Mathematics - 2017-12-09 (page 3 of 3)

Semiclassical

9:00 PM

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

anon

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

Semiclassical

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

anon

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

Semiclassical

ah

Tobias Kildetoft

@Semiclassical What heathen labelling is that?

2

Semiclassical

9:02 PM

so triplit $\oplus$ singlet

lol

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

Tobias Kildetoft

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

Kevin Driscoll

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

Tobias Kildetoft

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

Semiclassical

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

Tobias Kildetoft

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

Semiclassical

9:13 PM

depends.

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

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

Leaky Nun

Fundamental theorem of group theory @TobiasKildetoft

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

Semiclassical

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

Tobias Kildetoft

@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

Semiclassical

uh

we don't label by dimension

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

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

Tobias Kildetoft

@Semiclassical Ahh, good

silly half-integral weights :)

Semiclassical

9:18 PM

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

Tobias Kildetoft

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

Semiclassical

hmm. am I remembering things wrong? I probably am

Tobias Kildetoft

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)

Semiclassical

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

Tobias Kildetoft

@Semiclassical $-1$

Semiclassical

9:22 PM

ah, okay

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

Tobias Kildetoft

with vector being the other standard basis vector

Semiclassical

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

Tobias Kildetoft

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

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

Semiclassical

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?

Tobias Kildetoft

yeah

Semiclassical

9:23 PM

okay

then yeah, factor of two

physics would have 1,0,-1

Tobias Kildetoft

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

Semiclassical

yup

Kevin Driscoll

@BalarkaSen You around?

Tobias Kildetoft

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

Semiclassical

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.

Balarka Sen

9:26 PM

@Kevin Yep

Tobias Kildetoft

@Semiclassical So what would negative spin mean? :)

Balarka Sen

at least semi-around

Semiclassical

z-component of spin points up instead of down

Tobias Kildetoft

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

Semiclassical

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

Tobias Kildetoft

9:27 PM

(when considered as highest weight)

Semiclassical

oh

Tobias Kildetoft

maybe the connection in this case is just a coincidence really

Semiclassical

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 $s<0$

Kevin Driscoll

@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 $0 \to \mathbb{R} \to \mathbb{R}^2 \to \mathbb{R}^2 \to H^1(T^2) \to \mathbb{R}^2 \to \mathbb{R}^2 \to H^2(T^2) \to 0$

Semiclassical

to some extent this really comes down to the experiments

Kevin Driscoll

9:34 PM

@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

Balarka Sen

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

Semiclassical

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 $\hbar\sqrt{\frac{3}{4}}$

Kevin Driscoll

@BalarkaSen Yep

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

Semiclassical

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 $s,m_s$

Tobias Kildetoft

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

Semiclassical

9:36 PM

\hbar

Tobias Kildetoft

whatever

Semiclassical

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

Balarka Sen

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

Semiclassical

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

Tobias Kildetoft

@Semiclassical probably a fine approximation

Semiclassical

9:38 PM

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

Balarka Sen

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)

Kevin Driscoll

Thats the right direction

Balarka Sen

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?

Kevin Driscoll

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

Balarka Sen

Right. Hm.

Semiclassical

9:40 PM

@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

Tobias Kildetoft

@Semiclassical Would have been strange if it was I guess

Kevin Driscoll

@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

Semiclassical

yeah

Balarka Sen

Right, this is not just a chase game

Kevin Driscoll

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

Balarka Sen

9:42 PM

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

Tobias Kildetoft

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

Semiclassical

quite

Balarka Sen

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?

Tobias Kildetoft

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

Semiclassical

it's more a shorthand imo

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

Kevin Driscoll

9:44 PM

@BalarkaSen I wouldve said cohomology, but ya

Tobias Kildetoft

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

That just does not look right

Balarka Sen

@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 $U \cap V$

Semiclassical

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)

Balarka Sen

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 $H^1(L) \oplus H^1(K) = H^1(U \cap V)$

Semiclassical

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

Balarka Sen

9:47 PM

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

Semiclassical

~~doh, forgot~~ fixed

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

Balarka Sen

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

Vrouvrou

Hello@LeakyNun

Leaky Nun

hi

Semiclassical

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

Tobias Kildetoft

9:51 PM

@Semiclassical Right

Semiclassical

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

Tobias Kildetoft

@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

Semiclassical

hmm

Tobias Kildetoft

which is rather important in some cases

Semiclassical

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

Tobias Kildetoft

9:54 PM

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

Semiclassical

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

Balarka Sen

@KevinDriscoll Was the last thing I said sensible?

Semiclassical

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

Tobias Kildetoft

@Semiclassical I never did learn that bra-ket notation

Semiclassical

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

Kevin Driscoll

10:00 PM

@BalarkaSen I think so, but I not sure

Semiclassical

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

Balarka Sen

@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

Semiclassical

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

Balarka Sen

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

Semiclassical

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

Balarka Sen

10:05 PM

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

Semiclassical

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.

Balarka Sen

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

Tobias Kildetoft

@Semiclassical Right

Semiclassical

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

Tobias Kildetoft

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

Balarka Sen

10:06 PM

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

Daminark

Just landed in Texas

Kevin Driscoll

@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

Semiclassical

i mean, i'd have (using lazy notation since I can't be arsed to latex it all out)
L(1)^2 ~ L(2)+L(0),
L(1)^3 ~ L(3)+L(1)+L(1),
L(1)^4 ~ L(4)+L(2)+L(2)+L(0)+L(2)+L(0)

but I can't really see a pattern

Balarka Sen

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

Kevin Driscoll

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

Balarka Sen

10:09 PM

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

Kevin Driscoll

Oh okay, cool. Then ya I am with you

Balarka Sen

I have a meridian loop in $A$

and a meridian loop in $B$

Semiclassical

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

Balarka Sen

No longitudinal curves that goes around the hole

Tobias Kildetoft

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

Balarka Sen

10:10 PM

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

Semiclassical

niiice. upvoted

Kevin Driscoll

Ya I noticed that, though not in so many words

Tobias Kildetoft

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

Balarka Sen

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!

