Hello

Afternoon

@Jacobadtr how are you?

@Charlie I'm ok thanks, how are you?

@Jacobadtr fine, fine, thanks.

@Charlie I have to say, i'm very glad to have found this website

@TobiasKildetoft @JulianKuelshammer Hey guys

Do you know any references for geometry representation theory?

Atm I find Ginzburg and Criss too high level

@Jacobadtr really, why?

@Charlie I've changed courses at uni and I need to do a lot of maths revision. This site seems to attract lots of people who know what they're talking about, and the formatting for writing out equations is much easier to use than others i've seen before.

@Jacobadtr Hello

@BenjaLim Hi there, how are you?

not bad

Something I would like to throw out there for anyone who might read this - can anyone recommend a good book for an introduction to set theory?

@Jacobadtr TBH I never really learned set theory

just kinda picked it up along the way

I had a flick through some of my stepdads engineering reference books and a fair bit of the set theory stuff looks very familiar. It just wasn't called set theory in school.

Right but tbh I think having an understanding of naive set theory in most cases is probably enough @Jacobadtr

@Jacobadtr Halmos Naive Set Theory?

I am intrigued to learn a bit more about it after reading about Russel's paradox

@GustavoBandeira Is that a name of a book?

Halmos is the name of the author

@Jacobadtr Halmos is the name of the author.

@BenjaLim Yep. I'm reading Halmos' book by now.

@GustavoBandeira Thank you, what's the significance of 'naive'?

@Jacobadtr It means "the set theory that we all kinda need to understand in order to do mathematics"

@Jacobadtr It's a first step, doesn't have all rigor.

Ok great, that sounds like a good place for me to start :)

@BenjaLim What you mean? I tried to express that I am currently reading it.

nvm

@Jacobadtr But don't keep on recommendations, it's cool to look other books and see what book "fits" your mind.

@Jacobadtr There is always books.google.com.au/…

Don't claim to have read it but know someone who is

@BenjaLim That's nice to have available online

@GustavoBandeira I'll take a trip to the library before I go and pay any money for the book

@Jacobadtr That's a good idea.

@Jacobadtr I don't really know so much about set theory, if you want to ask for references on algebra that is another thing

@BenjaLim What are you studying in algebra?

@JulianKuelshammer @TobiasKildetoft Any of you studied things like Borel - Weil - Bott before? I got to say when someone was explaining it to me the other day there were like 15 terms I did not understand

@MarianoSuárez-Alvarez Hey

@MarianoSuárez-Alvarez Do you have any references for geometric representation theory?

@MarianoSuárez-Alvarez I find Ginzburg and Criss too high level atm

@BenjaLim yeah, but only in the algebraic groups setting

@TobiasKildetoft ok

@TobiasKildetoft You recommend springer's book?

@BenjaLim for an introduction, yeah

@MarianoSuárez-Alvarez Do you know where I can find out about line bundles over $\Bbb{P}^1$/ classification of them? @TobiasKildetoft

possibly supplemented by Humphreys

that book is crazy

I have heard a lot of good things about Borel, but that sems a bit more advanced

(and of course, at some point one needs to read Jantzen)

yea

@TobiasKildetoft I can understand the first chapter of springer's book

for a more "big picture" look, my advisor has written some nice notes

though I am not sure how easy those are to find, as they are part of the notes from asummer school

@TobiasKildetoft link?

@TobiasKildetoft you have a link of those notes?

@BenjaLim No, sorry.

@JulianKuelshammer It's ok then

got them?

yea thanks @TobiasKildetoft

@TobiasKildetoft @JulianKuelshammer Do you know of any flavour of representation theory that is like homological?

@BenjaLim what do you mean "like"?

FUCK SHIT DO I NEED TO KNOW SERRE DUALITY

@TobiasKildetoft I mean homological in flavour

@BenjaLim hmm, not sure precisely what you mean by that

@TobiasKildetoft ok perhaps I'm not being precise

obviously, a lot of homological algebra goes into the representation theory

@TobiasKildetoft yea I see resolutions in your document

chapter 1 in Hartshorne is generally held to be suboptimal as an introduction to varieties

Is it maths that you study?

how come? @TobiasKildetoft

@Jacobadtr ?

