Mathematics - 2017-11-30 (page 5 of 8)

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5:01 PM

Also @Mathein the only thing I'd try to do instead of Galois would be a reading course on Galois theory, one in which we'd probably go faster and then maybe do other stuff later on. 99% chance I'm not even doing that, but I def wanna learn the material one way or another

If you're not doing Galois theory y don't u join us

I mean I'm almost definitely doing it, chances are what'll happen in the spring is that I'll have harmonic analysis as an audit

ah

Anyway, what's up?

hey guys

Can I throw a question here and ask for help?

5:13 PM

@Daminark Not much. Fiddling with some elementary number theoretic things

Sick

1

Q: When $f(x)$ divides $d$ $f(x)=d(c+2ax+dx^2)\mod{N}$

Given $f(1)$ divides $d$ and $f(2)$ not, how to find other $x$ values that make $f(x)$ divisible by $d$? $$f(x)=d(c+2ax+dx^2)\mod{N}$$ $a,c,d,x,N$ are positive integers $c$ is a small number $d$ is a big number $c< a < d < N$ $gcd(d,n) = 1$ $N$ is a number that is hard to factor The questio...

number-theory elementary-number-theory divisibility prime-factorization

MatheinBoulomenos

Let $V$ be a volume bounded by a closed, positively curved surface $S$ in $\Bbb R^3$. Fix $x \in V$ and consider the set $U$ of points which may be connected to $x$ by a straight line that lies entirely in $V$. $U$ is open: fix $u \in U$, then the straight line that connects $x$ and $u$ is compact, so we may choose a finite cover of the straight line that connects $x$ and $u$ by balls that lie entirely in $V$, then after some fiddling with the triangle inequality, we see that we can find a ball around $u$ that is contained in $U$.

@BalarkaSen @Semiclassical

@MatheinBoulomenos Ah, that seems to be a sound argument

Yes, OK, this works. The important point is that the line joining $y$ to $x$ in the second paragraph can't be tangential to $S$

It has to intersect $S$ transversely. Once you know that, use genericity of transversality

MatheinBoulomenos

yup

5:23 PM

I was thinking about that but couldn't incorporate that into an argument.

Thanks a lot.

MatheinBoulomenos

glad to help

It takes an algerbaist to come up with that open-closed argument :D I would never have thought of that

We must cherish these times when algebraists are appreciated by topologists

Hi @Ted

MatheinBoulomenos

Hey @Ted

5:32 PM

I solved your cube

Hey @Ted!

Hi @Balarka, @Mathei, Demonark

It isn't mine, Balarka, but good.

There's actually a clever move using three corner squares

For me there's a one-word clue.

Google? /s

5:35 PM

lol

Um, no. I'm not a cheat.

lol

So was all this to-do about showing that a compact convex surface (where convexity means $K>0$ or that the surface lies entirely on one side of each tangent plane) bounds a convex region?

MatheinBoulomenos

yeah

yeah

5:38 PM

@Ted Right. I noticed that it's bounded on one side by all the tangent spaces

MatheinBoulomenos

sniped

But Mathei finished the argument

I managed to show that my volume was convex by linear algebra, which was cute

but I suspect I'm a broken record at this point

I haven't looked at this argument. I guess I would do a contradiction proof, assuming some line segment joining a pair of points left the region and getting the surface on two sides of some tangent plane.

Come on guys can't you put down your sniper rifles for just a few minutes when we're talking?

5:40 PM

@Semiclassical Not at all. Your surface is most interesting.

mmkay

I learnt a lot from thinking about it the past few days

I'm fine with this kind of rifle, Demonark, but not the kinds all over the US that are killing hundreds/thousands.

Anonymous

I'm trying to show that this series converges uniformly: $\sum \frac{x}{n+n^2x^2}$. So my method was to show that for $n\geq N$, $|S-S_n|<\epsilon$ for any $\epsilon>0$. But this method looks too difficult. Is there any other method for this ?

@Semiclassic: Particularly your algebro-geometric discovery is beautiful classic stuff.

