Mathematics - 2018-01-19 (page 3 of 4)

Once again: Good luck.

4:38 PM

@Semiclassical thanks :D

Don't you think that its possible with few constructions and calculations?

No.

hmmmmm

@AkivaWeinberger That is right.

@AkivaWeinberger I wondered that. Closed examples are certain Lens spaces

L(5, 1) and L(5, 2)

@BalarkaSen What dimension are these, again?

@Semiclassical So, should I start with something else? from scratch?

4:42 PM

3

for that?

those are quotients of S^3

Mhm

I think you should use trig and accept that that's how these problems work.

Weird

That'd be the smallest dimension in which that's possible though I guess

Wait, actually

4:43 PM

Closed 2-manifolds are determined by their homotopy type, yeah

Mobius strip and cylinder ($[0,1]\times S^1$)

Well, with boundary

Closed is the right condition here

They have different products with $\Bbb R$ anyway (orientability)

Mhm

Hmm

How about the double annulus (twice punctured D^2) and a neighborhood of the 1-skeleton of the torus?

Oh, right

(AKA punctured Tory)

4:46 PM

@AbhasKumarSinha the trouble is that, while you know sin(30 degrees)=1/2

(torus)

Yeah

(Not punctured UKIP)

you need to divide that angle by 2, 3, and 5

first isn't too bad. second also isn't horrible, though it's hardly trivial.

Wait, does that actually work??

Oh, wait

4:48 PM

but the last one is baaaad

It might if we were to multiply by $[0,1]$… but if we multiply by $\Bbb R$ then they have a different number of boundary components

(for comparison, note that it's provable that trisection of angles can in general not be done with compass and straightedge. 5-section of angles is worse.)

@Semiclassical sine(72 degrees) has a closed form, so that brings you down to 3 degrees

hmm

3 degrees not nine?

Oh whoops

9, yeah

cos(36 degrees) is half the golden ratio

4:51 PM

Oh hey: math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf

@AkivaWeinberger Boundary components?

That's a nice table

They look the same to me. Am I missing something?

@Semiclassical Oh, wait:

36 is doable, hence 72 is doable. 60 is also doable; use the difference formula to get 72-60=12

and then 6 and 3.

Hey @MikeMiller

4:52 PM

@akiva yeah, see the last two pages of what I just linked

Meh, you looked it up faster than I could figure it out from memory

lol

@BalarkaSen Double annulus has three boundary components; punctured torus has one.

Multiplying by $\Bbb R$ shouldn't change that

($\partial\Bbb R=\emptyset$)

hi

@AbhasKumarSinha check out what I just linked for the details of how you get exact values of sin(n degrees) for various values of n

it isn't nice.

Adeek

4:54 PM

@MikeMiller hey

@AkivaWeinberger Oh, right, I was taking the topological boundary to account

So x [0, 1]

Adeek

right @BalarkaSen

user8469759

is there a reason why a differential solid angle is usually represented by a cone?

Was thinking of an $\epsilon$-thickening

for instance, $$\sin(3^\circ)=\frac{1}{4}\sqrt{8-\sqrt{3}-\sqrt{15}-\sqrt{10-2\sqrt{5}}}$$

4:56 PM

@MikeMiller Is there a way to prove $M \times \Bbb R \cong S^n \times \Bbb R$ implies $M \cong S^n$ (topologically) without invoking the Poincare conjecture?

For closed manifolds, mind

Thanks, Akiva

And maybe even closed manifolds with boundary, we haven't gotten a proper counterexample yet

S^n x R has no boundary, so M cannot have boundary

$\partial(M \times \Bbb R) \cong \partial M \times \Bbb R$

Oh, $S^n$, whoops. I thought you had two arbitrary manifolds there.

4:59 PM

Ah, no. But I do want examples of compact (without boundary, ideally) manifolds $M, N$ such that $M \times \Bbb R \cong N \times \Bbb R$ but $M, N$ are not homeomorphic

by comparison, if you note that that's 3pi/180=pi/60 radians, which is small, you get the estimate sin(3 deg) ~ pi/60 = 0.05235987... when the exact value is 0.052335956

@Eric do you also happen to be familiar with the usage of the axiom of choice in functional analysis? Specifically is it possible to prove that the weak topology is Hausdorff without any form of $\sf AC$?

so the seemingly 'obscure' sine representation yields a very useful approximation

I suppose if we had an example with boundary, we could cap off the boundary components and get an example without boundary

@BalarkaSen No, I don't think so

5:02 PM

weak topology on a Banach space*

@MikeMiller Hm, weird

@Semiclassical There are much nicer approximations, such as the aforementioned $\pi\approx32\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}$

ehhh

again, definition of 'nice'...

