Mathematics - 2017-04-06 (page 6 of 6)

These bootstrapping arguments starting from Lie algebras are always tricky. I found a book that gives a proposition without proof, and refers to a considerably more difficult book, which only mentions the words "flag manifold" a few times near the end with no explicit statement of the result I'm looking for.

Mike Miller

What are you trying to do?

And you've checked Besse?

Danu

Yeah, Besse doesn't care about flag manifolds. Barely mentions them. No generalized flag manifolds at all.

What I'm trying to do is the following:

Mike Miller

You'll survive

<3

Ted Shifrin

@Eric: The definition actually comes a few pages earlier. He's working with Lorentz 5-space and the conformal Gauss map maps to the "unit sphere" $\langle x,x\rangle = 1$ in $\mathbf L^5$. He discusses conformal geometry of surfaces in $S^3$ in this light. But, yeah, if you've never done moving frames or the Cartan game, this is going to be a serious introduction.

Danu

11:08 PM

Two definitions: A generalized flag manifold is
(i) $G/Z(T)$ where $G$ is a compact Lie group and $Z(T)$ the centralizer of some torus (not necessarily maximal).
(ii) an adjoint orbit.

I can prove (i) implies (ii). The converse, idk.

Ted Shifrin

Generalized flag manifolds started after Besse. It became a big game starting in the 80s or 90s.

Danu

Hmm... What do people mostly do with them? I find mostly papers on Einstein metrics so far.

Ted Shifrin

There's lots of representation theorists playing with them.

Mike Miller

Crazy.

Danu

Borel & Hirzebruch gives lots of things to do

Ted Shifrin

11:09 PM

My (ex-) colleagues Graham and Boe have done stuff.

Danu

(which is exactly what my thesis is about avoiding :P)

Ted Shifrin

Trying to generalize Schubert stuff from Grassmannians to generalized flag manifolds ...

Danu

His papers look more fancy than the stuff I'm looking at

I'm doing pretty down-to-earth things.

Ted Shifrin

or even below-the-earth :D

Danu

even better

but I'm not "digging roots and lifting weights" ;-)

Ted Shifrin

11:12 PM

well, they are more algebraists, after all :P

Danu

I'm really happy about my topic :-)

Ted Shifrin

That's good, since you're engaged to it full-time for several months :)

Danu

I'm really pretty close to being done with all the computations. Then it's just a lot of understanding that needs to be done.

But I'll also have an exam for my symplectic geometry course in 12 days, and application deadlines not soon after that :\

That's why I'm trying to write a chapter of my thesis now

Ted Shifrin

Yeah, but you're learning a ton and enjoying what you're doing. I'd say you shouldn't complain.

Danu

Indeed. By the way, if $e^X$ lies in a torus then $e^{tX}$ lies in it too for any $t\in \Bbb R$, right?

Has to be true

Ted Shifrin

11:17 PM

Sure.

It is a group, after all. So mumble mumble powers of 2 ... continuity ... mumble mumble.

Mike Miller

I see no reason that works.

Danu

No?

Ted Shifrin

Hmm, no, I guess $X\in\mathfrak a$ doesn't tell us that the subspace spanned by $X$ has to.

Wait.

Duh. $\mathfrak a$ is a subspace.

Danu

Yeah, right?

If it lies in the Lie algebra of the torus, then its multiples do too

somethingsomething

Ted Shifrin

Right.

Danu

11:20 PM

Plus it'd just be perverted if not

Mike Miller

Why does $X$ lie in the Lie algebra of the torus?

Danu

The picture makes too much sense

arctic tern

e^X can be I without e^(tX) being where you want

Mike Miller

Clearly $e^{tX}$ is a 1-dimensional abelian subgroup. If it's not a closed subgroup, take its closure, and the closure is a torus. Fine. But if it's a closed subgroup, it's a copy of R, and has nothing to do with the torus containing $e^X$.

Ted Shifrin

Yes, Mike makes a good point. Non-injectivity of exp again.

Hi again, tern.

Danu

11:22 PM

OK, sigh

Mike Miller

Anyway, your previous question was pretty interesting, about adjoint orbits

Let's do that at the level of Lie algebras

Danu

What I really want is the following. Take $h\in T$ (a torus) and $g\in Z(T)$. Then I want to show that $h=e^X$ gives $\operatorname{Ad}(g) X=X$.