@BenjaLim it tends to be lacking in motivation compared to most other texts

@TobiasKildetoft I agree. But if you know commutative algebra quite a few of the problems can be reduced to commutative algebra.

ok

@BenjaLim Just wondering if it's maths you study, or some other subject that relies heavily on maths

So I had a go at concrete calculations

@Jacobadtr No I study straight maths and nothing else

no physics nothing

@BenjaLim At what level?

@Jacobadtr Still an undergrad

@TobiasKildetoft thanks

@BenjaLim At the moment I do something related to the A-infinity structure on the Ext-algebra of a module. I think that counts as homologically flavored :-)

it certainly does :-)

what is an A-infinity structure? @JulianKuelshammer

hah

I want to see this

/me grabs the popcorn

@TobiasKildetoft Informally speaking an A-infinity algebra is a graded algebra (with a differential) that is not associative, but associative up to homotopy. The difference is controlled by a map $m_3: A\otimes A\otimes A\to A$ (and there are maps $m_4,m_5,\dots$ controlling the lower ones)

neat

Keller has a nice survey paper: arxiv.org/abs/math/9910179

I will have to take a look at that at some point

are they in any way related to cluster algebras of type $A_{\infty}$?

not really :-)

ok

if $A$ is a differential graded algebra, $B$ a complex and $f:A\to B$ an homotopy-equivalence, then you cannot turn $B$ into an algebra using $f$ (as you could if $f$ were an isomorphim of graded vector spaces) An $A_\infty$-algebra structure is what you can construct on $B$.

Also a nice way of saying it. Has somehow the flavour of Karoubian envelope.

hi

@GustavoBandeira how are you?

@somaye Hello! I'm fine, you?

@GustavoBandeira FEINE :) COME TO YM OK?

@somaye Oky.

@GustavoBandeira MY NET IS LOW

@somaye =/

@GustavoBandeira TALK HERE?

@somaye Yep.

How are things?

ALMOST GOOD

@GustavoBandeira

@somaye You disappeared these days.

@GustavoBandeira RECENTLY I AM TOO BUSY

WITH MY JOB

@somaye Oh, got it.

FROM 8AM UNTILL 7 PM

ARE YOU THERE?

@somaye Yeah

Just looking some stuff.

Is @skull here?

@GustavoBandeira oK I AM READING AN ARTICLE TOO

:)

@somaye The same you told me last time?

@GustavoBandeira NO ANOTHER ONE

Morning

@Arkamis HI ED

@Charlie Allo! I live! sort of

@Arkamis we missed you, long time no see

i've been busy

Still am. Just taking a break to eat a burrito

@Arkamis aaah :)

@GustavoBandeira I DOUBTH

@somaye ?

@somaye Be back later. I'm gonna take a nap

See you.

@GustavoBandeira OK

@Charlie I am now.

@skullpatrol chilling in the chill

@skullpatrol how are you?

@Charlie Fine thanks.

@skullpatrol good :)

@Charlie My wifi seems to be acting up again >8(

@skullpatrol >8(

I read that as "My wife seems to be acting up again"

@Arkamis hahaha

ERMAGHERD.

@PeterTamaroff hi Pedro

@Charlie Hello, M.

Hi @Charlie

@PeterTamaroff how is it going, Pitt?

@skullpatrol hi skullpatrol

@Charlie Rainy. You?

@PeterTamaroff same

a rainy night

Hi @somaye how are you?

@somaye I'm back

Hi rob

Been a long time

@robjohn HI! HOW WAS THE MOVIE, ROB?

@Srivatsan hey there.. how've you been?

I am doing ok.

Been a bit busy. :)

@Charlie why the shouting? The movie was excellent!

