Mathematics - 2021-12-21 (page 2 of 3)

Ted Shifrin

5:00 PM

I wonder if the wrong lab was copied from somewhere/someone.

Balarka Sen

@Ted: got an A+ in analysis

leslie townes

only an A+?

Balarka Sen

dodged the guillotine this time but have to study all of this stuff

Alex

Hi!

leslie townes

ted's wonder is my wonder too. in the times i have seen clear instructions disregarded so completely (i.e., not just missing some aspect of the task, but doing the entirely wrong task), there was copying involved.

Semiclassical

5:03 PM

i doubt it tbh, this isn't a common lab

Ted Shifrin

@BalarkaSen deplorable! 😍

Hi Alex

leslie townes

i dunno. if i were the prof and had discretion to act, in the absence of clear evidence of copying or other gamesmanship, i would probably let it slide. that's the outcome at the university of leslie townes.

in the middle of the semester it is one thing, if you let stuff slide it creates a kind of 'precedent' that you then might have to follow in an unknown number of similar cases that might pop up later in the very same class. if it's just a one-off thing, who really cares.

Semiclassical

"take your passing grade and GTFO"

napstablook

@Semiclassical XD

leslie townes

i should start Leslie University. our first action would be legal action against Lesley University. we'd try to cancel all of their trademark registrations.

Alex

5:08 PM

I have this problem: Let $A\in M_{n}(\mathbb{R})$ such that $ A^3=4I-3A. $. Prove that $\det(I+A)=2^{n}$. By Calyey-Hamilton we know that there exists $p(A)=0$ such that $p(\lambda)=\lambda^{3}+4\lambda-4=(\lambda-1)(\lambda^{2}+\lambda+4)$. Hence all the roots are given by $1,-\frac 12\pm\frac{\sqrt{15}}{2}i$. How can I continue from here?

Ted Shifrin

Maybe you can take all of Tromp’s dissatisfied customers.

How is that determinant related to the characteristic polynomial?

leslie townes

alex: some of the quantifiers are garbled there and you're not really using cayley-hamilton, are you?

Semiclassical

the rest of their scores are good enough that even a 50% will get them a passing lab score

so i'll probably just do that

Xander Henderson

@leslietownes Will I be able to pay for classes at Leslie U with lesliecoin?

Semiclassical

@XanderHenderson also, i decided to ask the prof to look at the submission comment i mentioned. better him than me to deal with it

leslie townes

5:15 PM

xander: i'm glad you asked. yes.

Xander Henderson

@Semiclassical Oh, I didn't catch that you weren't the instructor of record. When I was a TA, I met nearly every complaint with "Go ask the prof."

Semiclassical

i've been taking that tack to some extent

i'm the one who decided how the problems were graded, though

there wasn't a rubric or anything

Xander Henderson

I taught for one guy who explicitly said "Tell the students that you want to give them credit, and that you will go to bat for them with the prof. Then send them to me. I'll say no. End of story, but your teaching evals will look great!"

Wolgwang

Is it ethical to start a bounty on some question so that it gets more votes?

Xander Henderson

@Wolgwang No.

Semiclassical

5:20 PM

well

bounties are intended to gain attention

Xander Henderson

Generally speaking, the goal of a bounty should either be to (1) reward an existing answer which you think is particularly good, or (2) to encourage users to post answers.

Semiclassical

but the amount of rep used to create them generally means that you can't expect to get a positive return on that

Xander Henderson

@Semiclassical I am not assuming that @Wolgwang is asking about putting a bounty on one of their own questions (or answers).

leslie townes

i don't understand how putting a bounty on a question would get the question more votes. it might get more views, but whether it gets more votes would seem to be more up to the people who do the voting than the person who places the bounty?

Xander Henderson

Of course, in practice, there is no way to distinguish between giving a bounty for a "good" reason, and giving a bounty for a "bad" reason.

Wolgwang

5:24 PM

@XanderHenderson Ugh! This post was substantially improved from a PSQ but hasn't received much votes and the OP is new. :-/

That's rare IMO.

