Mathematics - 2015-05-19 (page 4 of 6)

I just threw out my qualifying exam prep folders from 1974-75, @MikeM. I thought about scanning you some of the topology questions, but they're beneath you now. :P

iwriteonbananas

in the mayer vietoris sequence, the map $H_n(X_1) \oplus H_n(X_2) \to H_n(X)$ is $i_* + j_*$, where i,j are the inclusions, right?

Balarka Sen

You should have sent some of them to me, @Ted :( But I'll find them online anyways.

yeah, @iwriteonbananas

iwriteonbananas

kk

Mike Miller

Don't be silly, @Ted, they hadn't invented topology back then.

Ted Shifrin

No you won't, @Balarka. I made up all sorts of concrete alg top/diff top questions.

I'll tell Rob Kirby you said so, @MikeM. He was one of my examiners.

Balarka Sen

3:24 PM

ah, you mean it was made by you.

I see.

user147690

@octatoan Hahahahahaha

octatoan

What?

Paul Plummer

Click the little arrow

user147690

@octatoan I am so tired that I found that really funny

user147690

On that note I should sleep

user147690

3:26 PM

Ta ta

octatoan

Oh. G'night Alex, ta ta :)

See yah

Night, @AlexC.

See yah

Surprisingly, @Ted, Kirby will not be in my mathematical genealogy, despite his overarching influence.

Ted Shifrin

3:27 PM

No question of the latter.

Mike Miller

Morally we are all Kirby's children.

Balarka Sen

ps, @Mike, did you see what I pinged you about homology of infinite cyclic cover of knot complements and cech homology of "knotted" solenoidal spaces?

d'you think anything can be done with it?

No to the former.

okay.

@MikeMiller Not I.

3:30 PM

Fair enough, son of Chern

Balarka Sen

I wonder if something more general can be said about homology of universal cover of spaces and cech homology of solenoidal spaces obtained from taking inverse limit of all the (branched) covers.

Mike Miller

I don't know what those words mean.

Balarka Sen

which word, in particular?

Mike Miller

"Solenoidal space". Do you just mean an inverse limit of spaces?

Balarka Sen

yeah, inverse limit of covering spaces with bonding maps being covering maps.

Mike Miller

3:36 PM

OK, I read the thing you linked. I don't see much reason it should have anything to do with the Alexander polynomial and I vehemently disagree that Cech homology is easier to handle.

It behaves well for a wider class of spaces, but that does not make it easy for a human being to handle.

Balarka Sen

well, ok.

I guess I am just interested as it relates to the kind of things I like to think about :P

Mike Miller

Sure.

octatoan

@BalarkaSen do you know any category theory?

Balarka Sen

a bit

Vrouvrou

@TedShifrin hello, Are these two norms equivalent : $||u||_{L^p}+||\nabla u||_{L^p}$ and $||u||^p_{L^p}+||\nabla u||^p_{L^p}$ please thank you

picaposo

3:46 PM

Guys simple question: Is a lattice graph of an arbitrary number of sides n, always possible? I know triangular, square, pentagonal and hexagonal lattice graphs exist along with all polygons that can be constructed from a combination of them. But what about polygons with a prime number of sides? Can a lattice graph still be constructed in that case? I don't need a proof, just a confirmation.

Ted Shifrin

@Vrouvrou: The second isn't a norm, is it?

octatoan

Rudin chapter 1 done.

Winter is coming.

Does anyone know of any normal-looking non-archimedean subsets of $\mathbb{R}$?

Of the same cardinality?

Mike Miller

what's a non-archimedean subset

octatoan

A subset $S$ for which it is not true that there always exists integer $n>0$ such that $nx>y$ for $x,y \in S$ and $x, y > 0$

It's true for $\mathbb{R}$ and it's called the archimedean property.

Mike Miller

I don't understand. there are no non-archimedian subsets of $\Bbb R$. do you mean such that $nx$ is also in $S$?

Vrouvrou

4:00 PM

@TedShifrin right i'm sorry the norm on W^{1,p} is $||u||_{L^p}+||\nabla u||_{L^p}$ and not $||u||^p_{L^p}+||\nabla u||^p_{L^p}$

octatoan

Yes.

Mike Miller

gotcha

octatoan

Is there such a subset?

Paul Plummer

If you require $nx \in S$ then singetons are pretty normal looking

octatoan

"Of the same cardinality"

Yes

Paul Plummer

4:01 PM

Of the same cardinality of $\mathbb R$?

Mike Miller

I wouldn't be surprised if such subsets existed; I wouldn't be surprised if it was independent of ZFC

octatoan

Yes

Vrouvrou

@TedShifrin please see this if you have an idea math.stackexchange.com/questions/1284540/a-counter-example

Mike Miller

i would be a little surprised if you could actually prove non-existence

octatoan

Hm.

