just not topology

@TedShifrin those meant as Review , but ill Watch yours as well :D

@LeakyNun because they are not shrinking to 0 but are just having their maximum value at same value of y

Strang is just computations and he's ridiculously rambling ...

I already told you that you need linear algebra some time ago

Is $\sin(2)$ in closed form? What about $J_0(2)$?

in fact ill just Watch yours ><

@Semiclassical by "closed form" we mean elementary functions

I don't believe it, @Leaky.

@TedShifrin I gave you a graph...

trust the computation

Yeah Kasmir you'll want to learn serious linear algebra. And I mean for example, the bit that I saw from having audited the class used Jordan Normal Form

Okay. What defines elementary?

I'm doing calculus. The hell with the graph.

@MatheinBoulomenos yes I know , did not forget that, i asked you the question about what Products between the ones you mentioned, leaky took it as general question

Oh, you stated the problem incorrectly, Leaky. spank hard

@MatheinBoulomenos so now Ted told me direct Product is important, and i asked what else " between the Products" :D

You said the minimum for $f(x)=a^x-x$ was at $0$.

That is NOT what your picture is doing.

shit

That is crap, Leaky.

y=0.... not x=0...

sorry

@TedShifrin Ted do I need to Watch the integral parts too ? or i jump to linear algebra directly?

@Kasmir: Just watch linear algebra stuff.

because you mixed those up in a hard way :D

ok ill find where they start on the list

linear algebra is an integral part of the course

@Mathein kek

integral part means?

(My point just being that the division of special functions into elementary and non-elementary is usually just a matter of convention)

The easy stuff at the beginning you can skip, Kasmir. You want mostly chapter 4 and 9.

thanks Ted! :D

am looking now

@Ted do I need to learn symplectic geometry first before doing Kähler manifolds?

No, @Mathei. I sure didn't.

(I say mostly because there’s certainly a difference between, say, e^x vs erf(x) vs some Painleve transcendent)

@Ted I see, thanks

@TedShifrin just to be sure, i should start at lecture 31 right?

blah blah blah special functions blah blah

@TedShifrin titled rank and consistency

minimum occurs when $a^x \ln a - 1 = 0$, i.e. $x = \ln_a\left(\frac{1}{\ln a}\right)$

You should figure out what you know and what you don't. Well, echelon form stuff and general solutions of inhomogeneous equations came before that.

There will be more stuff way at the end (eigenvalues, vectors, etc.).

@TedShifrin are those enough?

And some stuff on quadratic forms in chapter 5.

Is the Kähler metric on a smooth projective complex variety canonical or it just some existence result? It's really intriguing that an algebraically defined object has such a rich geometric structure

If you say the, we presume it's the one coming from $\Bbb P^n$, @Mathei.

There are Kähler manifolds that are not, however, projective.

And the embedding might or might not be isometric. But if you don't specify, we assume it's induced ...

$\frac1{\ln a} = \ln_a\left(\frac{1}{\ln a}\right)$

@TedShifrin I got them :D the videos dont show chapters >< but ill look at them all and see what is needed, btw they can be watched without loss of continuity right? like if i skip the analysis part

$\ln \ln a = -1$

$\ln a = -1/e$

$a = e^{-1/e}$

aha

Oh of course, duh. We just need a Kähler metric on projective space and then restrict that. Okay that makes it less mysterious

Yeah, so, @Leaky, it is correct, but I don't see any way to solve for $a$. I think Desmos just did it numerically.

@TedShifrin I just did it

$\ln_a\left(\frac{1}{\ln a}\right) = \frac1{\ln a}\ln\left(\frac{1}{\ln a}\right)$ was the key observation

The critical point is $x=-\log_a(\log a)$.

does one need spectral theorem for rep theory?

or complex inner product for that matter

@KasmirKhaan how strong are you with complex numbers?

Complex inner products, I know are important for a certain "averaging trick" you do

So $a^x = 1/\log a$.

