Mathematics - 2015-05-29 (page 3 of 3)

Soham Chowdhury

14:04

They aren't, apparently.

@Paul, what have you been doing?

Paul Plummer

Nothing much mathematically, other than getting my next post figured out, and starting to read a paper. @SohamChowdhury He just faked passing out studying

And I am brushing my teeth right now...

@SohamChowdhury How about you?

Soham Chowdhury

I was away at an event all day. (Read back if you're interested.)

Gonna do a few exercises from Aluffi now.

Paul Plummer

In my next serving of salad (which by the way the title does not come from the food) look forward to inequalities, that I make absolutely no effort to make as good as possible :D

Why do Indians know German?

Soham Chowdhury

I believe $\omega = |\Bbb{N}|$?

@PaulPlummer Optional no-credit after-school course which I took for three years.

@PaulPlummer Where from, then?

Paul Plummer

Sort of, and depending on your construction of the naturals $\omega = \mathbb{N}$ even

Soham Chowdhury

14:14

Oh, yes.

That was confusing me a bit.

Although I'm certainly not part of your intended audience.

Paul Plummer

I am from America @SohamChowdhury

Why do you say that? @SohamChowdhury

Soham Chowdhury

It's way too advanced.

"The Strong Measure Zero Game"

Paul Plummer

Well my next post should be all right, I am sure I will be using some stuff you don't know, but I will give a decent intuition for what is going on to "justify" the ideas. @SohamChowdhury

Soham Chowdhury

Also, what was "I am from America" for?

I didn't understand.

Paul Plummer

What did you mean by "Where from then?" (I looked at the arrow and still not sure)

Where is the salad from?

Soham Chowdhury

14:19

Where does the name come from?

That's what I meant.

Paul Plummer

(I thought you maybe thought I should know why Indians learn German because of where I was from)

Soham Chowdhury

Oh, no, no.

Paul Plummer

Oh

Have you heard the phrase "word salad"

Soham Chowdhury

I wanted to know what the rationale behind "Math Salad" was. You said it wasn't the food, which confused me.

Yes.

Oh, I see.

Paul Plummer

It is symptom of disorders, basically of people having garbled words.

Hence the "tagline" Incoherent mathematical writings

pjs36

14:21

Too soon Paul, Gary Busey is still alive :(

Paul Plummer

Is something happening to Gary Busey?

pjs36

No, it was just a bad joke

Soham Chowdhury

Again.

Paul Plummer

:)

Was there anything in particular that felt too advanced (when I wrote it it felt like it was at a suitable level, that someone just knowing what an open cover is would do alright, not necessarily easy though) @SohamChowdhury

Soham Chowdhury

Well, I think it was written well enough for that "suitable level". (i.e. not highbrow "this is trivial, that is trivial, in fact everything on this page follows trivially from the commutativity of addition in $\Bbb{R}$" stuff).

I just need to know a little more. Like about open covers and order types.

Isn't an order type like the difference between $\Bbb{N}$ and $\Bbb{Z}$? Infinite in one direction vs. infinite both ways?

Paul Plummer

14:30

Yes

Soham Chowdhury

Or is that called something else? I've forgotten, it was a long time ago.

Oh.

Paul Plummer

Well $\mathbb{Z}$ does not have order type of an ordinal

(it does not have a least element for every subset, hence not an ordinal by definition)

Your intuition is correct

Soham Chowdhury

I read the Wiki page, I think. Or might've been The Emperor's New Mind, or GEB, or something.

All those "collection of cool things" books. (GEB isn't just that, though, imo)

@PaulPlummer is that because of subsets like $\{ x : x < 0\}$?

Paul Plummer

Yes

Soham Chowdhury

So if we disallow infinite ones . . .

Paul Plummer

14:35

Disallow infinite sets?

Soham Chowdhury

then it's no longer $\Bbb{Z}$.

Yeah.

Rambling, never mind.

I need to do a little work, I've done nothing today yet.

Paul Plummer

Been thinking about set theory recently (judging by that question)?

Soham Chowdhury

Well, every undergrad math book starts with set theory . . .

:P

Paul Plummer

It sure seems that way

Soham Chowdhury

And I learned a little about Schroder-Bernstein and Cantor's theorems.

Paul Plummer

14:38

I actually like set theory a lot, thought about becoming a set theorist for a while (I don't know much set theory though)

Soham Chowdhury

I'm secretly afraid of turning into a set theorist. Then logician, then intuitionist / axiom-of-choice-rejector / Kronecker / etc.
Just kidding, don't mind.

Paul Plummer

Haha, I think you are just afraid of becoming a philosopher

Soham Chowdhury

Well, maybe.

Anyway, I'll hit the book. You planning to stay online?

Paul Plummer

Yah, I will be around, although I might get logged out, they seemed to have changed things so you get logged even if the window is still open @SohamChowdhury

Soham Chowdhury

Oh, okay.

Mats Granvik

14:58

There is no point in adding a point when the point has been made...

Balarka Sen

@SohamChowdhury Just be a mathematician :P

you're alive, @PaulPlummer!

Paul Plummer

I am alive

Balarka Sen

to @AlexClark's great disappointment.

Paul Plummer

:P

1 hour ago, by Alex Clark

@PAUL!

