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12:02 AM
@Mike Miller I'd imagine what you want is in the appendix of the Big Mcduff-Salamon book, which I do not have in front of me. Is it in there, or is this not basic enough? I haven't ever worked through the Fredholm (or elliptic operators) parts of the places where I've seen it used.
 
@PVAL: I haven't taken a substantial look at that book yet. I'll try it. I could probably axiomatize lots of the analysis used wherever it shows up, but I enjoy the analytic details, and would like to understand them.
 
Perhaps Chapter 10 of this people.math.umass.edu/~sullivan/797SG/oh-floer.pdf is even better. (I've read other parts of this book and really like it).
 
Is it kinder than the big blue book? Admittedly I'm a little terrified of it.
 
Is the big blue book Fukaya et all? Oh's book is easier to understand than the big Mcduff-Salamon in my opinion.
Which means its relatively kind.
 
I meant McDuff-Salamon, Quantum Cohomology. I've never seen Fukaya's book. Sorry for the imprecision.
 
12:10 AM
I think its nicer.
 
Oh, I guess that's not blue, and they changed the name for the second edition? Anyway, thanks for the recommendation.
 
Easily the least kind big blue book in the area is Seidel's "Fukaya Categories and Picard Lefschetz Theory" though. I don't think I can understand more than a paragraph of that book (apparently its blue amazon.com/…)
 
Its description amusingly claims the emphasis is on simplicity.
 
@jeff just ot be a contrarian, I care quite a lot about divergent series! They're really quite common and sueful
 
@KevinDriscoll Useful for what? (just to play along with your contrarianism).
 
12:14 AM
I think the printed edition of McDuff Salamon is a lot better than the notes with Quantum Cohomology in the title by the way (though I still won't claim to understand most of this book).
 
@Jeff Usually when solving a difficult physics problem, it's impossible to find a closed-form answer. One could do numerics, but then you learn very little. If you do some kind of perturbation theory though, you find that quite often the sequence you get out of that does not converge. In fact they very often have a radius of convergence of 0.
 
@PVAL: The Quantum Cohomology notes are in printed form, I think it's the first edition of the J-holomorphic curves book. I heard the second edition was because the section on gluing was just wrong.
 
@Jeff Thankfully this doesnt matter because there are lots of ways of converting a naively divergent series into something meaningful
 
@KevinDriscoll Well, that's news to me. But you said the 'sequence' diverges. I'm asking about series.
 
@MikeMiller The book is much better than the notes available for free on say McDuff's website. I am not sure if that's the same as the first edition of the book though.
 
12:20 AM
I think it is. Size roughly doubled between editions.
 
@Jeff Same thing. I mean the sequence of partial sums diverges, so the series does as well
 
@KevinDriscoll I guess I'll have to stick with the advice here and say it fails the divergence test (i'll just rename it to the $n$th term test).
 
@robjohn @RandomVariable I made an update
4
Q: About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $

Chris's sis the artistHere is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known nice real analysis tools for calculating them (including here all possible sources in literature that are publicaly available). At some point I'll add my solution. It's a...

 
@Jeff Ya, sadly I dont have any pedagogical advice aboutt teaching this stuff to high school or college students. I learned all these tests in 11th grade. Promptly forgot all of them. And then looked them up when it became relevant again.
 
@Mike Miller I remember there being a proof in Gromov's paper which just says "apply Atiyah-Singer" and another which just says "use theorema egregium". So I wouldn't recommend starting there.
 
12:26 AM
@KevinDriscoll Sounds similar to my teaching strategy :D (except that I go through the textbook in order, reading it, and writing lesson plans).
 
@PVAL: Similar funny story. In a survey paper on contact geometry Eliashberg said that one can prove Cerf's theorem by reducing the the case of contact structures and then applying the theory of filling by holomorphic discs. A few years ago a couple people wrote a 50-page paper justifying this.
 
I am reading that paper
 
Oh, let me know how it is. It's on my infinitely large backburner
I read somewhere online that it should be possible to extend the technique to reprove Hatcher's theorem. Let me see if I can find the comment.
Oh, it was this. Still it'd be interesting to see.
 
