If someone is comfy with elementary Lie alg-stuff, please check to which degree I am lost here: math.stackexchange.com/questions/1343500/…

OK, this improves my mood $$\int_0^1 \frac{2 \text{Li}_3\left(1+\frac{1}{\sqrt{x}-1}\right)-2 \text{Li}_3\left(2+\frac{2}{\sqrt{x}-1}\right)-2 \text{Li}_3\left(1-\frac{1}{\sqrt{x}+1}\right)+2 \text{Li}_3\left(2-\frac{2}{\sqrt{x}+1}\right)+\log (4) \left(\text{Li}_2\left(2+\frac{2}{\sqrt{x}-1}\right)-\text{Li}_2\left(2-\frac{2}‌​{\sqrt{x}+1}\right)\right)+\log ^2(2) \left(\log \left(-\frac{2}{\sqrt{x}-1}-1\right)-\log \left(\frac{2}{\sqrt{x}+1}-1\right)\right)}{2 \sqrt{x}} \, dx$$

because it can be nicely done and related to that question.

@Chris'ssis, you messed up my screen.

@SohamChowdhury Oh, sorry, the integrals is a bit longer.

@BalarkaSen brinjal*

what're you doing?

the computer guy came and fixed everything up. :)

$$2\int_0^1 \left(\frac{\text{Li}_3(x)}{x} -\frac{\text{Li}_3 \left(\frac{ x}{x-1}\right)}{x}-\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (1-x)}{x}-\frac{\text{Li}_2(x) \log (1-x)}{x}- \frac{\log ^3(1-x)}{3 x}+\frac{\pi ^2 \log (1-x)}{6 x}\right) \, dx$$$$=\zeta (4)$$

@SohamChowdhury Nothing interesting.

@Balarka do you think "2 isotopy classes of embeddings of the open unit disk" is a reasonable definition of orientability (and 1 isotopy class a definition of nonorientability) of a surface, if we are forced to work in the absence of homology or differential structure?

I am trying to find a definition that I like, and I just concocted that one.

No.

oh, wait, I don't understand that.

What d'you mean by embeddings of the open unit disk?

well, the open unit disk is a topological space, and there is a notion of embeddings of topological spaces (continuous + injective)

basically, if you're placing a disk onto the surface, but allow yourself to stretch and warp the image continuously and move it around, then there are fundamentally two ways to do this (if it's orientable)

so you just mean $X$ is orientable if and only if there are two isotopy classes of maps $D^2 \to X$?

yes

originally I was thinking isotopy classes of embeddings of S^1 which have contractible images (so, little oriented loops), but then that's essentially embedding a disk

hmm

and actually the S^1 thing doesn't work because there's only one such isotopy class of embedding into S^2, say, even though S^2 is orientable. the reason is the loop doesn't know which connected component it splits the surface into is "inside" and which is "outside."

well, every map $D^2 \to X$ can be isotoped to a map from the point, right? so how does that work?

homotoped, not isotoped

isotopies must preserve injectivity

oh, i see. sorry.

i have never worked with isotopies much.

hmm

ok, what are the two maps representing the isotopy classes you mention, again?

the usual embedding $D^2 \hookrightarrow M$ is one. i don't see what's the other.

the usual embedding? wut?

hi @anon

$M$ is a 2-manifold. pick a point, take a nbhd homeo to $D^2$ around it.

there aren't canonical names for the two. pick one embedding $f:D^2\to S$ arbitrarily, then another one is $f\circ \rho$ where $\rho:D^2\to D^2$ reverses the orientation of the disk, and any two are isotopic to one of those two.

that's your embedded $D^2$.

@BalarkaSen yes, but there's two ways to do that...

There are infinitely many ways ...

isotopy classes, @Ted

for orientable manifolds, @Ted

But ones with the same orientation are isotopic :P

Goes back into his hole and departs ...

@TedShifrin no!

@anon i don't believe that the second one is not isotopic to the first one.

