@KevinDriscoll I don't run into rep theory in physics much, so I'll take your word on that

can I write $2\otimes 2=3\oplus1$ if I like dimensions as labels?

no. but you can write $2\otimes 2=4\oplus 3\oplus 2\oplus 1.$ :)

I can write that in the event that e.g. $2$ denotes the rep of dimension $2$?

ah

@Semiclassical What heathen labelling is that?

so triplit $\oplus$ singlet

lol

suppose you've got two particles with angular momenta $\mathbf{L}_1,\mathbf{L}_2$.

Certainy $L(1)\otimes L(1)\cong L(2)\oplus L(0)$

@anon If you want ot label your irreps by dimensions, sure

@KevinDriscoll But you should not label by dimensions, because what will you then do when there are multiple of each dimension?

the labelling in physics is usually so that a spin-$\mathcal{l}$ particle can have z-component $m_{\mathcal{l}}\hbar$ where $m_{\mathcal{l}}=\mathcal{l},\mathcal{l}-1,l-2,\cdots,-\mathcal{l}$

so that'll be $2l+1$-dimensional

so if you take two spin-1/2 particles and couple them, the combined system can have spin-1 or spin-0

i'm forgetting, how does one make a curly lowercase L

oh, right. $\ell$

\ell

@Semiclassical Do physicists never consider reps of higher rank groups?

depends.

in intro level quantum, it's always SU(2)

if you do particle physics, you'll do QCD and that uses SU(3)

Fundamental theorem of group theory @TobiasKildetoft

$$\Bbb{PSL}_2(7) \cong \Bbb{GL}_3(2)$$

if you work on standard model stuff / unification more generally, you certainly run into higher rank groups

but that's a fraction of a fraction

if you do condensed matter, you'll never do anything beyond SU(2)

so i guess i'd say that while some physicists do consider higher ranks, most don't

@Semiclassical It just seems like such a big potential for silly mistakes to get used to labelling by dimensions when there are other nice things to label by

uh

we don't label by dimension

we label by the highest weight in the su(2) rep

hence $\frac12 \otimes \frac12 =1\oplus 0$.

@Semiclassical Ahh, good

silly half-integral weights :)

lol

it's really the insistence that the steps between states always be 1

hence if you want two states separated by 1 you'd better do 1/2 and -1/2

Wait, how do you define the weight? The highest weight of the $2$-dimensional rep is $1$

hmm. am I remembering things wrong? I probably am

Because that is the defining rep, and the highest weight vector is just the one where the torus acts by multiplication by the upper element (i.e. the first standard basis vector)

well, what's the lowest weight of the 2-dimensional rep for you?

I suspect we're not actually labelling by the weights, though I think it's nearly so

@Semiclassical $-1$

ah, okay

for us it'd be 1/2 and -1/2

with vector being the other standard basis vector

so that 1/2 - (-1/2)=1

but those are not actually weights of the action of the torus

so there is some scaling being done, probably for physics reasons :)

the other reason is that, experimentally, the possible outcomes of when you measure angular momentum are $\pm \frac{\hbar}{2}$

do you do $+2,0,-2$ for the dimension 3 case?

yeah

okay

then yeah, factor of two

physics would have 1,0,-1

What a nightmare to try to translate between if you are not aware of it

yup

@BalarkaSen You around?

For me highest weight $1/2$ would give an infinite dimensional rep

i think part of the reasons physicists like that is because it lets them say that fermions have half-integer spin and bosons have integer spin

but by the same token a math person could just say fermions have odd weights and bosons have even weights

so I dunno.

@Kevin Yep

@Semiclassical So what would negative spin mean? :)

at least semi-around

z-component of spin points up instead of down

@Semiclassical but that is another thing that would result in an infinite dimensional rep

-1/2 is just saying that you'd measure the z-component as hbar/2 pointing down

(when considered as highest weight)

oh

maybe the connection in this case is just a coincidence really

highest weight corresponds to s in this case

so s=0,1/2,1,3/2,...

as in, if you measure the magnitude of the angular momentum, you'll get $\sqrt{s(s+1)\hbar^2}$

but a magnitude can never be negative, so you never do $s<0$

@BalarkaSen Im having a problem completing my Mayers-Vietoris sequence for this derham cohomology of the torus. For $T^2$ starting with $H^{-1}(U \cap V)$ and going all the way to $H^2(U) \oplus H^2(V)$ I've got $0 \to \mathbb{R} \to \mathbb{R}^2 \to \mathbb{R}^2 \to H^1(T^2) \to \mathbb{R}^2 \to \mathbb{R}^2 \to H^2(T^2) \to 0$

to some extent this really comes down to the experiments

@BalarkaSen I happen to know already that $H^{top}_{DR}(M)$ for $M$ compact is $\mathbb{R}$ already, but Im trying not ot use that theorem

