hey Guys,I answered This question (math.stackexchange.com/questions/2090665/…) and then Barto gave a link of an exact question, but it do not have any answer.. Should I answer that question too

??

Please guide me

I think that would be seen as trying to hog reputation. The question without an answer should preferrably be closed as a duplicate, I guess...

Yeah.

So, I think marking that one as duplicate is legitimate

Usually one references a duplicate that was asked and answered, not one that never received one.

I don't know what the standard line is, but I wouldn't be surprised if the older one would be the one to close on the grounds that it never got an answer.

But I dunno. The Math Mod's office might have a more definitive standard.

I don't know for sure either.

If you find both questions without an answer, you'd probably answer the older one and close the newer as a duplicate.

So, What I am supposed to do??

I'm still in favour for closing the older as a duplicate.

Perhaps this situation is worth asking about on meta, though?

Meta or the Math Mod's office.

Looks like DanielFischer weighed in and marked it duplicate

Hmm... I'm personally more in favour of meta, as 1) this isn't exactly an urgent question, and 2) it's more clearly documented for anyone with the same question at a later time.

Either way, guess it's resolved :)

Yeah, some documentation would be nice.

I think the key is that when a question is closed as a duplicate, the link does not say "This question has already been asked here" but rather "This question already has an answer here"

Hi chat

@GFauxPas: chat.stackexchange.com/transcript/36?m=34655807#34655807

Salut, @Astyx.

hi @ted, @Astyx

Hi @Semi @Ted

How are you ?

Doing just fine, thanks. Waiting to get my mail from the two weeks I've been gone.

@Ted Hirzebruch meant contravariant = usual tangent bundle.

The differential geometry language comes from you f***ed-up physicists ... looking at transformation of components. For mathematicians, $T$ is covariant and $T^*$ is contravariant :P

But one never knows ...

It doesn't come from physics, but sure :-)

I'm pretty sure it's the physicists who started that language ages ago.

I think Spivak made that claim in his 5-volume text, but I'm not going to go look.

Physicists didn't do much tensor notation before Einstein.

Levi-Civita etc...

Maybe you count those as physicists?

Woo, physicists bashing time?

I'm talking terminology. Of course, the tensors came about earlier.

In other news, help me understand the following maybe:

$i$ is better than $j$ !

^wtf

aint no engineers here

Lol

Anyways @Ted so I ahve the following

Someone was using $j$ for something totally bizarre in here a week or so ago, totally confounded me.

Consider $\Bbb P^3$ and projectivize its cotangent bundle

I thought $i = -j$

KIDDING

kind of

@TedShifrin I tend to just avoid the -variant words.

Well, at least $i^2 = j^2$ and $ij + ji = 0$.

If you think categorically/functorially, you can't avoid them, @MikeM :D

That's the "joke", its that it wouldnt make a difference

Then a point in $\Bbb P(T^*\Bbb P^3)$ corresponds to a line $\Bbb C^4$, plus a plane (namely the kernel of $\Bbb C\alpha$)

Just stick an ^op above your first category.

The $j$ use was not quaternionic when it occurred here a while ago. I've forgotten.

I think that 'general covariance' goes back to Einstein?

But how is the line "inside" the plane?

But when he did it it originally was about the content of physical laws rather than a mathematical statement.

hey Ted how come the extensions of the idea of complex numbers, quaternion and octonions, are powers of two? why cant you have 3 imaginary units?

$[\alpha]$ is the point downstairs, @Danu?

No, $\alpha$ is the covector

or rather

3 units

Oh yeah, duh @Danu.

I didn't give the line a name

I meant the point in the projectivized fiber.

yeah, right

@GFauxPas: For the quaternions you do have $i,j,k$?

so one real unit and 3 imaginary units

I want $\Bbb P(T^*\Bbb P^3)$ to be $U(3)/U(1)\times U(2)\times U(1)$ given by the flags $(1)\subset (3)\subset (4)$

and then one real unit and 7 imaginary units for octonions

Hey Guys Can anyone tell me what happened to this guy?? math.stackexchange.com/users/401635/kanwaljit-singh?tab=profile

8 units

Hirzebruch says that a point in $\Bbb P(T^*\Bbb P^3)$ gives the inclusion $(1)\subset (3)$. I don't quite see it.