@BalarkaSen yeah, I was fixing that

$x\mapsto -x$ keeps the orientation of $D^2$ (since $e^{i\pi}$ is just rotation by $\pi$ radians), but you can pick a self-homeomorphism $\rho:D^2\to D^2$ that reverses orientation, like $(x,y)\mapsto (-x,y)$

yes, true.

@Ted so what do you think about this definition (see above)?

hmm. i don't know. it looks okay, but you'd better check in with someone who has expertise in isotopies.

i guess a counterexample should come from Smale's construction.

i don't know, really.

you mean sphere eversion?

yeah

@TedShifrin so, we proved borsuk ulam today using that if there is no odd map $S^n\to S^m$ if $n>m$

it was a one-line proof. the work was in proving that preliminary fact

@BalarkaSen that creates an orientation-reversing automorphism of the sphere, it doesn't mean the sphere can't be equipped with an orientation (any more than the existence of an orientation-reversing automorphism of the disk means the disk can't be equipped with an orientation)

that morphs the codomain itself within an ambient space, rather than images of the disk inside the sphere intrinsically. different concepts. imagine keeping the sphere fixed and just moving a disk-about-a-point around the sphere.

@anon yes, i understand that.

it seems to me that if there were any isotopy of an embedding to its reverse (i.e. precomposing the a reversing of the disk), then unioning the image of all of the embeddings that make up the isotopy would yield a mobius band within the surface.

but i don't believe your definition is the right one

what definition do you use?

homological one

bleh

it's useful

true, homology is designed for computations.

one thing I don't like about most topological invariants is that they're first defined on representatives of things (like CW complexes on spaces, or planar diagrams of knots) and then shown the be the same on representatives of the same thing

it feels "indirectly intrinsic" instead of "directly intrinsic"

i would retort at this saying "there is more theory to homology than people think", but i guess i rather wouldn't.

Hi @Ted

@anon i don't know in what sense homology is indirectly intristic. it exploits the piece-wise linear structure on your space.

so i'd say it's definitely intristic.

@BalarkaSen there are multiple different ways of drawing inequivalent polygonal meshes on a given space. one then proves the homological stuff coming from your CW complex does not vary with your choice of representative.

of course it's intrinsic since it doesn't vary with choice, but you have to actually prove that by using some kind of relational structure on the set of representatives of a given space or knot or link or whatever (e.g. common refinements of complexes, or adding/removing parts of the complex, or reidemeister moves for knots, etc.).

i'll be back in a moment. have to eat.

...the hell?

well, I shall go eat and do yardwork

the first 1/4 seems alright, but it's certainly bad formatting

does it get crazy?

Seems so

I mean, a summation method is essentially an extension of the lim functional from the space of convergence sequences to a functional on a larger space. That much is true.

Well, scan the whole thing and tell me if it remains sensical :/

I can't make sense of the tuple-with-a-number-and-a-set, but it strikes me as the sort of thing that has a good idea behind it but the person doesn't know jack about communicating the idea.

I can't be bothered to figure out what the true idea is.

Ok

I have a hard time believing one can understand Ramanujan sums from set theory alone, which is what it sounds like he's saying

In any case, the terrible formatting/explanations are enough for me to flag it as low-quality

With regards to this question (math.stackexchange.com/questions/1343500/…), pondering some more I might have resolved my issue; spoonfeeding we consider the composite $$\frak h \to \frak gl(h) \to \frak gl(g) \to \frak gl(g/h)$$ the maps being the same as before.

By the induction hypothesis, there is a nonzero $x \in \frak g$ such that $\text{ad}_y(x) = 0$ for all $y \in \frak h$. As $\frak h$ is a proper subalgebra, so is $\frak g / \frak h$ and there is a nonzero element of $\frak g /\frak h$ that vanishes in the composition, again by the indiction hypothesis.

Does that seem about right?

