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19:00
'orning.
@DanielFischer Indeed. It does work though, which is sort of the point of the limit procedure :D
@DanielSank Yes, distributions tend to work as they should, they are nice in that way.
Morning @Mike.
Hey @Mike
Huy
Huy
@Mike: Do you know how to prove that a hyperbolic metric with totally geodesic boundary on $\Sigma_{0,3}$ is uniquely determined by the boundary components?
Don't think so, no.
Huy
Huy
19:04
my prof gave it as an "Exercise: (hard)"
=(
g'night @Mike @Huy
Maybe knowing that each of the boundary curves is a geodesic determines the exponential map in a small nbhd of each boundary point? And then you bootstrap from there?
Huy
Huy
morning @Ted
Evening @Ted.
evening, @DanielF
19:05
Pretty frustrated.
Yeah, I think it's something like that. Knowing that the boundary is geodesic plus the curvature should tell you a perpendicular geodesic, at least locally.
Evening? Morning? What is this?! I thought you guys were all US.
Huy
Huy
not me.
@Danu Scotland, presently.
Come up and have a beer with me.
@Frank's in France, @DanielF is in Germany, ... @Danu, you just want to be special? :P
Huy
Huy
19:06
I've been to Glasgow once. Not a very beautiful city, in my opinion.
@TedShifrin Ya.
@DanielSank Cool, that ain't so far.
How about you come over to Amsterdam?
I'll be back home in the afternoon :3
@Danu Can't. Busy.
Also no transportation.
I'll go to Germany for Christmas.
Usually mathematics is frustrating.
I agree with that
Especially when it's claimed easy on the book but I don't have any idea on that.
19:14
@Clarinetist ok so the tricky question seems to be, can we actually prove convergence to multivariate Gaussian? It seems to depend on the ratio of the number of columns to rows in a way I don't understand.
@dorothy I'm not sure how this ratio comes into play, unfortunately. Wish I knew this material better, ugh
using double angle formula show that sin^2(x) = 0.5 - 0.5 * cos2(x) are equal
help pls ^
What's the double angle formula?
cos2x = cos^2(x) - sin^2(x) , 1 - 2sin^2(x) , 2cos^2(x) - 1
sin2x = 2sinxcosx
Okay, so did you try to plug these into the right hand side?
19:22
yes
What did you get?
i just wont get seomthing like sin^2(x) = sin^2(x)
the things i get just goes back to the original equation
Plug $\cos2x=2\cos^2x-1$ into $(1-\cos2x)/2$, what did you get?
@KinsonChan Don't just come here to make people solve your homework questions.
@Danu im revising for a test and im only asking ones that im stuck on
19:26
The problem is that what we think easy may not be easy for those beginners.
So maybe he worked hard but still didn't get what's involved.
Maybe. I don't see a lot of attempts, though.
So I didn't give the complete solution, but indications to see what he really did.
*he/she
1/2 + sin^2x -1
ive tried this one
wait nvm
19:30
@Anubhav.K hi hi
One easy way to show all trig identities: Use exponentials.
3
@Danu +1
@FrankScience ive gotten sin^2(x) = 0.5 + sin^2(x)
@Danu how does that work?
@KinsonChan that's not true since that would be equivalent to 0 = 0.5
19:37
@KinsonChan Write out the procedure step by step, please.
please, please, write the procedures step by step please
sin^2(x) = 0.5 - (1-cos2(x)/2
@idonutunderstand using exponential definition of cosine, sine, tangent, like you know cos x = 1/2(e^(x)-e^(-x))
sin^2(x) = 0.5 - (1+1-2cos^2(x))/2
sin^2(x) = 0.5 - (2-2cos^2(x))/2
@user153330 nice!
19:39
sin^2(x) = 0.5 - (1-cos^2(x))
@user153330: As written that is not cos on the RHS.
sin^2(x) = 0.5 - 1 - cos^2(x)
sin^2(x) = 0.5 - sin^2(x)
@FrankScience ^
@MikeMiller oh my god, yes you are right, cos x = 1/2(e^(ix)+e^(-ix))
@MikeMiller that was the definition of sinh(x)
19:42
I think what you had is still cosh.