Mike Miller

(BTW, seems feasible you could have a picture that looks like $\phi$, where the bottom interesection point is 1 and the top is $e^X$

Danu

So I just wanted to do $h(t)=e^{tX}$ and claim that $g e^{tX}g^{-1}=e^{tX}$.

Meow Mix

@Ted Oops I fell asleep

arctic tern

11:24 PM

do I misunderstand what you mean by torus, or does $Z(T)$ mean $C_G(T)$ here?

Mike Miller

the latter

Danu

Sure

I don't know the notation. Is $C$ more standard? Then why does everybody use $Z$ for the center of a group?!

Meow Mix

So, when I changed it to $x^3 + y^3 = 1$, it looked the same around the first quadrant, but a little more square

So i changed it to $|x|^3 + |y|^3 = 1$

Mike Miller

I use $Z$

Meow Mix

And now it became a little square-y circle

And so in general I found $|x|^m + |y|^m = 1$ has the property where if you decrease $m$ it collapses into a cross and when it increases it becomes more like the unit square

Danu

11:26 PM

Oh lol

What I said before is much better

Just naturality of the exponential

Mike Miller

huh?

Danu

Meh, failure of injectivity is being lame again

$g e^X g^{-1}=e^{\operatorname{Ad}(g) X}=e^X$

I should be able to rule out multiples of $X$

arctic tern

@Danu center of T is just T, since T is abelian, and will not include anything outside of T (like a centralizer does)

at least in my conventions

Mike Miller

ah my b

i guess you usually do use c

Danu

@arctictern Ah, right. So $Z(T)$ is the center, not the centralizer. Thanks for setting me straight.

Akiva Weinberger

11:33 PM

@MeowMix Squircle | Superellipse

Meow Mix

@AkivaWeinberger Could we define a norm like this?

Ted Shifrin

LOL, Zach, @oops

Danu

@MikeMiller can I at least conclude that $\operatorname{Ad}(g) X=k X$ where $k$ takes on discrete values? Yes, right?

Mike Miller

I don't know or see how.

Danu

The kernel of exp is discrete, isn't it?

Ted Shifrin

11:36 PM

Zach: Be careful. What do you have when $m=1$?

Meow Mix

You have a square thing

wait

Mike Miller

@Danu Kernel of exp doesn't even make sense and also no.

Meow Mix

do you?

Mike Miller

I mean, I guess it makes sense as a set.

Meow Mix

Yeah, you have a square

and the metric you get is the taxicab metric

Akiva Weinberger

11:37 PM

A rotated square, yeah?

Ted Shifrin

But it's not the square you have as $m\to\infty$ :)

Mike Miller

For $SU(2)$ the exponential map is a diffeomorphism for norm up to $\pi$, and then sends all of $\mathfrak{su}(2)$ with norm $\pi$ to the identity.

Meow Mix

yeah, it's rotated by 45 degrees

Mike Miller

Wait, to the negative identity? Whatever, something like that.

Maybe you need norm 2pi. But what I said is more or less corerct.

Ted Shifrin

@MikeM, negative, I think.

Mike Miller

11:38 PM

Sounds right

Danu

@MikeMiller I'm pretty sure its kernel is a discrete subgroup of the Lie algebra.

Mike Miller

Then it loops back

I gave you an example where it's not.

Danu

(my representation theory book uses this)

Mike Miller

That's a shame.

Danu

Bröcker & Tom Dieck's book.

Mike Miller

11:39 PM

Well, I gave you a counterexample to what you claimed, so

Meow Mix

That's weird. So it's kind of like as you increase the power, the closer you get to containing $(1,1)$. And since the powers are so high, $0.99998$ or something very close to $1$ quickly becomes something like $0.5$

Danu

Hmm, is it true for Abelian?

Mike Miller

Yes.

Danu

(probably the case I want to work with in any case)

arctic tern

the inverse image of the identity element of G under exp is not in general closed under addition, nor is it discrete

Mike Miller

11:41 PM

That's the only case in which the exponential is a group homomorphism.

Danu

Sorry, for Abelian

Mike Miller

sure but that's still not relevant to you

Danu

Oh well, I guess it's not easy to prove.

Mike Miller

The thing about vector fields and adjoint reps?

Danu

hmm?

Akiva Weinberger

11:42 PM

@MeowMix What do you mean

Mike Miller

"I guess it's not easy to prove"

Ted Shifrin

DogAteMy: For what you're working on now, look particularly at Homeworks 6 and 7 for some good additional problems.

Danu

I don't know what thing you mean, Mike

Akiva Weinberger

As for "could we define a norm like this" — yep!

