@BernardoMeurer hallo

@heather Hey

wow, points for funny picture

@JohnRennie :p

It's old :P

@MartianCactus I do actually. It took me like five minutes to master that >_>

@G.Bergeron You'll do since ACM is a dead meme it seems

@BernardoMeurer you're gonna get stars every time you post that

\o/

o people her know what memes are>

@G.Bergeron Can you explain me a stupid primitive that is giving me brain cancer because I'm stupid?

@JohnRennie Me too, that's not so far away...

@BernardoMeurer what?

@G.Bergeron Solve $\int\frac{t}{t^2+1}$ plox

@BernardoMeurer what do you need?

Hey @ACuriousMind I learned a little bit about characters of compact Lie groups just now and I'm left wondering why they are the things that show up in partition functions of CFTs.

I can see why $\frac{1}{2} \log{|t^2+1|}$ is right, but idk how to compute this crap

Those are the same "characters", right? I only used to know the physics side.

@BernardoMeurer the answer seems to suggest that $t^2+1=u$ is a useful change of varaibles

try that

Why not a (hyperbolic) trig function?

Perhaps $\cosh^2-\sinh^2=1$ will help you (i.e. try $t=\sinh$)

I don't know about characters in CFT partition functions (I don't think I actually know any CFT partition function except for that of the torus, and I don't recall characters there)

@Danu I cannot use hyperbolics on the exam coming up :)

@ACuriousMind You know the free fermion on a torus?

@Danu yes

@AccidentalFourierTransform Yeah but then what do I do with the $t$?

This seems so simple

@BernardoMeurer So no contour integration either?

@G.Bergeron Nope

@BernardoMeurer well, if $t^2+1=u$, then $t=\sqrt{u-1}$ and $\mathrm dt=\frac{\mathrm du}{2\sqrt{u-1}}$

It's a freshman course in real analysis, calm your nipples folks

@BernardoMeurer it goes in the differential

@heather Very good.

@ACuriousMind That partition function is of the form $\sum_{i=0}^3 |\vartheta_i|^2$, right?

@BernardoMeurer May I point out that there's a $\mathrm{d}t$ missing from the integral?

@heather :D

@ACuriousMind I am having an emotional breakdown and this is what you tell me?

@BernardoMeurer But why no hyperbolic trig functions though? Those are pretty straightforward?

@Kaumudi.H That's a silly statement. Of course we're interested.

@Danu Uh, my version looks very different, and I don't know what the $\vartheta_i$ of yours are

@Danu Because the fat bald man said we can't use those spells, only spells with easier casts are allowed

@DanielSank =D like, really happy. I went from in agony to done with what I was doing!

@ACuriousMind Sigh... These shitty modular forms. Some special cases of some famous family...

@AccidentalFourierTransform AH

GOT IT

Makes sense

@BernardoMeurer you can also use partial fractions, $\frac{t}{t^2+1}=\frac{1}{2} \left(\frac{1}{t+i}+\frac{1}{t-i}\right)$

which are easy to integrate

@ACuriousMind So what does it look like for you?

I miss linear algebra :(

It was so much prettier than this

honey, we all do

good ol' days :-)

@Danu Characters as in traces of representations?

@G.Bergeron Yes

@Danu My version of the NS partition function is $Z(\tau,\bar\tau) = (q\bar q)^{-1/48}\prod_{i,j > 0}\left( 1 + q^{i+1/2}\right)\left( 1 + \bar q^{j+1/2}\right)$ with $q = \exp(2\pi\mathrm{i}\tau)$.

With no 1/2 and a 24 in place of the 48 for the Ramond version

@ACuriousMind Okay, so that's close enough. Those products of the form $(1+q)(1+\bar q)$ can be writen as some modular forms

So you need a Ramond AND NS sector

and then one more terms for modular invariance

That gives you the terms I mentioned (The fourth is identically zero)

@BernardoMeurer Everything is linear algebra in the end

It's the only thing we know to solve

@Danu Isn't the theta some kind of Jacobi elliptic functions?