At least over $\Bbb C$. :P

5:41 PM

for reference: "In 2013, there were 73,505 nonfatal firearm injuries (23.2 injuries per 100,000 U.S. citizens), and 33,636 deaths due to "injury by firearms" (10.6 deaths per 100,000 U.S. citizens)."

@Blue: Pointwise, uniformly, what?

@TedShifrin ya

Semiclassical should switch fields and come back to academics with an algebraic geometry mask

Anyone know how dirac bundle is defined over a riemannian manifold M?

@TedShifrin yeah, it's unfortunate. Such if life, I guess

5:41 PM

though the connection to semidefinite programming is esp nice

Anonymous

@TedShifrin uniformly :)

@quallenjäger: I did once. Spinor bundles and dirac operator, I assume. See Lawson & Michelsohn's book.

hmm, what's the symbol for positive semi-definite?

Anonymous

I'm unable to choose a $n\geq N$

@Blue: So have you considered the Weierstrass M-test?

5:42 PM

it's like $\geq$ but with a different >

Anonymous

@TedShifrin Oh, not yet. Checking

If you're taking a course, you should have studied this, @Blue!

@TedShifrin Is it also explained how to construct the clifford algebra?

Yes, @quallenjäger. It underlies everything.

ah, $\succeq$

5:43 PM

@TedShifrin Thanks!

@TedShifrin Are you familiar with Cartan Development?

Not sure to what you're referring.

@TedShifrin i.e. rolling a manifold over the other one along a given curve

Oh, I've thought about it, yes. Do you have something specific?

Anonymous

Okay, so Weirstrass M-test says that I need to choose $M_i>|u_i(x)|$ and $\sum|M_i|$ is convergent

Right, Blue, so find the maximum value of $|u_i|$, for starters.

5:45 PM

@TedShifrin mathoverflow.net/questions/286969/… Any idea?

My set of interest is equivalent (up to a linear change of variables) to $\begin{pmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1\end{pmatrix}\succeq 0$

that's a pretty cool statement

so it corresponds to the set of positive semi-definite 3-by-3 matrices whose diagonal is the identity matrix

ramble ramble ramble

@quallenjäger: For starters, the two surfaces have to be tangent along the curve you're rolling along, so you could roll to get a curve in $T_oM_1 = T_oM_2$ and then "unroll" that curve into $M_2$.

Oh no ... hides from @orbit

orbit-stabilizer

Haha, don't worry @TedShifrin. Just came on to look at the chat. I'm just studying for my existentialism midterm right now.

LOL, @orbit. OK. I'll give you more hint on that series question if you totally have given up.

Anonymous

@TedShifrin I can just think of $\frac{x}{n+n^2x^2}<\frac{x}{n}$ or $\frac{x}{n+n^2x^2}<\sqrt{n+n^2x^2}$ for each of the terms

Anonymous

5:51 PM

I don't know if that helps

orbit-stabilizer

@TedShifrin, no not yet. I'm going to reattempt it later.

@TedShifrin I am working a on problem rolling a curve $\alpha$ in $\Bbb R^3$ into a hyperboloid embedded in $\Bbb R^4$.

@Blue: Did you use calculus to actually find the maximum value of $u_n(x)$?

What they did is the following

You can't roll one surface along another if they live in different spaces, @quallenjäger!!

5:52 PM

Is there any algorithm that can help find a minimum of polynomial mod some big composite $N$?

@orbit: OK, just remember that I told you to write down actual numbers (or numerals).

@Ilya_Gazman: I'm not sure I even know what that means. You mean minimum in $[0,N-1]$? But I don't know this stuff at all. Computational number theorists do.

@TedShifrin $\Bbb R^3$ is embedded in $\Bbb R^4$ and the hyperboloid is also embedded in $\Bbb R^4$ or did I miss something?

Are the surfaces tangent along whatever curve you're talking about? How are you embedding $\Bbb R^3$ in $\Bbb R^4$? Is there a way in which this all makes sense?

Anonymous

@TedShifrin Oh, got it. At $x=\sqrt{1/n}$ it is maximum. The maximum value is $\frac{1}{2n^{3/2}}$

Let $M$ be a locally symmetric space of finite volume and negative curvature.
Is there some geodesic in $M$ which intersects itself?