Just wondering aloud, is all.

Back to work.

Paul Plummer

4:02 PM

Nah, pretty sure it shouldn't be that difficult

Mike Miller

well, actually, no, you're going to want something like density

$[0,1]$ doesn't satisfy it eg

Paul Plummer

Subtract all rational multiples of $\pi$, except $\pi$ from $\Bbb R$, then take $y=4$

Vrouvrou

@TedShifrin i want to find for example a sequence which converge in $C_{\theta}$ but not in $L^{p^*}_1$

Mike Miller

oh

meh

just need that integral multiples don't exist

octatoan

Density.

Mike Miller

4:05 PM

that's dense

octatoan

I've heard that word a lot.

Paul Plummer

True,

Ted Shifrin

@Vrouvrou: I don't think enough about this sort of stuff to be helpful, sorry. I'm working on cleaning out my office now.

Mike Miller

I think the above example has every property you could possibly want

octatoan

Yeah.

Thanks.

Vrouvrou

4:06 PM

@TedShifrin ok , do you know someone who can help me ?

Paul Plummer

Or you can just consider $\Bbb Q \cup \{ \pi \}$, I guess

(square root 2, anything that is not rational

Ted Shifrin

@Vrouvrou: That's why you post on Main. You get all sorts of people who might be able to respond.

Paul Plummer

Or consider $\Bbb R / \Bbb Q$, choose one element from each coset, so that it ends up dense (this uses choice). This is an instance of $G_1 (D,D)$, where $D$ is every dense subset of $\Bbb R$ :) @MikeMiller

Then you have no two elements elements have this "archimedian" property @octatoan

even $x,x$ since $nx$ is not in the set when $n>1$

(well it is an instance of the game if you choose from only countably many, but you can extend the game to continuum many innings)

octatoan

4:24 PM

I don't know that much group theory yet. :(

Vrouvrou

@TedShifrin i'm sorry but can you explain me this in french :Your first space contains only continuous functions by definitions while your second doesn't. Take u to be the indicator of some subset of Ω and you have your counterexample

octatoan

u is the counterexample

@Vrouvrou

does that help?

Paul Plummer

You don't need group theory, this is a quotient construction and it basically says that $x,y$ are in the same coset(equivalence class) if and only if $x-y$ is rational. @octatoan

octatoan

Oh, this was an exercise in the first chapter of Aluffi!

Paul Plummer

That sounds about right

octatoan

4:29 PM

Yeah, about equivalence classes and everything. Thanks.

Vrouvrou

@octatoan i juste don't clearly understand it

Paul Plummer

Although you should change "don't know much group theory"... :D

octatoan

I haven't done all of Aluffi.

.Just the first chapter.

Paul Plummer

I know, just saying you should learn more group theory (although just learn whatever you want)

octatoan

I'll do Artin now, then Aluffi.

Yeah, I like algebra.

Vrouvrou

4:30 PM

please +1 for this question if i have only -1 i will can't ask a question again :mathoverflow.net/questions/207047/…

octatoan

@Vrouvrou that is not a research level question I think

Paul Plummer

Is that research level? I am not going to upvote a question that should not be on MO

octatoan

Looks like an exercise.

Vrouvrou

no it is a research level question there are new spaces and there is no this in papers

octatoan

Oh, well

Sawarnik

4:33 PM

@robjohn You here?

octatoan

I'm more concerned about why the integral doesn't have the curl at the top on MO

(No pun intended.)

Vrouvrou

@octatoan the person who gives me the counter example "Hachino " delete it because it is not good

octatoan

Okay, but this is getting sort of spammy now.

Vrouvrou

it means that this question is not easy

r9m

@Chris'ssis okay ,, I have a way .. no gamma functions nothing ... 1st year calculus and induction :)

3

Vrouvrou

4:36 PM

i deleted it on Mo i recived -2

pfff

octatoan

"spammy"

Chris's sis

@r9m Great. :-) I think I also have a pretty elementary way.

Vrouvrou

why "spammy" i don't understand @octatoan ?