Balarka Sen

he did that out of shock, you know it, right?

hello, @AlexWertheim

Alex Wertheim

15:09

Hello @Balarka. How are you? :)

Paul Plummer

Yah, that is why he left, could not stand being in the same chatroom

Balarka Sen

@AlexWertheim almost alright. what about you?

I am thinking about a lot of things I have listed down in my notebook, but not getting any of them done.

Alex Wertheim

Just almost, @Balarka? I'm not too bad. Been a little more productive lately.

Haha, don't I know that. feeling =P

What's on your list?

Balarka Sen

@AlexWertheim lefschetz fixed point theorem and linear algebra of chain complexes, billinear forms on homology, fundamental group of linear groups, homology of a wacky chain complex and bordism homology.

Hippalectryon

$\square \!\!\! \checkmark$ Procastinate

Alex Wertheim

15:14

Cool, good stuff. A lot of homology :)

Balarka Sen

the first of all these is bothering me too much.

Paul Plummer

Missing a `\` @Hippalectryon

Balarka Sen

but I also want to think about the last one, because it's cooler.

so I'm all confus.

Alex Wertheim

Lol, I see. What's bothering you about the first?

Balarka Sen

Well, I believe that there must be a totally general nonsense proof of the Lefschetz fixed point theorem which essentially will boil down to doing linear algebra over $\mathsf{Ch}_\bullet$, category of chain complexes of $R$-modules.

but I can't seem to extract it out from Hatcher's proof of Lefschetz.

there is a basic algebra lemma over there, but it doesn't seem to be a lot of linear algebra.

Alex Wertheim

15:20

Mmm, I see. Well, I hope you can find what you're looking for!

Balarka Sen

are you familiar with a bit of homology?

Alex Wertheim

Only very basic principle of homology, and not much else, I'm afraid.

Balarka Sen

ah. well, if you want, i can tell you about the last need-to-think-about problem from the list, which is cool (i bet Mike knows about it, but I guess he's busy)

i am sure you'll be able to understand it.

Alex Wertheim

Sure, I'd be happy to hear it, though I am certain I can't contribute anything.

Balarka Sen

OK. Let's call maps $\Sigma \to X$ for $n$-manifolds $\Sigma$ as "singular $n$-manifolds" in $X$. $C_n(X; R)$ be the group of $R$-linear combination of singular $n$-manifolds in $X$, $R$ being some ring.

Alex Wertheim

15:28

I have to go for about 30 minutes, but I will read it when I come back, so keep typing if you'd care to do so. :)

Soham Chowdhury

Hey, @Balarka.

Balarka Sen

Consider the chain of homomorphisms $\cdots C_{n+1}(X;R) \to C_n(X;R) \to C_{n-1}(X;R) \to \cdots$, where the homomorphisms $\partial : C_n(X; R) \to C_{n-1}(X; R)$ are defined by sending the singular $n$-manifold $\Sigma \to X$ to singular $(n-1)$-manifold $\partial \Sigma \to X$.

Remember me

Hello@SohamChowdhury@PaulPlummer@BalarkaSen

Paul Plummer

@Rememberme Hello

Balarka Sen

(Forgot to mention that I am working with manifold with boundaries, so add that in anyway)

Soham Chowdhury

15:31

How many generators does the symmetry group of a cube have?
I think six: three rotations, three reflections.

Balarka Sen

OK, @AlexWertheim. Note that boundary of manifolds are closed and compact, so the composition of the homomorphisms $\partial \circ \partial$ is the zero homomorphism (which sends a singular $n$-manifold to the image of the empty manifold in $X$, which is the identity in $C_{n-2}$).

Soham Chowdhury

@Balarka, can you help me with that?

Balarka Sen

Thus, this defines a chain complex, so I can define it's homology $H_\bullet(X; R) = \text{im} \partial/\ker \partial$

This homology is suspiciously similar to unoriented bordism theory. So, my question is, can there be a serious connection with, say, $H_\bullet(X; \Bbb F_2)$ and unoriented bordism homology? Surely they are not isomorphic homology theories (bordism doesn't satisfy the dimension axiom).

It's cool that this thing is actually a functor $H_\bullet : \mathsf{Top} \to \mathsf{RMod}$. I am not sure if anything can be done with it though.

pjs36

@Soham I think you can get away with fewer than $6$. You need at most one reflection.

Soham Chowdhury

Oh, yes.

Balarka Sen

15:36

@SohamChowdhury You mean the isometry group of a cube?

pjs36

It helps if you know what the symmetry group is, as an abstract group.

Balarka Sen

It's $S_4 \times \Bbb Z/2$, if I recall correctly.

That $S_4$ copy corresponds to rotations.

pjs36

It is indeed, the standard trick for the $S_4$ part is to consider the $4$ diagonals, connecting opposite pairs of vertices.

Balarka Sen

right.

so, yeah, $4$ generators.

pjs36

Well, Balarka, I'd assumed you're familiar with all of this :)

Balarka Sen

15:39

i have done a bit of symmetry groups from Artin, yes

Paul Plummer

@BalarkaSen Not sure if it is that, I think it is probably a semidirect product of some sort

Balarka Sen

i haven't checked, @Paul

Paul Plummer

I have not checked either, I would just be surprised if it was a direct product

Balarka Sen

that one was from sheer memory.

pjs36

It's direct, every isometry is the composition of a rotation and the antipode map $x \mapsto -x$.