Anonymous

5:55 PM

So that looks like a p-series

Anonymous

It should converge

Aha. That's how you should immediate approach such problems.

Anonymous

And thus $\sum|u_i|$ should converge too (uniformly)

@TedShifrin Well, I got your point. That is what I want to ask in the following.

Anonymous

@TedShifrin Got it! :D Learnt something new today :)

5:56 PM

Do there a sequence (an)n≥0 such that, for all n, a0+a1X+...+anX^n has exactly n distinct real roots ?

:)

Ordinarily that should have been done in class, @Blue :)

@TedShifrin The problem is just, I have a curve $\Bbb R^3$ and want a canonical projection onto a hyperboloid.

Anonymous

@TedShifrin I'm actually learning this myself

Anonymous

:P

Oh. Well, good, then :P

Anonymous

5:57 PM

From Arfken

There are trickier problems where that approach gives something that doesn't work on all of $\Bbb R$. But probably physicists (now that you told me) wouldn't care about those.

@quallenjäger: I doubt there's anything canonical in what you're describing. But see if you can do what I said if you set it up right.

@abenthy: That sounds like a hard question in that generality. Are you looking at surfaces or arbitrary dimension? I don't think enough about Riemannian geometry to know offhand.

Anonymous

@TedShifrin Do you recommend any book for this stuff? It probably falls in the domain of real analysis.

@TedShifrin What the author of the paper did is the following: first they find a mapping from $\Bbb R^3$ into the lie algebra $so(3,1)$ by $x \rightarrow \begin{bmatrix}0 & x\\x & 0\end{bmatrix}$

Anonymous

Arfken doesn't have all the details but gives a brief overview

sup duds

5:59 PM

Hello, may I have a question? What is the maximum count of non-zero elements, that can a linear convolution of discrete signals of "lengths" 5 and 7 have? Tnahk you very much for any hint
question from signal analysis

Akiva Weinberger

Hi

Tomorrow's my birthday

Then this curve can be rolled onto the lie group $SO(3,1)$ by using cartan development

sorry there is a mistake in the matrix

Happy birthday in advance!

Anonymous

@AkivaWeinberger Happy birthday!

@Blue I like Rudin chapter 7 for uniform convergence

@AkivaWeinberger happy early birthday!

Akiva Weinberger

6:00 PM

@BalarkaSen @Daminark Thanks, you too!

$x\rightarrow $\begin{bmatrix}0 & x\\x^T & 0\end{bmatrix}$$

Anonymous

@Daminark Ah, I'll check. Thanks

Yo @Eric!

Anonymous

I bought the Stein-Shakarchi books. Must be there too

@Akiva I'll send you a meme as birthday present

6:01 PM

I'd guess that those more or less assume you know Rudin 1-7

@TedShifrin It seems to me like a hard question too, but I found a paper stating that its true and is a well known fact. I'm not only looking at surfaces.

Happy almost birthday, DogAteMy!! :)

The differential equation of the Cartan development onto Lie group along the curve $\alpha$ is then $d\Gamma=F(d\alpha)\Gamma$, where $\Gamma\in SO(3,1)$ and $F:=\begin{bmatrix}0 & d\alpha\\d\alpha & 0\end{bmatrix}$

@Blue: For starters you can look at Spivak's chapter on this stuff (24, I think) in his Calculus. I wrote some of the exercises, and there are some good ones.

@AkivaWeinberger lel

6:03 PM

@quallenjäger: I have no idea what you mean by $F(d\alpha)$. But this is intrinsic to one manifold.

By solving this differential equation we will find a $\Gamma$ and then apply it to the origin $(0,0,0,1)$ of the upper sheet hyperboloid, they claim that the obtained curve is then the hyperboloid development of curve $\alpha$

@TedShifrin Basically $\Gamma\cdot o$, where o is the origin, should be a hyperbolic development

@abenthy: You might ask @EricSilva if he's encountered this. You might look at Joe Wolf's papers or book (Spaces of Constant Curvature). Not sure if Petersen does this in his Riemannian Geometry book.