Semiclassical

@leslietownes well, it gets more eyes on the question, and more eyes = more opportunities to get upvotes

(also more opportunities to get downvotes ofc)

Xander Henderson

@Wolgwang If you feel that the question has been improved and deserves a good answer, go ahead and put a bounty on it saying as much. But putting a bounty on a question just because you feel sorry for the asker is not really okay.

leslie townes

yeah, that's the thing. i don't understand how the goal (whether ethical or not) would be implemented by the bounty.

one way of congratulating someone for improving their question is to answer it :). or to vote to reopen if it is closed.

Wolgwang

There was an answer to that post O_o

Ted Shifrin

@Alex Did you see my question to you?

leslie townes

5:28 PM

alex: in case my remark above was vague, follow ted's hint. (you don't know that the given polynomial is the characteristic polynomial, but you do know what any factors of the characteristic polynomial must be.)

wolg: it appears to have been deleted.

Ted Shifrin

Surprisingly, @leslie, there's enough info. I was not expecting that.

Xander Henderson

@leslietownes There is a dedicated thread on meta for getting reopen votes. And a chatroom, too.

@Wolgwang There was a hint answer. It is now a comment.

leslie townes

ted: yeah, it's a very fortuitous accident of numerology.

Ted Shifrin

Yeah, certainly it's going to work very rarely.

Semiclassical

is it an accident if the problem proposer arranged for it?

leslie townes

5:29 PM

ted: maybe we should worship this polynomial?

Wolgwang

Thanks for the inputs :)

Xander Henderson

@leslietownes All Hail!

(what polynomial?)

leslie townes

i guess, this family of polynomials.

Ted Shifrin

@Semiclassical Arranged, sorta like most antidifferentiation problems :P

leslie townes

xander: characteristic polynomials of real matrices with minimal polynomial dividing x^3 - 4x - 4.

they're what i believe in now.

Ted Shifrin

5:31 PM

Wrong polynomial. Alex had a typo. It's $x^3+3x-4$.

leslie townes

oh, yeah.

Xander Henderson

O_o

leslie townes

anyway when we figure it out, we'll worship it.

Ted Shifrin

snores

leslie townes

nurse, mr. shifrin has fallen asleep again. be sure to wake him in time for his sandwich at lunch.

Alex

5:33 PM

@leslietownes I recognise that this part of my reasoning does not reassure me. I know that if $A\in M_{n}(\mathbb{R})$ such that the characteristic polynomial is given by $p(\lambda)=\lambda^{3}+3\lambda-4$ so by Calyley-Hamilton we have $A^{3}+3A-4I=0\iff A^{3}=4I-3A$. But, I know I can't go back on the implication. However, I can ensure the existence of the polynomial $p(A)=0$ such that $p(\lambda)=\lambda^{3}+3\lambda-$?

leslie townes

you don't know what the characteristic polynomial is. A is nxn and its characteristic polynomial is degree n.

you know that A satisfies that degree-three polynomial.

this is the beginning of the miracle that ted and i were talking about.

Semiclassical

any zero of $p(x)$ is also a zero of $p(x)q(x)$

Alex

@TedShifrin Yes, I know that if $A\in M_{n}(\mathbb{R})$ so the characteristic polynomial for $A$ is given by $p(\lambda)=\det(A-\lambda I)$. In this case we have $p(\lambda)=\det((A+I)-\lambda I)$.

Semiclassical

no

knowing that A satisfies a certain equation does -not- tell you the characteristic polynomial

leslie townes

you're being told about the roots of the characteristic polynomial (but not their multiplicities). this tells you a bit about how the characteristic polynomial factors but does not determine the characteristic polynomial.

Alex

5:36 PM

ok

Semiclassical

consider the matrices A = diag(1,-1) and B = diag(1,0,-1)

Ted Shifrin

@Alex Don't use $p$ for both things. No, I'm asking about the characteristic polynomial of $A$. NOT $A+I$.

leslie townes

ted's original hint/question is still pending.

ted's annoyingly good at the giving of hints.

Semiclassical

both of them satisfy the equation x^3-x=0, but that's only the the characteristic polynomial for B

Alex

@TedShifrin The characteristic polynomial of $A$ is $p(A)=\det(A-\lambda I)$.

Ted Shifrin

5:40 PM

Not the same $p$, right?

So the original question is asking for what in terms of this?

