@JasperLoy You'll jinx us all!

@JasperLoy Don't make me sad!

I do not have a copy, I just used one from the library when I lived in Nevada.

@JasperLoy Have you heard of American Horror Story?

@JasperLoy A twisted horror series.

Pretty kickass.

@JasperLoy Timid is not the word there!

@JasperLoy People can learn fancy names, but not all the colours out there.

"Hey man, I got the "Emotional Skyblue Theme". "Oh, how nice".

"Hey man, I got the carmin color theme." "What is that. Isn't it a lipstick?"

@anon You there?

yes

@anon I have the following task:

Characterize the set $$\{A\in K^{n\times n}:A\cdot B=B\cdot A\;\;\forall B\in K^{n\times n}\}$$

i.e. $Z(M_{n\times n})$

@anon $Z$?

The "center" of an algebra is the subalgebra of elements that commute with all of the others.

@anon I'm talking about the vector space $K^{n\times n}$

you should just get scalar multiples of the identity

@anon Yes, I thought that. Actually, I thought

"Isn't it $=\{I_n\}$?"

But multiples makes sense.

@PeterTamaroff Even more, you're talking about the matrix algebras $M_{n\times n}(K)$. (Mere vector spaces don't have multiplication, after all.)

@anon Hehe, true.

@anon I should have said "the ring $(K^{n\times n},+,\cdot)$"

Hint. Consider plugging in the canonical basis elements $e_{nm}$ for $B$.

What makes you think that?

Yeah, that's what I thought.

@anon Nothing, I got something messed up.

@anon =)

:toughguyface:

@anon I denote them $E^{ij}$

cool notation bro

@anon Man.

I got that $A_{ij}=A_{ji}$ for each $j,i$

Oh, waitttt.

I am beginning to form the impression that quantum mechanics is in the process of being categorified.

@anon What might be the advantage in that?

Conceptual understanding.

Mathematical laxatives.

Span(Set): a categorification of the concept of sums-over-histories. mind=blown.

@anon Man, can I show you my work?

go ahead

it'd be easier with tensor calculus notation and the einstein summation convention methinks

@anon OK. So, considering the canonical matrices $E^{\ell m}$ for $1\leq m,\ell\leq n$, we have that

please use kronecker deltas :crossesfingers:

So

@anon =P

$${\left( {A \cdot {E^{\ell m}}} \right)_{ij}} = {\left( {{E^{\ell m}} \cdot A} \right)_{ij}}$$

Now, if $m\neq j$

$${\left( {A \cdot {E^{\ell m}}} \right)_{ij}}=0$$

And if $m=j$; then$${\left( {A \cdot {E^{\ell m}}} \right)_{ij}} = A{ _{i\ell }}$$

Similarily if $i\neq \ell$, then

$${\left( {{E^{\ell m}} \cdot A} \right)_{ij}} = 0$$

And if $i=\ell$ then

$${\left( {A \cdot {E^{\ell m}}} \right)_{ij}} = A{ _{mj}}$$

So our matrix is diagonal.

Just to let you know, I'm not going to bother reading that. :)

But if $m=j$ $i=\ell$ then $A_{mm}=A_{\ell \ell}$ so all the diagonal entries coincide.

@anon How would you do it?

@anon How do you use Kroenecker deltas for the canonical matrices?

Pretty much the same, I believe. Set $e_{nm}=[\delta_{in}\delta_{jm}]_{ij}$ so that $Ae_{nm}=e_{nm}A$ reads $A^{ij}\delta_{jn}\delta_{km}=\delta_{in}\delta_{jm}A^{jk}$ simplifies to $A^{in}\delta_{km}=\delta_{in}A^{mk}$.

If $k\ne m$ then setting $i=n$ we get $A^{mk}=0$, and if $i\ne n$ then setting $k=m$ we get $A^{in}=0$, which tells us all off-diagonals are $0$. Then setting $i=n,k=m$, we get $A^{nn}=A^{mm}$ regardless of $n,m$, so all diagonals are equal.

Which is essentially what you said (probably; I didn't read it).

@anon Well, but aren't you used to the notation $(A\cdot B)_{ij}$ fior the entries of the multiplication matrix?

The index notation and summation convention speed things up considerably, and is (arguably) occasionally invaluable in doing tensor calculations in say differential geometry. (Try deriving the geodesic formula or the Christoeffel symbols without them.) Arguably though coordinate-free stuff is more conceptual and easier to handle.

@PeterTamaroff what about it?

@anon Dunno, maybe we have a notation impass.

Well, I don't feel that I have any sort of notation impasse here. At any rate, since an element is central iff it commutes with a basis, from this we conclude central elements of the matrix algebra are scalar multiples of the identity, and conversely it is clear that scalar multiples of the identity are central, hence $Z(M_{n}(K))=KI_n$.