@anon when you say "most" then this is certainly false! perhaps you only know the euler characteristic for polyhedrons lol

I am referring to "one thing I don't like about most topological invariants is that they're first defined on representatives of things (like CW complexes on spaces, or planar diagrams of knots) and then shown the be the same on representatives of the same thing." Please make sure that you get the proper point of view, for otherwise you will be sorry later on

Is there a way to prove a bijection between $\{\mathbf{S}:\mathbf{S} \subseteq {1,2...50}, \sum \mathbf{S} \text{ is odd} \}$ and $\mathcal{P}(\{1,2,...,49\})$?

@Chris'ssistheartist Knuth's problem solution Version 2.0 complete .. I plan to add it to my blog soon :D

@r9m Great! I'm curious what tools you used. :-)

@Chris'ssistheartist I'm calling it 'hooga hooga hooga hooga' (as in cannibals dancing around a bonfire :P lol)

@r9m When it's available? You make me curious. :-))

@Chris'ssistheartist I am writing it now (slowly) .. should be done by morning :-) (4-5 hrs more)

@r9m KK :-)

@r9m by series manipulation only?

hello again

@Semiclassical hello

@Chris'ssistheartist nope .. but I think it's more general in terms of approach .. my initial approach was too bent on $H_{2n} - H_n$ .. this one is bent on the central binomial coefficient in general :)

@r9m Manipulations with series involving the central binomial coefficient? Anyway, I don't plan to work on that anymore, that solution I got is enough to me.

@Chris'ssistheartist yes .. and we can replace $(H_{2n} - H_n)$ with anything else, that is if they satisfy a few conditions and a few things about them is known to us :) ..

^ I think that'd be the record vaguest thing I have said since morning

@r9m Yeah, indeed! :-)

@r9m btw, it's not a joke the thing I told you today I think. All those particular series from Flajolet are done (including the worst ones and even more). Then I realized that problem you asked about is harder than many of those there.

ahh, flajolet

his analytic combinatorics text is great, though i've never succeeded it in reading it in depth the whole way through

@r9m The thing is that I worked on them for a while, so I stayed focused in a certain direction. Anyway, an elegant solution to those integrals would be a great gain.

@Chris'ssistheartist Did that ruin your mood? :P REVENGE EXACTED!! (for asking the questions on main, that I wanted to invest my time on .. ) now we are fair and square! :P

@r9m lolllllllllll :-)

Hi guys

was sick last couple of days :S

@KarimMansour what happened?

I just got bad cold so just was in bed last 2 days

which sucked

@r9m did I ever ruin your day? :-) lol, you're really funny! :-)

@KarimMansour ah! cold is my natural enemy .. I get bad headaches when I catch cold and I can't do anything in that state .. which sucks!

yeah I had fever and caughing and sneezing which in turn caused headaces so I couldn't do anything

I feel conscious too when I just waste 2 full days without doing anything

@Chris'ssistheartist can't remember .. :P I usually forget such boring trivia :P lol (joking)

@KarimMansour yeah ! that sucks ... rapid season change gives me cold

yeh me 2

bbl .. gotta finish a few stuff first!

@Semiclassical do you wanna see a nice integral?

@Chris'ssistheartist Now that's a pickup line if I ever saw one! :P

sorry, was distracted

sure

@Semiclassical $$\int_0^1 \frac{1}{x}\text{Li}_{5}\left(\frac{1}{1-\frac{1}{x}}\right) \, dx$$

Take the Cantor set on [0,1]. Its complement is open, and the indicator function of this open set is not Riemann integrable, correct?

@El'endiaStarman nah. for that it'd need to be a surface integral

hums innocently

@Semiclassical ;)

interesting. I've noticed you doing a lot of polylog integrals on $[0,1]$, though unlike some of the others i see no reflection symmetry

hi @Semiclassic

hmm. my immediate feeling would be to substitute $z=e^{-x}$ (that's more for taste than anything), write out the polylog as either a sum or a series, and integrate term by term

afternoon @ted

@TedShifrin Salut

salut @Gato

je suis sur le point de partir :)

@TedShifrin Ah dommage ! :P

qqch de vite? :)

@TedShifrin For english grammar , that could explain why people tend or tends ?

tend ... people = plural

in English the s verb ending is singular (don't ask me why)

ok, always ?