@KevinDriscoll Where $U, V$ are two annuli?

if you measure the z-component of an electron's angular momentum, you get $+\frac{\hbar}{2}$ or $+\frac{\hbar}{2}$. if you measure its magnitude, you get $\hbar\sqrt{\frac{3}{4}}$

@BalarkaSen Yep

I just don't know how to get the next step.

if you take the particle to be spin-s instead, those numbers would be $m_s\hbar$ where $m_s=s,s-1,\cdots -s$ and $\hbar\sqrt{s(s+1)}$ respectively

so physicists choose to label them by $s,m_s$

@Semiclassical Just set $h=2$ and everything works out :)

\hbar

whatever

and no. we're too used to doing $\hbar=1$ everywhere else :P

@KevinDriscoll Right, there's some busywork that needs to be done. Identifying the penultimate map $\Bbb R^2 \to \Bbb R^2$ is crucial

amusingly, $\hbar=h/2\pi$. so $\hbar=2$ would mean $h=4\pi$

@Semiclassical probably a fine approximation

the funny thing is that I'm pretty sure this same 1/2 factor is whats the cause of a different headache i'm dealing with lately

lol

That's the $H^1(U) \oplus H^1(V) \to H^1(U \cap V)$ map, right (check this; I have a tendency to confuse the homology/cohomology M-V, so I don't know if that direction is correct)

Thats the right direction

Cool

So, what is that map? It's just, pullback of the 1-form on $U$ to $U \cap V$ on the first component and pullback of the 1-form on $V$ to $U \cap V$ on the second component, right?

Ya you have $i:U \cap V \to U$ and $j: U \cap V \to V$ and then that map is $i^* - j^*$

Right. Hm.

@TobiasKildetoft I should point out that this issue is indicated at the start of Wikipedia's article on SU(2) reps: "As shown below, the finite-dimensional irreducible representations of SU(2) are indexed by a non-negative integer $m$ and have dimension $m+1$. In the physics literature, the representations are labeled by the quantity $l$, where $l$ is then either an integer or a half-integer, and the dimension is $2l+1$."

so this isn't an unknown issue

@Semiclassical Would have been strange if it was I guess

@BalarkaSen When we did this fro $S^1$ we had to chase around kernels and images of maps. But I tried that here and I found more than 1 thing seemed to be consistent

yeah

Right, this is not just a chase game

Oh, really? I guess I have to think harder then.

So I want to start by picking a basis for $H^1(U)$ and $H^1(V)$

But that does not mean it can not often lead to confusion, especially when people from physics post here asking about this stuff

quite

Take the loops in $U$ and $V$ which goes around the torus meridianally

And let $\omega$ and $\omega'$ be the forms which integrate to $2\pi$ around those loops

The cohomology classes of these forms generate those groups, right?

@Semiclassical I still really dislike the idea of labelling purely using the number though. At least add a bit of extra notation

it's more a shorthand imo

if you were to write it in terms of wavefunctions, you'd do stuff like $|s=\frac12 ,m_s=-\frac12\rangle\equiv |\frac12,-\frac12\rangle$

@BalarkaSen I wouldve said cohomology, but ya

I mean, consider the canonical map $\Delta(1)\to 1$

That just does not look right

@KevinDriscoll Thanks, I did mean cohomology haha

So, $(\omega, \omega')$ is a basis of $H^1(U) \oplus H^1(V)$. Let's say $L$ and $K$ are the annuli components of $U \cap V$

it reflects the fact that physicists love to omit stuff when they think it won't be confusing to do so

(for better or worse)

We can similarly choose basis for $H^1(L)$ and $H^1(K)$ by taking right-handed meridianal loops and setting $\rho, \rho'$ as the forms which integrate to $2\pi$ along those

So $(\rho, \rho')$ is a basis of $H^1(L) \oplus H^1(K) = H^1(U \cap V)$

main reason I like the notation is because I can do BS like this: $1\otimes 1\otimes 1=(2\oplus 1\oplus 0)\otimes 1=(2\otimes 1)\oplus(1\otimes 1)\oplus (0\otimes 1)=3\oplus 2\oplus 1\oplus 2\oplus 1\oplus 0 \oplus 1$

I want to understand the matrix of the $i^* - j^*$ map in these bases chosen for the domain and range space

doh, forgot fixed

(check for myself: 3^3 = 27 = 7+5+3+5+3+1+3)

It seems $\omega$ (or rather, $(\omega, 0)$) gets sent to $(\rho, -\rho')$ and $\omega'$ (or rather, $(0, \omega')$) gets sent to $(-\rho, \rho')$.