Mike Miller

Just tell me what you meant by "it".

Akiva Weinberger

11:43 PM

https://en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_dimensions

Danu

I wanted to get the second definition I gave from the first

Akiva Weinberger

$\|x\|_p=\sqrt[{}^{\Large p}]{|x_1|^p+\dotsb+|x_n|^p}$

Danu

So basically prove that $C(T)$ is the isotropy subgroup of the adjoint action

Mike Miller

Oh, sure, I was trying to do that, but I have no idea how $e^{tX}$ was relevant to that

Danu

Well IF

21 mins ago, by Danu

So I just wanted to do $h(t)=e^{tX}$ and claim that $g e^{tX}g^{-1}=e^{tX}$.

Then it's easy

Mike Miller

11:45 PM

Do you only care about compact groups?

Danu

yea, take exp to be surjective

Mike Miller

That's not the same as what I asked

Danu

Yea

Ted Shifrin

Until @Danu starts doing Lorentzian versions of his stuff ... :P

Danu

lmao

Akiva Weinberger

11:46 PM

> Of course the absolute value bars are unnecessary when p is a rational number and, in reduced form, has an even numerator.

That sounds like one person wrote "Of course the absolute value bars are unnecessary when p is even," and someone else edited it to make it more general, but the second person didn't notice the "Of course" at the start of the sentence.

Ted Shifrin

Wiki does have errors, DogAteMy.

Mike Miller

Wait

Meow Mix

@AkivaWeinberger Oh I've heard people talk about that before

Mike Miller

Did you ever actually need it to be a torus?

Why couldn't it just live in a $\Bbb R$ subgroup?

Meow Mix

$\ell_\infty$ would be the norm where $||(x,y)||$ is the greater of the two coordinates?

Akiva Weinberger

11:47 PM

It is factually correct. It's just rude to call things like that "obvious".

Danu

@MikeMiller The definitions I gave are the ones I want, yes.

Ted Shifrin

Absolute value, Zach.

Danu

No modifications

Akiva Weinberger

Yeah @MeowMix

Danu

Or is that not what you're asking?

Meow Mix

11:48 PM

Weird, is $\ell_0$ degenerate?

Akiva Weinberger

$\|x\|_\infty=\max\{|x_1|,\dots,|x_n|\}$

@MeowMix Wikipedia says it takes $p\ge1$

Ted Shifrin

Good exercise for you, Zach: Can you see that $\lim\limits_{p\to\infty} \|x\|_p = \max\{|x_i|\}$?

Mike Miller

@Danu Ah, sure, I see your point.

Akiva Weinberger

I wonder what fails when $p=$, say, $\frac12$

Meow Mix

Oh, boy.

Ted Shifrin

11:49 PM

Yeah, you don't want $p<1$. Convexity gets all messed up ...

Akiva Weinberger

@TedShifrin Does the triangle inequality fail, maybe?

Danu

I (am supposed to be reproducing the standard way to) try to generalize the standard flag manifolds

Mike Miller

@Danu Your argument that a point in $\mathfrak g$ has stabilizer $C_G(T)$. Clearly the relevant torus is the maximal torus living inside the stabilizer, and you want to show that its centralizer is the stabilizer. It's also easy to see that the centralizer of the maximal torus of the stabilizer fixes the relevant point. The hard part is to see why the whole centralizer acts trivially.

Meow Mix

@Ted Should I use the formal definition of a limit here?

Ted Shifrin

DogAteMy: I'm guessing that because convexity flipped it does.

Meow Mix

11:49 PM

I mean, intuitively it's kind of obvious

Ted Shifrin

Nah, Zach. Just figure out why it's true. Why is it obvious?

Adeek

@MikeMiller maybe you would like to discuss this with me

Mike Miller

Pick an element, I dunno, $g$ inside the centralizer of this maximal torus of the stabilizer of $x \in \mathfrak g$.

Adeek

@MikeMiller have an idea about maybe connecting algebraic tensor product and a geoemetrical way to get some information about it.

Akiva Weinberger

Hm. $\|\langle1,1\rangle\|_{1/2}=4$, but $\|\langle0,1\rangle\|_{1/2}=\|\langle1,0\rangle\|_{1/2}=1$

Mike Miller

11:50 PM

Hm, let's make it live inside the identity component of the centralizer.

Adeek

would you like to discuss ?

