}\left ( 1+\frac{a}{n}-1\right ) \Rightarrow a_n\geq \frac{a}{n^{1-\alpha}}$$
If $1-\alpha>0 \Rightarrow \alpha<1$ then $a_n$ converges to $0$. If $1-\alpha<0 \Rightarrow \alpha>1$ then $a_n$ goes to $+\infty$ so it diverges.
@robjohn Ah do you mean that when $a\leq 0$ or $a\geq 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\geq 1+\frac{a}{n}$ and when $0\leq a\leq 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\leq 1+\frac{a}{n}$ ? Or do you mean an other inequality?
Wolfram states: https://www.wolframalpha.com/input/?i=integrate+1%2F%28e%5E%282it%29%2B2e%5E%28it%29%29+from+t+%3D+0+to+2pi But if I take the change of variable u = e^(it), then we get integral limits 1 and 1, since e^(0) = e^(i 2pi) = 1 Is this change of variable illegal or am I wrong to assume that the integral is 0 if the limits are the same?
I'm saying that you should not be parametrizing in your problem. You should do algebra first. Don't ever try to do integrals explicitly when you don't have to.
Simplify the algebra and then factor the denominator and what does that suggest?
When $a\leq 0$ or $a\geq 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\geq 1+\frac{a}{n}$ and then $$ n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \frac{a}{n^{1-\alpha}} $$
When $0\leq a\leq 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\leq 1+\frac{a}{n}$ and then $$ n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\leq \frac{a}{n^{1-\alpha}} $$ @robjohn
You do not have a well-defined branch of log if you go all the way around the origin. Similarly for $\log(z-1)$ if you go all the way around the point $z=1$. So pay attention to what I said two hours ago and look at the explicit integration.
@robjohn no problem! So at each inequality we take the limit when n goea to infinity. At the case when $a\leq 0$ or $a\geq 1$ we get that the limit of the sequence is greater than 0 if $1-\alpha>0$ otherwise greater than infinity, or not? But we cannot say anything about the convergence, can we? At the other case we get than the limit of the sequence is less than infinity, which means that it converges. Or not?
@Ted I'm trying to figure out what's going on with plain $S^{n-1}$, but I don't really understand what the actual claim even is. Which forms precisely are supposed to be linearly independent?
The deluxe version, as I've said several times, is that we have $e_1$ as the position vector with $e_2,\dots,e_n$ spanning the tangent space to the sphere at $e_1$, and then it becomes $de_1\cdot e_j = \omega_1^j$. But identical except for notation.
This question is an old question from mathstackexchange.
Let $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $
And let
$ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $
It appears that
$$\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $$
Why is that so ?
Notice
$$\int_0^{2 \pi} \ln(\sin(x) + \f...