@Srivatsan good busy, I hope

Well, all kinds of busy :).

So how are you doing?

@robjohn AMAZING

@Srivatsan pretty well. Lilly passed away last month from torsion, and we have a new dog Dolly.

Oh, sorry to hear that

How old was she?

@Srivatsan I was quite depressed for several weeks.

@Srivatsan she was 13, but she was chasing rabbits on her last day. She was amazingly healthy for 13.

@robjohn Depressed or sad?

It must be hard for you.

Wow, I didn't know they (could) live so long.

@GustavoBandeira both.

@Srivatsan small dogs live longer, usually. Lilly was not a small dog, though

Did you know for 13 years as well?

know her*

@Srivatsan we had her for 12 years. she was just over a year when we got her.

I just saw her photo, with Smoky. She's beautiful.

@Srivatsan Yeah, she was Smoky's mom

I know you have mentioned Lilly many times, but I hadn't known about Smoky.

@Srivatsan we found smoky at the side of a road struggling with a broken pelvis.

Oh, I do know her(?). I was around when you found her and nursed her.

..back to health, I mean.

Is she doing ok?

@Srivatsan She is doing very well, but she misses Lilly

@robjohn Man, I am not sure what I'll do when my dog passes away.

Sorry to hear that, too. Hope you're well now.

@PeterTamaroff It was quite a blow, and very unexpected, due to her amazing health.

Greetings

@robjohn What happened?

@Chris'swisesister "...outlanders?"

@PeterTamaroff once in a while I still get a bit depressed.

@robjohn Well, 13 years is a long time. =/

@Chris'swisesister hies

@PeterTamaroff she got gastric torsion and it was fairly quick

@robjohn Oh.

I kinda saved my dog once, she had this infection in her uterus. We had her operated on and all her uterus and stuff and to be removed.

guys, why is that when I enter 2^(i) in my calculator , it doesnt give me an answer :( i feel depressed

@robjohn If you have the guts, google "Pyometra". =D

@PeterTamaroff it's an automatism. Usually, when the teachers from the school near my house see me, all together shout "Greetings!". I don't know them very well, but they know me pretty well. So, I said to myself to proceed the same way as they do to me.

@Charlie hello!!! How is it going?

clearly

@PeterTamaroff we had a large dog before and we caught the torsion in time, so he survived that. Lilly's was not caught in time.

@anon Oh, show that to Brian.

@robjohn How does one catch that?

k, @BrianMScott

@Charlie hello! How is it going? :D

hey, haven't seen @sri in a while

@PeterTamaroff Jake never whimpered, so when he whimpered, we took him in immediately

@anon XD

Thank god for the red circle.

@vvavepacket Try $\exp(i\log 2)$

My calculator gives $0.769238901363972 + i 0.638961276313635$ for $2^i$

@robjohn Oh, I see. I found my dog laying down, and she had some bloody discharge... well, you know. Pretty awful.

tnx peter

@PeterTamaroff yeah. Lilly had been throwing up, but she sometimes got into things that she shouldn't, so it didn't trigger an emergency response. Not until I saw the bloat, and by then it was pretty much too late

@robjohn Why doesn't the function $\phi(k) =e^{i\pi k} \sin \left(\frac{\pi}{2}k\right)$ seem to work for Ramanujan's Master Theorem: mathworld.wolfram.com/RamanujansMasterTheorem.html ?

@robjohn Anyways, I read a nice proof that $\varphi(a)\varphi(b)=\varphi(a\cdot b)$ when $(a,b)=1$.

@Chris'swisesister fine fine, you?

@Argon "Master Theorem". LAWL

hi @anon

hello

@Charlie Not in the best shape, but it's OK. I'm creating some limits involving hyperfactorial.

@Chris'swisesister interesting

Hi @robjohn, is there any way I could ask you a quick question in private?

@PeterTamaroff $\varphi(a)$ being the number of relatively primes below $a$

@robjohn Yeah.