Like when you study partition functions for spin chains

These things are all closely related

I don't recall the exact names etc ( around pages 133-136 or so in Blumenhagen's CFT book is one reference)

@Danu Last time I explicitly calculated stuff with that it was a total mess

But I think this appearance of the characters is what ties together string theory <-> representation theory <-> number theory

where the last connection is given by the magic formulas that we can find for characters, using physics

you forgot <-> pseudoscience

#haha

Ah, I think I know what you mean by the "characters" - the partition function is a trace over Hilbert space, i.e. we evaluate the character of the Virasoro representation on a special operator

(that joke is getting real boring)

(you are boring)

@ACuriousMind So a character in math is a function $G\to \Bbb C$ given by $\chi_V(g)=\operatorname{tr}(l_g)$ where $l_g$ is the action of $g$ on a fixed representation $V$.

Why is that what we are doing when computing a partition function?

Because of the trace

Okay, so the partition function is just computing a single value of a character? Namely for $g=$"the operator that gives the energy"?

@Danu Because the partition function is the trace over the Hilbert space $\mathrm{e}^{\tau H}$ for $H$ the Hamiltonian?

Not necessarily the energy

In CFT, the Hamiltonian is $L_0 + \bar L_0$, so you get (after adjusting for the difference between the plane and the torus) that you have to compute the trace over $q^{L_0 - c/24} \bar q^{\bar L_0 + c/24}$

Yeah, right.

the trace is nothing but the character of the Virasoro algebra in the representation of the free boson/free fermion/whatever

So the partition function computes the value of the character on a single element of Vir. What about the other ones?

I forgot the physics... what about the traces over $q^{L_j-...}$ for $j\neq 0$?

@Danu Well, an element of the Virasoro group, really. I don't think there's much meaning to a trace that does not involve $L_0$. However, using $1 = q^{-L_0}q^{L_0}$, we can say that the trace of $q^X$ is the expectation value of $q^X q^{-L_0}$.

Why you would be looking at that expectation value is anyone's guess, though ;)

@ACuriousMind Ohhhhh, now I understand.

:P

@ACuriousMind I'm just trying to see if there is a meaning to the other values of $\chi$

In math, I think you care about the values $\chi$ takes on the entire group, not just on one element

As I said, these are expectation values of operators, whether those are physically interesting operators is another question

@ACuriousMind So is the exponential shows up to get to a group element?

@Danu You care about it's value on the conjugacy class

@DanielSank It's very helpful :D

@G.Bergeron That's the same.

@Danu Yes

Since it's constant on conjugacy classes

Alas, I have to go

Aw

I guess we'll continue later (hope you'll be around too @G.Bergeron)

I approached those ideas coming from another perspective

It's funny to see the relation with other topics

I've never seen characters used for partition functions except in CFTs. Have you seen it used in other places?

I guess that it makes sense in any context

But nobody mentioned it in any of my other courses

Anyone good with cosmology?

@Danu Yes, but I need a little teasing mixed with self-mockery.

@Danu Well, I saw that first under the guise of integrability in the context of lattice models. But then again, these are related to statistical mechanics models and then to CFTs, but this was not the angle I worked. In the end the trace comes out because you are compactifying the space with periodic boundary conditions.

OK, anyone good with astronomy?

;-;

@SirCumference, no good, but curious

Well I'm wondering how the cosmic scale factor can reach infinity within a finite time in the Big Rip scenario.

@Danu The most interesting part for me, though, is the relation with ''quantized'' symmetries. So the operator your taking the trace of can be seen as endomorphisms of a state space. In that way, they relate to the dynamical algebra of the system. Now the point is that those dynamical algebra are given by these quantized symmetries which are obtained by much the same procedures as quantization, but on the level of the Lie group manifold instead of phase space.

...I am still wrapping my head around a proper interpretation

@SirCumference Wait, nvm

@ACuriousMind Sorry, potentially dumb (and late) question

What do $n$ and $\Omega_n$ represent?

@SirCumference That's more cosmology

@G.Bergeron Cosmology is a branch of astronomy

meh

:P

@SirCumference I've met some cosmologists with strong feelings on the distinction.

i spend way too much time on xkcd

@HDE226868 Really?

Yes

@SirCumference Yep.

To me astronomy is looking at stars and mapping known planets

Explaining the dynamics of a star or something like interstellar dust is astrophysics

Obviously, that doesn't cover all cosmologists - I won't even attempt to make a blanket statement there - but it's a very distinct field. Yes, there's some crossover, but only at certain points.