Hello@LeakyNun

hi

which in your language would be: $L(2)\otimes L(2)\otimes L(2)\cong L(6)\oplus L(4)\oplus L(4)\oplus L(2)\oplus L(2)\oplus L(2)\oplus L(1)$

@Semiclassical Right

the physicist in me just goes: ugh, so much repetition in there

@Semiclassical Yeah, but that means you make sure to keep track of the fact that these are simples and not some other reps with those highest weights

hmm

which is rather important in some cases

I should point out that, while a physicist might write $\frac12 \otimes \frac12 = 1\oplus 0$ as shorthand

And of course, some of these formulas would just end up looking dumb, like $\lambda \otimes p\mu = \lambda + p\mu$

the expression for the actual states would be

$|s_1,s_2,s,m\rangle = \sum_{m_1,m_2} C^{s,m}_{s_1,s_2,s,m}|s_1,m_1\rangle \otimes |s_2,m_2\rangle$

where that C is a Clebsch-Gordan coefficient

@KevinDriscoll Was the last thing I said sensible?

and the thing that keeps you honest is that states on both sides contain four quantum numbers, just not the same ones. you'll often times see physicists omit $s_1,s_2$ on the LHS, but that's bad practice

@Semiclassical I never did learn that bra-ket notation

I should've had that sum as $\sum_{m=m_1+m_2}$, woops

I actually ran into that issue a week ago, when I tried to use the standard table of Clebsch-Gordan coeffs to write $|\frac12,+\frac12\rangle\otimes |\frac12,+\frac12\rangle\otimes |\frac12,+\frac12\rangle$ in terms of $|s,m\rangle$ states

@BalarkaSen I think so, but I not sure

problem being that there's too few numbers in the result. the total states should really be of the form $|s_1,s_2,s_{12},m_{12},s,m\rangle$

@KevinDriscoll The thing I am doing is I am homotoping the meridian loops I chose around $U$ and $V$ to the meridian loops to $L$ and $K$

The integral of the form over that remains the same, upto a sign caused by change of orientation

where $m=m_1+m_2+m_3=\frac12+\frac12+\frac12=\frac32$ and $m_{12}=m_1+m_2=1$. if you omit those, you end up thinking that two states are the same when they really shouldn't be

So the pullback of the form gets multiplied by that sign accordingly

@TobiasKildetoft here's a question I came up with, though, and I'll write it in proper form

I really should write it as $T(\omega) = \rho + \rho'$ and $T(\omega') = -\rho - \rho'$ I guess

Consider $L(1)^{\otimes n}$. This should be isomorphic to a direct sum of $L(k)$'s from $k=1$ to $n$ with some set of multiplicities.

So that the matrix is [1, 1; -1, -1]?

@Semiclassical Right

Is there a name for these multiplicities? I feel like it should be something standard but I don't actually know.

@Semiclassical I don't know a name for them, though they have a fairly nice form

@KevinDriscoll In which case, this matrix has rank 1, so the image space is R.

Just landed in Texas

@BalarkaSen The thing that is weird to me is you can homotope the one meridian loop to be in my $L$ or $K$, but the other one gets cut into 2 disjoint pieces and and I dont see how you homotope that one to be in either $L$ or $K$

So I think I am misunderstanding something

but I can't really see a pattern

@KevinDriscoll Wait, by the other one, you mean the longitude? That's not playing the game

But I think the same kind of arguemnt works if you just say "take the forms that intergrate to $2\pi$ and are 0 on the other component

I have a meridian loop in $U$ and $V$, both

Oh okay, cool. Then ya I am with you

I have a meridian loop in $A$

and a meridian loop in $B$

this is definitely known, but I'm ignorant of it

No longitudinal curves that goes around the hole

@Semiclassical See my answer at math.stackexchange.com/questions/155589/…

See, the point is, $A$ and $B$ both deformation retract to a circle - their "core meridian loop"

niiice. upvoted

Ya I noticed that, though not in so many words

@Semiclassical I don't really recall where I got that formula from though.