In Milnor's Notes on the h-cobordism theorem, he says Cerf's theorem was hard. I think that Eliashberg's proof is probably more elegant than the original (and it isn't in French), but there's still probably a large difficulty gap between proving Hatcher and Cerf.
 
Fair enough
 
12:41 AM
Goodnight @MikeM
 
Morning
how's the move
 
 
1 hour later…
1:46 AM
Slow so far; movers Friday, I hope @MikeM .... I'm planning to visit you guys the end of the week of Labor Day ... So make escape plans :)
 
2:06 AM
@Ted: OK, I'll keep that in mind.
 
 
5 hours later…
7:08 AM
hi @Kaj
 
Hey @BalarkaSen
Haven't been around much in here recently, haha
 
been busy?
 
Of course I'll be back once the semester starts in a couple weeks
Not really. I've just been enjoying the end of the summer for the most part. Watching movies, reading a book, that sort of thing.
 
ah, I see. so not much math?
oh hey, you look like Thurston in your avatar, lol.
 
Just a little. I do a few problems from Hungerford here and there, and preparing for the subject GRE
Thurston?
 
7:12 AM
you've never heard of Bill Thurston?
 
Name sounds familiar, but I can't put my finger on it
 
he's, like, the one of the greatest 3-manifold topologists of the century.
 
GRE is like a graduate school entrance exam for students in the US. Not sure how familiar you are with how things work here, haha
 
yeah, I know.
 
Yeah, I can kind of see the semblance :P
 
7:16 AM
what're you taking the next semester?
 
@BalarkaSen, for math I'm taking cryptography and an algebra course using Hungerford. I think I might also be doing some sort of TA'ing for a pre-calculus course.
 
ok, I see. haven't you already taken algebra before? so what kind of course is this one?
 
Yeah, I've taken two undergraduate courses. This is designed to help prepare grad students for their quals.
 
ah.
 
I'm fairly interested in algebra, so I'm trying to absorb as much as I can before grad school. With any luck, I might be able to take commutative algebra from Pete in the spring.
 
7:22 AM
that's nice. commutative algebra is good, I'm doing that stuff again from Atiyah-Macdonald.
 
8:05 AM
@BalarkaSen do you know the homology/degree theory proof of the FTA?
good morning btw
 
8:17 AM
@iwriteonbananas I don't know what you mean by that. You can just mimick the usual proof by replacing "$\pi_1(S^1) = \Bbb Z$" by "$H_1(S^1) = \Bbb Z$" and winding number of $S^1 \to S^1$ by degree of the map. Those are the same things.
Would you consider this "degree theory" proof of FTA?
The gist of the usual topological proof of FTA is that the map $z \mapsto z^n$ in $\Bbb C - \{0\}$ is not homotopic to the constant map. You can also use the the first map has degree $n$, while the constant map has degree $0$.
Degree is homotopy invariant, so this is impossible.
 
@BalarkaSen yeah, i guess they're the same
@BalarkaSen ok i think i understood the main idea now
basically the point is that on large circles, a polynomial $p(z)=z^n+a_{n-1}z^{n-1} +...+a_0$ is dominated by its leading term $z^n$
 
8:39 AM
right. all you need to show is that for a polynomial $p$, $|p(z)|$ for large $|z|$ is almost like $|z|^n$. thus for sufficiently large $|z|$, the graph of $p(z)$ is homotopic to $z^n$ in $\Bbb C^n$. varying $z$ continuously gives a homotopy between $p(z)$ and $p(0)$.
this in turn gives a homotopy between the graph of $z^n$ and $p(0)$.
in $\Bbb C^n - \{0\}$, such a homotopy is impossible.
so $p(0)$ has to be $0$.
huh?
oh, whoops, I mean $p(z)$ must be $0$ at some point of time.
 
if $p(z) \neq 0$ for all $z$, then the homotopy is done all inside $\Bbb C - \{0\}$.
and apologies for the typo : I meant $\Bbb C$ instead of $\Bbb C^n$
 
no worries
just to make sure i got it right: let's define $$f_{R,t}(z) = \frac{ Rz^n+t(a_{n-1}Rz^{n-1} +...+a_0)}{Rz^n+t(a_{n-1}Rz^{n-1} +...+a_0)}$$ then $f_{R,1}$ is homotopic to a constant map by letting $R$ go to $0$. on the other hand, letting $t$ go to $0$, $f_{R,1}$ is homotopic to $f_{R,0}$ which is the map $z\mapsto z^n$
oh and $R$ is so big that $z^n$ is larger than the rest of the polynomial for $|z|=R$
no wait
 
your $f_{R, t}(z)$ is equal to $1$, as you've wrote it.
it's still $1$ :P
 
sigh
too late to edit
meant to put absolute value in the bottom
 
8:49 AM
what is $t$? what is $R$?
 