Soham Chowdhury

15:41

Well, I just took one of my cubes out and rotated it while keeping one pair of diagonally opposite vertices fixed. That rotation has order 4, it seems.

pjs36

Err, the antipode map, if necessary, not necessarily

Balarka Sen

@pjs36 OK, but that doesn't imply it has to be the direct product.

Soham Chowdhury

Does my little experiment mean anything?

Balarka Sen

It's clearly an $S_4$ extension by $\Bbb Z_2$ (or whatever), and they're just two of them, so I guess it's not hard to check anyway

Soham Chowdhury

You're losing me.

"Extension"?

Balarka Sen

15:42

It was a reply to @pjs36

pjs36

Good point Balarka, I didn't realize that wasn't good enough.

Soham Chowdhury

See, the text hasn't introduced products and all yet. How do I find out the order of the isometry group?

pjs36

Have you covered orbit-stabilizer?

Soham Chowdhury

Nope!

Paul Plummer

Well wiki does say it is iso to what @pjs36 and @BalarkaSen said

Soham Chowdhury

15:43

That's the second last section of the chapter.

Now, how do I find the order of the group?

pjs36

Is that a "yes" to orbit-stabilizer, @Soham? :)

Soham Chowdhury

No.

I really don't know anything about those things.

Paul Plummer

Well you will learn

Soham Chowdhury

Yet.

But the exercise is at the beginning of the chapter!

Paul Plummer

I thought you said you read the whole chapter and just going back to do some exercises you hadn't done?

Soham Chowdhury

15:47

@PaulPlummer I skimmed it once, didn't really understand the last two sections well. Just wanted to get an overview of what was there.

Paul Plummer

I do the same when reading textbooks @SohamChowdhury

Soham Chowdhury

I wanted to see if Lagrange and Cauchy were there.

pjs36

Without orbit-stabilizer or even group actions... the order is pretty hard to compute!

Soham Chowdhury

All five platonic solids. gulp

pjs36

Oh, and also no homomorphisms either, if you're going strictly by the book.

Soham Chowdhury

15:51

And, contrary to the first exercise, which I needed a lot of help with, the second one feels far easier.

@pjs36 Yeah.

pjs36

But you do have symmetry groups $S_n$, and you'll find $S_4$ by labeling the four diagonals of a cube

So just convince yourself that the rotational symmetry group does indeed permute those diagonals, just like $S_4$. I'm not sure how you can deal with reflections, at this point.

Soham Chowdhury

Yeah.

I'll just think about it for a while and skip that question for now.

Paul Plummer

@SohamChowdhury It seems like it would be a pain. I think the idea is to use the same ideas in the section on dihedral groups, you should be able to apply your intuition (which ends up being pretty close to the ideas introduced in orbit-stabilizer, so that may be the point, to get you ready)

For example, you have rotational symmetry of faces, and then you can choose where you place the face.

Soham Chowdhury

13 mins ago, by Soham Chowdhury

Well, I just took one of my cubes out and rotated it while keeping one pair of diagonally opposite vertices fixed. That rotation has order 4, it seems.

Paul Plummer

Which might be good enough to count the orders, havn't really though tabout it

Balarka Sen

15:55

It's swelteringly hot in here. Can't concentrate on work.

Soham Chowdhury

Same here.

Paul Plummer

Are you Jasper @SohamChowdhury

Balarka Sen

oh look, we got a new Jasper

pjs36

That rotation should have order $3$, it cyclically permutes the edges connected to one of the fixed vertices

Balarka Sen

lol, @Paul

Soham Chowdhury

15:56

@BalarkaSen Don't know who that is, but I suppose I must wear his robes with pride.

Paul Plummer

I guess he was before your time (by a suspiciously short "before your time")

Soham Chowdhury

An illustrious ancestor, I surmise.

:P

So I'm skipping the exercise, with your blessings.

Paul Plummer

You should be able to figure out the order you basically count the symmetries of the faces and where the faces can go

Balarka Sen

I +1ed just for the "not even wrong".

love that phrase

Soham Chowdhury

Under what conditions is $[a^n]_p = [a]_p^n$?

a coprime to p?

Paul Plummer

16:03

I don't really think math.se is even an appropriate place for those sorts of questions (basically asking for peer review). I will flag the question for closure. (A more appropriate question would be to ask about specific parts of the paper...)

Alex Wertheim

Neat question @Balarka. As I expected, I have nothing to contribute =P

user112495

I've been given the ellipsoid $\epsilon := \{ (x, y, z) \in \mathbb{R}^3 : x^2 + \frac{y^2}{4} + \frac{z^2}{9} = 1\}$. I need to find its tangent plane, $T_{(p, q, r)}\epsilon $assuming that it's a smooth 2-dimensional manifold of $\mathbb{R}^3$. Am I right in thinking that this is just $2xp+\frac{1}{2}qy + \frac{2}{9}zr=0$?

I also need the find the point on this ellipsoid closest to the point $(0, 1, 2)$ which I assume I'm supposed to use lagrange multipliers with. But I keep getting a horrible quartic which doesn't seem to have any nice roots.