So yeah once you understand all the computations the image/kernel of the map $\Bbb R^2 \to \Bbb R^2$ in the penultimate position falls out to be $\Bbb R$ both

Now you can play the game a little

$H^2(T^2) \cong \Bbb R$ falls out easily eg

@Daminark Woo!

you mentioned Littlewood-Richardson, and that seems on point

especially since, looking at the wiki article for that, there seems to be a nice description for that in terms of young tableaux (which would make sense)

@Semiclassical Yeah, the first part is easy enough. I just don't recall how I came by the formula in terms of binomial coefficients

ah

@BalarkaSen Ya once I have that that image in 1 dim, $H^2(T^2) = \mathbb{R}$ follows immediately from what I already have

i'm no help there

yup

Littlewood-Richardson is precisely the way to go in general, once you adjust for working with $sl_n$ instead of $gl_n$

hm

my few forays into schur functions have convinced me that I don't have a clue what's going on there

but for $sl_2$ the formula is especially nice when you have just two factors

right

i could believe that for L(2) instead of L(1) it's also known but it'd definitely be harder

@Semiclassical Yeah, the general theory is pretty deep actually, and there is a reason the Littlewood-richardson rule is often prefaced by "celebrated"

right

right, $L(2)$ would make sense to have been studied a lot too as that is the adjoint rep

@TobiasKildetoft Hi Tobias :D

@KasmirKhaan Hi

where are you from if you dont mind me asking? :)

@KasmirKhaan Denmark

vaaaaah :D

är du dansk :D

ja

najs ju :o

trodde du var tysk eller nåt =p

And you are Norwegian?

At some point I gotta sue Denmark for copyright

@BalarkaSen So more generally, we can say that if our $U$ and $V$ are differmorphic then $dim(im(H^K(U) \oplus H^k(V) \to H^k(U \cap V))) = dim(H^k(U))= dim(H^k(V))$

nah living in sweden =p

but dont let my name fool you , it is not my real name =p

@KasmirKhaan I see. I thought my Swedish was good enough to recognize when people write in Swedish

(especially the nåt did not look familiar)

@Daminark lmao

@daminark You mean trademark?

would have expected nånting

@TobiasKildetoft haha , well because that is how that Word is shortned ><

so you were right =p

någonting

@KevinDriscoll Wait, how are you concluding that?

hmm

Ahh, I am just not properly down with how young people use their text-speak in Swedish

haha >< that part was my bad =p

Anyway :o I came here to ask ya about representation theory _

we using the book of serr

all i know are swedish curses

"Pieri's formula, which is the special case of the Littlewood–Richardson rule in the case when one of the partitions has only one part, states that $S_{\mu}S_{n}=\sum _{\lambda }S_{\lambda }$ where Sn is the Schur function of a partition with one row and the sum is over all partitions $\lambda$ obtained from $\mu$ by adding $n$ elements to its Ferrers diagram, no two in the same column."

I started Reading it without any lectures

but was very hard , did not get anythin gfrom chapter 1 is that normal ?

:O

@TobiasKildetoft would that be related to the case of $L(1)$?

@BalarkaSen where do ppl allways learn that first ?

@BalarkaSen Ya that was too hasty. Because $U \cap V$ maight have more than 2 connected components.

I'm probably gasping at straws.

@KasmirKhaan Hard to say. I never did read Serre, but it is supposedly quite good

@KevinDriscoll Right, you have to be careful

This is a very specific case

@TobiasKildetoft Do you Think that i should just wait untill january when the lectures start on class, then take Another go ? ( I wanted to be preperared in a way, but so far cant do it on my own :/ )

@Semiclassical that would be related to all the rank 1 cases

hmm

@KasmirKhaan Depends a lot on how used you are to reading textbooks on your own. It is a skill that takes a lot of practice

@BalarkaSen But I was thnking that if it does have 2 connected components then the forms that form the basis for $U$ are isomorphic to those that are a basis for $V$, so then the $i^* - j^8$ is going to have rank that half the sum of the dimensions, which is just the dimension of $H^k(U)$ or $H^k(V)$

alright

@TobiasKildetoft tbh dont have that skll yet :/ still dependant on lectures

I wish ill get there this year :D

@KasmirKhaan Then it is probably better to wait, but start practicing with some easier material

@KevinDriscoll yeah yeah that

@BalarkaSen Oh but I dont want $U$ either I need ot be talking about $L$ and $K$

@Semiclassical We just need to do an identification where whenever we get a partition with two parts, we remove the second part and the same amount from the first part

@TobiasKildetoft well I started with abstract algebra from 0 , almost done with Groups part now and gonna do rings this week or next one =p also planning to revisit some of linear algebra, is that good?

ah

@KevinDriscoll The point here is not that $U$ and $V$ are diffeomorphic, but the components of $U \cap V$ are naturally related to $U$ and $V$... right

We just happened to have $U \cong L$

Well, just being diffeomorphic does not suffice

I mean

@KasmirKhaan Yes, that is good. But that also means that going into representation theory using something like Serre will be a bit of a jump in difficulty