@anon Not you man, I mean we use different notations.

I use your notation and more. :)

Unless you just mean which choice of letters we are using for the indices; that can be paved over if desired.

@anon Well, you are the algebra overlord here.

So long as Mariano and Dubuque and Jack and ... aren't here.

@anon Bah! You're one of the algebra overlords

@anon Who's Jack?

Jack Schmidt, into groups and representations.

this is blowing my mind

@anon What does Mariano do?

algebra

(like, actual algebras)

@anon ?

"Degroupoidication."

LAWL

I believe you're missing a couple letters.

@anon Odd. I copied and pasted.

Degroupoidification.

It's yummy like alphabet soup.

@anon Man, at least I somehow understand the $\bf FinVec$ and $\bf FinSet$ with $\mathbb N$ part.

@anon I find it interesting that we can represent and work with any linear transformation as a matrix.

That is like the first grasp of abstract algebra I got, I guess.

:deadpan:

Hey guys....So I'm in the process of applying to graduate schools (doctoral programs), and I'm collecting my letters of recommendation. Problem: One of my references asked me to write my own letter, and I have no idea where to begin. Can anyone give any pointers or references?

Please no "0x821F3D" pointers either =)

Look at examples of other letters. Think about any sort of details about yourself and stuff you've done that could be spun in ways they sound relevant to people reading your letter.

Sell your soul.

@DanM.Katz "Why should you take me in? Bitch, I'm fabolous".

Perfect, ship it.

@JasperLoy So you carry nuts?

Uhh pretty sure everybody going for a PhD in math is something of a nutcase

@JasperLoy I prefer almonds.

@JasperLoy You're making progress with all that social awkwardness of yours

@DanM.Katz There are mods around...

Lol sorry

Apologies =) I don't post here often; I should probably take a look at the guidelines

@DanM.Katz Just delete it.

I can't look at your name without thinking of Dean Koontz.

He is a suspense-thriller novelist. I read Velocity.

Like 6 or 7 years ago.

@JasperLoy You should comment this.

I like differential operators

@belisarius Hello!

@GustavoBandeira hi

@GustavoBandeira Portugolisidelian.

math.stackexchange.com/questions/217203/… - please someone make a comment that more eloquently (and politely) than I'm capable of explains to the OP that category theory is not a suitable topic for students unfamiliar with the motivations for it

@kahen Sometimes, you should let the children get burnt with the stove, man.

Else, they don't learn.

George Carlin said that, in other words.

@anon Do you know about the symmetries of the weight diagram of a representation of sl_3?

it's beautiful

nope.

it's like a hexagon

It has $S_3$ symmetry group

don't know what a weight diagram is. although I believe according to them I have been teetering on the verge of unhealthily underweight for most of my life.

@anon how tall are you?

your weight?

I'm about 185

weight 64

Ah, hexagons are my favorite tesselation, possibly because of a minigame in mario party 2 and an SA2 advertisement in nintendo power in my childhood.

I am 130 and do not know my height (average).

your weight is 64? what kind of units are you using?!

kg

not pounds

this is australia

this is the internet, which is part of 'mericuh.

wow you're quite underweight

@anon Quick question

why do we care about characters of say $GL(V)$?@anon

Is it true that I have two representations of $GL(V)$ @anon

they have the same characters @anon

then they are isomorphic? @anon

did you just edit every single one of your lines to put a ping in

that is hilariously you written all over

what do you mean

anyway I dunno, I don't know much about representations of infinite groups

right

are you at uni @anon ?

yes, I started my freshman undergrad year this semester

oh wow

am I right in saying that you are going to uni late?

by three years yes

I mean like when did you finish high school?

@anon How come you're so pro?

I do math more than I do video games

I don't do video games too

And I'm not even pro

Guys

Well, I do video games, but only my old ones I have nostalgia for, and a few new ones.

@anon which uni are you going to?

I wouldn't consider myself pro.

Could you make some justice?

University of Nebraska at Omaha

hmm, it seems pedro has been downvoted

@anon nice.

@BenjaLim You're not serious I guess....

why?

@PeterTamaroff what in the world are $\phi$ and $\xi$ and how did you get the solution? I mean, I know how, but you pulled it out of a hat in your answer it seems like.

@anon Keep on reading...

you should put \underbraces around the implicit definitions of $\varphi$ and $\xi$

@PeterTamaroff I hate DEs

@BenjaLim Why?

sucks like fuck

they aren't all that beautiful, more like a giant handbag of symbolic tricks, but applying ideas of spaces of solutions and algebraic operations with operators are pretty cool imo. the Euler-Maclaurin expansion with the adjoint actions IIRC can be used to get the BCH formula, for a Lie-flavored example.