You prefer $15\sqrt{8-\sqrt3-\sqrt{15}-\sqrt{10-2\sqrt5}}$?

I prefer to accept that $\pi$ is a useful constant absent approximations :P

5:08 PM

which is a (slightly) worse approximation

i.e. sin(3 deg) is to me a far more useful answer than the above mess of square roots

It's sine 45/8 I think

(The bit after the 32)

sure

Cute question:
Consider a set of triples, each member appears in 6 triples exactly. Show you can two-colour the triples, such that each member appears in both colours

I mostly just think it's silly to prefer the exact forms of something like sin(3 deg). there's a reason why we use trig functions---a closed-form expression for a sine value is by and large pretty useless.

to the extent that you get anything useful out of it, it's by crunching out the resulting value. but in that case there's easier approximations which you might as well do instead

(blah blah soapbox blah)

5:12 PM

@AlessandroCodenotti i think you need hahn banach which can be proven without full choice but im not certain about that

@Studentmath We don't know how many members in total there are?

@EricSilva I see, thanks

@Akiva nope. You can think of it as a 3-uniform 6-regular hypergraph if you want (I think), and show that there are two disjoint hyper-edge covers of it

@Akiva you can assume they are finite, though

i think all you need is the ultrafilter lemma? idk

Now I find myself wanting to make up a story for this setting :P

5:16 PM

something weaker than choice in any event

@EricSilva That's enough for Hahn-Banach, yes

cool

@Studentmath And the triplets don't care about order?

@AlessandroCodenotti @EricSilva I don't think you need any form of choice. The weak topology on $V^*$ is induced by the familiy of seminorms $\|f\|_v := |f(v)|$ for $v \in V$. Since we trivially have $(\forall v \in V \|f\|_v=0 )\Rightarrow (f=0)$, the topology is Hausdorff

@Akiva precisely, that's why 3-uniform 6-regular hypergraph is a good representation (well, for me at least it helps)

5:17 PM

Maybe something like

I don't quite know what 3-uniform 6-regular means

Hypergraph is like edges have three vertices instead of two?

Yeah precisley

That's 3-unform hypegraph

And 6-regular just means each vertex is in 6 hyperedges

Ah

I want to think of it as a 6-dimensional simplicial complex.

@MatheinBoulomenos What do you mean with "is induced by a family of seminorms"?

5:19 PM

You've got a bunch of people practicing group presentations. Each presentation must involve three people, and each person must participate in six presentations.

The other type of "group presentation"

lol

hadn't even thought of that

@BalarkaSen Closely related but not precisely the same, I think

And then everyone participates in at least one morning presentation and one afternoon presentation. @Semiclassical

hmm

that may work, yeah

5:20 PM

It's really cute though. Haven't yet thought of a complete proof

Or, rather, it can be scheduled like that.

I think to each hypergraph you can associate a simplicial complex, by saying that {v_i} form a simplex if they're connected by some hyperedge (which may connect more things)

Is what we need to prove.

@MikeMiller Maybe it won't be a simplicial complex. I was thinking, like, let $T_i$ be the triples. If $T_i$ and $T_j$ intersects, put an edge. If $T_i \cap T_j \cap T_k$ is non-null put a 2-simplex.

@AlessandroCodenotti if you have a family of seminorms $\| - \|_i$ on a vector space, then they induce a topology. You define your open balls by $B_{\varepsilon}(x)=\{y \in V \mid \forall i \|y-x\|_i < \varepsilon \}$

5:21 PM

Hypergraphs are cool, since they can represent more systems than regular graphs

Another way to do it would be to have two different observers

And so on

@MatheinBoulomenos I'm thinking about the weak topology on $V$ rather than $V^*$ as well. But I have to be away for a while now, I'll ask you more about this later probably

And anything with "hyper" in its name gets a lot more exciting

And you want to schedule the talks so that each observer sees each student at least once

5:21 PM

Except for maybe hyperplanes

though that sounds like a really good airplane

Haha

I'll think about this for a bit.

anyhow

@AlessandroCodenotti oh yeah I think you need Hahn-Banach for that

Concord 2.0: Hyperplane

5:23 PM

@MatheinBoulomenos yeah it shows up somewhere at some point

Can you draw hypergraphs?