let's go to the algebraic topology chat room
 
right
 
Huy
Bye.
@MikeMiller: We have defined geodesic polar coordinates by placing polar coordinates on some $T_pM$ and transfering them to $U \subset M$ via $(\exp_p)^{-1}$. On wikipedia it too says that "Composition of $(r,\varphi)$ with the inverse of the exponential map at $p$ is a polar coordinate system." Why the inverse? $\exp_p: T_pM \to M$, so shouldn't it just be the exponential map itself?
 
I agree.
 
 
2 hours later…
10:51 AM
hey @Balarka
 
hi. how's it going?
 
oh, real well.
 
hi guys :)
 
hello.
I am finally going to get a 50/50 on my math first term internals.
that's a first
"eto onko parish, school-e ki hoy?" etc :P
did you go to school, @Balarka?
 
I always leave out the 15 or so marks corresponding to the arithmetic (i.e., confusing word problems)
@SohamChowdhury nope. exams in less then 5 days.
 
10:55 AM
your math teacher loves you, doesn't he
ok.
good luck. no math now then?
 
Hey @Robjohn. Are you here?
 
@SohamChowdhury not much, other than fairy-tale math.
 
what book?
 
fairy-tale math means random reading/learning from different branches. story-book math is reading stuff from pop-math books, on the other hand.
3
 
good.
I gotta go. Have done no math in a while.
 
10:59 AM
What will you do?
as in, what kind of math? algebra?
 
@robjohn Are you sre that $ \left( 1 + p/n \right)^n = \sum_{k=0}^{n} \frac {|p|^k}{n^k} \binom{n}{k} \leq \sum_{k=0}^{n} \frac {|p|^k}{k!}$??? In your answer here, I think you meant $ \left( 1 + p/n \right)^n \leq \sum_{k=0}^{ \infty } \frac {|p|^k}{k!}$ did you?
 
9
Q: About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $

Chris's sis the artistHere is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known nice real analysis tools for calculating them (including here all possible sources in literature that are publicaly available). At some point I'll add my solution. It's a...

 
5
Q: Show that $(1+p/n)^n$ is a Cauchy sequence for arbitrary $p$

Valery SaharovIt is a generalization of this question. I am looking for a similar derivation as in here. Can we prove that $(1+p/n)^n$ is a Cauchy sequence for any $p \in [a, b]$ by showing that $$ \Bigg| \left( 1 + \frac{p}{n}\right)^n - \left( 1 + \frac{p}{m}\right)^m \Bigg| \leq f(n)$$ where $f(n)$ is so...

 
11:26 AM
@BalarkaSen indeed.
modules are cool stuff.
 
ADG
hi
 
@SohamChowdhury I dont know why that happens :p
 
ADG
i m back after a long time
 
Even happens to me
 
ADG
just entered iit
 
11:32 AM
Hey @Balarka
 
ADG
were being taught real analysis
 
@SohamChowdhury why d'you think they are cool?
 
Not to me anymore.
Probably the calculator! :P
@BalarkaSen Because of the examples. Every ring is a module over itself, every ab-grp is a Z-module . . .
Plus they generalize vec. spaces, which is a pretty natural thing to do.
 
@Balarka I feel cyclic groups are the most fanciest stuff I have encountered till know in groups
 
Yeah, plus the CRT tells you everything about their structure possible (I think?).
 
11:34 AM
@SohamChowdhury can you classify vector spaces of dimension n?
 
oh, shut up
I will be able to in a while.
 
if you could, i could have said cooler things about modules
 
Okay. When I can, I'll pester you :P
 
one of the drawbacks of not knowing linear algebra before doing modules.
 