Is anyone able to help me with this?

pjs36

16:25

Yeah, summer classes! I get to teach at 8:00am, two sections of Intermediate (read: remedial) Algebra.

user112495

I get the two cases: $\lambda = 1$ and $x=0$. $\lambda = 1$ leads to imaginary values for x, and so we must have $x=0$. But then this gives me a quartic in $\lambda$ which I am unable to solve. Would someone be able to suggest where I might be going wrong?

Paul Plummer

16:50

@SohamChowdhury Asaf gave an interesting answer to your question, working in $\mathsf{ZF}$

Semiclassical

17:22

@Chris'ssis: just in case you might be interested, i've got another integral i'm working on (related to the one i showed you a while back): $$\int_0^1 \frac{\ln(1+t)}{t}\frac{dt}{1+t^{r}}$$ for $r<0$

it's calculable at $r=0$, and appears to decrease monotonically as $r\to -\infty$. but i've been unable to find a closed-form

Soham Chowdhury

@PaulPlummer Yeah.

Chris's sis

17:58

@Semiclassical Very interesting question

@Semiclassical This reduces to $$\frac{1}{r}\sum _{k=1}^{\infty } \frac{(-1)^{k-1} \left(\psi ^{(0)}\left(\frac{k}{2 r}\right)-\psi ^{(0)}\left(\frac{k}{r}\right)\right)}{k}+\frac{\log ^2(2)}{r}+\frac{\pi ^2}{12}$$

Semiclassical

hmm, interesting

Chris's sis

@Semiclassical Now, there is an interesting connection to make ... (just a bit to find the link)

Semiclassical

kk

actually, while i said $r<0$, it also converges for $r>0$, and (aside from an overall constant shift) seems to be an odd function of $r$:

Chris's sis

@Semiclassical The idea is that one may try to connect the whole thing to $$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{H_{kn}}{n}=\frac{k^2+1}{24 k} \pi^2-\frac{1}{2}\sum_{j=0}^{k-1} \log^2\left(2\sin\frac{(2j+1)\pi}{2k}\right)$$ by carefully choosing the values of $r$ where in my formula is positive.

Semiclassical

hmm, that does look interesting

Chris's sis

18:09

Also see arxiv.org/pdf/1010.1842.pdf for a proof.

Semiclassical

that's a nice link, thanks

Chris's sis

@Semiclassical You can get infinitely closed forms for certain values of $r$.

Semiclassical

yeah. integer values of $r$ in particular seem to work well

Chris's sis

@Semiclassical I think you need $r=1/(2p)$

Semiclassical

that's probablyr ight

Chris's sis

18:17

@Semiclassical I think you wanna get rid of that 2 in denominator inside the first digamma $$\psi ^{(0)}\left(\frac{k}{2 r}\right)-\psi ^{(0)}\left(\frac{k}{r}\right)$$

Semiclassical

right. when $r=1/(2p)$ then both of those arguments are integers

Chris's sis

@Semiclassical Yeah. Then you can prepare what you have and use the above formula. :-)

Semiclassical

ahh

heh

Chris's sis

I love analysis because of the amazingly nice connections one can do. Every time I solve a problem I do my best to use it further not to let that there to be caught by dust.

Semiclassical

what i find striking myself is how tanh-like the graph is

Chris's sis

18:21

@Semiclassical Yeap, I noticed.

Semiclassical

though that may be a bit deceptive: the tails go to 1 slower than one would expect (like $\pm 1-a/r$, I think)

Chris's sis

@Semiclassical Did your problem arise in some physical problems?

Semiclassical

yeah

Chris's sis

@Semiclassical Nice physics you do! :D

Semiclassical

it's related to that integral I showed you before, but with a change of variables $(x=e^{-t})$ and generalized in a different direction

Chris's sis

18:25

I see.

Semiclassical

namely, it shows how a certain quantity (evaluated at finite temperature) departs from one of the limiting cases from the earlier integral (which was done at zero temperature)

namely, the one limiting case that you could compute easily by hand :)

so yeah, there is motivation for it :)

Chris's sis

As I previously said, many very nice questions in terms of integrals (also series and limits) come from physics.

Semiclassical

indeed

especially since, if one has an integral arising from physics which one knows how to do, then one can often find natural generalizations by considering how to generalize the physical problem

Chris's sis

@Semiclassical Yeap.

@Semiclassical This might be a nice question to give in a contest (based upon your question) $$\int_0^1 \frac{\ln(1+t)}{t+\sqrt{t}} \ dt$$ :-))))))

Semiclassical

hmm!

Hippalectryon

18:31

@Chris'ssis It feels weird written that way though. $\int_0^1\dfrac{\ln(1+t)}{t+\sqrt{t}}\mathrm{d}t$ would be more natural

Chris's sis

@Hippalectryon Right! I just noticed that

Semiclassical

that's the $r=-1/2$ case, which as you noted before should be in the calculable class

Chris's sis

@Hippalectryon see above now.

Semiclassical

so yeah, a nice special case

Chris's sis

@Semiclassical May I have your permission to add it to my book? That particular case?