Semiclassical

it's probably best to give symbols for two polynomials: the one which you're told A satisfies, and the characteristic polynomial

otherwise it'll continue to be ambiguous which polynomial you're talking about

Ted Shifrin

@Semiclassic Not that I didn't complain :P

Xander Henderson

Sisyphus is rolling a boulder up a hill towards Hilbert's Hotel. The hotel does have an infinite number of rooms, but is full, so it may (or may not) have room for Sisyphus or his boulder. There is a side path which leads down the hill to Theseus' ship. The boulder will surely gain momentum on the way down the hill and break the ship to flinders, but all of the constituent parts of the ship have been replaced over time, so it may not even be Theseus' ship.

So, a question:

Alex

@Semiclassical $p(A)=\det(A-\lambda I)$ is characteristic polynomial of $A$ and $q(\lambda)=\det((A+I)-\lambda I)=\det(A-(\lambda-1)I)$ is the characteristic polynomial of $A+I$ is it correct now? I was planning to use the fact: the determinant of a matrix is the product of its eigenvalues. So my goal was to obtain the eigenvalues of $A+I$ from the eigenvalues of $A$.

Xander Henderson

Is Sisyphus happy?

Ted Shifrin

5:49 PM

@Alex Why do you keep writing $q(\lambda)$? I told you we care ONLY about the characteristic polynomial of $A$. You need to focus on that.

Semiclassical

@XanderHenderson one must imagine so

Alex

@TedShifrin :-(

Ted Shifrin

@Xander It depends only on whether Sisyphus is sleepy.

leslie townes

alex: is there a number you could put in for lambda to relate det(A - lambda I) to the thing the question is asking about

Ted Shifrin

glares at leslie

Semiclassical

5:58 PM

myself i'd approach this as: what can we say about the eigenvalues of A?

Ted Shifrin

Nah.

Semiclassical

we aren't in a position to say what their multiplicities are, but we do know what the only options are

Alex

@leslietownes Nice. Yes, we can put $\lambda=-1$ so $\det(A-\lambda I)=\det(A+I)$.

Ted Shifrin

Well, if you don't know multiplicities, you're SOL.

@Alex YES!

Semiclassical

no, you're not

well

Alex

5:59 PM

:-)

Semiclassical

you know that two of the three multiplicities have to be equal :)

leslie townes

my apologies, ted :) i went to the store to get some oats and arrived home to see things in the same position as before, and like my daughter, decided to kick the thing that wasn't moving

Ted Shifrin

Go on and finish the problem your way, @Semiclassic. I'll wait.

Semiclassical

already have

or at least i think i have :P

Ted Shifrin

OK. I think the approach I was pushing is an important one for Alex to learn/master.

Semiclassical

6:00 PM

without saying it out loud it's hard for me to be sure

leslie townes

semi, you've probably got it

has there ever been a more starred comment than ted's string of emojis?

Ted Shifrin

Yeah, it probably works with similar exponent observations. It's just a lot more of a pain.

Semiclassical

nah

Ted Shifrin

I don't even remember why I emotively emojied.

Semiclassical

not painful to notice the pattern with the magnitudes

leslie townes

6:01 PM

he's a physicist. he loves eigenvalues. they are the opposite of pain. or maybe pain and pleasure are the same thing to him.

Semiclassical

you're not wrong there

leslie townes

have extradimensional beings appeared after you solved a wooden box puzzle, semi?

Semiclassical

not yet

Ted Shifrin

I'm a geometer. I love eigenvalues, too, but the symmetric functions of eigenvalues are what I really love.

Semiclassical

but this kind of stuff is something i do use a lot. e.g., if A^3 = A then determine exp(itA) in terms of A^2, A, I

(obviously one can use Taylor series for that, but it's kinda boring)

leslie townes

6:03 PM

that's a similar spirit, i think? you can work directly with the polynomials without going down to their roots.

Semiclassical

there's a variety of approaches

leslie townes

without opening the hellraiser box.

Ted Shifrin

Yup, having to find roots of polynomials is arduous and often impossible!

Semiclassical

my preference is to write $I=P_+ + P_0 + P_-$, $A=P_+ - P_-$, $A^2=P_++P_-$

Ted Shifrin

I remember a theorem along those lines... vaguely.

Semiclassical

6:05 PM

and then $e^{itA}=e^{it}P_+ + P_0 + e^{-it}P_{-1}$

Ted Shifrin

You're stipulating that all the P's commute.