I have an inch there .. HEHAHAHA
 
11:35 AM
I'll do Artin in the evening.
@Rem, can you?
 
yes, he can
 
Hey @Balarka What have you been doing today?
3 mins ago, by Remember me
I have an inch there .. HEHAHAHA
Did you guys take that seriously?
Everything became suddenly abnormally quiet so I got a bit worried
 
@SohamChowdhury more interesting examples : every k[x]-module is a k-vector space with a linear transformation. every k[G]-module is a k-representation of G
@Rememberme nothing much
 
@Balarka Do you know anywhere I can get a pdf of artin ? I want to look at that book..
 
11:42 AM
it's not available online
 
Oh...
 
12:04 PM
1
Q: Find a constant $C$ such that $ \Bigg| \frac{\prod_{i=0}^{k-1} (n-i) }{n^k} - 1 \Bigg| \leq \frac{C}{n}, \forall k \leq n$

Valery SaharovConsider the following: $$ \Bigg| \dfrac{\prod_{i=0}^{k-1} (n-i) }{n^k} - 1 \Bigg| \leq \frac{C}{n}, \forall k \leq n $$ How to find an expression for $C$ independent of $k$ and thus $n$? It arises in estimating the speed of convergence of exponential series and was mentioned in here for examp...

Anyone to help out? I'm close to a very useful result
 
12:30 PM
After looking how naive I was here when I first came , I really feel like digging up a hole and jumping into it :p
 
1:20 PM
HEYY GUISE HALP SOLV REEMON HAIPOTHSES PLZZZ!!!!1!!!one!
@Balarka:
229
Q: What does "coalgebra" mean in the context of programming?

missingfaktorI have heard the term "coalgebras" several times in functional programming and PLT circles, especially when the discussion is about objects, comonads, lenses, and such. Googling this term gives pages that give mathematical description of these structures which is pretty much incomprehensible to m...

 
why are you linking me that
 
The "factoring out" thing is cool.
Have you encountered F-algebras before?
 
of course I have
they are very basic concept in algebra
 
> very basic
right
 
they really are.
 
1:25 PM
ok
 
and as much as that guy thinks the mathematical concept is incomprehensible, I find the answer incomprehensible. :P
@SohamChowdhury if you think they're hard, you don't understand them. think of an R-algebra A as something which has natural left/right R-module structures coming from the morphism R --> A which takes the image of R inside the center of A.
equivalently, an R-algebra can also be thought off as a generalization of a vector space with a billinear form (cross product to you). But you won't understand it, once again, as you don't know linear algebra.
I don't think you'll get much intuition about things in module theory without knowing linear algebra.
 
F-algebras are not algebras in that sense!
In mathematics, specifically in category theory, F-algebras generalize algebraic structure. Rewriting the algebraic laws in terms of morphisms eliminates all references to quantified elements from the axioms, and these algebraic laws may then be glued together in terms of a single functor F, the signature. F-algebras can also be used to represent data structures used in programming, such as lists and trees. The main related concepts are initial F-algebras which may serve to encapsulate the induction principle, and the dual construction F-coalgebras. == Definition == If C is a category, and F: C...
 
1:41 PM
silly name
 
perhaps
 
The name is so because it comes from a functor F
it makes sense
 
i.e., because it already exists.
 
ok, back to work. this stuff is interesting.
 
an "co" is a usual thing in category theory
 
1:43 PM
@SohamChowdhury what are you learning?
 
Hi :)
 
@ValerySaharov "co" is used in a lot of other places than category theory.
but yeah, it's a terminology with it's origin in general nonsense
 
 
1 hour later…
2:52 PM
@DanielFischer I didn't immediately understand what you were talking about yesterday. But then I realized you were simply saying

$$ \sum_{n=0}^{\infty} \text{Res} \left[ \frac{\tanh^{2m} z}{z^{2}}, i \pi (n + 1/2 ) \right] = \sum_{n=0}^{\infty} \text{Res} \left[ \frac{\tanh^{2m} (z+ i \pi n)}{(z+ i \pi n)^{2}}, i \pi /2 \right] = \text{Res} \left[ \tanh^{2m} (z) \sum_{n=0}^{\infty} \frac{1}{(z+ i \pi n)^{2}}, i \pi /2 \right].$$
 
3:08 PM
Yup.
 
