Mathematics - 2018-11-13 (page 6 of 6)

Ted Shifrin

11:00 PM

@Jake: If you fix an $x$, say $0$, what $\xi$ contribute to the integral?

Jake Rose

@TedShifrin yeah I should, but I’m getting quite confused and I’m not sure exactly how to tackle it

it doesn’t seem like a hard problem which makes it all the more frustrating

Ted Shifrin

So I just asked a question.

CaptainAmerica16

@TedShifrin Ok, pay attention to what's necessary and what isn't.

Daminark

I never ended up doing Hatcher 1.3 before the class started but figuring things out for that first pset helped me get something of a grip on covering spaces. Still probably gonna eventually go through things more systematically

CaptainAmerica16

I suppose I shall attempt #5.ii now.

Jake Rose

11:02 PM

@TedShifrin -a to a?

Ted Shifrin

Right, so the integral is very easy.

Jake Rose

But if x isn’t 0 won’t these limits change?

Ted Shifrin

Sure, sure. Now do the fiddling with algebra. What do you need $\xi$ to satisfy for a general $x$?

What happens if $|x|$ is huge compared to $a$?

Jake Rose

$x-\xi \geq a$

Leaky Nun

@TedShifrin I'm still fascinated by the generalization of the fact that every matrix has an eigenvalue...

maths is beautiful

s.harp

11:06 PM

@Leaky what generalisation do you mean?

Jake Rose

G = 0 for large x

Ted Shifrin

Right, @Jake. Can you tell me how large?

Leaky Nun

@s.harp the spectrum of every element in a unital Banach algebra is nonempty

s.harp

did you read that in Folland?

Leaky Nun

yes

Jake Rose

11:08 PM

Bigger than $a+\xi$

Ted Shifrin

No $\xi$ in this answer, @Jake. In terms of $a$, how large an $x$ guarantees $G(x)=0$?

MatheinBoulomenos

I like that you combine complex analysis and functional analysis to prove it

CaptainAmerica16

@TedShifrin Is $a-b = -b$ a true statement?

Ted Shifrin

Ask yourself.

CaptainAmerica16

Why you do this to me

but fine

@CaptainAmerica16 Is $a-b = -b$ a true statement?

Jake Rose

11:16 PM

@TedShifrin I’m not sure

Ted Shifrin

Write down the two inequalities you need $\xi$ to satisfy to get a nonzero thing in the integral, @Jake.

chandx

k, so i did next task, is it done k?
check if there exists linear isometry $F$ such that $F\binom{1}{1} = \binom{1}{-1}$ and $F\binom{1}{-1} = \binom{-1}{1}$
so let $m(F) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} = A$ and I did liek
$A\binom{1}{1} = \binom{1}{-1}$ and $A\binom{1}{-1} = \binom{-1}{1}$ so it led to $a+b=1$, $a-b=-1$, $c+d=-1$ and $c-d=1$, summing stuff up and i got $a=0$, $b=1$, $c=0$, $d=-1$, so $A = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}$, but then it can't be isometry, cause matrix of linear isometry has to be invertible, and $\det A = 0$, also from another rea…

Ted Shifrin

@chandx: An easier way to think about finding the matrix is to deduce from linearity what $F(1,0)$ and $F(0,1)$ must be.

But, yes, you're right, because if $F$ maps two linearly independent vectors to linearly dependent vectors, then $F$ is singular.

Jake Rose

@TedShifrin yep done that

CaptainAmerica16

@TedShifrin Ok here is what I came up with for #5.ii Prove that if $a<b$, then $-b < -a$: Given $a<b$, we can create the statement $a+(-b)<b+(-a)$. Then we have $a-b<b-a$. Therefore, $-b < -a$.

Assuming it's even correct, my wording still seems kind of bad.

Leaky Nun

11:28 PM

man when I write $\mathcal L$ it looks like a weird $h$

Fargle

At least you're not the default TeX font writing $\mathfrak{R}$ and $\mathfrak{N}$.

That still irks the heck out of me.

Daminark

So I'm not on computer and thus can't see what you guys wrote, but I am fairly annoyed by mathscr

Leaky Nun

So $\mathcal L(C(X))$ is unital iff $X$ is compact?

that's L(C(X)) for Demonark

no, that makes no sense

Daminark

What's L here?