Semiclassical

more than that, i'm stipulating that they're projection operators

Xander Henderson

Claim: $\log(a) + \log(b) + \log(c) = \log(a + b + c)$. Example: $\log(1) + \log(2) + \log(3) = \log(1+2+3)$.

Semiclassical

b/c i'm assuming A is hermitian and therefore nice

leslie townes

ted: i think A is self-adjoint. like R being a big number.

Semiclassical

6:05 PM

@XanderHenderson perfect

Xander Henderson

@Semiclassical :D

Semiclassical

amusingly, there's an HTML5 page which simulates spin measurement stuff

Ted Shifrin

Unless $R$ is curvature.

Good assessment there, Semiclassic.

Semiclassical

and as part of the javascript code which computes everything, you can find the following

// Function to calculate the propagator matrix for a magnet/rotator component.
// The formula is exp(-iHt) = 1 - i*sin(theta)*H + (cos(theta)-1)*H^2,
// and apparently it works as long as the eigenvalues of H are all 1, -1, or 0.
// (I pulled this from my old Pascal code and I have no recollection of whether
// I derived the formula at that time, or how I even learned of its existence.)

i think t=theta here, though that's a bit confusing

anyways. the proof i sketched above does that, even if the eigenvalues have multiplicity since the projection operators are still valid

Alex

@TedShifrin Well, so if $p(\lambda)$ is the characteristic polynomial of $A$, so I'm finding the value of $p(-1)$. How can I relation it with $A^{3}=4I-3A$?

Semiclassical

6:10 PM

i'm not sure how else you'd prove it

leslie townes

eventually you reach enlightenment when it is all just star-algebra and there might not be an underlying space or eigenvalues

Semiclassical

yeah, i'm approaching that

Ted Shifrin

That is indeed the question. So you have $f(\lambda) = \lambda^3+3\lambda-4$, which you did factor. What do you know about $p(\lambda)$?

Semiclassical

though in physics you tend to talk about C* algebras specifically

Ted Shifrin

I assume you know about minimal polynomials, etc., @Alex.

leslie townes

6:13 PM

my daughter announced this morning that she wants to shave her head. "i don't want this," she said, while running her hands through her hair. the look on my wife's face was something else.

Semiclassical

the bit i want to properly understand at some point is the GNS construction

i sorta know it but not completely

leslie townes

my observation that it would shorten bath time considerably was not well received.

Semiclassical

the basic story i know is that you pick a reference state in your C* algebra, and try to use it to define an inner product

in general that won't work

Xander Henderson

@leslietownes I encourage the saving of heads.

Semiclassical

but if you pass to an appropriate quotient space, it does work and you get a Hilbert space representation

Ted Shifrin

6:17 PM

Where is munchkin learning about shaved heads?

leslie townes

semi, there are lots of good mathy treatments. kadison and ringrose have a good development. arveson's spectral theory. fillmore's guide to operator theory.

Xander Henderson

My daughter has had a nearly shaved head recently.

Semiclassical

kk

Ted Shifrin

In the olden days that was cuz lice.

leslie townes

ted: i have a nearly shaved head. i think i made it look cool.

it's actually surprising she hasn't demanded it earlier, given that she's around me all the time, and given how cool i am

Semiclassical

6:18 PM

what i'm ultimately interested in is the following. start from a C* algebra, pick a state and get a Hilbert space representation

Ted Shifrin

spits up in mouth

Semiclassical

one can then define the Hilbert-Schmidt space of operators on said space

is there a way to go directly from the C* algebra to the Hilbert-Schmidt space of operators?

it'll still be relative to the reference state, of course

but some more direct relationship than having to go via the underlying Hilbert space representation

leslie townes

semi my knowledge of this is too hazy and out of date. note that some properties of HS operators are intrinsic to the C* algebra, such as the spectrum.

Semiclassical

yeah

i feel like there should be a good story here

leslie townes

maybe also being a norm limit of finite rank operators. you could probably define x in a general algebra A to be of finite rank if xAx is finite dimensional as a complex vector space, irrespective of any representation of A.

