@AlexClark thanks, that's weird

Not sure how it would suddenly need an additional package that it did not previously

what package does it ask for?

hmm, I can't seem to find any info on that package

very strange indeed

cool

@AlexClark Well, the first part of my summary of the latest referee report was directly transcribed. So good

Just finishing up a paper on classification of simple transitive $2$-reps of certain types of Soergel bimodules

we ended up being able to almost fully classify the ones we considered, with the exceptions being the possible existence of one additional one for each of the dihedral groups of order $24$, $36$ and $60$

which we just do not see how to rule out (or prove exists)

one major problem being that the hypothetical reps do exist if we forget the $2$-structure (i.e. work with just group representations)

@AlexClark good question. We were almost ready to publish this paper several times when we got ideas on how to extend the results. But this time we are completely stuck and have no idea how to even start going further

I am also almost done writing up a paper based on my dissertation. Still debating with myself whether it is worthwhile to include specific calculations for the cases $SL_2$ and $SL_3$

@AlexClark not so much I would say, at least in this case.

@AlexClark it takes time to write up nicely

and it makes the paper longer which may reduce the possible journals that it can be submitted to

plus of course, often the calculations could be performed easily enough by any reader interested in them

I don't think the $SL_2$ and $SL_3$ examples I would include would help illustrate the concepts in any way. They would just be to have the results for those cases completely written up (and I am not sure if the calculations are really doable in any larger cases)

Yeah, for some thing examples can really help. But remember that the main audience for a research paper is not students (even if it is nice if it can be read by them)

In my case, the specific calculations are just a matter of computing some Littlewood-Richardson coefficients in some special cases and doing a bit of tensoring stuff.

@AlexClark No idea. They show up in various places

@AlexClark Well, they show up when tensoring reps in type $A$

@AlexClark Right, I can see in the abstract that they are related to Schur functions, which is where the coefficients come from originally (the Schur functions are the characters of certain reps in type $A$)

Every time I write a private message to someone through forums or emails, why am I so careful and take a large amount of time thinking and perfecting my response?

because your Vietnamese parents taught you to always be a perfectionist

@BalarkaSen: let $S_g$ be a closed genus $g$ surface. $\pi_1(S_g)$ is finitely generated. is the minimal number of generators always $2g$?

I think it has to do with the phonomonology of writing, specifically that it is so permanent, and that it is asymmetric. Whenever you write a message you nessisarily have to anticipate the response of the other person, which is uncomfterable, and because writing itsn't ephemeral as the spoken word, then it only makes that pressure more accute.

@Huy haha, that might be true after thinking about it... what if I'm insecure?

Phenomenology wasn't the right word... I meant the characteristics.

Anyway, good morning, guys.

@JesterTran: again, because your Vietnamese parents would always compare you to the "better" even if you were already one of the best, which leads to insecurity. in other words, it's all your parents' fault. :P

@BalarkaSen we should continue our conversation from yesterday.

Thanks for the response, Juan. I have noticed that if time is not a concern, I will spend as much time as I need to write up a satisfactory message

@Huy That is plausible and thought-provoking. Cảm ơn Anh :)

em khoe khong?

khoe ma can thang

can thang?

stress

have I spelt it wrong? lol

maybe, I don't think I know the word for stress, actually

I use a lot of non-Vietnamese words when I talk with Vietnamese people

haha

why stressed?

third year mathematics is getting tough, especially analysis and algebra

I see, good luck with that!

thanks, I'm going to get back to studying

good bye ^_^

wikipedia

Math on Call, works, I think, along with the related books Algebra to Go and Geometry to Go (in the "Customers Who Bought This Item Also Bought" section on the bottom)

Might be too basic.

@samayo

@Huy yes, because the generators will also generate the abelianization.

HI

I need to find the dimension of the space V, which subspace of R4

V is also al linear span of the set:

imgur.com/eQzjbJv

how to start ?

with puting into matrix or... ?

morning chat

Hey all. I can't see how by differentiating under the integral sign, they got what they did. I would've thought that you'd need something like $\partial(y+uh)$ in the partial derivative instead of $\partial y$ and so on.

Hey, @Semiclassical!

Afternoon here and a lovely one at that (for a change!) :-)

pretty nice here as well, though right now i'm in one of the computer labs preparing to do some calculations

I took my stuff outside. Inside the math building was too stuffy :-b

It's a wondrous thing that they have wi-fi literally everywhere :-0

ahh. there's a big set of windows in the conversational space on the same floor as my desk, so i tend to wrok there

but that requires using my laptop, and that's a bit too slow for what i want to do

Yep, you're doing some heavy stuff as far as I can recall

yeah, pde solving stuff

morning SemiC

whereas you, from what i see up there, are doing calculus of variations stuff?

dear all:

Yep, just started a very short module in the stuff, @Semiclassical

It's pretty cool so far with extremisers and Euler-Lagrange etc.