Sure

Hypergraph

In mathematics, a hypergraph is a generalization of a graph in which an edge can join any number of vertices. Formally, a hypergraph H {\displaystyle H} is a pair H = ( X , E ) {\displaystyle H=(X,E)} where X {\displaystyle X} is a set of elements called nodes or vertices, and E {\displaystyle E} is a set of non-empty subsets of X {\displaystyle...

right, i see it now

yeah

Considering 3-uniform hypergraphs, you can try to imagine it as some triangles-sort-of-thing. I never really found it too helpful, but you can then throw some theorems about triangles in Graphs

right

so it's sorta like a vertex in a polyhedra being the point of contact for multiple faces.

5:26 PM

The point of the proof, is just like v2 in the above picture being in two different colours - for triples such that each member appears in 6 triples, you can 2-colour the triples (hyperedges) and cover all members s.t. they have two colours on them

it's not a perfect analogy, though

Similar in a way, I'd say

I wonder if thinking of it as a bipartite graph with one subset of vertices corresponding to the triplets and the other subset corresponding to members help. Each member is connected to 6 triplets and each triplet is connected to 3 members

You want a 2-coloring on the subset of triplet vertices

Such that each member makes it to two colors

It seems more efficient than thinking about hypergraphs

I can see a particularly simple example: Form a triangular lattice, with each vertex corresponding to a member and the faces corresponding to groups

If you want finitely many people you'll have to take a lattice with periodic boundary conditions and the right proportion of people, but that's not too bad

@Balarka That's nice

5:32 PM

and then it's rather plausible that you can color the faces so that adjacent faces have opposite colors

in which case you're done.

Buuuuut that's an easy version.

Maybe turning it the other way around would make it easier.

I mean, consider vertex colouring and not edge colouring

But then the structure won't be that neat.. nevermind

The complete thingy with five people satisfies this, right?

Each object would be in $\binom42=6$ triplets

@Akiva not sure I follow

@Studentmath We want a bipartite subgraph of the thing I said such that it looks like 2n "triplet" vertices on one side and n "memeber" vertices on the other and each member vertices is joined by two hands to one black "triplet" vertex and one white "triplet" vertex

@Studentmath Say there are five people, and all possible groups of three of them gives a presentation

Then each person will be in 6 presentations, if you work it out

Person A has six ways to complete his triplet out by choosing two people out of {B,C,D,E}.

And so for everyone.

5:48 PM

@Balarka Sounds like that, preciesley
@Akiva right

@Akiva you can see the "theorem" is true in that case

@Studentmath It cannot happen that two distinct member vertex do not have a pair of edge coming out of them ending at different triplet vertex, yes? By counting

Sure. Each member vertex has 6 edges going to the other side

Well what if it looks like the complete bipartite graph K_{2, 6}?

I guess it can happen and I am wrong

Actually, there is a cute short embedding lemma for this. For such a bipartite graph, we can embed every 6-vertex 3-degenerate graph

(Not that it helps)

I can embed K_{3, 6} as a bipartite subgraph but not K_{4, 6}. OK, that's the limit.

5:54 PM

I don't think it can look like the complete bipartite graph K_{2,6}

Why not?

You mean the subgraph?

Or the entire structure?

A bipartite subgraph

Ah, sure then

K_{3,6} is the "certain" subgraph

Yeah

5:55 PM

I think

But you can't get K_{4, 6}, is the point

Okay okay okay

Anderson Felipe Viveiros

Anybody?

https://math.stackexchange.com/questions/2612101/conditions-on-a-lipschitz-function-fu-subset-bbb-rm-to-bbb-rn-which-guara

I think I found out where my friend took the question from
http://math.mit.edu/classes/18.095/lect6/notes.pdf

First exercise, think it's pretty much the case (maybe a coincidence, will ask him)

Aha

mercio

is it possible to ignore some users so you don't see their questions anymore ?

6:02 PM

Put black tape on your monitor over the questions of that users that show up in your feeds

Simple

does

Oh no too much black tape

Let's say, for contradiction, no such colouring is possible.

Hrm

It means there is at least one member..

More Kepl than ever

6:06 PM

@Akiva that's cool

Coolio

@Studentmath I considered that but it gets complicated quickly

It means there is at least one member, such that in every two colouring of the triples, it will be in one colour only (stating the obvious here)

@Balarka Well yeah..