@KinsonChan Check this.
my mathjax isnt working rip
@KinsonChan sniff :'(
i know cosx = 2cos^2(x) - 1
but her it is -cosx
19:44
so shouldnt -cosx = 1 - 2cos^2(x) ?
LHS should be $-\cos2x$.
@KinsonChan try x=pi/2
@Danu: So what are you up to today?
@MikeMiller Just finished writing up my topology notes.
Tonight I'll pack my bags and get ready to go to Amsterdam tomorrow :)
@FrankScience it is same as cosx
19:47
I'm also reading the CFT textbook.
@KinsonChan No, and please learn how to write math in TeX.
Furthermore, I'm spending a lot of time debating about what to do for my (MSc.) thesis project
Conformal Field Theory (physics)
@Clarinetist hi
@Clarinetist are you about?
19:48
With the amount of effort I'm putting into typing up these (algebraic) topology notes, I am thinking about what I can do with them once I'm done; It'll be a ~200-250 page manuscript with very high quality pictures etc.
@Clarinetist the two issues are a) the correction factor because differential entropy is not scale invariant and b) does it actually converge?
b) is more interesting :)
Ah, physics is all Greek to me. Even quantum mechanics is difficult to me.
A simple question I've on this paper: ams.org/journals/mcom/1983-40-162/S0025-5718-1983-0689471-1/… -- what are the primed x's in the second page, and why is their sum equal to 1?
@FrankScience "even" haha
What's wrong?
19:50
QM is not the simplest thing.
@Clarinetist I think that the literature on multivariate CLT should tell us but I don't understand it properly sadly
@FrankScience can u remind me how s a "developpement limité" translated in english ?
limited development ?
Taylor expansion?
@Danu: There's only one CFT book?
@Agawa001 developpement limité as in finite expansion
19:51
@user276387 : aren't they dx/du
@FrankScience yes thank u
@Agawa001 no thank u for me?
@user153330 thk u , thats also said
@MikeMiller No, but there aren't many.
In my case, it's Blumenhagen's book (he's doing the lecture I'm attending, yay!)
Bell's inequality, etc.
I'm confused
19:53
and Bell's inequality is not basic QM either.
What do you mean by basic QM? Fundamental postulates of QM?
@CarlMummert thanks for the reply. How come that's dx/du?
No, solving Schroedinger's equation for a simple potential, for instance.
It's not how QM is taught here.
Well, the prime usually means a derivative, and dx / du is what you need if you want to do a change of variables from x to u in the integral, and it gives the correct equation (equal to 1) based on the previous equation in the paper @user276387
19:55
All reference books here start with fundamental postulates.
That's one approach.
Such as Tannoudji, or Le Bellac.
@FrankScience or Zettili, or Griffiths
Griffiths does not start with the fundamental postulates.
@Danu yeah but he just gives you like a very small discussion and then immediately provide you with schrodinger equation, i don't remember when he talks about the fundamental postulates
but because that's what i hate about those textbooks, they provide those postulates without giving any motivation
19:58
Sakurai is a classic (and does provide motivation).
@FrankScience do you know a good intuitive (ie gives motivation) introductory QM textbook with full historical context?
I am math-oriented and I come directly into formalisms.
@FrankScience yeah but even with formalisms one needs a bit of intuition and motivation, coming up with it on ones own costs tons of time and most of the time your intuition is flawed
What I was confused about is the composite system.
Thanks @CarlMummert If I may trouble you more, a bit above that when they say "equation (3) reduces to an nth degree equation for x of the form..." how did they reduce it?
20:01
For example, if $H_1$ is the state space of one system and $H_2$ is that of the other.
@user153330 Like I said, Sakurai does this stuff.
@Danu okay thanks i'll look into it
$H_1\otimes H_2$ gives the state space of the composite system.
I believe it comes from combining the fractions to get a common denominator, then multiplying that by both sides, then rearranging the resulting polynomial equation to put 0 on one side. Try it out in a specific case (e.g. n = 3 or 4 and make up values for the constants) and see what happens. @user276387
@FrankScience so...