Is there a way to visualize how $\eta(\Sigma\eta)\in\pi_4(S^2)$ has torsion? Where $\eta:S^3\to S^2$ is the Hopf map and $\Sigma\eta:S^4\to S^3$ is its suspension.

I just skipped ahead in Hatcher and found this. Apparently almost all of the higher homotopy groups of spheres have torsion, and I can't visualize how that's possible.

you wrote the group it lives in wrong

I did?

By the way, to guess at the answer to my question: "No."

oops

and in any case why do you think there should be a good picture of a mapnfrom $S^4$ to $S^2$

Wishful thinking?

k

@Akiva Good luck that.

Good luck visualizing higher homotopy groups in general.

@Akiva: chat.stackexchange.com/transcript/message/26553598#26553598.

If you want to read that.

There are more geometric ways of understanding all of this that I am not going to write down.

I'm guessing Thom-Pontryagin, but I don't know anything about that.

Hello all!

@PVAL Did you email the guy?

I'm not using one, @WillHunting.

Can someone help with explicity computation of bases from projection matrices?

in principle. go on?

(remember: ask; don't ask to ask)

@Juan I saw your message above. Still want to continue that conversation?

@Semiclassical nice to see I received a 50 points bounty for the answer below

My answer wasn't chosen, but this is the part that I care about less. As long as the quality of the math is awesome, all is perfect.

Not to put it right, that problem is pretty easy compared to what I prepared in my book (and also pretty short).

Hope some professors will consider such problems during their exams in uni.

@BalarkaSen definitely!

@JuanSebastianLozano Let's make a summary of what we know so far.

$X : U \to \Bbb R^2$, a vector field on some open subset $U$ of $\Bbb R^2$, is conservative if $X = \nabla f$ for some smooth function $f$ on $U$. This is equivalent, as we proved, to asking $\int_\gamma X = 0$ for any loop $\gamma$ on $U$. Equivalently, $\int_\gamma X$ is independent of what the path $\gamma$ is as long as the endpoints are fixed.

Smooth vector field?

@MikeMiller Familiar with the contour integral computation of projectors?

$X$ is irrotational if $\partial X^2/\partial x = \partial X^1/\partial y$. We understood that this mean $X$ is not "swirly", which is what curl = $\partial X^2/\partial x - \partial X^1/\partial y$ measures.

No, you've lost me now.

Of course, since, $f$ is smooth.

Sure, whenever I say function or vector field I will mean smooth.

Yes, alright, and I think that's as far as we got.

Right. We didn't discuss the correlation between irrotational and conservative vector fields, which we will do now.

First, can you prove that any conservative vector field is irrotational?

@Semiclassical almost forgot to ask you if you're done with $$\int_0^1 \operatorname{li}(x) \ dx$$

@BalarkaSen you can also try it

(it's so simple ...)

Okay, so if $X$ is conservative, then it's curl is equal to $\frac{\partial}{\partial x}\frac{\partial f}{\partial y} -\frac{\partial}{\partial y}\frac{\partial f}{\partial x}$, for some $f \in C^\infty$, and since $f$ is smooth, then its mixed partials are equal, so the curl has to equal $0$.

That's it.

The converse is true by Stokes' theorem.

No.

You're half-right. Figure out the wrong half.

or the right half

@user1618033 haven't, but i may take a few minutes to try while i wait for my mathematica code to run

which half are you huy

@MikeMiller: I'm half-half

@MikeMiller: thanks for your response btw

what did i respond to oh that ok yes you're welcome

@MikeMiller: I think I finally understood the proof for Dehn-Nielsen-Baer

do you want to teach me

very creative proof

oh that's not the one i was thinking of

i want you to teach me why the mapping class group is generated by dehn twists

ok, later

eating my bolognese atm

dehn-nielsen-baer holds for a wide class of 3-manifolds, too, called Haken manifolds

the proof is similar but a bit more complicated

I've seen those mentioned before, here, by you

oh

I was actually gonna look at the punctured case later and then check out the two other proofs given in Farb & Margalit

one seems to be working with topology exclusively

or almost exclusively

and the other one uses analysis and is like 1 line

but of course with heavy machinery required

i guess that's something about quasiconformal whatevers

someone explained this to me once and i cried

the analysis one is about harmonic maps and the energy functional

i dunno what a Haken manifold is.