@quallenjäger: So it seems they're using the idea I suggested, more or less. Roll along a curve in one, then unroll in the other, using the fact that $SO(3,1)$ acts on both, presumably.

@TedShifrin $F$ is just the defined mapping $F:x\rightarrow \begin{bmatrix}0 & x\\x^T & 0\end{bmatrix}$. $d\alpha$ is the differential of the curve $\alpha$ in $\Bbb R^3$

@TedShifrin Thanks for the hint, Wolfs book is a good idea!

@quallenjäger: You actually mean $\alpha'$, not $d\alpha$, and my point was that you used $d\alpha$ in your original definition of $F$.

6:07 PM

May I ask for something from signal analysis? -> What is the maximum count of non-zero elements, that can a linear convolution of discrete signals of "lengths" 5 and 7 have?

@TedShifrin Can you specify a little bit more which is rolled on which and how they proceed?

@quallenjäger. No. :) I'm not willing to engage my brain that hard to work on this for an hour.

@FilipCZ: Offhand, I'm not sure anyone here knows signal processing. I know people that do, but that doesn't help. :P

@TedShifrin Alright, I think I can figure it out by myself.

@TedShifrin Thank you!

Good :P

@Blue: This topic is undergraduate real analysis, which doesn't quite fit the Stein/Stakarchi series. They assume it.

what's up i was tabbed out

6:10 PM

@TedShifrin hi, long time no see

@Akiva: I did wish you happy birthday, but I never tag you (for obvious reasons) :P

Anonymous

@TedShifrin Ah, okay. Got it. I bought those books mostly with the intention to self-learn them. But okay, I'll look at Spivak and Rudin 1-7 too

@EricSilva: abenthy asked about geodesics on locally symmetric spaces of negative curvature (and finite volume).

Hello. I would like to know if the Ecole Normale Supérieure is a recognized school abroad?

@TedShifrin Always a great help to talk to you. You always point out the point I am missing. Now I understand it.

6:11 PM

This was my question: chat.stackexchange.com/transcript/message/41460322#41460322

I'm glad I could help, @quallenjäger. I just made an educated guess and I hope it works out for you! :)

Yes, @Lucas.

@TedShifrin Well, I was confused why they need to roll it on the Lie group. Now I see, they are not living in the same space and I need to find a connection.

@TedShifrin A connection between the both space, not the connection of a manifold.

@quallenjäger: Yeah, for homogeneous spaces with the same group, this should work out in an abstract way, rather than a concrete physical way.

Yes, yes. :P

i have not encountered this so I am of no use

@TedShifrin Very sophisticated idea. I was restrict myself on the physical way and I could not get my head around it.

@TedShifrin Again thank you for pointing this out.

6:15 PM

Eric, I was useless, also. The symmetric spaces in which I know geodesics (other than hyperbolic space) aren't of negative curvature.

Sure, sure, @quallenjäger. I'm glad to help.

I think every negatively curved locally symmetric space has rank $1$, right?

I usually have much more reason to care about things of nonnegative curvature so I end up knowing embarassingly little about negative curvature

So there are basically only 4 kinds of those spaces and I guess one checks it for every one of them

I actually think negative curvature is quite interesting ... But ...

I actually don't even know that, @abenthy. Is that true? Why?

I just read it in some paper, I'm gonna look it up.

6:17 PM

Well negative curvature is quite interesting. I am working on this topic right now.

@Ted me too but the geometric analysts here do variational things which are almost always more interesting in pos curvature

Hmm ... so rank $\ge 2$ forces nonnegative curvature. That's plausible, actually.

This is where I say I'm more of a complex algebraic geometer :P

something about the fact that negative curvature forces the behavior of critical points of functionals to be too tame to be interesting

I've taught graduate courses which covered some Riemannian stuff (including some symmetric spaces stuff) ... but it's not my major bed of knowledge.

Well, Eric, you'll expand your view eventually :P

I mean compact (say) quotients of hyperbolic space are interesting!

ya ofc

i just mean what I've been exposed to a lot over the recent period has skewed things a bit

6:20 PM

Is this true if I rescale the curve large enough on a hyperboloid, it will become close to a geodesic?