CaptainAmerica16

Ooh, I think I left out some important stuff looking back at what I have.

Leaky Nun

11:30 PM

$C_0(X)$ is unital iff $X$ is compact

Daminark

Ah, yeah I think that's right

Leaky Nun

@Daminark For a Hilbert space V, I hope, L(V) is the unital Banach algebra of bounded linear endomorphisms of V

(yes, those are certainly all words)

s.harp

@Leaky $C_0(X)$ is unital iff $X$ is compacat, $\mathcal L(V)$ is always unital

Leaky Nun

If I ever teach group rep, I'll make sure to do it over an arbitrary Hilbert space instead of a f.d. C-v.s. :P

Daminark

Why is Hilbert relevant?

Leaky Nun

11:32 PM

good question

maybe it isn't

@s.harp teach me

chandx

well, $F$ is additive, so $F(X+Y) = F(X) + F(Y)$, therefore $F\binom{1}{1} = F\binom{1}{0} + F\binom{0}{1}$, but that's just $a+b=1$ and $c+d=-1$, ye, bit simpler

Leaky Nun

@Daminark at least a Banach space, right

s.harp

further the correspondence $X \to C_0(X)$ can be turned around, if you have an abelian C* algebra $A$ you can construct a locally compact Hausdorff space $X$ with $A=C_0(X)$

this space is unique up to homeomorphism

Leaky Nun

so V is a Banach space, and L(V) the unital Banach algebra of bounded linear endomorphisms

s.harp

(this is all done in the first chapter of Folland)

Leaky Nun

11:34 PM

A representation of a locally compact T2 abelian group G is a continuous group homomorphism G -> L(V)*

Daminark

I feel like Banach would kick in because you want L(V) to be a Banach algebra.

Leaky Nun

"kick in"?

@s.harp is it the spectrum of A?

Daminark

As in, it'd be a condition on the space that we want

s.harp

what is interesting is now this: If $X$ is not compact there are a few standard procedures to compactify it, like one-point or Stone-Chech, this will translate on the algebra level to a unitisation of $C_0(X)$

Leaky Nun

as in, what does that phrase mean

Fargle

11:35 PM

"kick in" = "turn on"/"go into effect"

s.harp

different compactifications give different unitisations, ie stone-chech gives the multiplier algebra of $C_0(X)$

Leaky Nun

@s.harp and what is the result of the translation?

Fargle

roughly

Daminark

Hmm, there are distinct unitizations of a Banach algebra? How does that look algebraically?

Leaky Nun

thanks @Fargle

@s.harp what is the multiplier algebra?

lol @Daminark is way too late :P

Daminark

11:35 PM

Does adding the unit to other vectors look differently?

Leaky Nun

vectors?

Daminark

A Banach algebra is still a vector space, you know what I mean

Leaky Nun

hi @Semiclassical

Semiclassical

hi

s.harp

@Leaky the space $X$ is indeed the spectrum of the algebra. If you did algebraic geometry I can tell you that the points of $X$ are the maximal ideals of $A$.

Leaky Nun

11:37 PM

A group representation is a continuous group homomorphism from a locally compact Hausdorff abelian group to the unit group of the unital Banach algebra of bounded linear endomorphisms of a Banach space.

@s.harp that's exercise 1.26 of A.M. and I do know some algebraic geometry

@s.harp is that right? ^^

ninja hatori

@MatheinBoulomenos still in that case why this is true? it must come from {\mathfrak{m}^t \cap I \otimes M} for all {t}. Thus it belongs to {\mathfrak{m}^t(I \otimes M)} for all {t}; my guess is due to intersection commute with tensor so

s.harp

@Leaky the elements of the multiplier algebra are pairs $(L,R)$, where $L, R$ are linear maps $A\to A$ so that $L(ab)=L(a)b$, $R(ab)=aR(b)$ and $aL(b)=R(a)b$

Leaky Nun

oh man

that's way too nested

s.harp

here L is a generalisation of the map "multiply on the left" and R a generalisation of "multiply on the right"

Leaky Nun

I mean...