Semiclassical

6:26 PM

like, what conditions have to be fulfilled for a particular subspace of the C* algebra to be the Hilbert-Schmidt operators for some Hilbert space representation?

robjohn

1. Sisyphus is rolling a boulder up a hill 2. Sisyphus has to live with being called Sisyphus his whole life.

Sisyphus is not happy.

leslie townes

well you say 'some' representation. if you don't require the representation to be faithful you can make anything hilbert schmidt.

Semiclassical

does he make a phus about it

@leslietownes hmm. gross

leslie townes

if it is faithful, i'm just wondering if there's anything about HS-ness that can't be recovered from *-algebraic properties.

Jam

I need to understand what $\mathcal{O}(1)$ and $\mathcal{O}(-1)$

Ted Shifrin

6:27 PM

@Semiclassical Certainly if you call him Sisy.

I prefer $\mathscr O(-1)$ some days, @Jam.

Jam

haha hey Ted

how did you type that

Ted Shifrin

mathscr

Jam

$\mathscr{O}(1)$

robjohn

@Jam right click on the rendered MathJax

Semiclassical

now here's the real question: how do you write \mathscr{O} by hand

Jam

6:29 PM

ohh helpfull

script O?

right

show as TeX commands

$\mathscr{O}$

@Semiclassical trust me i practiced to write it by hand hahah

Ted Shifrin

6:30 PM

So what is your actual question? And what are you learning? Complex geometry? Algebraic geometry? Sheaves? Line bundles?

Semiclassical

though the one i really struggle to write nicely by hand is $\mathcal{N}$

mine always come out wacky

Jam

Ofc let me explain

Semiclassical

also $\mathcal{D}$, though to a lesser extent

Ted Shifrin

I did a lovely script Q when we were doing quadratic forms. My students made a whole meme about it years ago.

Jam

im learning alge geometry. Specifically intersection theory i need to understand Chern Classes and it seems this particular $\mathcal{O}(1)$ and -1 plays a big role as a bundle . But the definition seems to be using divisors

Ted Shifrin

6:32 PM

You need to learn basic stuff before doing advanced stuff.

Jam

actually the book 1324 im using by harris doesnt have a clear definition

Ted Shifrin

So you're on $\Bbb P^n$?

Jam

yeap

Ted Shifrin

I suspect Joe Harris assumes you have background in basic stuff (e.g., Griffiths/Harris or his own introductory algebraic geometry book).

So there is the tautological line bundle (which turns into the sheaf $\mathscr O(-1)$) and its dual, which is called the hyperplane section bundle ($\mathscr O(1)$), whose sections have hyperplanes as their zero divisor.

robjohn

Was @Alex able to answer their matrix question?

Ted Shifrin

6:35 PM

The tautological line bundle has as its fiber over $\ell\in\Bbb P^n$ the actual line $\ell$ in $\Bbb C^{n+1}$.

Jam

ok nice

Ted Shifrin

@robjohn Still working on it, as far as we know. I didn't see an answer to my last question to him.

Jam

i think i understand it a line bundle

but not as a sheaf

as a*

robjohn

@TedShifrin He's no longer here. Okay, thanks.

Ted Shifrin

Well, there's a standard yoga for going back and forth between vector bundles and locally free sheaves. You just look at germs of sections. But I think that for intersection theory (at least at the beginning) you'd rather think about bundles and divisors.

Jam

6:38 PM

yes exactly

Ted Shifrin

You need to look at some standard books (as I said) for background. Wells's book on complex manifolds might be helpful, too.

Jam

ok ive seen that divisors are formal sums of points on a curve with coefficients the multiplicity

so i guess the divf of a polynomial is the sum of the multiplicity of tis zeroes?

Ted Shifrin

That's on curves. On higher-dimensional varieties (manifolds), divisors are integer linear combinations of hypersurfaces.

Jam

so they are sums of codimension 1

Ted Shifrin

Divisors of meromorphic functions are zeroes - poles. Polynomials, when you think of them on $\Bbb P^1$ (or any algebraic curve) have poles, too, at infinity.

You need to read background. Look at Chapter 1 of Griffiths/Harris.

Don't try to understand all the fancy stuff, but learn the vocabulary and how divisors and line bundles work.

Jam

6:43 PM

Griffiths Hariis?