Hello, I asked a question but I messed it up when I asked it so now i have some answers on the question that answer the question I asked but not my problem. Should i make a new question?
 
@crossboy007 Considering that some people have taken the time and effort to answer your first question, it is best not to pull the rug out from under them by changing the question. It is best to ask a new question and reference the previous question if needed.
 
Ok, thank you. I shall do that then :)
 
3:39 PM
Hello!! Could someone of you take a look at my question:
1
Q: Show that $t^n-1 \mid t^m-1 \Leftrightarrow n\mid m$

Mary StarI want to prove the following lemma: $t^n-1$ divides $t^m-1$ in $F[t, t^{-1}$ if and only if $n$ divides $m$ in $\mathbb{Z}$. I have done the following: $\Leftarrow $ : $n\mid m \Rightarrow n=km, k \in \mathbb{Z}$ $t^n-1=t^{km}-1=(t^m)^k-1=(t^m-1)(t^{m(k-1)}+\dots +1)$ So, $t^n-1\mid t...

?
 
3:59 PM
@MaryStar It's already been closed.
You can think of it this way $\frac{x^m-1}{x-1}$ and $\frac{x^n-1}{x-1}$ are polynomials and at $x=1$ they evaluate to $m$ and $n$ respectively. That proves that if the polynomials divide, then the orders divide. The other way around is simple.
 
4:23 PM
Hello@BalarkaSen
Let us say that if a subgroup of a group is cyclic then does that necessarily mean the subgroup has to be abelian
@BalarkaSen ^^
 
@Rememberme A cyclic group is abelian, period. It does not matter whether the group lives inside a bigger group as a subgroup. Of course, the subgroup being abelian does not imply that the bigger group is abelian.
 
Yes I know that ... The subgroup being abelian not implying the bigger group being abelian (I know this one)
 
@Rememberme Then, I am not sure where the question came from.
 
4:59 PM
@MaryStar That is not really a duplicate of the question it is supposed to duplicate. However, lab bhattacharje has a gold badge in elementary-number-theory, so his vote closed the question unilaterally.
 
5:16 PM
I think we should also pay out respects to Dr kalam
RIP : Dr APJ Abdul Kalam
 
@Rememberme the cyclic group is abelian, so obviously.
what Boni said answers your question
 
Ohkay ... I figured it
Did you watch A Kalam's burial today @BalarkaSen
 
Tom
Anyone here familiar with calculating MTBF?
 
5:34 PM
It seems there are still people that downvote my post here
10
Q: About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $

Chris's sis the artistHere is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known nice real analysis tools for calculating them (including here all possible sources in literature that are publicaly available). At some point I'll add my real analysis so...

I would like see those that only downvote how they perform in a real math job at these integrals.
Of course, the system allows you to hit anyone anonymously, but a strong person would never do that anonymously. Just my opinion.
I mean when you have somehting to say, say it face to face, assume an identity as a username, whatever.
 
Tom
@Chris'ssistheartist Where do you live?
 
@Tom Romania
 
Tom
Too far. I can't give a face to face that far away
 
@Tom lol, well, it was about assuming an identity as a username at least since one can downvote a lot anonymously, you don't even know the user to have a talk to him/her. :-)
 
Tom
@Chris'ssistheartist It's kind of the point of anonymous downvoting that they don't have to give an identity...
 
5:43 PM
@Tom Yes, that's perfect, but by MSE identity I meant username here.
 
Tom
@Chris'ssistheartist It's still the point. It's like the evaluations at the end of a college class. If I review the professor. It should be anonymous. The feedback I leave is valid, but leaving bad feedback for a bad professor could be dangerous if they know something about who did it.
Do you expect the person who upvotes your post to say something about it?
 
@Tom The point is that I don't know what the negativity is good for, because it's not about a single downvote, but looking back, I note that there are some that downvote me again and again.
I have doubts there are different persons at every post.
@Tom It's more important to know the resons when you get downvotes, maybe you may improve something.
 
Tom
@Chris'ssistheartist If someone had to post a comment every time they downvoted something, users would be less likely to downvote. That would artificially increase the number of votes on a question. Do you explain everything you downvote?
 