And as long as one doesn't care about actually finding a solution, there are plenty of beautiful existence theorems :)

Also Lie theory is applicable to PDE theory, but I don't know anything about that. Also, PDEs are critical to understanding differential geometry and modern theoretical physics.

I always thought existence theorems were very nice to have handy, but not beautiful in themselves, and just plain tedious to prove. Kinda signifies I'm not a career analyst.

@PeterTamaroff Dividing out differential operators. Reminds me of Euler and infinite sums.

@anon In what sense?

@Peter So you have learned some operator theory since Truesdell?

@BillDubuque I studied some "operator theory" from Spiegel.

Really liked it.

@PeterTamaroff Polynomials are ubiquitous.

@PeterTamaroff In the sense of "hey the symbols behave nicely but we'll ignore the question of if this means anything..." It could use some more clarification methinks but I liked it so I will upvote.

@anon Well, study the Laplace transform and it will make sense.

@BillDubuque =D

It does make sense, but you give no indication of why or how (or even if) to uninitiated readers.

@BillDubuque anon tells me you're an Algebra demi God.

@anon Well, that'd be a long story...

Which I can't probably answer yet!

I have some little insights, that's all.

@Peter No, I worship them while standing on some big shoulders! Your answer would be easier to comprehend if you made it clear that those even and odd part functions are defined later on.

@BillDubuque Hehehe, true, and, yeah, guess so.

@anon The Lie theory of symmetries underlying separation of variables is quite beautiful and has many applications. It's part of the group-theoretical approach to special functions. See e.g. work of Willard Miller cited in some of my posts.

@PeterTamaroff lol

@belisarius I'm back.

@belisarius This question is amazing: mathematica.stackexchange.com/questions/5676/…

Hi, I have a question.

If $E$ is a closed subset of a metric space $X$ and for any closed subsets $E\supset E_1\supset E_2\supset\cdots$, $\bigcap_{n=1}^\infty E_n\neq\emptyset$, does it manifest that $E$ is compact?

A space is compact if and only in every system of closed sets which has finite intersection property has a non-empty intersection.

I don't think that in general countable decreasing systems suffice.

Oh, now I noticed that you want metric spaces, not arbitrary topological spaces.

Yeah, it's just part of mathematical analysis, not so abstract.

Oh, sorry.

$E_n\neq\emptyset$

These closed subsets are nonempty.

Ah, it's implied. I'm confused with these concepts.

For metrizable spaces, countable compactness, sequential compactness, limit point compactness and compactness are all equivalent. Says Wiki.

I hope I did not miss something.

But your property seems to me as characterization of countable compactness.

The proof would be the same as characterization of compactness which I mentioned above.

For metric spaces, sequential compactness is equivalent to compactness.

Rudin's Principles of Mathematical Analysis Theorem 2.41.

I think this blog post is exactly what you need: The Finite Intersection Property in Compact Spaces and Countably Compact Spaces

See Theorem 2 there.

Sorry, that website is not accessible for me.

How come?

Is that related?

What about [google cached version}(webcache.googleusercontent.com/…)?

That's okay.

Yes, the PlanetMath link you posted is related - it is a very similar proof.

It seems that it's just the inference of de Morgan's law, therefore I want to think it over first.

Yes, exactly.

This was weird - a minute ago MSE looked as there were no questions on the site: screenshot.

Too many new concepts.

Hausdorff

@FrankScience If you are only interested in metric space, you can ignore that. All metric space have this property.

What's $\bigcap_{n<\omega}$?

Just a different notation for $\bigcap_{n\in\mathbb N}$.

Or $\bigcap_{n=0}^\infty$

Um, I see.

$X$ is separable $\iff X$ is countably compact. Is that right?

In metric space.

$\mathbb R$ is separable.

$\{(k,k+3/2); k\in\mathbb Z\}$ is a countable cover with no finite subcover.

countably compact

Or $[n,\infty)$ is a decreasing system of closed sets with empty intersection.

The cover I suggested is countable.

It seems that I misunderstood the concept. Let me reread the wiki page.

The condition that every open cover has countable subcover is called Lindeloff space.

This one is equivalent to separability (in metric spaces).

Countable compactness: Countable cover has a finite subcover.

Obviously: Lindelöf+countably compact $\implies$ compact.

Yeah, I confused these two concepts.

Let me try some examples.

No. Every metric space is Hausdorff.

Sorry if I provided you with too many new things and confused you....

No, it's just strengthening my ability of comprehension.

That was quite ELUoquent.

Just this trick: $\bigcap_{n=0}^\infty G_n=\bigcap_{n=0}^\infty\bigcap_{m=0}^n G_m$.

I have to go afk for an hour or two.

I hope you'll manage that somehow, Frank. (You're doing fine I'd say - you suggested the claim of that result yourself.)

See you later.

Thank you, but unfortunately maybe I'll be away.

da fuq

@anon

hey

eh

I don't understand what is happening here

You know pieri's formula?

nope

Do you have a copy of fulton and harris?