Ted Shifrin

Same Harris.

yearning4pi

or Hartshorne II.6 :P

Ted Shifrin

Griffiths/Harris is the standard text for complex algebraic geometry. There are others, but since you're reading Harris, this makes sense.

Do NOT look at Hartshorne.

Jam

hahaha

Ted Shifrin

Unless you know a year of commutative algebra and plan to spend 2 years on Hartshorne's book.

Jam

6:44 PM

ok i think ill listen to Ted

yearning4pi

I find that to be one of the most difficult sections if one does not know about divisors already

Jam

my commutative algebra is really basic haha

Ted Shifrin

I taught a semester-long course many years ago out of parts of Griffiths/Harris (mostly).

Jam

Tho from what ive looked

i prefer the more geometric approach

of the algebraic geometry

more classical

i havent been introduced to scheme theory alot

Ted Shifrin

Yes, you mean not all the formal algebra. That's why I'm recommending what I am. Also look at Harris's own introductory algebraic geometry book.

For fancy intersection theory, though, it is about scheme structure.

I even had an embedded component show up in one of my little papers. I was stunned.

Jam

6:47 PM

which chapter will help with my question on griffiths harris book?

Ted Shifrin

I told you.

Jam

ohh found it

my bad

you had an embedded component show up?

means you had to use theory from those things?

Ted Shifrin

I'm saying that in very natural geometric constructions, schemes do show up. This was the Nash blow-up of the Whitney umbrella. (Nash blow-up is one way to make sense of a tangent bundle when you have singularities.)

leslie townes

i hope you weren't indoors when you nash blew-up the whitney umbrella. that's bad luck.

Jam

oh yes ive seen that in gathmans notes about blow ups

So Ted have you taught Intersection Theory?

or just studied it?

yearning4pi

6:56 PM

The nash blow-up is different from blowing up a closed subscheme (which is what Gathman seems to be covering).

Jam

yeah i was talking about regular blow ups haha

Ted Shifrin

No, no fancy intersection theory. Lots of Chern class stuff in my research and teaching.

Jam

ohh thats perfect brace yourself

hahaha

Ted Shifrin

Yeah, Nash blow-up is looking at the Gauss map of the subvariety, not the variety itself.

Jam

ill have a lot of questions

Semiclassical

6:57 PM

insert the old anecdote about not talking about Nash at airports

Ted Shifrin

Chern-Mather classes instead, @Semiclassic. Mum's the word.

Semiclassical

clever

the other one i always liked was getting a grant proposal in the 70s by emphasizing how one is working on annihilating radical left ideals

Ted Shifrin

That's more of that commutative algebra stuff. Ugh.

Jam

hahahahaha

Xander Henderson

@TedShifrin Who's mum?

Ted Shifrin

7:07 PM

You mean whose or who's? If the latter, word is.

leslie townes

my daughter brought up shaving her head again. this might escalate to the level of a problem that i have to deal with.

Ted Shifrin

Sounds like it's already escalated.

leslie townes

she's never had a haircut, so this is the first time she's actively thought about her hair as something that can be changed.

but we're not taking her to get a haircut with all of this coronavirus floating around.

Ted Shifrin

Well, if you're just going to shave her head, that doesn't take so much talent.

leslie townes

yes. it's almost like the universe wants us to shave her head.

Xander Henderson

7:10 PM

@TedShifrin I meant "who's". As in "who is?"

@leslietownes SHAVE HER HEAD! SHAVE HER HEAD! SHAVE HER HEAD!

Ted Shifrin

OK, I thought you were being brit and asking whose mum.

So, if who is ... I already said. The word.

leslie townes

in normal times, a bargaining position would be, we can schedule a haircut and go from there, but it won't be shaving your head.

at home, all we're equipped to do is shave her head.

Balarka Sen

@TedShifrin Speaking of Chern classes: I can prove $S^{4n}$ has no almost complex structure by studying its $p_n$, but I do not know why this magic works.

leslie townes

maybe i could bargain her down to having her hair shaved to the length of our cat's fur. i have an attachment on my clippers for that.

studying its what?

sorry, adolescent humor.

Ted Shifrin

@Balarka I do not remember the argument.

Balarka Sen

7:14 PM

Pontryagin class

copper.hat

when my daughter was around 10 she was on a 'swim team' and when we would get home from practice i would detangle her hair. one day i said, you know if you cut your hair short it would save 15 mins every time we do this? her reply, cut it off.

