@math101 hmm.no

its been solved already

@N3buchadnezzar wassup man?

Aaron!

@Charlie Hey!

Its finally Friday!!

@Argon Finally!

Got to get down on Friday

Everybody's looking forward to the weekend

@Argon stop it,please

here we go...

@JasperLoy I know

Can surface integrals find surface area?

@JasperLoy What happens when you integrate a function over a surface physically?

@Charlie Funny!!!

@JasperLoy what do you think about that?I can write too!

@JasperLoy Area of what?

@JasperLoy I see. That is very strange!

@JasperLoy It makes sense, it's just really cool!

@JasperLoy No doubt.

Things get even cooler with vector calculus.

@anon I need to learn vectors better!!

Yo

@OldJohn Yo!

@OldJohn Oldjohn!

@Charlie Hi!

@OldJohn wassup?

@Charlie Not much here - just chillin :)

@JasperLoy agaaainn!

@JasperLoy i only know 3 menaings for XXX...

...que o pariu...

@JasperLoy not the whole thing

@JasperLoy less than 9 billion

7 billion guess

@JasperLoy yeah

I once read that 90% of all scientists who have ever lived are still alive

Is that different from the ratio of people today to people ever?

@JasperLoy Neptune will be annoyed

@JasperLoy Pluto

@JasperLoy mickey's dog...

Chatanooga Choo choo, won't you choo choo me home

hey

@anon How's your representation theory coming along

@BenjaLim hey!

@Charlie hey

@Charlie where are you in the world?

@BenjaLim I'm a nowhere man

kidding!

south america

right.

@OldJohn are you watching tottenham vs chelsea tomorrow?

It's tonight for me

@BenjaLim Not sure - I think I have too much to do tomorrow, unfortunately

lots of things to catch up on after a holiday

@BenjaLim pretty meh

I am spending too much time reading things I don't understand than actually learning the fundamentals.

@anon meh?

@OldJohn back from abu dhabi?

Mackey functors, plethysms, ...

right.

@anon Do you know much about the schur functor in any case?

I know what it is in a bare sense.

@BenjaLim No - back from Slovakia - Abu Dhabi is next month :)))

You would probably like it it's very combinatorial

@OldJohn So how now john senior the explorer

Slovakia was great - very interesting place - and great beer

Yes, the trace formulas for the effect schur functors have on reps, when viewed as polynomials in the eigenvalues, ubergeneralize identities involving symmetric polynomials (namely elementary and complete homogeneous ones in the newton-girard identities).

yeah my god so combinatorial

But it's not so bad I managed to prove pieri's formula after a little help :D @anon

I wonder if the generating functions $$\sum_{\lambda\in M}\mathrm{tr}(\Bbb S_\lambda V)q^{|\lambda|}$$ have nice closed forms for maximal chains $M$ of the Young lattice other than just $(1^n)$ and $(n)$.

@anon Well the trace of an element of $g \in GL(V)$ on the schur functor is the schur polynomial evaluated at the eigenvalues of $g$

but what is a maximal chain ?

young lattice?

A maximal chain of a poset $P$ is a linearly ordered subset that has no linearly ordered oversets in $P$.

woaaah what is the young lattice

The Young lattice is the lattice (a poset with sups and infs) of Young diagrams (which are in bijection with integer partitions obviously) with the $\le$ relation given by iteratively adding squares; see wikipedia.

right

@anon This shit is really combinatorial

and I have to tell ya I hated combinatorics at school

@anon @OldJohn Right I'm going to make a milkshake now

banana smoothie rather

bye guys!

@BenjaLim bye for now

So there are maximal chains $M=\{(1),(1,1),(1,1,1),\cdots\}$ and $N=\{(1),(2),(3),\cdots\}$ (viewing diagrams as partitions). The former corresponds to the sign representations and the latter to trivial representations of $S_n$, which under the effect of the schur functor and taking traces correspond to $e_n$ and $h_n$ respectively.