V is a Banach space, L(V) is the endomorphisms, and then if you unitize it...

s.harp

11:39 PM

L(V) always has a unit

Leaky Nun

oh

I'm stupid

@s.harp is this right? A group representation is a continuous group homomorphism from a locally compact Hausdorff abelian group to the unit group of the unital Banach algebra of bounded linear endomorphisms of a Banach space.

s.harp

it is too restrictive what you say

Leaky Nun

then how should I say it?

s.harp

if $V$ is a topological vector space, $G$ a topolgical group, then a representation of $G$ on $V$ is a continuous action $V\times G\to V$ where $G$ acts by linear maps

Leaky Nun

I don't like your directions

but does Maschke apply?

s.harp

11:41 PM

if you work with that you can reformulate: a representation is a homomorphism $G\to L(V)$ that is continuous when $L(V)$ is given the topology of pointwise convergence

Leaky Nun

how general can you be, and still have Maschke?

MatheinBoulomenos

@LeakyNun Maschke applies for unitary representations of compact groups, since you can do the averaging trick that you know from finite groups with a Haar measure

s.harp

we dont have maschke here @LeakyNun

Leaky Nun

@MatheinBoulomenos what is unitary representation of compact group?

s.harp

@Mathein unitary, not unital :)

Leaky Nun

11:43 PM

ouch

MatheinBoulomenos

@s.harp thanks

s.harp

@LeakyNun its more ouch than you think

a QFT is a unitary representation of $\widetilde{SL_2(\Bbb C)}$, irreducible representations correspond to the particle content of a QFT, but because that group is not compact we do not have a decomposition into irreps.

ie there are QFTs without any particles

MatheinBoulomenos

also Maschke applies to admissable smooth represenations of profinite groups

Leaky Nun

@MatheinBoulomenos let's not go that way

I want to focus on analysis for now :)

so what is a unitary representation?

Daminark

@LeakyNun :(

Leaky Nun

11:47 PM

@Daminark hey Daminark

A group representation is a continuous group homomorphism from a compact Hausdorff abelian group to the unit group of the unital Banach algebra of bounded linear endomorphisms of a Hilbert space.

s.harp

if your target space is a hilbert space and the action of the group is by unitary maps, you call the representation unitary

Leaky Nun

what is a unitary map?

and why do we need hilbert instead of just banach?

s.harp

well, $A$ is unitary if $A^*=A^{-1}$, it turns out this is the same as $A$ being an isometry

MatheinBoulomenos

I'm too tired I'm talking nonsense, what I meant is that you can always make a representation on a Hilbert space unitary with the averaging trick

Leaky Nun

heh?

Semiclassical

11:48 PM

$A^*$ being both transposition and conjugation (if we're talking about A as a matrix)

Daminark

Uh, I wouldn't define group representations like so if only because I know from number theory that representations over $\mathbb{Q}_p$ are a thing

Leaky Nun

@Daminark let's not go that way

I want to focus on analysis for now :)

s.harp

@Daminark those representations are not continuous (or possibly they are, I cannot remember right now)

MatheinBoulomenos

@LeakyNun you're repeating yourself

@Daminark let's go that way

Leaky Nun

@MatheinBoulomenos oh

quite a tautology, @Nobodyrecognizeable

Daminark

11:51 PM

I mean I definitely don't define a representation to be that. I'd define a representation generally and in a given context say to restrict. Also I'm mildly nervous that you're mapping to linear endomorphisms instead of automorphisms

Leaky Nun

aha, I forgot the word

I said "unit group of ... endomorphism"

Nobody recognizeable

@LeakyNun what ?

Leaky Nun

@s.harp I was about to ask you how to determine if a map is unitary, lol

s.harp

@Daminark since its a group action the endomorphisms that are hit must automatically be invertible. And your qualms are justified, I am talking about "continuous representations", since these are the only ones I have ever cared about I drop the word "continuous"

Daminark

Also there is a topology on Q_p as a profinite group

Nobody recognizeable

11:53 PM

@LeakyNun do you know which surfaces should be used as closed surfaces to y divergence theorem?

Daminark

While you probably mostly care about R and C, there is continuity at work for sure

s.harp

there is, but the vector spaces are not usually given a topology in those represnetations

Leaky Nun

@Daminark I mean, Q_p is locally compact, so there's that

also Q_p is just locally profinite