So $e_1,...,e_{n-1}$ is an orthonormal frame of $S^{n-1}$ with coframe $\omega_1,...,\omega_{n-1}$. We identify $T_xS^{n-1}\cong\{x\}^{\perp}\subseteq\mathbb{R}^n$ and $dx$ is the vector-valued $1$-form with entries $dx^1,...,dx^n$. So the claim reads as $\sum_{i=1}^ndx^ie_j^i=\omega_j$ ($e_j^i$ being the $i$-th entry of $e_j$), which we can check on a basis: $(\sum_{i=1}^ndx^ixe_j^i)(e_k)=\sum_{i=1}^ndx^i(e_k)e_j^i=\sum_{i=1}^ne_k^ie_j^i=\langle e_k,e_j\rangle=\delta_{ij}=\omega_j(e_k)$.
I don't see where the connection forms come into play
I'm new to manifolds. I'm reading "An Intro to Manifolds" by Tu, when I can. A diffeomorphism is a conformal map if the pullback of the metric results in a conformally equivalent metric.
Are there any references about diffeomorphisms that permute angles (not conformal)? (If this is impossible, p...
Let $a_n=x^n+1/x^n$. Then $\sqrt3a_n=\left(x+1/x\right)\left(x^n+1/x^n\right)=x^{n+1}+1/x^{n+1}+x^{n-1}+1/x^{n-1}=a_{n+1}+a_{n-1}$ thus $a_{n+1}=\sqrt3a_n-a_{n-1}$
Or it might be better to let $b_n=x^{2n}+1/x^{2n}$ and derive a recursion for that
let me see....
Then $b_n=b_{n+1}+b_{n-1}$, that is $b_{n+1}=b_n-b_{n-1}$
How do you get the Poincare Dual of a closed/compact $n$-form, and does it always exist ?
By poincare dual of an $m$-form $\eta$ on a $n$-manifold $M$, I mean a $n-m$-dimensional submanifold $N$ (possibly with boundary) such that for any $n-m$-form $\omega$, we have $\int_{N} i^* \omega = \int_{M} \omega \wedge \eta$
Also, is there a nice way to explicitly write down Poincare duals for $1$ dimensional submanifolds of surfaces ? And possibly unrelated, but it seems if you take a=an embedded one dimensional submanifold in a surface, take the dual of it, and integrate that along the submanifold, the result will be an integer ! I only checked it for toy cases, how do show this (or is this false ?) in general ? Can this be used to give explicit duals ?
@Alessandro hahaha I thought so; I've also been looking at Münster for a PhD. There are two profs there who do things that I am potentially interested in
Oh nice :D It looks quite streamlined lol, when does one have to apply? Like a year in advance or ? (E.g., if I wanted to start in October 2022 or smth)
is there someone here who has failed midway in a PhD course and records of that failure have badly affected their chances of getting a job in academia or the industries?
I took a look at some sane dynamical systems stuff @Balarka, in particular I saw that there's a nifty dynamical proof of Van der Waerden's theorem. To get that you need Birkhoff's recurrence theorem, turns out that the topological dynamics book had a more abstract nonsense version of Birkhoff's theorem as an unnamed lemma...