20:19
Hey everyone, what is the correct way to take the exterior derivative of the product of two functions? i.e $d(fg)=?$ Here is my attempt, but I am left with a result that doesn't look too good imo... $d(fg)=d(f\wedge g)=dg\wedge f+g\wedge df $
@CarlMummert Wonderful, thanks again.
20:34
@AngusTheMan Since functions are $0$-forms, hence of even degree, your result turns out to be correct. Generally, one has $d(\alpha \wedge \beta) = (d\alpha) \wedge \beta + (-1)^{\deg \alpha} \alpha \wedge (d\beta)$. So $d(fg) = d(f \wedge g) = (df)\wedge g + (-1)^0 f\wedge dg$. But since $f,g$ have degree $0$, that is equal to $dg \wedge f + g \wedge df$.
(This is usually taught in first-year calculus courses with the name "the product rule".)
Huy
Huy
ö_ö
Don't give me that look. :(
Huy
Huy
.___.'
@DanielFischer Thank you for the reassurance! Sorry if it is obvious, I am a chemist after all :p
20:44
$$\huge{\text{What an amazing result I got!!!}}$$
Where is @robjohn?
Could someone of you take a look at my question:
0
Q: Reparametrization - First fundamental form

Mary StarSuppose that a surface patch $\tilde{\sigma}(\tilde{u}, \tilde{v})$ is a reparametrization of a surface patch $\sigma (u, v)$, and let $\tilde{E}d\tilde{u}^2+2\tilde{F}d\tilde{u}d\tilde{v}+\tilde{G}d\tilde{v}^2$ and $Edu^2 + 2F dudv + Gdv^2$ be their first fundamental forms. I want to show that...

?
@dorothy
@offsharing what kind've result?
@Semiclassical A series involving the reciprocal of Fibonacci numbers. Most probably unknown to this date.
well, here's hoping :)
20:56
look my hat
had same panador hat, looks awsome on ma mountain
Me and this paper again > ams.org/journals/mcom/1983-40-162/S0025-5718-1983-0689471-1/… -- at the very end of page one, in the sum can we have Cj = aj?
21:25
So I'm a complete noob to Topology. Define the ball of radius $\delta$ centered at $c \in \mathbb{R}$ by $$B(a, \delta) = \{x \in \mathbb{R} : |x-a| < \delta\}$$
A set $U \subset \mathbb{R}$ is open if for all $c \in U$ there is a $\delta > 0$ such that $B(a, \delta) \subset U$.
I would like to show that an open interval $(a, b)$ is open. I understand what I'm reading, but I'm wondering (like with $\delta$-$\epsilon$ proofs) if there's a brute-force way to find $\delta$.
@user276387: according the the text at the top of the second page, yes
@Clarinetist: for a given point x, the delta to take is the distance from x to the closer end of the open interval
(sorry, the definition above is wrong. It should be the ball centered at $a$)
@Carl Is that always going to be the case?
I mean, with an open rectangle in $\mathbb{R}^n$, sure, that's easy
But I'm wondering if there's a brute-force way to find a $\delta$ that will satisfy the above
what do you mean by "always". In the case of an open interval, certainly. In general, the largest delta you can take, given x, is the distance from x to the complement of the open set (which will be positive if the set is open - basically, you are trying to show it is positive if you are trying to show that the set is open)
@Clarinetist: By definition, yes. Set delta to be the infimum of the distances to points outside the set. Then by definition, points within delta of x are inside your open. The question is whether this is not zero.
Thanks @Carl, @Mike
21:31
In the case of an open interval, the distance is the minimum of two positive numbers, which will always be positive. One number for each end.
@CarlMummert thanks.
21:45
@Clarinetist Thank you. where is that from?
@Clarinetist I plotted a graph which might show some of the problem. imgur.com/PdYUzb4 is a graph of 2^17 6 by 12 matrices whose entries are 0 or 1. The x-axis is the entropy with random $\pm 1$ vector $x$ and the $y$ axis is the log of the determinant. Sorry about mixing M and B as the matrix names
Guys, consider an open square say $A= (-1,1) \times (-1,1)$, and $A \setminus \{(0,0)\}$ - I'd guess they are not homeomorphic, but how should I show this? Didn't know how to search for this, so a hint on what to search for would be good too.