Like, the condition for not being able to change the color of the triple

But I think a constructive proof will be even uglier

That means it's connected to something which is connected to everything of opposite color

(except one)

And ugh

6:10 PM

It could be even worse, could be a chain of connections that ends up ruining all up

Yeah

And it could be more than one member... maybe contradiction isn't the best idea here, yeah you are right

I'll think about it. It doesn't look like a super easy problem to me

But I do think the bipartite formalism helps

It's definitely tougher than it seemed to me at first

Ideally you could write down an actual algorithm

6:25 PM

@Semi That was where I started. Somehow it just got to too many options to cover, but ideally I'd guess that's the idea

Hey, the intersection of compact sets is compact, right?

Wait

Dammit

In Hausdorff spaces, the intersection of compact sets is compact, right?

@Akiva yep

Iirc in non-Hausdorff spaces that's not always true

Silent

If $p>0$, then $\lim_{n\to\infty}\frac{1}{n^p}=0$.

*Proof.* Take $n>(1/\varepsilon)^{1/p}$. (Note that the archimedean property of the real number system is used here.)

Where is archimedean property used in this proof?

Is it used to assure that there is an $n$ with $n>(1/\varepsilon)^{1/p}$?

@Studentmath Yeah, line with two origins. Take the two versions of $[-1,1]$ in it and intersect them

My reaction to this problem: math.stackexchange.com/q/2612497/137524

be praying

Eric Auld

7:02 PM

Suppose I take the singular value decomposition of a matrix like $\mathbf{I} + \mathbf{u}\mathbf{v}^*$.

My temptation is to take my right singular vectors to be $\mathbf{v}$ along with a basis for $\mathbf{v}^\perp$.

I know in general I can't choose my right singular vectors, but here what doesn't work? If $\mathbf{A} + \mathbf{I} + \mathbf{u}\mathbf{v}^*$, then either $\|Av\|\geq 1$ or $\|Av\|<1$.

anonra

Heya. I've been asked this question by a friend, but as a poor CS student my knowledge of probability isn't good enough to crack it. Could someone point me in the right direction?
A band whose concerts are usually 80% booked wants to book a concert hall for their next concert. How many seats should the place offer so that there's a 90% chance of only 100 seats remaining empty but also not more than 100 people not being able to get a seat.
I tried playing with the CDF of the binomial distribution and confidence intervals, but that got me nowhere and I haven't been able to find any similar qu…

Hey @EricAuld. Haven't seen you around in this chat before.

Eric Auld

In the former case $v$ is a good first singular vector. In the latter it is not, but why can't I choose a bunch of singular vectors corresponding to $\sigma = 1$ in $v^\perp$?

Hi @BalarkaSen ! Thanks for all your good answers.

Haha, the proportion of good answers I posted is not very high, but I'll take it :) I did remember enjoying answering/learning from the answers to a few of the questions you posted.

Eric Auld

Oh, I'm starting to think I can just do this the way I want. I was confused. The matrix is singular iff $\langle v, u \rangle \neq -1$, and that jives with $v + \|v\|^2 u = 0$.

7:12 PM

@Balarka remind me where your avatar comes from

Oh yeah that's from "Tale of Tales" by Yuri Norshteyn

thanks

@Anonra I would indeed go for $p=0.8$ for a seat being taken, and play with that

@Balarka I think there are some theorems about matchings/1-factorisations in bi-regular graphs. Maybe something there could help us, with your formalisation

Mm, I don't know much graph theory. I'd like to see those theorems

7:42 PM

@Balarka we should retitle my analysis class "Ergodic theory" at this point

Sounds fun

Last class we did Krein-Milman, generalized it slightly today from normed spaces to locally convex spaces

You're going to end up doing it for the space of probability measures on a compact space eventually, I suppose

Then we said alright, so if $K$ is a compact metric space and $\phi:K\to K$ is a homeomorphism, then if you look at the space of $\phi$-invariant probability measures on $K$ (which sits in $C(K)^*$), call that $M(\phi)$, then $\mu$ is ergodic iff it's an extremal point of $M(\phi)$

Mhm

7:46 PM

We did this modulo the statement that

$\frac{1}{n}\sum_{k=0}^{n-1} f\circ \phi^k \to \int_K fd\mu$ in $L^2$ if $\mu$ is an ergodic measure

That's right

This we will prove next class using an even stronger ergodic theorem

philmcole

Hello. Does anybody have some recommendations for reading material on ODEs? I'm searching for a comprehensive and well structured book which doesn't lack of detail and explains the concepts well (not for some loose Wikipedia articles).

That's the same as saying $\mu_n \to \mu$ in the weak topology, I think, where $\mu_n$ is the average delta measure along a length $n$ orbit of $\phi$

Yeah that should be right

7:52 PM

I need to get some chemistry done

5

Faust

$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2y)}{x^2+y^2}$

Oh come on

l m a o

Faust

L seems to equal 0 yes?