@LeakyNun i dont understand the question , but i took complex analysis

@KasmirKhaan ah

Setting that equal to $x=-\dfrac{\log\log a}{\log a}$ yields $\log\log a = -1$.

And that, in turn, yields $a=e^{1/e}$. That's easier than the stuff you're doing, Leaky.

There's no hermitian metric on the sphere, @Mathei. Watch out.

@TedShifrin thanks

Oh I remember now @Leaky

oh yeah, of course

@Daminark what is it

@KasmirKhaan find a maximal open subset of $\Bbb C$ wherein $z \mapsto \ln z$ is continuous

Let b=log(a). Then the above becomes e^(bx)=1/x —> bx e^(bx) = b -> bx= W(b)...bleh

So let's say $\rho:G\to GL(V)$ is a representation, you call it unitary if $V$ has an inner product such that $\langle v,w\rangle = \langle \rho(g)v, \rho(g)w \rangle$ for any $g\in G$ and any $v,w\in V$

@LeakyNun leaky I got algebra stuff to do -.-

@LeakyNun we use Log z not ln

I shouldn’t do algebra on an empty brain

@LeakyNun and that need to be defined where it make sense :D

Now for any finite dim rep of $G$, you can find such an inner product by letting $(v,w) = \frac{1}{|G|} \sum_{g\in G} \langle \rho(g)v, \rho(g)w\rangle$

@Semiclassical take something sweet , it will boost ur energy :D

also Brains need glucos

Where $\langle | \rangle$ is any inner product on $V$

It’ll have to settle for fructose, I’ve got some fruit

haha

let G be a Group s.t all subgroups of G are normal in G, for a,b in G prove that ba = a^j b for some j

this is easy unless am missing soemthing

@Daminark I see

bab' = a^j , because of the normality of all subgroups

a^j has to be in a normal subgroup call it A

and we conjugating by b

This is useful because if $V$ is a unitary G-rep and $W$ is some invariant subspace, then $W^{\perp}$ is also invariant

@Daminark you have heavy notation problems

you used 3 notations for the same thing

any handsome person saw what i just wrote?

Leaky, you're in no position to complain, after confusing domain and range.

@TedShifrin :'(

Wait what did I do? Just for general reference?

@Daminark you used $\langle , \rangle$, $(,)$, and $\langle \mid \rangle$ for inner product

Well they're different inner products

Yeah, it looks like Demonark was being quite careful. Not that I wish to defend him.

I use $\langle | \rangle$ for $\langle v,w\rangle$ when I'm just denoting the inner product without any input because the comma just doesn't look too nice

It looks fine to me.

$\langle \cdot , \cdot \rangle$

Your notation looks too much like bra-ket notation.

now it looks like a penguin

I'm not too familiar with bra-ket notation but yeah I dunno, some vague shit sense of aesthetic I guess?

@Daminark nice trick

Whatever the case, the thing about using parentheses was a very deliberate choice, since you started with one and got another

Yeah I think it's called Weyl's averaging trick

You apparently can do it on compact groups by integration but I'm gonna let someone who actually knows math answer that

why is it necessary to divide by $|G|$?

to get the right scaling properties

it won't be an inner product otherwise

That's not right. Any positive multiple of an inner product is an inner product.

exactly

Oh whoops

That's actually a good point then, :thinking:

oh yeah, brainfart

Damn. What's the world coming to. First Mathei defends me. Then I defend Demonark. Then I defend Leaky. I resign.

what's wrong with a world of mutual-defence?

@TedShifrin you cant quit now Before you defend or be defended by kasmir

@Daminark more like, :thonking:

@MatheinBoulomenos what's a spectrum?

I might guess that it has to do with when you try to generalize this trick to Haar measure? Otherwise nothing comes to mind

in algebraic topology or in algebraic geometry or in functional analysis?

spectrum can mean different things

@MatheinBoulomenos linear algebra

just the set of eigenvalues

wat

@Daminark or else the sum will not converge when you take it to infinity, I guess

the $\frac1{|G|}$ is just $\mathrm dx$

if the group happens to be a manifold

No, it's $dx/\text{vol}$.