Ohhhh

As I have gone through a three days suspension recentaly.. I m interested in it

but i've never seen for example $i^2 = j^2 = k^2 = \ell^2 = ijk\ell = -1$

First of all, let's do dimensions.

The kernel is just a plane

octonions are weird, though. not even associative :/

So you add the line

to get the (3)

Well, at least they're alternative

they're not semi? Weird

@Danu: I'm not keeping up with you.

@THELONEWOLF. It says voting irregularities

If you don't need me, I won't try.

@TedShifrin So I want to understand how a point in that space gives $(1)\subset (3)$ where $(n)$ denotes "subspace of dimension n"

And there's that drawing of the octonions as the fano plane to help remember how to use them

yes @Danu, But Is there anyone who knows when a user is suspended due to such Voting irregularities

The base point is the (1) and the kernel of $\alpha$ gives a (2); then you add the (1) to get a (3) in which the (1) is included

@THELONEWOLF. Voting irregularities means sock puppet, usually. Your interest in this case casts some suspicion on yourself.

Does that make sense, Ted?

Sock puppet??

So you have to remember the interpretation of the tangent space (similarly, cotangent space) of $\Bbb P^3$ at a point $x$. $x$ is a line, and any tangent vector corresponds to point in the orthogonal complement of that line.

@Danu, What is sock puppet??

Having a dummy account to upvote your posts

Use Google

So a $1$-form will kill any vector along that line.

So $\ker\alpha$ contains the line.

oh...

Oh, but that's not so nice, since I want a (3)

Whoa ... A dummy account to upvote your posts sounds a lot like our corrupt lying president-to-be.

Hmm... I think I'm starting to see...

@Ted Who ?

The orthogonal complement of the line is a $3$-plane in $\Bbb C^4$.

@TedShifrin, President??

president or president-elect? not that said distinction will matter for much longer ...

Yeah, wait, I'm not thinking ....

Danu can see my recent facebook post re that.

So, a tangent vector to $\Bbb P^3$ actually should be corresponding to a (2), not a (3).

@Danu: At any rate, you have to remember what the tangent/cotangent plane are.

If you don't like Trump, you spent the majority of the year praying it to end---and then the rest of it praying from 2017 to never start :/

A tangent vector is an element of $\text{Hom}(\ell,\Bbb C^4/\ell)$, so it's (non-canonically) a point in $\ell^\perp$, which is $3$-dimensional.

@TedShifrin I'm weirded out by this, since how is this idea of yours going to distinguish the tangent and cotangent bundles?

Well, we need the dual bundle, of course.

But my brain is not working on this now. I'll come back later.

\o

Salut @Hippa

@Astyx bon jour

Salut @Null, comment vas tu ?

@Astyx pourquoi ne pas "cava"? Com ci com ca ;)

Not quite sure I got that :p

@Astyx i never heard comment vas tu hehe

but i am honored :)

It's more "formal" than "Ça va ?", which is casual, at least that's how I see it

are french keyboards armed with the swirly "c"?

And you forgot two m's and two e's, and two cédilles, but else that's okay :p

I'm using a qwerty for my part, but yes they are

alt-c on my keyboard

ķ

ļ

And so on

@Null Italian ones too and I don't know why

Polish has ę

Wait, the swirly is pointing the wrong way

i get a spascism when i see $\delta$, so I'm biased I guess

wait... is there a math-keyboard available?

It should look like this : fr.wikipedia.org/wiki/C%C3%A9dille

@Astyx i learned french, it's just that I don't have a french keyboard :/

Neither do I

Well actually I do, but I don't use it

I'm typing in qwerty on an azerty

Which is confusing to everyone else trying to type on my keyboard

@Astyx hehe

i need help for this : math.stackexchange.com/questions/2090674/…

Any ideas why someone downvoted me with nary a comment?

Because your answer is bad and you should feel bad

You're o so helpful.

@TedShifrin So I still don't understand what you're saying.

Let's think about covectors for now.