@Charlie maybe you like this question: math.stackexchange.com/questions/162142/…

It is: $(a,b)=1$ then $$(a,y)=(b,x)=1\implies (ax+by,ab)=1$$ and the fact that if $(a,b)=1$ then $$x\equiv x'\mod b,y\equiv y'\mod a\iff ax+by\equiv ax'+by'\mod ab$$

@PeterTamaroff I wouldn't call CRT quick, but that is the way I usually see it.

@Chris'swisesister cool

@robjohn Yes, I know.

@PeterTamaroff which is pretty much what you have there

@robjohn Heh, kinda, yes. =)

So if we let $x,y$ range over a reduced set of residues $\mod b\;\mod a$, $ax+by$ ranges over a reduced set of residues $\mod ab$. $\blacktriangle$.

@Charlie Well, I'm really glad to "hear" you say "cool". :-)

@Chris'swisesister :D hehe why?

@Charlie there are not too many that like these questions. :D

@Charlie btw, I also received a downvote in the past.

@Chris'swisesister it doesn't mean it's not good

@robjohn How often have you used Euler's $f(\lambda x)=\lambda^p f(x)\implies x\cdot \nabla f(x)=pf(x)$?

Seems to come in handy when using Lagrange Multipliers.

I am learning about those now! =) @robjohn

@Charlie yeah, that's right.

Now, I have $0=\nabla f(x)+\lambda g(x)$, and $f$ is homogenous of degree $2n$ and $g$ is homogeneous of degree $2$, so I can change that to $0=2nf(x)+2\lambda g(x)$, yes?@robjohn

:9641045 Woosh! Five seconds left!

@PeterTamaroff ":) phew!

A cute question from a math contest: Test for convergence $$\sqrt{1!\sqrt{2!\cdots\sqrt{n!}}}$$

hey @anon

hey

How have you been?

pretty good

Sorry, had to step out. I just dropped in to see what this chatroom looks like. :)

@Chris'swisesister Since $\sum \frac{n\log(n)}{2^n}$ converges, I'd say that converges.

@Srivatsan ?

@robjohn yeah. :-)

@Chris'swisesister That's not hard, man. You know better!

@Chris'swisesister So even replacing $n!$ by $n^n$ it should still converge

@PeterTamaroff is it important to be hard? It's important the way (imo).

@robjohn didn't get your question (mark)..

@Chris'swisesister Don't get cheesy now.

@Srivatsan you don't know what this chatroom looks like?

Well, I was wondering more like: Are the same people around? etc.

@PeterTamaroff all questions must be appreciated. :-)

@Srivatsan some are; sadly, some are not.

@Chris'swisesister exactly :)

@Charlie :D

@Chris'swisesister Really?

@robjohn Um, I miss this place too, but real life intervenes at the moment. :)

@Srivatsan Are you saying this is not real? snif

:)

@PeterTamaroff what sorts of thing are you using them for?

@robjohn "Them"?

command[recall]: failed

@PeterTamaroff That was linked back to the Lagrange Multipliers comment

By the way, @rob, I have a geometry question related to a paper I am writing. (If I know the answer to the question, it'll simplify the explanation somewhat. Otherwise, we have a backup plan that kind of gets around it.)

Question:

Could you take a look at it for me?

@Srivatsan what question?

@robjohn Give me a sec, let me type it up.

@robjohn do you think the limit might have a nice closed form?

Mathematica doesn't help.

@Chris'swisesister I don't see that it does, right off hand

ok

Take a (hollow) d-dimensional cylinder of unit radius & infinite height (extends in both directions), and let x be a point on its (curved) surface. What can we say about the surface area of the intersection of the ball B(x, r) and the cylinder? Think of r approaching 0; I want the just the first two terms in its taylor expansion.

@Chris'swisesister The log of the limit would be $\sum\limits_{k=1}^\infty\frac{\log(k)}{2^{k-1}}$

@Srivatsan You mean the $n$-sphere $\times\mathbb{R}$?