But yeah so if $V$ is a unitary G-rep and $W$ is an invariant subspace, then $W^{\perp}$ is as well, so to prove that every finite-dim complex $G$-rep (where $G$ is a finite group) is completely reducible, all you'd need to establish the existence of an invariants subspace $(V^G?)$, do this averaging trick to show that this is a unitary $G$-rep under the right inner product, and then induct

hi @TedShifrin @Daminark

@MatheinBoulomenos

Hi @Adeek

Demonark, @Mathei: I think the normalization really only matters when you want to average things and make sure they're in the same "homotopy" (or cohomology) class.

hi Karim

@TedShifrin I was visiting with my analysis prof he told me I have pretty good ideas but I tend to overcomplicate things lol

You sometimes get confuzled and don't see the forest for the trees, yes, Karim.

yeah

@Daminark I won't try to imply that I know math, but yeah it works out for compact groups in the same way, you just replace the sum by an integral and do the same thing

@Mathei, Demonark: Continuing ... there's always a unique left-invariant metric on a Lie group ... unique up to positive scalars.

Does Haar measure fit in this somewhere?

Let $\rho : G \to GL(V)$ be a unitary representation. Let $W$ be a subspace of $V$ such that $\rho(g)W = W$ for all $g \in G$. Let $w' \in W^\perp$ and $g \in G$. Then, for all $w \in W$, $\langle \rho(g)w', w \rangle = \langle w', \rho(g)^{-1} w \rangle = 0$, so $W^\perp$ is also invariant.

Haar measure = left- (or right-) invariant measure on a Lie group ... but it makes sense more generally

Does this look right? @Daminark @TedShifrin

Kk

@Ted I'm not too familiar with how cohomology comes up here, but it sounds reasonable that it's a topological question if anything? Since I'm not seeing why the algebra forces that. Also that's nifty about the Lie group metric

@Mathein so I guess my question would then be, why do we require compact instead of finite measure?

You can run into that kind of stuff in physics, such as when you want to do generalized coherent states

compact = finite measure for the Haar measure

Dumb question: If $V$ is a vector space that is a topological vector space with respect to some topology $\tau$, will $V$ also be a TVS with respect to any topology finer than $\tau$?

Oh... Okay that makes sense. I need to learn harmonic analysis tbh

Yeah, that looks right, Leaky.

@TedShifrin thanks

Though it’s not a standard topic

@user193319 no. $\Bbb R$ with the discrete topology is not a topological vector space, since scalar multiplication is not continuous

@TedShifrin for $\int_{-a}^a [\sqrt{(a^2 - x^2)}]$, I should set $a=sin x$ then?

that's the best method

@MatheinBoulomenos Shite! Thanks for the response.

@Trey: You can — but you mean $x=a\sin\theta$. And if you need an antiderivative, you should do that. But if you want that integral, you should use your brain instead, as I suggested an hour ago.

> you should use your brain instead

damnnnn savage af

Sometimes thinking is harder than proceeding mechanically.

But I don't like being ignored. :P

That's why Alessandro says hi to me, whether he wants to or not.

what if he doesn't

k e k

Hi @Ted

why does herstein uses A(G) instead of S_G

the Group of permutations?

does the letter A stand for something ?

@KasmirKhaan because it's the automorphism group of the set G with no structure

automorphism

Something's not right there, Kasmir. There's no reference to $n$.

oh right ><

@TedShifrin lol come on it's clearly understood

Oh yeah, sure.

hmm so the Group of permutations is the Group of automprphisms?

Automorphisms of the set ... yes.

but i thought aut (G) is a subgroup of that

aha

hmm

@KasmirKhaan that's when G is a group

There were zillions of reasons I never liked Herstein. Although there were a few good exercises (and some ridiculous ones).

iam doing it for exercies ><

what do you mean ridiculous?

DF had no hard ones

I mean hard and not (to me) interesting in any way.