So what Hirzebruch says is the following

So I agree that you get a $3$-plane containing the line.

Am I not always ?

@TedShifrin I'm not sure anymore.

@Astyx: I liked you better when you were sickly.

Hirzebruch says

Hehe :p

@TedShifrin w/e, upvoted out of spite

@Null: Unless it's my long-ago arch-enemy, I'm curious how to improve it. The OP accepted the answer.

Maybe the OP misclicked when he accepted the answer ?

"A point of $\Bbb P(T^*\Bbb P^3)$ corresponds to a line $\ell$ and a plane through $\ell$ which, as a tangent plane in $\ell$, is annihilated by a (line's worth of) covector(s) in $T^*_\ell\Bbb P^3$." Let's call that covector $\alpha$. This makes perfect sense to me... Except that $\ker\alpha$, to my mind at least, naturaly lies in $T_\ell \Bbb P^3\cong \Bbb C^3$... Also, it doesn't contain $\ell$ in a natural way.

@Astyx meh, what i thought

He/she could always unaccept and ask a question, @Astyx.

No, I meant he meant to upvote, but clicked on the wrong arrow

@Astyx: No, that just happened, after the acceptance a while ago.

It seems like a good answer to me, some people just enjoy downvoting others I guess

I was trying to give intuition, not a rigorous proof. That's what he needed.

What am I doing wrong?

What does "tangent plane in $\ell$" mean, @Danu?

Well, people tend to downvote without giving a reason, sadly.

I don't know! The obvious meaning, to me, is a plane in $T_\ell \Bbb P^3$, i.e. a subspace isomorphic to $\Bbb C^2$. This is what's confusing me.

OK, @Danu, my point is that if you pull $\alpha$ back to $\Bbb C^4$, the kernel does contain $\ell$.

So we want to lift everything up to $\Bbb C^4$.

@TedShifrin Okay. Pullback. I should've thought of that (in some non-rigorous sense I was, but I was getting confused at the same time)

@Astyx I really think it's a missclick

So we can use a hermitian metric (ok, so we're cheating) to identify $\ell^\perp$ with $T_\ell\Bbb P^3$.

@Ted there are some very strange downvote things on the site sometimes, like this question has 5 downvotes for some reason: math.stackexchange.com/questions/2084047/…

@s.harp: I can't remember a time I downvoted without posting a comment to complain. And usually I give the person time to fix what I complain about. When no one does, then I downvote. But I am not a typical citizen.

I was actually asking you guys for input on how to make the answer better. If you have none, so be it.

@TedShifrin Like for a sphere. Okay, I'm fine with this. Now for the tangent bundle. Here we should get a 2-dimensional subspace (not a 3-dim one) containing $\ell$... I'm tempted to just say $\Bbb C\ell\oplus \Bbb C v$ where $v$ is a tangent vector.

No, the kernel will be a $3$-plane containing $\ell$, up in $\Bbb C^4$, @Danu.

@TedShifrin I'm telling you, Hirzebruch says we should understand the tangent bundle in terms of a flag $(1)\subset (2)\subset (4)$. So we're looking for a plane containing $\ell$ in this case.

Oh, no more kernel of $\alpha$? Yes, what you just said is correct.

Just direct sum it? It feels "cheap" but I guess there is nothing else (no pullback, pushforward does no good...)

No, $(\pi_*)^{-1}(\text{span}\,v)$, where $\pi\colon\Bbb C^4-\{0\}\to\Bbb P^3$.

This is not unrelated to our discussion months ago of the Euler sequence.

wonders if @Danu fell asleep

the proof I most enjoyed was $\sum \frac{1}{n^2}$ is convergent, via comparing the terms with terms from $\frac{1}{2^x}$. The diagonal argument from cantor is another I liked very much

@TedShifrin I played a 1 minute chess match ;) Sorry

@TedShifrin I realized that much :)

@Null: There is a cool way to consider $\sum n^{-p}$ using the Cauchy condensation test ...

We need to check that $(\pi_{*tx})^{-1}(\text{span}\,v) = (\pi_{*x})^{-1}(\text{span}\,v)$, @Danu.

I.e.., you need to check.