$L$ is a deformation retract of $U$. Right, that's what I meant to say

That's why you could slide the core circle in $U$ to the core circle in $L$

Don't try to generalize this. It's a very very specific situation

Im glad I did this exercise though. Because I thought for something simples like $T^2$ that is just $S^1 \times S^1$, known the deRham cohomology for $S^1$ and the MV, it would be enough to just chase around kernel and image dimensions to get everything

@TobiasKildetoft grrrr I noticed that, but hmm i mean , my teacher told me that after linear algebra and abstract algebra , we could take rep theory

Right, it's fun to see how that doesn't work out

Try to compute $H^1(T^2)$ now

@KasmirKhaan You could probably do okay with the first 31 pages. (Of Serre)

@KasmirKhaan Sure, it is doable. I had rep theory as the final topic in the algebra course I just finished lecturing

But for that I did have to adjust the approach a lot

What uni are you at btw?

Stockholmsuniversity =p @TobiasKildetoft

cool

@Narcissusjewel okay if you say so , ill give it Another go :D

@TobiasKildetoft You should come here one day and give seminar :D ill be the first to go :D

@KasmirKhaan That might be cool. But I don't really know anyone at Stockholm, since I was at Uppsala

@TobiasKildetoft damn it >< uppsala is not too far , when was this?

@KasmirKhaan fall 2014 to fall 2016

well first time i was still in high school and second was doing first courses

so tell me next time you are here :D

@BalarkaSen This actually assuages some worry I had previously about MV. Becuase the ideas and maps that go into it are almost trivial. So by conservation of difficulty I expected it to not directly gives us toooo much. But something like all the DeRham cohomologies of the torus was kind of a borderline case

I can sneak to that uni with eas i Think :D

@KasmirKhaan @Semiclassical Anyway, I need to go to bed. I really should have been sleeping for hours now

@KevinDriscoll Right, sometimes you need to explicitly understand the maps instead of just doing a diagram chase

night

Good night Tobias!

Night!

@TobiasKildetoft Okay ! thansk for the help as allways ! and good night :)

@BalarkaSen $H^1(T^2) = \mathbb{R}^2$ I can get now just by playing the game chasing images and kernels.

Right

Once you have a map or two written down, the rest becomes easy

I renew my thought from weeks ago this is is almost sudoku

True!

@KevinDriscoll never thought I'd see real coefficients

But I have

This'll be a story to tell my grandkids

@BalarkaSen I'd like to try to do another one, but I think I've run out of examples. We did $S^n$ and now I've done $T^2$. $T^3$ will work with essentially the same game because the open sets will be homotopic to $T^2$. Do you have a recommendation? Maybe $\mathbb{RP}^2$ or $\mathbb{CP}^2$? Im running out of manifolds!

(or at least ones I understand)

@Daminark Real coefficients of what!?

The projective spaces would be good for your health, I think, yes

@Daminark de Rham = real coefficients

@BalarkaSen I'm hoping Etnyre will ask a question about one of these on the exam. And I will thus nail it.

lol

He likes to ask about projective space on exams, it seems

And it was on each of the like first 3 homework assignments. But not ont he alst 3

so its a sneaky choice

I am not sure if Mayer-Vietoris is the right tool to deal with $H^*(\Bbb{RP}^n)$ though

I haven't thought about this. Hm

Ah, ok, no, it's not hard to do using MV

@Daminark sup dami :D

@Daminark how did it go with your exam?

@anon Hello Anon :D

If you use cellular you can just compute $H^*(\Bbb{RP}^n)$ directly

hello

@MatheinBoulomenos Mathein :D

He's doing de Rham cohomology, @Mathein

@anon long time no see! how are you ?

No cell structures

alright

just sauce

Hi @Kasmir

:D

this might be a silly question guys

but consider G = (Z,) a group with the operation ab =a-b

we do not have an identity here right?

so this set with that operation is not a group for 3 reasons, non associative, no identity and no inverses

well, it does have a right identity

@Balarka yeah I know, it's just the first time aside from that Schlag pset that I've seen De Rham, the only time we've ever actually talked about it with Peter we'd speak of integer or Z/2

the issue is the left identity (and the lack of existence thereof)

This is from the proof that the product of two diagonal matrices is a diagonal matrix, if it makes any difference .

what does n ∈ Z mean?

@james it means n is an integer

i.e. n belongs to the set of all integers

but z usually denotes complex numbers

that's the small letter z

ok

thanks

it should actually be the blackboard-bold ℤ

n ∈ ℤ