Ted Shifrin

Often swimmers prefer very short hair, regardless.

copper.hat

the pontryagin principle (optimal control) proof is still on my todo list

leslie townes

we may have reached the day that i thought would never come, in which i follow parenting advice from copper.hat

Balarka Sen

If $J$ was an ac structure on $TS^{4n}$, one could look at the complexification $J : T^{\Bbb C} S^{4n} \to T^{\Bbb C} S^{4n}$ and decompose $T^{\Bbb C} S^{4n}$ into two eigenbundles corresponding to eigenvalues $\pm i$.

copper.hat

7:16 PM

you might want to consult with my offspring (one very public, one very private) about that before taking any parenting advice from me

Ted Shifrin

Ah, but because $4n$, $p_n$ will be the same on those two bundles.

Balarka Sen

These eigenbundles are isomorphic to $E = TS^{4n}$ and $\overline{E}$ respectively. Then $E\otimes \Bbb C \cong E \oplus \overline{E}$, taking $c_{2n}$ gives $0 = c_{2n}(E) + c_{2n}(\overline{E}) = 2c_{2n}(E)$ but $c_{2n}(E)$ is the Euler characteristic of $S^{4n}$, $2$.

Ted Shifrin

$4n$, but no matter.

Balarka Sen

Thanks.

Seems a bit like magic. This is my first time actually using the Pontryagin class though.

copper.hat

an eigenbundle sounds like offspring of some linear operator

Ted Shifrin

7:19 PM

I'm confused.

Where did the $0$ come from?

@copper Wrap it in swaddling clothes.

Balarka Sen

Ah because $E \otimes \Bbb C = T^{\Bbb C} S^{4n}$ is stably trivial, so $p_n(E) = c_{2n}(E \otimes \Bbb C) = 0$

copper.hat

and feed it roots

Balarka Sen

Or rather, because $TS^{4n}$ is real stably trivial, i.e., $TS^{4n} \oplus \underline{\Bbb R} = \underline{\Bbb R}^{4n+1}$, tensoring everything with $\Bbb C$ we get $T^{\Bbb C} S^{4n}$ is complex stably trivial, so $c_{2n}$ vanishes.

Ted Shifrin

@Balarka Ah, I was misremembering my computation in my head of $p_1(S^4)$. Yes, of course, the Pontryagin form is actually dead $0$.

Jam

@leslietownes will happen embrace it haha

Balarka Sen

7:25 PM

Ah!

Pontryagin form is $\mathrm{tr}(\Omega^2)/8\pi^2$?

Ted Shifrin

Maybe with sign.

So, yeah, it's easy to see from the fact that $S^{4n}$ has a constant curvature metric that the top Pontryagin form always vanishes.

Astyx

good evening

Ted Shifrin

Salut @Astyx

leslie townes

wha?? it isn't even the afternoon.

Jam

damn that griffith harris book aint that easy

Ted Shifrin

7:31 PM

Add 8 or 9 hours.

leslie townes

ted: too confusing.

Astyx

add 1-2 day periods

Jam

what is $\mathcal{H}^{0}$

Astyx

In what context?

Ted Shifrin

You need to learn basics of cohomology for sure.

Just $H^0$. No script. Global sections.

You need to understand basics of Cech cohomology. Back to Chapter 0.

Jam

7:33 PM

it appears on exact sequences

of line bundles

or sheaves

Astyx

what Ted said

Balarka Sen

@TedShifrin I'm slow with this, but let me try. The diagonal terms of $\Omega^2$ are something like $\sum_j \Omega(e_i, e_j) \wedge \Omega(e_j, e_i)$, right?

Ted Shifrin

@Jam This is nuts. Even the title of Harris/Eisenbud says A SECOND COURSE in algebraic geometry.

Jam

hahaha

Ted Shifrin

I don't like that, @Balarka. You mean $\Omega_{ij}$? You're not evaluating on any vectors.

You have to take the first course, or at least read a lot of it, @Jam. Go do it.

It's like trying to study real analysis when you haven't learned what multiplication and derivatives are.

Jam

7:36 PM

So which book would be a good one for a first course?