@Ted Maybe I would emphasize more on the fact that convergence is an asymptotical property, thus if two series "tend to be the same" at $\infty$, they have similar nature (is that the terminology). Otherwise your answer seems absolutely fine to me.

Well, they aren't equal, @Astyx. But the decision about the convergence is identical.

Yeah, I realised the absurdity of what I had written a few seconds too late

I always taught it with the phrase "looks like" in comparing two series, meaning the limit comparison idea.

I guess I said we could ignore any finite number of the first terms, @Astyx. I guess I don't want to muddle it up by adding too many words.

@TedShifrin Yeah, I definitely think that works (you'll get a scaling factor, as we discussed for the Euler sequence, but that doesn't mess with the span)

Right @Danu.

So are we OK now?

Definitely!

Thank you so much

Is the convergence/divergence of a series called its nature in English ?

Flag manifolds are neat :)

Sure thing.

Or is that specific to french ?

Yes, my thesis is full of them. (Incidence correspondences galore.)

Now I need to do a more complicated case on my own

I don't think we say that in English, @Astyx.

$G_2/U(2)\cong \Bbb P(TS^6)$

Well, when you do that, @Danu, you can teach it to me.

Right, thanks for the info

I first need to understand the flag perspective for those $G_2$-quotients.

That'll take some time. In the meanwhile, I'm doing some other calculations.

Just so long as your interest doesn't flag :D

Rehi DogAteMy :)

Kotschick is on top of my stuff now---he came by my office again today (first day after the holidays!)

Hey chat

I might come back, bye

Wow, he's being great, @Danu.

Bye, @Astyx.

And gave me more stuff to compute

Yeah, very nice.

I hope I don't disappoint him too much

Does he want you to stay to do your Ph.D. with him?

I hope?!

Oh, I thought you wanted to go elsewhere.

Well

I mostly want ANY PhD spot

I don't give myself such great chances for finding one

Well, keep working hard and learning. I think you've learned a lot.

Anyhow, I look forward to a $G_2$ lesson from you.

We'll see when the time comes...

Now I'm going apply to this nice workshop/thing:

wwwf.imperial.ac.uk/~at515/bigworkshop.html

So, what exactly is a Higgs bundle? @Danu

@Semiclassical I will find out, won't I?

lol

A priori it's just a holomorphic bundle with some extra data...

Can someone pick a number 1 - 5?

4

Thanks

Lapsang Souchong, it is.

Eww

3smoky5me

("chosen by fair dice roll, guaranteed to be random"?)

I only like my whisky smoky, not my tea :P

I need something to keep me awake during this lecture :/

Smoky will do

tiny lemma: if $\sum_{k=1}^{2n+1}a_k$ converges for $n\to\infty$, then $\sum_{k=1}^{\infty}a_k$ converges too. Or can there be an easy counterexample?

Yes, there is an easy counterexample.

I've never heard that question before, but I like it.

Think of it in terms of the partial sums instead.

It's similar in flavor to a problem I gave DogAteMy a while ago ... but not quite.

it's just part of my unsatisfaction in one of my proofs @TedShifrin missing basicly the "even" stuff^^

but good to know it's no true

What were you trying to prove, if you don't mind me asking?

well, i try to extend my proof here math.stackexchange.com/questions/2077990/… without AM-GM or similar stuff, only extend mine to even $n$

@Null: You should take the time to make up your own counterexample so you understand what's going on.

(not that I dislike AM-GM, just curious :) )

@Null: Can you prove it's correct if $a_n\to 0$?

@AkivaWeinberger well, alternating series are the first thing I'd try ;)

Back at school now, DogAteMy?

Just ended for the day

Ok, just asking wether it is clear that i mean that the top value is $2n+1$, so 1,3,5...

Yes, of course we get that

mmh

But I'm asking if you can prove that the original series converges if $a_k\to 0$ as $k\to\infty$?

@TedShifrin no, easy example would be 1/n

(or 1/k)

With your assumption about odd partial sums?!

ah, ok, then I will think about it ;)

ok, just asking. Does it even make sense to say that we add an odd number of terms infinitly many times?

Are you adding partial sums?