Semiclassical

18:33

sure

Chris's sis

@Semiclassical Thanks!!! :-)

$\displaystyle \int \frac{\log (t+1)}{t+\sqrt{t}} \, dt$=2 (Log[-I+Sqrt[t]] Log[(1/2-I/2) (1+Sqrt[t])]+Log[I+Sqrt[t]] Log[(1/2+I/2) (1+Sqrt[t])]-Log[1+Sqrt[t]] (Log[-I+Sqrt[t]]+Log[I+Sqrt[t]]-Log[1+t])+PolyLog[2,(-(1/2)+I/2) (-I+Sqrt[t])]+PolyLog[2,(-(1/2)-I/2) (I+Sqrt[t])])

Hippalectryon

@Chris'ssis You know you can export it as LaTeX right ?

Semiclassical

oh what a joyful bunch of polylogs

Chris's sis

@Hippalectryon Yes, but it put me in trouble trying that.

@Semiclassical true

Hippalectryon

:(

Semiclassical

18:38

though, for me mathematica gives the simple result $-\frac{\pi^2}{24} + \frac{3}{2}(\log\,2)^2$

Chris's sis

:21910860 hahahaha :-)))))))))

@Semiclassical Yeap.

Hippalectryon

bolbteppa

19:26

"Non-commutativity of operations on a space may be expressed either using algebras or categories. Thus we may construct an algebra from a category, or vice versa, so let us consider obtaining an algebra from a category. Since composition of arrows in a category is associative we must construct an associative algebra. But an arrow may decompose into the composition of other arrows, usually in more than one way.

This forces us to sum over all possible decompositions of an arrow giving us an associative convolution algebra".
http://en.wikipedia.org/wiki/Convolution#Discrete_convolution
Why sum over all possible decompositions? How does this ensure we get an algebra? :(

Tobias Kildetoft

@bolbteppa You seem to have linked to a different part of the article than the quote

anyway, I don't see why we would need to do any summing to get an algebra from a category, but I suppose it depends on our precise needs

bolbteppa

Sorry, the quote is a statement of fact, the link is an example of a convolution algebra, the reason for summing is because an arrow can decompose in many ways

Tobias Kildetoft

but why would we need to sum because of that?

there is no reason to require unique factorization in our algebra

bolbteppa

It is described on pages 19 to 20 properly here its.caltech.edu/~matilde/SpinFoamCover.pdf They use convolution, the reason is this factorization of maps reason, but I don't see why :(

I mean how in the world would one think to sum over all maps as a means to get an algebra?

Tobias Kildetoft

@bolbteppa Well, because it then looks a lot like what you had with the group algebra (where the multiplication can also be defined in a more natural way, but giving the same thing)

They also mention that this is supposed to be a special case of categorification, but I didn't think too much on which one it would be

Possibly decategorifying via the trace (which I am only vaguely familiar with)

bolbteppa

19:35

But why use convolutions in the group algebra, they knew to use convolution in that context for the same reasons it works in this context, if you get me

Tobias Kildetoft

@bolbteppa In the group algebra, we know what the multiplication should be, and it just happens that it can be described as convolution

Still trying to think of a way that the group algebra would come about as a decategorification in a nice way

Balarka Sen

hey @TobiasKildetoft, I just remembered that Mike referred me to you in a discussion about quantum things and knot invariants. He said that combinatorial knot invariants crop up from representation theory. d'you have some time to elaborate a bit (in a topologists way, if possible : my rep. theory knowledge is near to null)

Tobias Kildetoft

@BalarkaSen Hmm, my knowledge of it is not that good

bolbteppa

Knot invariants show up in quantum groups because of the braid Yang-Baxter equation

Balarka Sen

blank stares at @bolbteppa

bolbteppa

19:40

This is like an insanely interesting topic haha, give me a moment to formulate what I know about it

I do not see how they knew what the multiplication should be, in the group algebra wiki they say it comes from the group ring multiplication
http://en.wikipedia.org/wiki/Group_ring
which pushes the question back, there should be an intuitive obvious way to know that one should use convolution as a means to eliminate the fact that maps can split up in many ways $s = s_1 s_2 = s_3 s_4$ and that this gives us an algebra :(

Semiclassical

mmm, quantum

Tobias Kildetoft

@bolbteppa I still don't see how the splitting of the maps presents any sort of obstacle to getting an algebra

Semiclassical

i don't actually know much about Yang-Baxter stuff, though i've run across the phrase in the context of math-physics re: solvable models in stat mech

i pretty much don't touch quantum groups, for better or worse

bolbteppa

@TobiasKildetoft perhaps it's not an obstacle, but we want is a non-commutative associative algebra so maybe that makes it more obvious?

Balarka Sen

i have no idea about any of the quantum stuff out there (except possibly topological quantum field theories, which i understand as a purely mathematical tool rather than something of physics)

Semiclassical

19:44

you'll get something, just not the something you want?