Herstein's got a soft spot in my heart, it was fun enough, though it did get me thinking about permutations in the $x(f)$ sense

well kasmir is lost by the aut of G busness as a set

Which people seem to not do anymore, so that threw me off a bit when I kept multiplying cycles and getting them wrong

Oh, gotta love functions on the right. NOT.

need to put that into my head

i also hate that

functions looks silly on the right

@KasmirKhaan do you know what automorphism means abstractly?

we should write from right to left that way

(I will say, I do see a case for replacing the way things currently work with that, it just feels more natural)

well i thought i knew leaky

(But I never heard of anyone other than Herstein do it like that so I'd recommend against learning it that way)

because what i had in mind was the bijectiive mappings

@KasmirKhaan all of them?

or in other Words what i mean is

be specific

an isomorphism from G to itself

1-1 onto hom, from G to G

@Daminark gotta love $f \circ g$ being $A \xrightarrow g B \xrightarrow f C$ in the "right" way

but without structure

@KasmirKhaan right, so an automorphism is a bijective map that preserves structure

how would we make sense of automorphism as sets?

@KasmirKhaan no!

-__-

did I say automorphism as set?

I said automorphism group

of the structureless G

automorphisms always form a group

and if there's no structure, then all bijective maps are permitted

that's why you get S_G

aha

neat

so we get all the permuations that way

of G

yes

@Kasmir: Someone just asked a question on main you should figure out. Suppose $A$ is an $n\times n$ matrix satisfying $A^2 = I$. Prove that $A$ is diagonalizable.

okay :D

@TedShifrin $\sqrt{A^2} = \sqrt{I} = I$, so $A=\pm I$ :P

Leaky should make sure he knows how to do it, too.

well I don't even know if I know how to do it

I'm amused that DonAntonio (who has almost 150K rep on here) posted a wrong answer.

high reps mean nothing these days

there are some high-rep users who post basically nonsense

CRUDE has seen many

It comes from doing stuff way too quickly. But there's no excuse for someone like that.

I'd like to believe my rep means something.

sup nerdoids

@TedShifrin sure it does

I just mean, not every high-rep user is good

@MatheinBoulomenos Okay. How about this. Suppose that $V$ is some vector space, and let $f : M_n(\Bbb{C}) \to V$ be some map, where $M_n(\Bbb{C})$ is the TVS of all $n \times n$ matrices. Define on $V$ the largest/finest topology with respect to which $f$ is continuous. Will $V$ also be a TVS with respect to this topology?

Ted not only it means something, it also underrated :D

@EricSilva inf nerd

because you should get upvotes for the answers u put ( more than u getting )

His excuse (after he removed his wrong answer) is that he was too busy in private chat. I have been known to be stooopid, too, so I'll give him a break.

donno why tho ppl dont get them upvoted ._.

Yeah that's my question, if you're a high rep user, presumably people saw the answer, read it over, felt it was correct, and then upvoted

well tbh i really forgot how to do it

._.

I think it's a lot like referring of journal articles and NSF grants. I think often big-name people get treated better than they should, just because of their big names.

@user193319 $f$ is any function? then I don't see why it should be TVS. if $f$ is just constant, then the finest topology which makes it continuous is the discrete topology

@Kasmir: I'm not sure you know how to do it. But this kind of thing is important for rep theory. You need to understand eigenspaces.

@MatheinBoulomenos Hmm...Okay. It isn't constant map, but it is a linear map.

the map which is constant $0$ is linear

But I am saying that $f$ isn't constant.

@TedShifrin i ll put that in my notes :D not i got linear algebra , eigenspaces and direct Product , thanks Ted ! il make sure to read them Before january

Add the hint for the problem: Show that the (two) eigenspaces of the matrix span $\Bbb R^n$. @Kasmir

Be careful about that though, there isn't much time and you can only go for so long on a superficial understanding, which you may get if you go a bit too quickly

I think the answer is yes iff $f$ is surjective

@TedShifrin Done ! thanks again :)