@G.Bergeron The astronomers I know rarely make the distinction

I've met many who say outright "astronomy = astrophysics"

And cosmology is mostly the application of GR on the whole universe, with a thickening side of QFT

@SirCumference What are they doing? It involves quite different skills

@G.Bergeron One of them was a planetary scientist, the other studied neutron stars, and I don't know what the last one specifically did

Also, in an astronomy course, my professor said "astrophysics and astronomy historically meant two different things, but in modern times they refer to the same thing"

@HDE226868 I know a lot of cosmology involves particle physics, etc., but isn't in mainly based on the evolution of our cosmos?

Especially considering how closely it and astronomy tie together

@SirCumference Define "cosmos".

@G.Bergeron Sorry, what do you mean? What are quantized symmetries?

@HDE226868 The universe in the large scale

Of course, particle physics comes into play, but most of it involves the large scale universe just like astronomy

@SirCumference Right, so that is cosmology. Astronomy and astrophysics deal with it on a slightly smaller scale.

Astronomy isn't really about large scale stuff I think, is it?

@HDE226868 Huh? You don't think studying voids and galaxy filaments counts as astronomy?

@SirCumference Not particularly. I mean maybe, but probably not.

Astronomy translates pretty literally to "study of stars"

@Danu you know, all this quantum groups stuff.

I don't know what a quantum group is.

@HDE226868 I don't know anyone who studies galaxy filaments, but I know an astronomer who's done work in studying voids

@Danu Historically, sure

But you need to remember that when the term "astronomy" was first coined, only stars and comets were really ever studied

No one knew about galaxies, black holes, etc. until the 20th century. That's when the field changed.

A nice quote from Chris White:

@HDE226868 Wikipedia:

> Historically, astronomy has included disciplines as diverse as astrometry, celestial navigation, observational astronomy and the making of calendars, but professional astronomy is now often considered to be synonymous with astrophysics.

I think you distinguished theoretical astronomers from observational ones

Also voids, which are on the same scales of galaxy filaments, seem to be considered astronomy

@SirCumference That's not too definitive.

I don't think there's one true answer here.

> However, since most modern astronomical research deals with subjects related to physics, modern astronomy could actually be called astrophysics. Few fields, such as astrometry, are purely astronomy rather than also astrophysics.

I think pure astronomy is pretty rare

. . . but existent.

> Based on strict dictionary definitions, "astronomy" refers to "the study of objects and matter outside the Earth's atmosphere and of their physical and chemical properties" and "astrophysics" refers to the branch of astronomy dealing with "the behavior, physical properties, and dynamic processes of celestial objects and phenomena"

In any case, let's drop this pointless debate :D

@HDE226868 I suppose, but it's also important to note that astronomy and physics have become so tied together, they're very often considered the same thing

That's why we differentiate theoretical astronomy from observational astronomy

Whatever. I'll concur with @Danu's suggestion.

I forgot what i was originally talking about...

Oh yeah

@Danu well you used loop algebras, no? They are a related type of deformations

@Danu It's all those ''groups'' full of q's

@JohnRennie Small question

I'm screwed! I spent too much time fiddling spheres and now I still have one big homework due now.

@JohnRennie Just so I understand, does the universe's overall temperature determine its fate?

More specifically, if the temperature was higher such that $w > 0$, would that affect the fate of the Universe?

@G.Bergeron Ehh... I'm not sure. I recall that Kac-Moody algebras are related to loop somethingsomething. But I don't quite remember.

@G.Bergeron ?

@Danu For instance the theta functions are special functions arising in the realization of quantum group symmetries. There is always a bunch of $q$'s everywhere. Like in what ACM posted.

q-Pochhammer, q-Gamma, etc.

They can all be found in the representation theory of quantum algebras

Oh wait, @Jim, can I ask you something?

@AccidentalFourierTransform By the way, you're #1 on the hat leaderboard for this site.

@SirCumference yes

@Jim Right, so this might sound pretty dumb

First, the equation of state $w$ for the Universe is equal to $\frac{p}{\rho} = \frac{C^2}{c^2}$, right?

@SirCumference something like that. It's not fresh in my mind

@Jim Ok, so how could $w$ possibly be $< -1$ in the Big Rip scenario?

Wouldn't $C$ have to be negative?

@SirCumference yeah, you'd have to check in a textbook. I'm pretty sure there's a good explanation, but it's been a year or so since I had to deal with it