the Dehn one is about quasi-isometries

I think the topology one uses pants decomposition

though i have heard it numerous times.

i guess i should look at that. i only understand the topology one.

anything to keep me from working in the two days when i need to work most

the quasi-isometry one is really cool. uses hyperbolic geometry in a very elegant way

no idea how to come up with it though, as Dehn

the topology is the one that generalizes to 3-manifolds

ic, so I should probably look into that

the quasi-isometry one is the one that generalizes to hyperbolic manifolds

so they're both valid viewpoints that are helpful in different directions

yeah, their proof of 8.9 is the only hard part in the way i know, and i would probably end up more or less writing the same proof they do

oki

lol this analysis proof is "We cite two big theorems that we do not prove"

exactly

:D

but the paper they're from is ~40 pages which seems assailable

"it is a well-known fact that..."

"one can readily check that..."

:(

omg looked at it no thanks

:D

2hard, i'll trick my officemate into reading it

2smart

he's real into harmonic analysis and quasi-whatever

well the quasi-isometry one was the original proof by Dehn

@JuanSebastianLozano Did you figure things out?

apparently

and it's really cool

if you want I can explain it to you at some point

yeah, i do

So given an irrotational vector field, you have $curl(V) = 0$, and therefore $\int\limits_S curl(v) \cdot dS = 0$ for any surface $S$ in the same space as the vector field. By Stokes' theorem (and a special case is Green's theorem for $\mathbb{R}^2$), this is the same as the line integral over the boundary of the surface $\int\limits_S curl(v) \cdot dS = \int\limits_{\gamma = \partial S} V\cdot d \gamma = 0$. Because the surface is arbitrary then the boundary is too

my officemate did and i did not understand him

, meaning over any loop the integral is $0$

but I'm a maths teacher, I'll make you understand

and if you don't it's homework

@JuanSebastianLozano You have made an implicit assumption here about the loop.

@Huy i think it's cool that you're teaching and also learning more math

i can only do one at a time

I'm trying my best to keep doing it

will teach more after summer

What Mike said.

because one needs monies to survive

but I hope I can do like a textbook every year or so

@Huy if I am ever in whatever dumb country you live in I'll buy you a drink

Switzerland my friend

most other maths teachers pretty much stopped learning, even the ones with a PhD :(

totally dumb country.

The loop has to be oriented and closed because of the surface has to have an orientable and closed boundery?

But, that should always exist in the case of $\mathbb{R}$

No, that's not the point.

Hint : We're working vector fields defined over open subsets of $\Bbb R^2$.

@Huy maybe one day I will be in Switzerland

and because your country is like two feet long I'll be able to find you ezpz

I'll likely be in the US again at some point. if I can get across the wall

fun fact: Los Angeles is bigger than Switzerland

I know

I have lots of relatives in CA

Also please don't call regions bounded by loops as "surfaces". I think they should just be called "regions".

Surfaces to me lives in R^3.

twitter.com/ironghazi/status/629725691685261312?lang=en

It is a surface in $\mathbb{R}^3$ bound by the condition $z=0$ :P

Meh.

You got me, but still.

I do not agree with your complaint @BalarkaSen

Anyway, I still don't see what assumption I made.

i'm with Juan on that point, actually. if you're going to be talking about curl, you might as well accept that you're working in 3D (albeit with everything uniform in $z$)

@JuanSebastianLozano Consider the unit circle in $\Bbb R^2 \setminus \{0\}$.

@JuanSebastianLozano You assumed the loop bounds a region.

@MikeMiller you've posted this before, last Friday

yeah but you didn't see it then

evidently he did

I did, otherwise I wouldn't know

no you didn't

I did

based on that map I personally am three times bigger than you

do you really want to mess with me

lol

no, please leave me alone

@JuanSebastianLozano Did you realize what goes wrong? Can you fix it?

Yeah, I can see what's wrong: I assumed the subsets were simply connected.

That's it.

some slight further subtleties but not worth bothering with atm

Note that it's not trivial that loop bounds regions in simply connected subsets, even. That's Jordan curve theorem. Just worth mentioning.

schoenflies

Ah, yikes, I always confuse which one's Jordan and which one's Schoenflies. Jordan says it separates

really hope that twitter picture was intended as being a joke.

@Semiclassical the USA is very large

yes, larger than itself even

@Juan Anyway, so for simply conencted open subsets $U$ of $\Bbb R^2$, we have that irrotational and conservative means the same thing.

Can you give an example of a irrotational field on $\Bbb R^2 - 0$ which is not conservative?