@Tom I don't even remember when I downvoted the last time. There must be something exceptionally bad, but I think there is no need for downvotes, it disturbs the peaceful environment.
 
Tom
I downvoted a post the other day because the answer didn't answer the question, just posted some random fact about the subject matter. I didn't put the reason down in a comment.
 
5:53 PM
I'm trying to prove the identity sin^{-2}(x) = \sum_-\infty ^\infty (x-n\pi)^{-2}. This definitely looks like a Fourier series-type thing. Can someone give me a hint?
 
@PhilipHoskins do you know about the residue theorem in complex analysis? that's the easiest route to that identity
 
@TedShifrin I found this concerning the difference between the Moore method and the Socratic method of teaching.
 
Semiclassical: I am, but I'm looking for a real analytic argument
 
hmm
@PhilipHoskins then i'd suggest you ask our resident guru on real-analysis methods for obtaining identities...
(i'm looking at you, @Chris'ssistheartist :) )
 
@Chris'ssistheartist I have never downvoted anything, not even on meta. I rather just be frugal with my upvotes.
 
5:59 PM
one way to derive it is from knowing something about infinite-product expansions, but that just pushes the problem back to understanding those (and that's once again really complex analysis)
 
@RandomVariable That's great! :-) The biggest number of downvotes I ever gave happened when I quarreled with Don Antonio. He downvoted me, and I did the same. It was a horrible mess at that time, but he started it and it was so hard for me to stay away. :-)
@Semiclassical lol, what you mean? :-)
 
Actually, I might have an idea
 
i mean that if anyone would know a purely real-analytic argument for something that's usually derived via the complex residue theorem, it'd be you :P
 
6:03 PM
i actually think you might be able to get there via fourier analysis directly, but i'm forgetting the approach
ah, yes. you should be able to derive it from the Poisson summation formula.
assuming the integrals are doable, of course
(i really mean eq. 1 of that link, not the more specific result quoted below)
 
don't tell me anything I think I almost have it
lol
 
heh, fair enough
 
@Semiclassical Doesn't it flow naturally from deriving $\cot(x)$ series once?
@skillpatrol It's an illusion as far as I can see, it's not a cube at all.
The illusion is a cube, indeed. :-)
 
@Chris'ssistheartist yes via photo shop
 
6:14 PM
that's probably right, but then one needs to derive that instead
 
Yeah I got it
Parseval's formula does the job
 
nice
ahah, nice
 
@Semiclassical Well, we use Euler product for $\sin(x)$, take log and differentiate once.
 
sure. but then, how do you get the Euler product in a sufficiently rigorous way? :P
 
You use the fourier series of e^{-i(t/pi)x}. So simple.
 
6:17 PM
obviously one can derive it like Euler did, i.e. product of all the zeroes of $\sin(x)$
but that's more a formal derivation than a rigorous one. i don't know how to do it rigorously without complex analysis.
 
See here
16
Q: Is there an elementary proof for Euler's product for Sine?

SuperbusI've been looking at proofs of this equation: $$\displaystyle \frac {\sin \pi x} {\pi x} = \displaystyle \prod_{k \mathop = 1}^\infty \left({1 - \frac{x^2}{k^2} }\right)$$ All the proofs seem to rely on complex analysis and fourier series. Is there a (more) elementary proof?

 
there we go
 
Quivers!
@Balarka, I am reminded of fundamental groups (although I am not legally allowed to know about them yet :P)
 
6:33 PM
@Soham how did that happen :P
 
Quivers are a bit similar.
All that composition of paths business.
 
Back in 30-60 min.
 
what even is a quiver
 
don't link me
i am too lazy to open links
explain them
 
nah
got better works to do
 
ok, wait
 
a quiver is a directed graph
 
it's basically a directed multigraphy thing
 
it is not really anything like the fundamental group
 
6:35 PM
that's it?
 
it's not basically, that's what it is
 
yes, you can make a vector space based on one
 
where does $\pi_1$ comes from, then?
 
what i hear are mostly phrases like 'quiver gauge theories' and i don't have a clue what that means
 
@Mike yeah, but I haven't really seen composition of paths anywhere else. I'm not suggesting they're related.
 