Semiclassical

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)

Tobias Kildetoft

10:14 PM

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

Semiclassical

ah

Kevin Driscoll

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

Semiclassical

i'm no help there

Balarka Sen

yup

Tobias Kildetoft

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

Semiclassical

10:15 PM

hm

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

Tobias Kildetoft

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

Semiclassical

right

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

Tobias Kildetoft

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

Semiclassical

right

Tobias Kildetoft

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

Kasmir Khaan

10:17 PM

@TobiasKildetoft Hi Tobias :D

Tobias Kildetoft

@KasmirKhaan Hi

Kasmir Khaan

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

Tobias Kildetoft

@KasmirKhaan Denmark

Kasmir Khaan

vaaaaah :D

är du dansk :D

Tobias Kildetoft

ja

Kasmir Khaan

10:18 PM

najs ju :o

trodde du var tysk eller nåt =p

Tobias Kildetoft

And you are Norwegian?

Daminark

At some point I gotta sue Denmark for copyright

Kevin Driscoll

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

Kasmir Khaan

nah living in sweden =p

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

Tobias Kildetoft

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

(especially the nåt did not look familiar)

Balarka Sen

10:20 PM

@Daminark lmao

Kevin Driscoll

@daminark You mean trademark?

Tobias Kildetoft

would have expected nånting

Kasmir Khaan

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

so you were right =p

någonting

Balarka Sen

@KevinDriscoll Wait, how are you concluding that?

Semiclassical

hmm

Tobias Kildetoft

10:21 PM

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

Kasmir Khaan

haha >< that part was my bad =p

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

we using the book of serr

Balarka Sen

all i know are swedish curses

Semiclassical

"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."

Kasmir Khaan

I started Reading it without any lectures

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

:O

Semiclassical

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

Kasmir Khaan

10:22 PM

@BalarkaSen where do ppl allways learn that first ?

Kevin Driscoll

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

Semiclassical

I'm probably gasping at straws.

Tobias Kildetoft

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

Balarka Sen

@KevinDriscoll Right, you have to be careful

This is a very specific case

Kasmir Khaan

@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 :/ )

Tobias Kildetoft

10:23 PM

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

Semiclassical

hmm

Tobias Kildetoft

@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

Kevin Driscoll

@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 $H^k(V)$

Tobias Kildetoft

@Semiclassical Since the relevant partitions always are like that for our cases here

Semiclassical

alright

Kasmir Khaan

10:25 PM

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

I wish ill get there this year :D

Tobias Kildetoft

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

Daminark

@KevinDriscoll yeah yeah that

Kevin Driscoll

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

Tobias Kildetoft

@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

Kasmir Khaan

@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?

Semiclassical

10:26 PM

ah

Balarka Sen

@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

Kevin Driscoll

We just happened to have $U \cong L$

Balarka Sen

Well, just being diffeomorphic does not suffice

I mean

Tobias Kildetoft

@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

Balarka Sen

$L$ 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

Kevin Driscoll

10:28 PM

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

Kasmir Khaan

@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

Balarka Sen

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

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

Narcissus jewel

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

Tobias Kildetoft

@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?

Kasmir Khaan

Stockholmsuniversity =p @TobiasKildetoft

Tobias Kildetoft

10:32 PM

cool

Kasmir Khaan

@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

Tobias Kildetoft

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

Kasmir Khaan

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

Tobias Kildetoft

@KasmirKhaan fall 2014 to fall 2016

Kasmir Khaan

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

so tell me next time you are here :D

Kevin Driscoll

10:35 PM

@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

Kasmir Khaan

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

Tobias Kildetoft

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

Balarka Sen

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

Semiclassical

night

Daminark

Good night Tobias!

Balarka Sen

10:37 PM

Night!

Kasmir Khaan

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

Kevin Driscoll

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

Balarka Sen

Right

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

Kevin Driscoll

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

Balarka Sen

True!

Daminark

10:42 PM

@KevinDriscoll never thought I'd see real coefficients

But I have

This'll be a story to tell my grandkids

Kevin Driscoll

@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!?

Balarka Sen

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

@Daminark de Rham = real coefficients

Kevin Driscoll

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

Balarka Sen

lol

Kevin Driscoll

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

Balarka Sen

10:46 PM

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

Kasmir Khaan

@Daminark sup dami :D

@Daminark how did it go with your exam?

@anon Hello Anon :D

MatheinBoulomenos

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

anon

hello

Kasmir Khaan

@MatheinBoulomenos Mathein :D

Balarka Sen

He's doing de Rham cohomology, @Mathein

Kasmir Khaan

10:50 PM

@anon long time no see! how are you ?

No cell structures

alright

just sauce

Hi @Kasmir

:D

11:02 PM

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

Semiclassical

well, it does have a right identity

Daminark

@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

Semiclassical

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

Bennett

Let $\displaystyle (AB)_{ij} = \sum_{k=1}^{n}a_{ik}b_{kj}$ $ ~~~~~\longleftarrow$ (this might not even be that relevant).

Why does $i \ne j \in \left\{1, \ldots, n \right\}$ implies $i \ne k$ or $k \ne j$?

If $i \ne j$ then $i \ne k$ because $k \in \left\{1, \ldots, n\right\}$ which contains $j$.

My question is where does the ***'or'*** bit come from?

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