@robjohn Oh, because in Lagrange I usually get $0=\nabla f(x)+\lambda \nabla g(x)$, but with homogeneous $f,g$ I can change it to something involving $f,g$ only, upon dot producting by $x$ =P

As I said, the answer to this question is desirable, but not super crucial for our paper, so please don't spend too much time on it (unless you find the problem interesting, of course). :)

@robjohn Yeah, your $n$ would be my $d$, but yes, it's the product of a $d$-sphere and $\mathbb R$. The infinite height is just to remove the boundary effects.

@robjohn For example, take $f(x_1,\dots,x_n)=(x_1\dots x_n)^2$, and let's find it's maxima and minima over $\Vert x\Vert =1$.

Hey guys, when you learn math, do you memorize all the theorems and formulae, or do you just memorize the main things, or do you not memorize at all, instead focusing on understanding from the bottom up?

Then I set up $\nabla f(x)=\lambda\nabla g(x)$. We have $\nabla g(x)=2(x_1,\dots,x_n)$ and since $f$ is $2n$-hom we get $$2nf(x)=2\lambda $$

nevermind, i found this math.stackexchange.com/questions/3782/…

@Raindrop If you understand, you learn and you thus memorize. But memorizing doesn't mean understanding.

@Chris'swisesister which has a PolyLog answer which is approximately 2.7612068419574980332

Instead of memorizing, I prefer "internalizing".

@robjohn according to Mathematica, log of that limit is $1.01567$.

@robjohn yes, it's correct.

Basically, @rob, I think I will be able to do the surface area calculation in small number of dimensions (say, $d=2$), but I don't know how to write the integral in higher dimensions.

Colleagues.

@Lord_Farin Hi!

@Charlie Hello Charlie.

@Lord_Farin how are you today?

@Charlie I hit rock bottom earlier today (no elaboration). Now that I can only look up, things are improving quickly.

How about you?

@Lord_Farin I'm fine

@Srivatsan looking at it now

Sure, thanks, rob.

@Lord_Farin Hello there.

Have you seen this?

@robjohn if you arrange all terms of the sum on a square and then complete the square with the missing terms, the sum is exactly 2 times the sum of the terms that are to be found on the main diagonal.

@PeterTamaroff Hello. I have now.

What's special about it?

It motivates the definition of continuity in closure spaces.

@PeterTamaroff I see. (Not that I know anything about closure spaces, but they seem to be a generalisation of top. spaces.)

@Lord_Farin Not really, there is a one one correspondence between topological spaces and closure spaces, methinks.

I noticed that IMO syllabus is upper-division undergraduate while IPhO syllabus is lower-division undergraduate. No wonder it is harder to prepare for IMO than for IPhO. see fiitjee.com/campus/olympiad.pdf for syllabus

@PeterTamaroff It would seem that a general closure space does not necessarily admit the analogues of the union and intersection properties.

@robjohn, but yes, double sums is lovely as well.

I need some sleep now.

@Lord_Farin Come again?

@PeterTamaroff That the union of two "closed" sets needn't be "closed".

Anyone know if there is a simple way to calculate sums of the form $$\sum_{n=1}^\infty \frac{\chi(n)e^{inx}}{n^k}$$ for various characters $\chi$ and integers $k$, reals $x$

@Chris'swisesister which problem you talking about?

@Lord_Farin Hmm, yes $\overline{(A\cup B)}=\overline A\cup\overline B$.

@Lord_Farin Eh?

@robjohn Closed with respect to a general closure operator.

Like if I wanted to calculate $$\sum_{n=1}^\infty\frac{\chi(n)\sin(2\pi nx)}{n}$$

Similarly $(A\cap B)^\circ=A^\circ \cap B^\circ$

@Ethan hi, Ethan

@Peter @robjohn I'm considering closed as "in the image of a closure operator".

@Lord_Farin Closed is defined as $\overline A=A$ in closure spaces, isn't it?

@PeterTamaroff Yes, that's the same.