Anyway, I have a mathematic confusion. I've been exploring hyperbolic geometry and polygonal tilings and the like. Right now, I'm working on understanding Coxeter diagrams, and one thing I don't get yet is how they're related to groups. The Wikipedia page here says (or seems to say) that $A_2 \equiv \bullet - \bullet$, but the associated graphical representation earlier on the page is this:

for regular verbs, for sure

thanks, ted, last question when we put a 's' at the end of the word ?

El'endia, I am no expert on Lie groups/algebras, but the thing you have drawn two things up is a Dynkin diagram for the group. The graphical representation is showing how the group is represented as a symmetry group.

My thinking is that $A_2$ contains only the identity because the only other permutation on $\{1,2\}$ is $\{2,1\}$, and I'm pretty sure that's an odd permutation, hence is not in $A_2$. So how can a group with only the identity be associated with something interesting (i.e. non-trivial)?

no, no, definitely not the identity. You need to do more reading. :)

These are not alternating groups. This is the classification of Lie groups according to $A_n$, $B_n$, $C_n$, $D_n$ ... nothing to do with symmetric groups.

@TedShifrin ....AH. That explains my confusion.

These are mostly in fact infinite Lie groups.

sorry, @Gato: "s" at the end of a noun makes it plural (most of the time). "s" at the end of a verb makes it singular (most of the time) :P Consistent, eh?

@TedShifrin Considering English's history of stealing - AHEM - incorporating all sorts of things from other languages, that's not bad. :P

@TedShifrin no problem, yep interesting approach, I need to improve my english :)

Linguistics is a fascinating field. I almost went into that. I love languages and grammars.

As do I, (don't as my why :P) I don't have this 'love' in English ^^

Well, my French many years ago was once as good as my English, but that's very much no longer true. :( I also studied German, Latin, and Russian. I'm going to work on Spanish when I finish moving.

@TedShifrin Impressive, I will start Russian in september, and Arabic ? :P

@Semiclassical You meant to be bring it into this form? $$\frac{1}{k^5}\int_0^{\infty } \text{Li}_5\left(\frac{1}{1-e^z}\right)\, dz$$

Sorry, don't know any Arabic :)

Morning.

that'd work, sure, though I also had in mind the fact that (if memory serves) polylogs are what a physicist would label as Fermi-Dirac or Bose-Einstein integrals

Good night, @MikeM. All done?

so insert the integral representation

@TedShifrin When I will visit you, I will '''''''teach'''''' you.

Hi @TedShifrin

LOL, ok, @Gato.

hi @Karim.

I'm long gone, @Ted. Lunch break, so trying to get some reading done.

Speaking of lunch break, I have dinner company coming over in a few hours. I'd best get busy.

Lost my phone and wallet last night, which is not so good. None of my financial instruments were in my wallet so it's not as bad as it could be.

Yikes, @MikeM ... I'm supposed to be that forgetful, not you.

You'll not survive without your phone. Where/how did you lose 'em?

@Semiclassical $$\int_0^{\infty } \text{Li}_5\left(\frac{1}{1-e^z}\right) \, dz$$

Think I left them in the cab but they said it's not in their lost and found. After they said that they immediately hung up.

right. then look for useful integral representation of Li-5

Oy ... You need to work on your typical mathematician scatterbrainedness. :(

The next person in the cab may not have turned them in :(

There was nothing in the wallet for them to steal. It was all IDs.

@Semiclassical Yeah, I see your point.

Do you have the "Find my iPhone" app on your phone?

Not in a foreign country where I don't have service.

Ohhh, you're there already ...

I guess they could have taken my wallet for my BevMo card... :)

You'd better contact the US consulate for help.

You still have your passport separately?

Yes. I don't really understand the need to contact the consulate.

Yeah, if you have the passport, you're ok. My bad.

IDs are an inconvenience and phone is a monetary inconvenience, but it's far from the end of the world. Couldn't use the phone while I was here anyway.