Okay the claim that negatively curved implies rank $1$ is on page 9 right before 3.2 in the article arxiv.org/pdf/0909.2899.pdf

What does that mean, @quallenjäger?

That seems totally wrong, @abenthy. Spheres are rank $1$ symmetric spaces.

@TedShifrin Well, what we are currently doing is, we develop a curve onto the hyperboloid using the method above and rescale it by multiplying with a constant $\lambda$.

I don't know what that means on an ambient manifold that's curved, @quallenjäger. I'm sure you can make sense of it, but I don't get it.

@TedShifrin But aren't spheres rank $1$ symmetric spaces?

6:22 PM

I just said that. Did I misread?

@TedShifrin Sorry, I mean we rescale it first. i.e. multiplying with $\lambda$ and roll it onto the hyperboloid

He said negatively curved iff rank 1. That's crap, I believe.

@TedShifrin If we choose $\lambda$ large enough, we can show that the curve becomes arbitrary close to a geodesic.

Oh, scaling in the fixed tangent space, @quallenjäger. I see. Seems reasonable since that makes it more linear in the tangent space.

@TedShifrin Is this reasonable observation?

6:24 PM

I posted it as a question now: math.stackexchange.com/questions/2544833/…

Well, you should check it out, but exponentiating lines gives geodesics.

@TedShifrin exponentiating lines in the tangent space gives geodesic on the hyperboloid?

Antonios-Alexandros Robotis

That works on more general (Riemannian) manifolds, right?

Yes, @Antonios. The issue, @quallenjäger, is to understand what developing and undeveloping (in the Lie group) actually does. Undeveloping isn't just exponentiating ... or is it?

Wow ... how did all this geometry end up in here!!

@TedShifrin Do you have reference recommendation where I can read the riemannian geometry of a hyperboloid?

@TedShifrin You are right! I will check this out! Thanks1

Antonios-Alexandros Robotis

6:28 PM

We're just talking about the exponential map, no?

(Sorry, I just started reading lol)

Antonios, no, we're not, I don't think. It's interesting geometry. You can look at quallenjäger's question that's linked up there somewhere.

Antonios-Alexandros Robotis

Will do.

@Antonios-AlexandrosRobotis We are not at the tangent space this is the problem. It could be anywhere.

Antonios-Alexandros Robotis

Hmm, neat. If I remember I'll check that out. For now, I suppose I should keep reading about ramification lol. Finals are approaching...

But you probably can think about what happens in the group before you map it down to the homogeneous space.

What are you ramifying, Antonios?

MatheinBoulomenos

6:33 PM

I hope prime ideals

Antonios-Alexandros Robotis

Prime ideals x D

MatheinBoulomenos

Nice

Damn. I hoped it was riemann surface theory. :P

lol

Antonios-Alexandros Robotis

hehe. Maybe soon

6:33 PM

@TedShifrin here is my question for better context. I want to know if there is an algorithm that can find the minimum of polynomial mod $N$ not for $0<x<N$ there is no any limit on x, but just in general when f(x) is a minimum, where $f(x)$ is a polynomial that $\mod N$ been applied on it.

MatheinBoulomenos

@Antonios-AlexandrosRobotis are you taking an algebraic number theory course?

Antonios-Alexandros Robotis

@MatheinBoulomenos basically, yeah.

What $x$ values solve it?

MatheinBoulomenos

wow awesome

@TedShifrin Will do, I am just not sure what lie group action does matrices like $\begin{bmatrix}0 & x\\x & 0\end{bmatrix}$ generates. Is it a rotation around all 3 axis?

6:39 PM

What group are you working with? Why is it just a $2\times 2$ matrix? I really haven't been paying attention to details.

Alessandro Codenotti

Let $\tau$ be the coarsest topology on $\Bbb C$ according to which every $f\in\text{Aut}(\Bbb C)$ is continuous, what can be said about it? Is it discrete?

@Ilya: I have no earthly idea.

@Alessandro: So what is $\text{Aut}(\Bbb C)$?