Fundamental group.
@TedShifrin Now that you're freshly retired, how do you spend most of your time? :)
Thanks Danu, haven't learned that in school yet, I'll look into it.
You're in high school? Or do you mean university?
@Jake: I don't know any other effectively different approach here unless you know some machinery. What have you guys learned?
21:54
For this one, can't you show that the homeomorphism would have to swap 0 and \infty, so to speak, and go about it that way?
Yes, I suppose one could show that one has one and and the other has two ends. But this seems like the kind of idea that wouldn't be covered in a first topology course.
I don't think any proof of that question is included in a first point-set course :) Definitely not mine.
My first had the fundamental group in it... but I spose whether or not that's true is the taste of the instructor.
Actually, I think you can do this:
A homeo will induce a homeo between space- points
finitely many points
and somehow reductio ad absurdum should work
or maybe there is nothing more absurd one can do :P
I can't take what you said and make an argument out of it.
22:05
On $A\setminus \{(0,0)\}$, there is a harmonic function without a harmonic conjugate ;)
Now prove that's a topological invariant, @DanielF :)
@MikeMiller Well, we have the Riemann mapping theorem after all. Another approach: $A\setminus \{(0,0)\}$ has a two-sheeted (unbranched unlimited) covering.
hehe
I like this... I wanted to study some complex analysis next semester break.
@DanielF: Ok, now let's prove that doesn't happen for the plane...
I guess that's simple enough.
@MikeMiller Once you know your fundamental groups, that one isn't so hard any more ;)
22:19
Grumble, had to do a buckload of work today.
@MikeMiller Not that much, first course in point set topology - basic definitions, projectie inductive generated spaces, embedding lemma, separation axioms with urysohn lemma,tietze ext theorem, stone weierstrass theorem, tychonoff theorem, compactifications, uniform spaces.
@Danu The problem with using exponentials to show the angle summation identity for trig functions, say, is that this might or mightn't be circular depending on your definition of trig functions. To derive the Euler's formula, you need to compute the derivative of $\sin(x)$ and (depending on defn) you'd need angle sum formula to do that.
@BalarkaSen Circularity? Ain't no physicist worried about that ;D
Fair enough :)
I'd never even seen a proof until about half a year ago.
22:23
Proof of what?
Euler's formula
How can we calculate the integral $\iint_{\{u^2+v^2<1\}}\sqrt{1+4(u^2+v^2)}dudv$ ?
Anything
22:25
I didn't know what a topology was about 15 months ago.
Me neither.
I still don't know what a topology is.
In what sense?
Maybe that is how your first course got to fundamental groups so quickly @Mike :)
hehe
22:31
@Balarka: The definition, of course
BS.
I would guess it would be completely possible to do a lot of topology without knowing the definition of a topology, if everything you study is a subset of R^n for some small n
Who needs anything more than metric spaces? I haven't.
One can just study metric spaces.
Mike beat me to it.
Of course, to take quotients under group actions, one needs a G-invariant metric. But any student of topology knows that if G is compact one can do this and get a metric space homeomorphic to the one you started with.
I guess I lose now, because I have wanted at times to quotient by the action of a noncompact group
22:40
While I agree that every reasonable space is metrizable, I am not sure if one can start off with a metric without much trouble. The one you mentioned about coset spaces is one (the solution to which you have also mentioned). I haven't thought about it, but I'll believe you.
There's also the implicit assumption that I only care about reasonable spaces.
from a certain definition of "reasonable". I have my own definition, of course...
Mostly, me too.
@CarlMummert The reasonable spaces, by definition, are precisely... the metrizable spaces :D
@Carl: "A reasonable space is a compact, connected, non-Hausdorff space..."
@user159870 Shoot it down by polar coordinates.
22:43
I have heard there are plenty of uses for Zariski topologies, although that isn't my area. I have studied completeness properties of spaces to some extent, and we have other spaces there that are of interest in that program
You mean to set $u=r \sin\theta$, $v=r \cos\theta$ ? @OFFSHARING
Zariski topology on varieties aren't quite used to study topology, though. They are there for the sheaf theory.