Dattier

Hi,

A sequence of real functions $f_n \in C^2(\mathbb R)$ with $\exists M>0, \forall n\in \mathbb N, f_n '' \leq M$, and the sequence simply converge to $g$.
Is-it true that $g$ is continuous ?

Dattier

8:14 PM

The answer is true, but why ?

Year of No Light - Perséphone

►

This is very good

8:36 PM

Hey @Mathein!

Hi @Daminark

How's it going?

Pretty good, thanks. I've got a bachelor thesis advisor I wanted, though we don't have a topic yet

And for you?

Nice, do you have some ideas for the topic already?

Hey @Alessandro! And I'm just waiting on one of my classes to upload the pset yet

8:45 PM

Hi @Dami

In the meantime I have no purpose in life

Really it's just like, our prof in functional is already a couple days late in uploading the pset and it's prob gonna be heavy so I want to get started on it as soon as it comes up

What are you doing now in functional analysis?

Hhey

I'm probably going to do something related to number theory. Probably something that involves both algebraic and analytic techniques (I'm taking a course on modular forms before I'll start on my thesis and a second course on algebraic number theory.) I'd like to do something with elliptic curves as well

hey hey hey

8:46 PM

@Alessandro right now we're doing ergodic theory

@Daminark your functional analysis course sounds pretty advanced

I want to look at number theory again at some point

I see, sounds cool @Mathei

I'll probably do some descriptive set theory for my thesis

@Mathein yeah it's doing good stuff. I found that one of the students in the class is TeXing notes for it

I know next to nothing on descriptive set theory, but what I've seen was pretty cool

8:47 PM

I have my copy of Iwaniec-Kowalski lying around gathering dust on my bookshelf

I feel bad about that book

(I have classmates doing it but like, I know a grad student who has a webpage and is uploading notes to it)

math.uchicago.edu/~dsli/notes/math313_w18_notes.pdf

Tobias Kildetoft

@MatheinBoulomenos Have you read mathoverflow.net/questions/290776/… ?

@Tobias no, thanks for the link

Hello, can someone explaine me why we have this: Let $X$ be a Banach space and $f,g\in X'$ then for every $\lambda\in\mathbb{R}$:$$ \sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f,y\rangle=\sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f-\lambda g,y\rangle\leq \sup_{\|y\|=1}\langle f-\lambda g,y\rangle,$$

Hasn't uploaded today's notes but today we briefly talked about a generalization of Klein-Milman from normed spaces to locally convex spaces, defined ergodic measures, and showed that a measure is ergodic iff it's extremal in the space of $\phi$-invariant probability measures. Except for one lemma, that $\frac{1}{n}\sum_{k=0}^{n-1} f\circ \phi^k$ converges in $L^2$ to $\int_K fd\mu$ for any $f\in L^2$ (e.g. any continuous function) if $\mu$ is an ergodic measure

@Vrouvrou so if $y\in \ker(g)$, you know that $\langle f,y\rangle = \langle f-\lambda g, y\rangle$, right?

8:52 PM

yes my problem is with $\leq$

and this: that is, $$\sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f,y\rangle\leq \|f-\lambda g\|,~\forall\lambda\in\mathbb{R},$$

@Daminark

So in one case we're trying to take a sup over the set of points in the kernel of $g$ such that $\|y\| = 1$

In the second case we're also allowing a larger set, namely anything with norm 1, without restricting to the kernel of $g$

So the sup is at least as large

i don't understand you

Okay so if $A\subset B$, then $\sup A \le \sup B$, right?

we have for $y\in\ker g$, $\langle f-\lambda g,y\rangle=\langle f,y\rangle$

@Daminark yes it is right

Okay, so take $\{\langle f-\lambda g, y\rangle : y\in \ker(g), \|y\| = 1\} \subset \{\langle f-\lambda g, y\rangle : \|y\| = 1\}$

This is true because every $y$ in the first set satisfies $\|y\| = 1$, so it is in the second set. We're removing conditions that we require so we get a larger set

9:01 PM

i understand

and how we get this $$\sup_{\substack{y\in\ker g\\\|y\|=1}}\langle f,y\rangle\leq \|f-\lambda g\|,~\forall\lambda\in\mathbb{R},$$

$\sup \{\langle f-\lambda g,y\rangle : \|y\| = 1\} = \|f-\lambda g\|$

So by the above inequality we're done

why $\sup \{\langle f-\lambda g,y\rangle : \|y\| = 1\} = \|f-\lambda g\|$