Chris's sis

@r9m that is using $2p$th derivative of the (cosh(x))^n at the point $x=0$.

r9m

@Chris'ssis EXACTLY SAME APPROACH we had then :D

Chris's sis

4:41 PM

@r9m :D:D:D:D:D:D:D

@r9m Take care about Knuth's problem, it is wrong. I corrected it and sent it to AMM. They accepted my solution. :-)

@r9m the story with the Catalan number generating function is wrong.

r9m

@Chris'ssis wow! okay! :)

Chris's sis

@r9m One page solution. (and 2 tiny rows I think one the other page):-)

r9m

@Chris'ssis omg! Nice :)

Chris's sis

@r9m :D

@r9m Don't forget to ask me to show you my solution in case they don't publish it (hope it's not that case).

r9m

@Chris'ssis sure! :-) I wait eagerly :D

Chris's sis

4:46 PM

(after they publish the solutions to this problem)

Remember me

Hello @r9m

Remember me?

r9m

@Rememberme hi

Remember me

@r9m you are still in CMI?

r9m

@Rememberme yes

Remember me

So where are you going to do your MSC@r9m

r9m

4:52 PM

@Rememberme Tifr Bangalore

Vrouvrou

@octatoan you do -1 for my question so give me an idea !

Remember me

You beauty!!! TIFR

So we might hope to meet some day :D@r9m

evinda

Hi @cirpis @DanielFischer There are given 3 vectors v_1, v_2, v_3, of dimension m, of which the elements are real numbers.
I tried to write a quicker than $O(m^2)$ algorithm that finds if there are three numbers , one from each of the vectors v_1, v_2, v_3, that have sum equal to 0.
Is the following idea right?

We look at the elements that are at the midpoint of our interval. If v_1[mid]+v_2[mid]+v_3[mid]=0, we are done. If v_1[mid]+v_2[mid]+v_3[mid]>0, we decrement the index of the last vector by one. If v_1[mid]+v_2[mid]+v_3[mid-1]>0, we decrement the index of the seond vector by 1. If v_1[mid]+v_2[mid-1]+v_3[mid-1]>0, we decrement the index of the first vector by 1. If v_1[mid-1]+v_2[mid-1]+v_3[mid-1]>0, we decrement the index of the third vector by 1 and so on.
And similarly if v_1[mid]+v_2[mid]+v_3[mid]<0.

r9m

@Rememberme sure :)

Remember me

Well where is tifr in blore I don't know the address sadly@r9m

octatoan

4:55 PM

@Vrouvrou I didn't.

r9m

@Rememberme It's in Yelahanka Satellite Town

Remember me

Ahh yelahanka pretty far away from my place a 2 hour drive from my house @r9m

Your course is in pure maths or applied maths @r9m

r9m

@Rememberme pure math .. mainly focusing on mathematical analysis

Remember me

Ahh.... Topology also??

Nice analysis you surely going to have fun

r9m

@Rememberme the main focus is on functional analysis and partial differential equations

LeGrandDODOM

5:01 PM

@Chris'ssis Can you compute $\sum_{n=1}^\infty\frac{(-1)^n}{n+1}H_n$ ?

evinda

@RickDecker Hi!!! Do you maybe have an idea about the following?

There are given 3 vectors v_1, v_2, v_3, of dimension m, of which the elements are real numbers.
I tried to write a quicker than $O(m^2)$ algorithm that finds if there are three numbers , one from each of the vectors v_1, v_2, v_3, that have sum equal to 0.
Is the following idea right?

We look at the elements that are at the midpoint of our interval. If v_1[mid]+v_2[mid]+v_3[mid]=0, we are done. If v_1[mid]+v_2[mid]+v_3[mid]>0, we decrement the index of the last vector by one. If v_1[mid]+v_2[mid]+v_3[mid-1]>0, we decrement …

Chris's sis

@LeGrandDODOM elementary. First arrange a bit the numerator and write $H_{n+1}-1/(n+1)$. Then use Cauchy product, cleverly. Q.E.D.

LeGrandDODOM

@Chris'ssis tell me more

Paul Plummer

Getting a -1 does not mean it is easy and just because it may be hard does not mean it is research level

@Vrouvrou

Vrouvrou

@PaulPlummer it is a research level

Chris's sis

5:04 PM

@LeGrandDODOM No need for that arrangement. See here, just Cauchy product math.stackexchange.com/questions/292973/…

@LeGrandDODOM I discovered it my research independently, but I knew it also appeared there.

Paul Plummer

How do you know a counter example exists? @Vrouvrou

Chris's sis

@LeGrandDODOM you should also see @robjohn's way. Not sure now where that link is.

Karim Mansour

hi guys

Chris's sis

@KarimMansour Hi

Paul Plummer

I upvoted your math.se question, maybe it will get more attention, but I don't know enough to say if it is a good fit for math.se, I have a feeling it is not, but idk @Vrouvrou

robjohn

5:07 PM

@Chris'ssis are you thinking of this answer?