Balarka Sen

Ah yeah $\Omega_{ij}$. The $(i, j)$-th component of the matrix, which is a 2-form, not feeding the form anything.

Ted Shifrin

Right.

Balarka Sen

It's $\langle \Omega(-, -) e_i, e_j \rangle$, rather.

Ted Shifrin

The point of constant sectional curvature $K$ (exercise) is that $\Omega_{ij} = -K\omega_i\wedge\omega_j$ [sign conventions to be ignored].

Right @Balarka

Balarka Sen

Easy enough, $\langle \Omega(e_i, e_j) e_j, e_i \rangle$ is the sectional curvature on the $i,j$-plane.

Ted Shifrin

7:37 PM

Yes, I know that. So finish the exercise :)

Minus, actually, by my conventions. But who knows.

Balarka Sen

I switched $j$ and $i$, so I think that's where minus comes from. But yeah

Ted Shifrin

I think that's the standard order, but I don't berember now.

And maybe you want $\theta_i$ instead of $\omega_i$. I just mean the coframe.

Balarka Sen

OK, yeah, so you're saying that if you have constant sectional curvature then the cross-terms like $\Omega_{ij}(e_k, e_l)$ of the tensor are actually zero.

Hm, one second.

Sounds like parallelogram law

Ted Shifrin

Well, even with $k=i$ or $j$. Etc.

Balarka Sen

Right, $\delta_{ik, jl}$.

Well, $i = l, k = j$ can happen, but I got you :)

Ted Shifrin

7:42 PM

You can do the case of the sphere directly using its embedding as a hypersurface and using the Codazzi equations, but this is still a good exercise.

Balarka Sen

Yeah this I ought to know. This is a triviality in 2D which is the only case I have dealt with moving frames.

But it seems like an easy corollary of "$\|x + y\|^2 - \|x\|^2 - \|y\|^2$", give me a moment.

Jam

So ive read Gathmans notes/ Fultons parts on an introduction to algebraic geometry but it does not have any of this. Where do i start?

Ted Shifrin

Go back and read parts of Griffiths/Harris Chapter 0 as needed. You cannot do anything without the basics of sheaf cohomology and long exact sequences. Harris's introductory book has lots of examples, but none of the technology to proceed.

Jam

So Chapter 0 will be my savior

Balarka Sen

@TedShifrin Yeah I know this one. If $(a, b, c, d)$ are multilinear operators which satisfy skew-symmetry if you switch $(a, b)$ and $(c, d)$ individually respectively and symmetry if you switch $(a, c)$ and $(b, d)$ simultaneously AND satisfies Bianchi identity, then it must be determined by expressions of the form $(a, b, a, b)$

Ted Shifrin

7:54 PM

There you go.

It's the uniqueness result.

Balarka Sen

Yeah

Ted Shifrin

Cool. :)

For practice, use the structure equations for $S^n\subset\Bbb R^{n+1}$ to get it, too. :D

Semiclassical

so, here's the argument i was getting at above

Balarka Sen

@Ted Extremely clean. So this tells me $\Omega^2$ has only off-diagonal terms.

Aka trace 0

Very nice. This is proper intuition.

Jam

isnt it funny when advanced books will write down the formula of the norm of a vector but not of the harder stuff hahaha.

Semiclassical

7:56 PM

since A^3+3A-4=0, the only possible eigenvalues are $1,-1/2\pm i\sqrt{15}/2$

Ted Shifrin

No $i$?

Semiclassical

oops

let the multiplicity of the $+i$ eigenvalue be $\nu$. since it's a real matrix, the multiplicity of the $-i$ eigenvalue is also $\nu$, and the multiplitiy of eigenvalue 1 is thus $n-2\nu$

therefore the eigenvalues of A+I are $2,1/2\pm i \sqrt{15}/2}$ with the same multiplicities

Ted Shifrin

Yes, I made the identical argument with the polynomial factored as linear times quadratic, without needing any roots.

Oh, right, I never wanted to think about eigenvalues of $A+I$, either.

Semiclassical

since the complex eigenvalues have modulus 2, the product of eigenvalues is $2^{n-2\nu} 2^{2\nu}=2^n$

Ted Shifrin

I will try to come up with a problem where my approach is needed.

Semiclassical

7:59 PM

kk

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$Mathematics$ Mathematics