I do..

@BillDubuque Hey, I have a question to ask you. Do you know of any references concerning the irreps of $\mathfrak{sl}_n$ as schur functors? I am looking at fulton and harris but I don't understans what they are saying.

@anon pg 455

A.7

I proved Pieri's formula the other day with a little help from someone on math.se

@anon but now in A.9 they say "formula (A.7) applied inductively yields"...

I get how $H_\lambda = s_{(\lambda_1)} \ldots s_{(\lambda_k)}$

That came from the fact that by Pieri's formula, we have for example that $s_{(3)} \cdot s_{(2)} = s_{(5)}$

this is because there is only one way to add 2 boxes to the young diagram that is just 3 boxes in a row.

@anon

what's the question?

I don't understand A.9 of Fulton and Harris

How is it that by applying A.7 inductively we get A.9

@anon

Hi

Filling up boxes with 1's corresponds to adding boxes in the first multiplication of $s_{(\cdot)}$'s, filling up boxes with 2's corresponds to adding boxes from the multiplication by the third term, and so on, right?

hello Chris's

hmmm kinda

but I don't get what they mean "applying formula A.7 inductively" gives...

Isn't "applying A.7 inductively" a succinct description of what I just wrote?

I'm kinda confused

applying the first time gives the filling by 1's, applying it the second time with 2's etc. (or something like that)

right but my understanding

for example $s_{(3)}s_{(2)} = s_{(3+2)} = s_{(5)}$

I don't get any sum at all by applying the pieri formula inductively @anon

or perhaps 1's label the boxes there from the start, and 2's label those added on after the first multiplication. I think that works better.

Is not $s_{(3)}s_{(2)}=s_{(3,2)}+s_{(5)}$?

no why?

You are looking at the sum over all $s_\nu$

where $\nu$ is a young diagram obtained from that of the young diagram with just 3 boxes in a row

there is only one way to add 2 boxes to a young diagram with 3 boxes in a row @anon

well, for example, $(x^3+y^3)(x^2+y^2)=x^5+x^3y^2+y^3x^2+y^5=(x^5+y^5)+(x^3y^2+x^2y^3)$ though, right?

DA FUQ................

@BenjaLim can't you start new rows though?

hmmmmmmmm

oh yeah da fuq

what da fuq

ahh crap

fuck i'm stupid

@anon Thankx

fucking hell

yup

@anon thanks man

thanks I owe ya one

@anon I should get ready for dinner

that was fun and entertaining

@anon Hey

I think we are very much alike

@anon about the same age

you're probably 2 years older than I am

but we have interests in rep theory

@anon One day we may well be collaborating together

doug is the only user on mse who can (and does) annoy me

doug spoonwood

@anon Sorry if I'm pingy

@anon bye man

later

"skullie" sounds like something anon would say.

@JasperLoy will see them. I am searching for the link jonas posted about his em for pedestrian.

@JasperLoy yup.

Hi

Where could I find a proof for the identity (35) here?

@JasperLoy dude, I was just interested in whether he was going to include second quantization in his presentation.

Also, I have the ten volumes by landau, however, thats not the point.

@Chris'ssister Interesting Identity. I do not know the proof though. Thinking on it.

For even, I think I can guess.

@JayeshBadwaik: yeah. Based upon on it one may create a lot of nice problems.

It can be just $\tan{(\pi/2 - x)}$ or something.

@Chris'ssister In here search for continued fractions, you get a formula for $\tan nx$ may be that can help. May be

@JayeshBadwaik: OK. Thanks!

I don't know how else to get a square root of $n$ from the tan. Or probably just use taylor series?

Well, don't bother, that continued fraction looks a lot like taylor series!

@JayeshBadwaik: have you met it before?

I would write it as the $\tanh$.

@Chris'ssister No.

@JonasTeuwen Hmm.

Will give you exponentials.

@JonasTeuwen Yup. I tried that. Though, I wrote $\tan$ directly in terms of $e^{ix}$ (which is the same thing actually)

Yes, of course.

And then you write it as $1 + \text{junk}$.

@JonasTeuwen Yup, and then bound that junk? (in case of even)

but that is not very promising for the odd case (at least I cannot see it) .

Hey

@N3buchadnezzar hi

Yea whatever, need food.

How to depict the least-upper-bound property in general metric space?

Good morning :)

Hello everyone, I see lots of formulae on Math.SE, and obviously it's all written in latex. But do you use some kind of editor (wysiwyg or something like that) to put down the formulae quickly?

Hi @JonasTeuwen, do you know how to estimate pointwise the difference of Fourier transforms $\hat{mu}(\xi) - \hat{f_{*}\mu}(\xi)$ of measure and it's pushforward without series expansions? I failed with the last...