Tobias Kildetoft

@bolbteppa Not really

Semiclassical

the word 'quantum' is really really broad, alas

Tobias Kildetoft

@bolbteppa I mean, if we decategorify any 2-category with just one object (via the Grothendieck group) then we get an associative algebra, and rarely a commutative one

Balarka Sen

indeed

Semiclassical

quantum mechanics, quantum information, quantum groups, quantum stat mech / quantum field theory

bolbteppa

19:46

They are all the same thing basically, at the core of it all is the simple idea of non-commutativity really

Semiclassical

eh, yes, but fulfilled in different ways

Tobias Kildetoft

@bolbteppa I suppose I might have a better idea of the why if I could figure out how they mean to make the examples be special cases of decategorification

Semiclassical

i should note that i don't grok category theory, so i don't have that perspective

Tobias Kildetoft

@bolbteppa Well, of both non-commutativity and non-cocommutativity

Balarka Sen

feels like he doesn't belong in this conversation

bolbteppa

19:48

Well yes Grothendieck groups will come up but you are focusing on 'objects' there, that is my end-goal alright, but the intuitive motivation is that they are taking the standard idea of convolution of maps (arrows) and just copying that idea for the domains/images of arrows, i.e. objects, and getting a predictable end result

Tobias Kildetoft

19:59

@bolbteppa Yeah, it does feel a little forced to me that they want to think of everything as "maps to $\mathbb{C}$" or something like that. But that does seem to be pretty common in some areas

bolbteppa

20:10

I guess I'll just accept it for now haha

Tobias Kildetoft

@bolbteppa I have certainly not seen those example before, and I have done quite a bit of categorification

bolbteppa

It comes up in the context of quiver categories math.uni-bonn.de/people/habecker/KacMoody.pdf

Paul Plummer

@BalarkaSen So I am typing up the winning strategy for player one in that group theory game. I decided against, for now, doing a series on small cancellation theory so I am just going to define some of the basic ideas, and state some fact for $C'(1/6)$ groups. There will be a few estimations on bounds of a couple things, but I don't think I will even attempt to make them good estimations, just make sure they are good enough for the result.

bolbteppa

That pdf might look insane but I've almost managed to reduce lots of it to baby-speak, I'd love a baby-speak way to see why one would know to use convolution

Balarka Sen

I don't even know what a $C'(1/6)$ group is, so that'd be pretty cool, @Paul

Tobias Kildetoft

20:15

@bolbteppa It seems that the idea is that the product in some way should reflect how many ways it is possible to get some morphism as a composition of two others

Balarka Sen

so, you already found out a strategy for the game for groups, did you?

Tobias Kildetoft

@bolbteppa Best bet is probably to look up the original papers (which are probably by Ringel I would assume given the name)

@bolbteppa Or you might be lucky to catch the author of the .pdf here on the site (Hanno Becker). I have seen him around occationally

bolbteppa

I think it might be obvious when viewed in terms of polynomials as on the group ring page en.wikipedia.org/wiki/Group_ring My question is now 'what is the motivation for the group ring multiplication' basically, progress!!! :D

Paul Plummer

@BalarkaSen Yah, in fact it is a "fixed strategy", so player one does not even need to "change" depending on what player two chooses. So there is a sequence $A_1,....,A_n,...$ that player one will not need to deviate from, so in every game, on the $n$th turn player one plays $A_n$, no matter what happens in the previous turns.

bolbteppa

Wow awesome

Tobias Kildetoft

20:22

@bolbteppa Well, the usual way to define it is pretty much the only way one could define it (when it is not viewed as a set of maps)

@bolbteppa Anyway, I need to go to bed now

bolbteppa

Thanks for the help man :D

@Semiclassical I know that when you decompose the partition function in statistical mechanics into a trace of products of transfer matrices, as in the Ising model, if they are commutative then the problem can be solved. The condition for commutativity is that they satisfy the quantum Yang-Baxter equation. Vaguely I think it's like finding an oasis of commutativity in a non-commutative algebra, i.e. a solvable subalgebra or something...

@Semiclassical so that is the Stat Mech link to the quantum Yang Baxter, and the quantum Yang Baxter is like a deformed Lie algebra Jacobi identity, so a quantum group is basically a deformed Lie algebra which reduces to a Lie algebra (modulo some technicalities). Also, note the partition function in stat mech can be set up for any integrable non-linear PDE, i.e. any non-lin integrable pde can be written as a partition fun, so you can basically set up a quantum group and Yang baxter for solitons

Semiclassical

funnily enough, i'm trying to learn about that connection to nonlinear DE's

mainly from knowledge of how it works in the 2D ising model, where stuff like diagonal correlation functions can be expressed in terms of solutions to the nonlinear Painleve equation

bolbteppa

@Semiclassical the insane thing is that this crazy formality even shows up in elementary quantum mechanics when you solve 1-D scattering problems
https://books.google.ie/books?id=db0dTvNKWFgC&lpg=PA211&ots=QoqznPIiZY&dq=Ising%20Model%20Yang-Baxter&pg=PA211#v=onepage&q=Ising%20Model%20Yang-Baxter&f=false
(The pages around that page are insane, they give intuition for needing Weierstrass/Jacobi elliptic and theta functions as being obvious!)