MatheinBoulomenos

@AlessandroCodenotti I'm not sure I understand the question. Every map whose codomain has the trivial topology is continuous

@TedShifrin On $SO(3,1)$ and its a $4x4$ matrix where $x$ is a three dimensional and its should read $\begin{bmatrix}0 & x\\x^T & 0\end{bmatrix}$

Ohhhh ... now it makes more sense.

Alessandro Codenotti

6:43 PM

@TedShifrin The group of field isomorphisms $\Bbb C\to\Bbb C$, so identity, conjugate and $2^{|\Bbb C|}$ nowhere continuous (in the standard topology) morphisms

Oh, you meant the discontinuous ones. Surely it must be the discrete topology.

Alessandro Codenotti

@MatheinBoulomenos sure, but I'm interested in the coarsest one with this property

@quallenjäger: I've worked this out before. It's not hard. Just do it. You should get the hyperbolic version of circles, so you get $\cosh$ $\sinh$ stuff that turns into hyperbolas.

MatheinBoulomenos

@AlessandroCodenotti the trivial topology is the coarsest topology on any set

Alessandro Codenotti

@MatheinBoulomenos yes, but no automorphism of $\Bbb C$ is continuous in the trivial topology

6:45 PM

@TedShifrin Well I have exactly the same result, sinh and cosh everywhere, can't get my head around how to interpret cosh and sinh on hyperboloid.

MatheinBoulomenos

every automorphim of $\Bbb C$ is continuous if you give both copies of $\Bbb C$ the domain and the codomain the trivial topology

It's intersections of the hyperboloid with planes through the origin.

Those are geodesics.

conic sections?

Alessandro Codenotti

@MatheinBoulomenos ah, sure, I was implicitely using the standard one on the second one, hmmm I'm not sure about what question I want to ask then

MatheinBoulomenos

@AlessandroCodenotti do you fix the topology on the codomain and are wondering about the domain?

6:48 PM

@quallenjäger: There aren't any cones here. Just hyperbolas running on the hyperboloid. Work it out! :)

Alright!

sounds like someone is doing a nice exercise

what is up dramalertnation

it's yo boi wikileaks here

Alessandro Codenotti

@MatheinBoulomenos hm, actually let's use the same topology both on the domain and codomain. Both the discrete and the trivial topology have the property that they make al automorphisms of $\Bbb C$ continuous. Are there more?

Hey everyone!

6:52 PM

hi Perturb.

MatheinBoulomenos

Hi @Perturbative

Hey! @TedShifrin It's nice to see you again :)

Hey @Perturbative

Hey @MatheinBoulomenos, @BalarkaSen, @AlessandroCodenotti, @EricSilva

The hellos take longer than most problem sets :P

6:54 PM

So what kind of geometry were we speaking of?

Alessandro Codenotti

Hi @perturbative

there should be a third button for jsut saying hi to people coming into chat

next to send and upload

Lots of Riemannian stuff, Balarka. Including something about rank 1 symmetric spaces which seemed totally wrong to me.

hmmm. elementary probability question

Eric, that would make it nice and personal.

6:55 PM

@EricSilva Don't cheat; lest you'll get sued by Epic Games

@TedShifrin Ah, I see

let them come at me @Balarka

I fear no man

suppose I've got two binary random variables X,Y which are not necessarily independent

is there a name for Pr(X=Y) - P(X!=Y) ?

If they always agree that's 1 and if they always disagree it's -1.

Isn't P(X = Y) + P(X ! = Y) = 1?

It is.

Then that's 2P(X = Y) - 1

6:58 PM

I know I'm overcomplicating it

Alessandro Codenotti

Actually @Mathei I have a more sensible (and interesting question) I wanted to ask you, what's a simple example of a polynomial $f\in\Bbb Q[X]$ whose roots can be expressed via radicals but whose splitting field is not a radical extension of $\Bbb Q$?

@Alessandro Wait, that's possible?

What I'm interested in is the fact that when they're perfectly correlated it's +1 and when they're perfectly anticorrelated it's -1

I want to say that it's literally just the correlation of the two random variables.

I mean if the roots are expressible by radicals the splitting field E/Q is obtained by adjoining radicals to Q successively... am I missing a subtlety?

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