(although I don't know much about algebraic geometry)
@user159870 I only gave you a hint, not telling more.
Ok. Thanks for the hint! @OFFSHARING
I wonder if I have ever seriously thought about spaces which do not have any CW-structure.
22:46
@Balarka: You claim to like topological manifolds.
@user159870 Welcome! ;)
Non-metric topologies are unavoidable if one wants to do analysis, where occasionally you have mapping spaces that want the point wise convergence topology.
@MikeMiller I haven't seriously thought about topological manifolds, but I am very excited to know there are topological manifolds with no CW structure. Is there an easy example?
@Balarka: No; but rather, it's open that they all do have a CW structure. I would be quite unsurprised if one failed to.
In analysis, even "topologies" are not general enough, so no surprise that metric topologies are not enough for the modes of convergence that are topological
22:50
This is fun. Thanks. Is it known if every topological manifold is homotopy equivalent to a CW-complex?
Yes.
@Carl: You're thinking of uniform structures?
Oh, fair point, I have given $Map(X, Y)$ some thought. They are certainly not CW-complexes.
convergence a.e. is not characterized by any topology
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2
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@MikeMiller Can you refer me to a proof?
22:54
@Balarka: Why are they not CW complexes?
No, the proof that topological manifolds are htpy eq to CW-complexes. I haven't seen the proof, but I think it should be significantly less harder.
PERFECTION IN MATHEMATICS ^^^
No, I'm asking you why Map(X,Y) is not a CW complex.
Anyway, see the appendix of Hatcher.
Anyway.
@Carl: Hm, fair enough.
22:59
@MikeMiller Hmm, I probably can prove that for some choice of X, Y, Map(X, Y) never admits a CW-structure with countable # of cells in each dimension by looking at fundamental group, but not sure about uncountable cell structures.
I don't know.
What are X and Y?
@dorothy See here, p.6
CW-complexes.
@Balarka: Then that has the homotopy type of a CW complex.
@MikeMiller That is helpful, thanks. The relevant proposition seems to tell me that every compact manifold has the homotopy type of a CW complex, though.
23:08
That was exactly what you asked for?
Oh, you want noncompact too.
Should work by a similar argument if you have a proper emvedding into Euclidean space.
Ok, I'd have to look at the argument carefully.
@MikeMiller Yikes.
If it has the homotopy type of a CW-complex, I can't expect $\pi_1$ to capture it, can I?
Huh? What does that mean?
I want to prove Map(X, Y) is not a CW-complex. I can't expect to prove this with $\pi_1$ if it's homotopy equivalent to a CW-complex.
So my idea can't really work.
23:15
You can't have ever expected to prove that with \pi_1 because you can find a CW complex with whatever \pi_1 you like.
Oops. I was thinking of CW-structure with countable many cells in each dimension.
I should go to sleep.
And you can disprove that with \pi_0. Take X to be the integers and Y to be two points.
But \pi_0 of that is \{+1, -1\}?
I think it just means it does not admit connected CW-structure. But then I am sleepy.
@Danu Haha, nice. That was the first thing I suggested and someone shot me down. Then you suggest it and it's starred. It's almost like different people have different opinions on the internet.
@MikeMiller Yikes, misread. Thought X was the reals. I follow your example, good one. Thanks.
23:32
If both X and Y are finite CW complexes this has the homotopy type of a countable complex.
oh, to completely toot my own horn
i think i mentioned my collaborators and i had a paper accepted for publication
and i only just realized that it shows up online as an accepted paper now :)
Grats!
Your first paper? @Semiclassical
23:47
Oh, that was a quick acceptance, @Semiclassic. Way to go!
Hi @Ted.
Hello @TedShifrin

I look again at an exercise in differential geometry and need some help.

The exercise is:
Show that the Weingarten map changes sign when the orientation of the surface changes.

The Weingarten map is defined by $W_{p,S}=-D_pG$, where $D_pG$ is the derivative of the Gauss map, the rate at which the unit normal $N$ varies across $S$.

So does the sign changes because of the direction of $N$?

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