Semiclassical

snerk. ah, the joy of reflectionless potentials

bolbteppa

20:38

Yeah so I think the real thing going on is that they all come from non-linear integrable pde's, i.e. the partition function is basically the 'time-ordered exponential solution' to a non-linear pde, and there are tons of conjectures saying non-linear integrable pde's are solvable in terms of Painleve functions,

Semiclassical

nod

hard to parse that stuff, though

bolbteppa

Yeah

@BalarkaSen I don't understand it well enough to sum it up sadly :D

Semiclassical

the direction i'm trying to understand is from hierarchies (KP, for example) of integrable equations

which isn't an easy nut to crack

bolbteppa

20:53

@Semiclassical that is not a good place to start tbh

@Semiclassical basically, view integrable non-linear pde's as though it was literally quantum mechanics, it has a Schrodinger formulation, a Heisenberg formulation, a Hamiltonian system formulation, a path integral formulation, a spinor formulation, (wtf)

nullgeppetto

Hi ppl! I have started a bounty for this question, if you like please take a look! Thanks and apologies for spamming the chatroom!

bolbteppa

@Semiclassical you have a non-linear PDE $F(V) = 0$ and assume you can associate some linear operator $L$ to it whose expected value does not change with time, i.e. $<L,L>$ is time-independent which means $L$ can be written as an eigenvalue equation with a time-independent eigenvalue. This is like writing the Schrodinger equation in implicit $F(V) = 0$ form (where $V$ is the potential btw!), and considering the position operator or something.

From this you re-write $L$ in the Hamiltonian formulation and set up the Heisenberg equations. In non-linear pde's the exact same process gives you the Lax-matrix method of solution. Usually this method is presented backwards in pde's books so you don't see the connection.

Semiclassical

the thing i should point out is that i'm not tryingt to understand the Painleve stuff from a terribly high-level perspective

bolbteppa

So the key to integrability is being able to take your non-linear pde (i.e. your 'Schrodinger equation') and re-writing it in Lax form (i.e. as 'Heisenberg's equations'). If you can do this you have an integrable pde.

Semiclassical

what i'm really after is to understand how the (Toeplitz) determinants which show up in the 2D ising model (and more generically under the rubric of Fisher-Hartwig singularities) can be understood as solutions to the KP hierarchy

bolbteppa

21:05

Stay with me for a moment, I am basically just telling you things you already know but with new words

Semiclassical

which as I understand it amounts to showing that those determinants are tau-functions

bolbteppa

Tau functions come up naturally in a moment if I can formulate this right, you already know what they are

You will be able to see it yourself if we do this right hha

@Semiclassical So to sum up again, in a non-linear pde you are trying to find $V$, so you try to assume it is like Schrodinger's equation where your $V$ is actually just the potential not the wave function, then rewrite your pde as a Heisenberg equation, if it can be done, and it can't always be done, you have an integrable pde, the only difficult thing is doing the factorization, for which no general method exists, so this is why it's difficult.

But since we have established the existence of a Hamiltonian, the non-linear pde can also be written as a Hamiltonian system, i.e. you can set up an analogue of Hamilton's equations for the pde too, but you get an infinite system of them

@Semiclassical In fact you get an infinite number of conserved charges too, it's like conservation of energy for each Hamilton's equations. So thus far from a non-linear PDE we have a Schrodinger formulation (the pde itself), a Heisenberg formulation (Lax), a Hamiltonian formulation (Hamiltonian system).

It can also be written like Dirac's equation, factored using spinors, to give the Zero-curvature formalism. It can also be written using 'time-ordered exponentials' (a standard ode/pde method) to give an analogue of the path integral solution of Schrodinger's equation. Now, the 2-D ising model partition function is just a path integral, so already you see we expect some integrability formulation,

The KP Heirarchy is just the Hamilton's equations formulation of a non-linear pde, the Kadomtsev-Petviashvili equation, which can be factored into Lax form

Semiclassical

21:24

if ii'm remembering right, in the case of KdV the point regarding Schrodinger's equation with $V$ as potential is actually literally true

bolbteppa

Yeah so KdV is the canonical example because the factorization into Lax form is so easy

Now my (not 100%) understanding is that Tau functions are just the non-linear pde analogue of the QM/QFT partition function, so when you say you want to see why Toeplitz determinants are solutions of the KP hierarchy I think you are equivalently asking why Toeplitz determinants give partition functions which are the path integral solution to the KP equation

To quote a book "Mikio Sato discovered that the totality of solutions of the KP equations form an infinite dimensional Grassmannian, and established the algebraic structure theory of completely integrable systems" (Don't see how this fits in yet)

It's crazy how Toeplitz matrices are connected to convolution (and fourier series), topics so closely related to this whole discussion...

Semiclassical

busy, sorry

bolbteppa

@Semiclassical so if we can figure out exactly how KP relates to Toeplitz and justify determinants of Toeplitz matrices give partition functions I think the whole thing makes sense and is conceptually unified... This pdf arxiv.org/pdf/hep-th/0110125v1.pdf and the first book in the references will give far more detail on what I've said

No problem, back in a while

and a clue is that you said the Ising model uses these ideas

pjs36

21:49

@bolbteppa I'm about as far from applied/physics as can possibly be, but I've come into contact with the combinatorics of some KP-related stuff in the past. How odd!

Mike Miller

@lenticcatachresis If you're ever online, I have a spectral sequence question I'd like to ask you.