Good positive attitude. Keep up the good work, but try not to be forgetful :P

Not nearly as scary as when I thought I didn't bring any shirts.

(They were zipped into a different part of the suitcase, not visible without unzipping.)

LOL

Have a great time at the conference!

OK.

There is such a crazy rain here! My ears hurt!

i suspect i'd be jealous, i like rain like that (when i'm indoors watching it, mind)

@Semiclassical I liked to do that and especially walking through the rain.

i dimly remember something i did for a high-school composition class

"Rain from my window / such harmonies of water / are playing the grass"

so yeah, i like rain :)

:-)

i feel like the second comment to this answer is missing the point. answers aren't just for the OP, they're for the site as a whole.

...now that I've done more reading, I found out that the Lie group $A_2$ corresponds to $SU(3)$, which seems far more complex than the graphical representation would suggest.

Dynkin diagrams?

@Semiclassical Coxeter diagrams, more like, but they're the same here. $\bullet - \bullet$

right. i had to employ some of that for an REU project like...6 years ago

@r9m @Hippalectryon ^^^ :-)

@Chris'ssistheartist I've seen that one already :P

I'm subbed to this channel

@Hippalectryon Also this one is amazing :D

@Chris'ssistheartist too short :(

@Hippalectryon extended versions? Let me see ...

Nah, it would just be the same thing looped

There's another cool recent one

@Hippalectryon nice, nice

@TedShifrin If I was doing some honours project with a prof and I don't want to do it with him anymore

can I back down and not work him anymore ?

@Semiclassical How would you explain Coxeter/Dynkin diagrams, then?

I mean, I sorta understand Wikipedia's explanation that uses mirrors, but if you have a different perspective, I'd love to hear it.

@Chris'ssistheartist Here goes my v2.0 on AMM 11832 :) .. hope you like it :-)

Lol I read it before you even posted it in the chat xD

@Hippalectryon :-) :-)

@Chris'ssistheartist you might also like this one :D then! :)

@r9m very nice, but it seems advanced machinery! :-)

@Chris'ssistheartist I refered to Boyadzheiv's paper .. :) he gives a very nice exposition to the application of the theorem to various similar harmonic number identities

@r9m Yeah, I know. :-)

@r9m Well, nice work, indeed!

@r9m I'll check tomorrow the details, too tired to think now.

@Chris'ssistheartist sure :-)

@r9m I wanna see what that paper contains.

@Chris'ssistheartist okay :D

@KarimMansour This is a touchy matter. Where I was, Karim, you had to actually sign up for a course with the professor with credit to do such a thing, so you can't just back out. Is this an optional thing that you're doing?

I agree with him, though, @Semiclassic. I don't want to write for an hour if no one is really interested. Even when I write just a few sentences, it annoys me when the OP doesn't respond one way or another.

@r9m btw, don't you post the solutions to the integrals on main? :D Or you plan to wait some more? :-) At the moment I don't have a pure real analysis solution!

@Chris'ssistheartist no! as a continuation of revenge :P ..

@r9m Anyway, I'm very glad people like you exist and attend such stuff! You have all my admiration! :-)

@Chris'ssistheartist LOL! wish my professors would say that to me :-) No one really encourages me to do these here (refering to my uni) ..

@r9m lol, why not? ::D

@Chris'ssistheartist I don't know .. they really don't care one way or the other

@r9m I see, they work on RH (maybe).

(joking)

:P rofl

:D

@Chris'ssistheartist good night! It's 3:40 am here .. need to sleep now! Bye :-)

@r9m Good night! :-)

Anyway :-)

@TedShifrin i can understand that at the level of "why should i put an answer together if the OP won't respond." but that still leaves the question of "if i choose to post something, is it appropriate as an answer or as a comment?"

@El'endiaStarman it's been a long time, so i'm not sure I can. what i recall is this: what dynkin diagrams are about are the root vectors of the lie algebra, and in particular what kinds of subspaces they form

link or something ???

i got it ^_^ ,i found derive of laplace inversion and understood the way,here is the link if any one has similar problem google.com/…

what's up everyone?