Birkhoff says that if $X$ is compact and $f$ an automorphism then there is an $x\in X$ and a sequence $n_k\in\Bbb Z$ with $f^{n_k}(x)\to x$ (there is a recurrent point in dynamics terminology)
And it's actually true whenever a topological group $G$ acts on a compact space $X$, there is an almost periodic point $x$, meaning that for all nbhds $U$ of $x$ there is a syndetic $A\subset G$ with $xA\subseteq U$ (syndetic is a generalization of quasi-dense in discrete groups, it means that there is a compact $K\subseteq G$ with $G=AK$)
The idea is that compact spaces always have a minimal subspace (this is an easy Zorn's lemma argument) and every point in a minimal subspace is almost periodic (this part is more painful but not too bad either)
tbh it seems like just some olympiad combo problem, use the $\delta$-hyperbolicity condition to prove these theorems and you don't need to use anything except triangle inequality
Oh that was just saying the following: Take any two geodesics which are Hausdorff close, then they are close in the pointwise sense, under the uniform norm, as well
So since in lemma we're dealing with local geodesics, his kind of example where you go forward and come back a lot shouldnt pop up
I mean $c$ be the local geodesic, we know $c$ is some $3\delta$-Hausdorff close to $[c(a), c(b)]$ (I think part $1$ says $\text{im}(c)$ is in the $2\delta$-nbhd of $[c(a), c(b)]$ and part $2$ says $[c(a), c(b)]$ is in the $3\delta$-nbhd of $\text{im}(c)$, but whatever). Now pick a point on $c$, look at the arc of length $k$ around that point -- that is an honest geodesic segment which is $3\delta$-close to some segment of $[c(a), c(b)]$
and Hausdorff-close geodesics track each other in the parametrized sense as well aka they are pointwise close... so
Anyway I don't know seems like an annoying fact
@Lelouch I'm really intrigued by the 4-point condition and I'm disappointed he doesn't like my idea. Here's what I really want to note: Fix a basepoint $o \in X$ in your $\delta$-hyperbolic space, and say the distance between two points $x, y \in X$ - imagine them as being very very far from $o$ - is $e^{-(x, y)_o}$.
Why? because $(x, y)_o$ is measuring how long the geodesics $[o, x]$ and $[o, y]$ stay close to each other (think of the tripod picture), if you do $e^{-that}$ that tells you if they are nearby as points at infinity
He mentioned something like you get an inner product structure even when there's no tangent space, but I don't understand what he meant by you get an inner product structure
@BalarkaSen Yes. So the rough picture to keep in mind is that $(v,w)_x$ is large will imply that the geodesics $[x,v]$ and $[x,w]$ are very close to each other for most of the time (and vice versa), right ?
I mean I get what he's saying; (1) there's no preferred basepoint (2) by sending the points to infinity I am not really understanding the $\delta$-hyperbolicity of the space, because the condition is for all $o, x, y, z$
Doesn't it irritate you that the inequality has to be algebraically manipulated to make sense of it using the tetrahedron condition?
it's probably instructive (and maybe trivial) to find a space for which the inner product condition holds for one particular base point, but it's not $\delta$-hyperblic
Too hard, maybe you can just take like a tree, $o$ be a basepoint, and then very very very very far from $o$ append a massive complete graph of a bunch of vertices
then its mostly delta-hyperbolic but very very very very far away its not
so at that basepoint the inner product thing should work out
But maybe that doesn't work
You need some substantial set of vertices I mean, otherwise its just quasi isometric to the tree
@Threnody yes, and you soon find out that the arctangent that comes from integrating $\frac1{1+x^2}$ can be written as $\frac1{2i}\log\left(\frac{x-i}{x+i}\right)+\frac\pi2$
@Knight Think about the inequality you just proved.
@Lelouch I asked Mahan this point, and his response was "what you are saying is right. and is the right perspective at infinity. but for finite points, better think tetrahedra" -- so maybe not entirely useless.
So we can maybe spend some time thinking about this interpretation and won't be a waste of time
@Thorgott right, I wrote for any ring of dim=1 in my removed messaged, but then I realized there might be many chains with different smallwr element switnessing dim=1. Of course if (0) is prime that doesn't happen
@abhas_RewCie Yes I found this by doing FullSimplify in Mathematica on nested limits that give derivatives of the reciprocal of the Riemann zeta function.
@robjohn Sir, high school algebra is much about : adding and subtracting the same thing to get a good expression, dividing both num and deno with same number to get something familiar. In high school, all the indefinite integrals are more about algebra than the Calculus.
My question is: Is higher mathematics similar to this? I don’t think so, because whenever I see @Thorgott discussing something of his level, he almost never writes any expression, they all discuss it like philosophy