Semiclassical

@bolbteppa: that bit about the infinite grassmannian is something i'd like to understand a lot better myself, since I think that should connect with the free fermions description of the tau-function

i've got some sources myself, but uh

heavy lifting, to put it lightly

@bolbteppa: the source which would probably resolve things, if i could actually digest it, is

arxiv.org/pdf/1212.6049v3.pdf

bolbteppa

22:41

@Semiclassical Page 33 of your pdf:
"As it has been established in the works of the Kyoto school, the expectation values of group-like elements are τ -functions of integrable hierarchies of nonlinear differential equations."
Hopefully you see how this makes sense from what I said. I'm glad they said this haha :D The grassmannian thing is interesting, I'm kind of shocked your paper is restricted to discussing Fermions, that's like an awesome sign this is so coherent

@pjs36 weird, my guess is that the link is via moment-generating functions which I think are like Taylor expansions of random variables which are basically operators acting on states in QM linking all this to integrability so in some perverted sense that might be it

pjs36

@bolbteppa Beats me, I just know I was supposed to read a section of this paper for a reading course on combinatorial polytopes.

bolbteppa

@pjs36 wow cool, the set up of that paper is exactly what we're discussing, there is a lot of this finite field grassmannian combinatorics swimming around this theory too for some reason

pjs36

It's a small world after all! :P I think the triangulations of $n$-gons were what I was supposed to focus on

The pictures of KP waves are crazy, by the way. Here, and here, and here

bolbteppa

23:08

@Semiclassical From the intro and chapter 9 of Babelon's integrability book, it says the KP equation comes up because it is the abstract formulation of an integrable PDE with one singularity, i.e. the Lax matrix will have one singularity. The way to avoid worrying about singularities is to go to projective space. Also, this makes sense because the path integral, representing probabilities, cannot be singular, so the KP path integral must be in projective space...

idk if you can see page 299 but they basically show a path integral books.google.ie/… I mean it's crazy

@pjs36 do you think the middle of the X in your second picture is the one singularity?

pjs36

@bolbteppa It certainly seems like a distinguished point! But unfortunately I don't know enough about any of this to have even a guess.

Semiclassical

i've got a copy of Babelon's book, but alas I find it pretty impenetrable

all in all, it's the developments of the kyoto school which i probably need to understand. but easier said than done

bolbteppa

23:23

@Semiclassical do you see that the motivation is to just set up the non-linear PDE integrability form of the path integral for the KP equation? However because the Lax matrix you get from KP has a singularity we cannot do this directly, because probabilities cannot be singular, so we must go to projective space to avoid worrying about singularities. So we have a path integral in projective space. Nobody mentioned Grassmannian so far, and your paper doesn't mention it

Semiclassical

that one doesn't, no, but it does talk about plucker coordinates. so while they don't talk in terms of grassmannians i think they're stil there

back later

bolbteppa

As Babelon says, "Grassmannians are projective algebraic varieties", so here is our first link

I'm guessing that grassmannians arise just because you want to take subspaces of the fermionic coordinates at a time in summing the path integral, that's all

So "if the tau functions are interpreted in terms of infinite Grassmanians (fermionic approach) then the Hirota relations sometimes boil down to Pluecker relations", ncatlab.org/nlab/show/Hirota+equation , which is just some condition, and Schur functions are like functions that come up in partitions, so I feel the whole paper you linked to is making some sense :D

@Semiclassical okay cool

This Hirota Bilinear actually has a physical analogue too, wtf... Vertex operators!???

Overall summary so far:
Non-linear PDE = Schrodinger equation in Schrodinger formalism
Lax matrix method = Heisenberg equation form of Schrodinger equation
Hamiltonian system = Hamilton's equations arising from existence of Hamiltonian in Lax method
Zero curvature formalism = Spinor decomposition of non-linear PDE mimicking Dirac equation coming from Klein-Gordon
Tau function = Path integral form of non-linear pde ('time-ordered exponential solution')
KP Equation = General Non-Linear PDE Coming From Assuming Lax Matrix Has One Singularity, Path Integral formulation must be on a projective s…

Chris's sis

23:52

@robjohn have you seen this one?

3

Q: Find this limits $\lim_{n\to\infty}n^2\bigl(n(H_{2n}-H_{n}-\ln{2})+\frac{1}{4}\bigr)$

Question1: Find this limits $$\lim_{n\to\infty}n^2\left(n(H_{2n}-H_{n}-\ln{2})+\dfrac{1}{4}\right)$$ where $$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$ Question 2: Can we obtain a higher asymptotic expansion? I know $$ \lim_{n\to\infty}n(H_{2n}-H_{n}-\ln{2})=-\...

It can be done pretty elementarily ... no need for special functions ... :-)

Anyway, it's too easy, forget that.

I also think here is too much machinery (well, that doesn't mean I don't appreciate the solutions)

19

Q: How can I find $\sum\limits_{n=0}^{\infty}\left(\frac{(-1)^n}{2n+1}\sum\limits_{k=0}^{2n}\frac{1}{2n+4k+3}\right)$?

prove that$$\sum_{n=0}^{\infty}\left(\frac{(-1)^n}{2n+1}\sum_{k=0}^{2n}\frac{1}{2n+4k+3}\right)=\frac{3\pi}{8}\log(\frac{1+\sqrt5}{2})-\frac{\pi}{16}\log5 $$ This problem, I think use $$\sum_{k=0}^{2n}\dfrac{1}{2n+4k+3}=H_{10n+3}-H_{2n+3}$$ Thank you everyone help

sequences-and-series closed-form

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