Chris's sis

@robjohn Yes, precisely! :-)

Vrouvrou

because i take a space $L^{p^*}_{\alpha}$ and i foud a condition on $\alpha$ to have a continuous embedding from C_{\theta} to L^{p^*}_{\alpha} it was $\alpha >1$ so i take $\alpha=1$ @PaulPlummer

Chris's sis

@LeGrandDODOM see the link above robjohn just showed.

Ted Shifrin

hi @Karim salut @LeGrandDODO

hi @robjohn

robjohn

@TedShifrin good morning/afternoon.

Ted Shifrin

5:09 PM

LOL ... something like that, @robjohn :)

Mike Miller

Morning @Ted, @robjohn

LeGrandDODOM

salut @Ted.

Ted Shifrin

We've been through this already, @MikeM. I said goodnight hours ago :P

Mike Miller

Was I there for that?

Paul Plummer

ta ta @TedShifrin as the British, and the Indians seem to say

Ted Shifrin

5:11 PM

How would I know, Mike? :D

robjohn

@MikeMiller I was at UCLA for a midterm on Friday. I was pretty tight-scheduled, and on the way home, a got a flat tire.

Ted Shifrin

oh no, @robjohn

I always worry that something like that will happen to me on the way to the airport at 6 AM

Karim Mansour

robjohn grad student?

Mike Miller

@robjohn Ah, that's awful.

robjohn

@KarimMansour nah... I was a grad student over 30 years ago.

Ted Shifrin

5:13 PM

he's just young compared to me, @Karim

octatoan

Anyone here good at German and reading Fraktur?

Mike Miller

@Ted: I remember now.

Karim Mansour

oh I see must be a prof or something

Paul Plummer

No one is good at reading Fraktur :D

Ted Shifrin

Who would want to read much fraktur?

Mike Miller

5:14 PM

The only German I know is "Arsch ersetzer".

octatoan

Artin's first chapter.

evinda

Hey @robjohn
Did you see my question?

Ted Shifrin

oh, his little quote?

octatoan

Yes.

Ted Shifrin

I don't have the book here, though.

Karim Mansour

5:14 PM

isn't that ass hole or something @MikeMiller?

Ted Shifrin

It's packed somewhere.

Mike Miller

No, @Karim, don't be crass.

octatoan

Endlich wird alles . . . ?

I know a little German, it won't suffice for all of that.

Ted Shifrin

yeah, mine either at this point

Karim Mansour

I learned german but I don't like it

Ted Shifrin

5:15 PM

DanielF is a German speaker.

Karim Mansour

weird language

octatoan

^ I went to Germany last year for a month for a school program. When I came back everyone was curious why I was ending my sentences with prepositions.

Karim Mansour

haha

LeGrandDODOM

5:31 PM

@Chris'ssis well that Cauchy product is non standard for none of the series converges absolutely

Chris's sis

@LeGrandDODOM You find the needed theorem in the Joseph Edwards's old book such that you don't need to be concerned at all.

Ted Shifrin

@octatoan No, you should have ended your sentences with past participles.

octatoan

Haha.

Instant Yoda?

Chris's sis

@LeGrandDODOM There are many ways to get there. For instance you can use the generating function of the harmonic numbers combined with a bit of integration and Taylor series ...

Karim Mansour

I want to learn like 3 languages more I already know 3 :D

octatoan

5:44 PM

Duolingo is your friend.

David Zhang

Hey everyone, quick question. When you have two vector spaces V and W, what's the standard way to denote the set of all linear maps V -> W?

Karim Mansour

hey @TedShifrin

so I was just looking at this example in DF

1 moment typing

Ted Shifrin

@DavidZ: $\text{Hom}(V,W)$ or $L(V,W)$.

David Zhang

@TedShifrin Is it usually a plain italic $L$, not a script or double-struck one?

Mike Miller

@Ted: $P$ is a complex polynomial. Assume $|P(z)| \leq M$ for all $z$ in the unit circle. What's an example of one such that $M < \sum |a_i|$, where these are the coefficients?

Ted Shifrin

5:51 PM

It can be whatever you want, @DavidZ. Some people use each of these.

Karim Mansour

We show that the center center of $D_8$ is subgroup ${1,r^2}$ we know that $r^3$ isn't in the center well they didn't show it but I did the following computation $sr^3 = r^{-3}s = r^{-2}r^{-1}s = r^5s = rs \neq sr^5 = sr$ but there is other way to show this easier way I guess

I can feel it in my gut

Mike Miller

Nevermind. $(z-1/2)^2$.

Ted Shifrin

I was about to say it shouldn't be hard, @MikeM, but I don't have one off the top of my head.

octatoan

The set of untimely June insects is dense in my room.

Mike Miller

No, that doesn't work.

Karim Mansour

5:52 PM

what do you think @TedShifrin?