But I wasn't done typing!

yes that is correct

Because the schur functor in those cases is either $\textrm{Sym}^n V$ or $\bigwedge^k V$ @anon

Which, when put in the generating function I gave, give $\prod (1+x_nt)$ or $\prod (1-x_nt)^{-1}$ (I may be mixing up signs and which one goes to which now, hard to keep track)

/rant

I always get mixed up too

@anon Great both of us deal with similar things, we should chat more often!

@BenjaLim Dawg.

yes

@BenjaLim Do you have any old linear algebra asessments I could do?

(from your uni)

About matrices

no man

That stuff I did like march last year

@BenjaLim But don't you have access to the page?

holy moley where did my drink go

@anon Man. Am I correct in saying that the dimesion of the subspace of all symetric matrices in $K^{n \times n}$ is $n(n+1)/2$?

Yes.

Now I have to prove it !

I can use the canonical matrices as a basis

dude that is obvious

Interesting, arxiv.org/pdf/1210.0601.pdf attributes the peculiar existence of extra polytopes in dimensions three and four to algebraic properties of symmetry groups and Lie algebras.

Actually that's totally unsurprising on second thought.

@BenjaLim My algebra instructor gets me on those things though. I once had 1/2 a bonus point not obtained because I didn't prove $\{x,x^{-1}\}$ had two elements when $x$ was order two in a group.

And I'm like, you just don't want me to score perfect do you bro.

@BenjaLim What is?

That the subspace has the dimension 1+2+...+n.

@anon Why?

I mean, it is easily reasoned out, but why say "obvious"?

The (say, lower) triangle of elements determine the strict upper triangle and each component of the lower triangle can be chosen independently, hence 1+2+...+n.

Pretty intuitive once you have enough familiarity.

@anon Yes, of course.

@anon Surely you mean something else than what you wrote.

How do you figure?

@anon ?

Oops, I mean not order 2.

Good catch.

@anon HOw would you prove that $\dim S_1=\dfrac{n(n+1)}2$?

By forming the basis matrices $e_{nm}$ of $M_{n\times n}$ and showing $\mathcal{B}=\{e_{nm}+e_{mn}:n\le m\}$ is a basis for the symmetric matrices (this would need to be tweaked slightly in characteristic $2$ along the diagonal) and has $n(n+1)/2$ elements.

@anon $e_{nm}$ are the canonical matrices?

right

@anon What about "induction" from one dimesion to the other?

That is, if I had one row and one column, the dimension increases by $n$...

You can do it that way too.

@anon OK

The linear independence is immediate from the fact that the set of canonical matrices is linearly independent, so induction might be overkill.

@peoplepower Overkill?

Bleh

@JasperLoy I know, Captain. I mean, why?

Overkill: doing two proofs using precisely the same argument.

@peoplepower Come again?

Initial case + Inductive case

@anon Maaaaaaaaaaaan

What have you done to me.

@anon Man

@anon For antisymmetryc matrices, we have the particular case $2\cdot a_{ii}=0$. But $a_{ii}=0$ only if $2\neq 0$, For example, in $\Bbb Z/2\Bbb Z$, this is not true.,

Over any ring of characteristic two, the symmetric and antisymmetric matrices are the same.

It also serves as an answer for basement-level fruit.

@anon I can't see it.

Oh.

$-1=1$

=)

@JasperLoy What does "characteristic $2$" mean? What is the characteristic of field?

The characteristic of a ring $R$ is the integer $n$ such that $\underbrace{x+x+\cdots+x}_m=0$ for all $x\in R$ if and only if $n\mid m$.

@PeterTamaroff Any simple ring (division rings and fields fall in this class) has the property that there is some prime number $p$ for which every nonzero element has additive order $p$.

The characteristic is this prime p or 0 if they are all infinite cyclic.

Hmm. The trivial ring is the only one of characteristic $1$.

Not simple

I was typing before you commented on simple rings.

Ah, it is a matter of convention that 0 is a field, right?

No, {0} is not a field. Fields need 1 and 0 distinct (by convention), for reasons I forget over and over again.

The lack of an $\Bbb F_1$ though creates an $\Bbb F_1$-shaped hole in mathematics, into which many cool things fall.

Among them a possible route to proving the Riemann Hypothesis.

A mathematical reader hearing these things for the first time could be forgiven for believing I am trolling.