6:36 PM
@SohamChowdhury Why is knowing about something illegal?
 
don't ask me, @semi-c
 
wasn't trying to :P
 
@Boni, I tend to skip ahead and Balarka doesn't like it. :P
 
nah, i think you can look up what a fundamental group is now
 
@SohamChowdhury Eh? Is he teaching you or something?
 
6:37 PM
No.
It was just a joke.
 
not really.
i just object to knowing what a fundamental group is without knowing what a topological space is. that's it.
 
But he tends to scold other people.
 
Oh.
 
I know what a topological space is, duh.
 
now you do
 
6:39 PM
yes.
 
@ted how goes the move?
 
I learnt about fundamental groups without a direct link to topological spaces.
 
anyway, if you ever feel interested in how you can make a vector space from a directed graph, that link is there.
 
that's not really knowing/learning
 
good night. :)
 
6:39 PM
i'll read up later, @Soham
'night
 
Sweet dreams.
 
later pal
 
It was a passing remark by my supervisor when he explained about knot theory.
 
hi
 
6:42 PM
Sorry, just tried. Just this time I get to have chat privilege
 
welcome :-)
 
Thank you :)
 
@Lulu in the bottom right hand corner of the screen there are "help | faq" links
 
6:58 PM
@Semiclassic: So far so good, but the movers haven't responded to my message asking them to confirm tomorrow delivery ....
 
that's unsettling
 
yikes
 
Well, high blood pressure is good for me :) How's probability going, @MikeM?
 
next week
 
Oh ... So late! UGA starts fall in just over 2 weeks.
 
7:01 PM
Quarter system
 
Yeah, I remember my Berkeley days, but I don't remember "late" summer quarter.
 
Hi @TedShifrin
 
Hi @Lucio
 
It's two sessions. Each are five weeks. I teach in the seond one.
 
Do most classes run through both, Mike?
 
7:05 PM
Heard you're about to evacuate the state @Ted, not a bad idea I think
 
Hi, Kevin. I flew out a week ago!
 
@TedShifrin Are you familiar with the result: Let $X$ be a partially ordered vector space with $A \subset X$. If $\lambda \geq 0$ then $\sup (\lambda A) = \lambda \sup A$ if the right hand side exists.
 
I put high spf on my upper back and nothing on my lower back. Wind up with my upper back much darker than my lower back. Any science buffs that can explain this to me?
 
The right-wing politics is indeed globally demoralizing, Kevin.
Did your lower back turn red and peel, anon?
 
Hello Professor @TedShifrin
 
7:07 PM
no, no apparent effect on my lower back
 
Hi, skull.
i've never thought about it, Lucio, but surely you can prove it.
 
@TedShifrin did you get my message?
 
@Ted: no classes run through both
 
Can't look at it, skull, until I'm back on a computer.
Ah, Berkeley's classes ran 10 weeks, Mike, I'm pretty sure. Maybe 8.
 
@TedShifrin let me know what you think...
 
7:10 PM
You 're an overall anomaly, @anon.
 
10 is a quarter class. Summer moves twice as fast
 
Just saying it was different 36 years ago, Mike ....
 
10 weeks classes. Madness!I've been confined almost exclusively to 15 week classes
 
15 seems too long for most students, Kevin.
 
our semester courses are 15 weeks. (the algebras I teach only meet 75min once a week though)
 
7:13 PM
"most"
 
is this TAing?
 
some sources say I'm a TA and my boss is the instructor, other sources say I'm the instructor/teacher (I lecture the material in class). maybe it's a legal or bureaucratic thing.
I'm amazed I spelled bureaucratic right on the first try
 
sounds like a cushy deal. you can't even teach as a grad student here unless you have 3 years of TAing experience
 
did you see the moore method teaching message I left for ted?
 
I'm the only undergrad in my position, and I happened to have about 2 or 3 yrs of TA experience (meaning, only tutoring the material in our computer lab) before anyway.
 
7:22 PM
@Semiclassical did you see my last question on main? math.stackexchange.com/questions/1378536/…
 
I figured as much
 
hadn't yet
 
well, bbl, bike ride
 
later pal
 
enjoy the bike
@Ted: so what do you do with your days now?
 
7:43 PM
orthogonal polynomials: how do i love/hate thee, let me (not) count the ways...
 
Hello
 
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