Mike Miller

I'm trying to think about Danu's problem right now

Akiva Weinberger

So $\|\langle1,1\rangle\|_{1/2}\not\le\|\langle0, 1\rangle\|_{1/2}+\|\langle1,0\rangle\|_{1/2}$

Ted Shifrin

Karim: Seriously. Interrupting people who are in the middle of something complicated isn't cool ...

Adeek

oh ok

Danu

11:51 PM

@MikeMiller Okay..

Mike Miller

@Danu I'm just fucking around. I think I got close earlier but not quite.

Ted Shifrin

DogAteMy: Triangle inequality is intimately related to convexity. The chord above the graph is good; the chord below is not.

Danu

Caveat: I found some people take some more assumptions. The alternative definition I found was $G/H$ where $G$ is semisimple, compact and connected and $H$ has equal rank and centralizes a torus.

Mike Miller

!!

Danu

No idea if that's a big simplification

Ted Shifrin

11:53 PM

Tern should probably be involved, Karim, but what's your geometric idea?

Mike Miller

Compact is reasonable, don't see the semisimple requirement (probably important for the geometry), connected doesn't matter.

Well, I guess it matters a little. But I don't really care much. You don't get interestingly distinct structures.

Ted Shifrin

semisimple is important for metric things usually

Mike Miller

I don't know what rank is.

Someone has told me once every two months for a few years.

Ted Shifrin

dim of maximal torus

Danu

(I think dimension of Cartan subalgebra?)

yeah

Mike Miller

11:54 PM

Aren't those different??

Ted Shifrin

or dimension of maximal abelian subalgebra

Danu

No, those are the same

Mike Miller

I thought Cartan subalgebra was a nilpotent thing, not abelian

Danu

I learned it is the Abelian thing

So let's not try to make these assumptions

I'd like to work without them

Meow Mix

@Ted well, do you see that $||(ax_1,ax_2,\dots)||_p = a||(x_1,x_2,\dots)||_p$? if so, i'm just going to generalize it to the unit "squircle"

Danu

11:55 PM

(I think they are just taken in that paper since the flag manifolds they actually need are extremely simple)

Ted Shifrin

I think Mike's right ... Cartan is nilpotent.

Mike Miller

Cartan subalgebra

In mathematics, a Cartan subalgebra, often abbreviated as CSA, is a nilpotent subalgebra h {\displaystyle {\mathfrak {h}}} of a Lie algebra g {\displaystyle {\mathfrak {g}}} that is self-normalising (if [ X , Y ] ∈ h {\displaystyle [X,Y]\in {\mathfrak {h}}} for all X...

properties is relevant.

Ted Shifrin

Zach, you need $|a|$.

Mike Miller

it's only abelian if your group is semisimple

Danu

Weirdest thing... Let me check the book I learned it from

Meow Mix

11:55 PM

oh, sorry, yes

Danu

Ah, I see. I only learned about semisimple.

Mike Miller

I should abandon ship in the next 15 minutes or so if I can't answer this then. Can you give me a reference that presents both equivalent definitions (without proof) in the generality you want?

Danu

I don't know anything about algebra, lol

Mike Miller

Me neither, but I'm not the one doing a thesis on it!

Danu

Little book by Arvanitoyeorgos

Ted Shifrin

11:56 PM

Gesundheit!

Danu

He even tells you what results are supposed to establish equivalence

Mike Miller

Beh, I'm not downloading that.

Meow Mix

Anyways so I claim that as $p$ increases, it approaches a unit square. Obviously, if it was a unit square, then the norm would be the maximum of the absolute values of the coordinates

Danu

(including stabilizer = centralizer)

Meow Mix

So the reason it approaches a unit square

Danu

11:56 PM

I can post it in 2 pictures

Mike Miller

Oh, I think I might have written a proof of this down a while back for SO(3)

Down Christopher

CMC: (chat mini challenge):
Do this with math (formula)
Check for all perfect license plates with length N

To check if a license plate is perfect:

1. Sum up the characters, with A = 1, B = 2, ... Z = 26.
2. Take each consecutive chunk of numerals, and sum them; multiply each of these sums together.

If the values in part 1 and part 2 are equal, congratulations! You've found a perfect license plate!

### Examples ###

License plate: AB3C4F

Digits -> 3 * 4
= 12
Chars -> A + B + C + F
= 1 + 2 + 3 + 6

Wow that one boxed

Meow Mix

Is because, well, consider the points like $(x,1)$ and $(1,y)$ for $x,y\in(-1,1)$

Danu