@Daminark hello

@ShaV: First of all, you messed up Pythagoras again. Secondly, I don't know what "$p'\ne p$, a linear combination of $v$" means. But your idea is correct: When $p'-p\ne 0$, the error $e'$ is longer.

Hi @Ted

Hi, Zach. Yes, your school is supposed to be that easy, given what you work on in here.

Just pay attention to details and don't be sloppy.

I, for one, highly endorse sloppiness

Jk precision is good

And hey @Vrouvrou

glares yet again at Demonark

Zach is already too sloppy. Don't be a bad influence.

I know, I know, I was kidding

wilts under glare

I can't believe this

A guy in San Diego I was helping with Calc II yesterday asked if my students liked me. How to respond ... :D

Why do all my headphones break in a month?!?!?!

Cuz you're a brute, Zach.

Or cuz you buy cheap s**t.

Actually, this was high quality usb wireless thing because the last "Cheap shit" headphones I bought got stuck in the jack and now its unusable

rip

Interestingly, a lightning bolt apparently fried my ethernet port on my old desktop. Of course, now I have a new computer and no ethernet jack nearby to connect it to.

How does that even happen?

Yours or mine?

Yours

@Ted oh right.. Well, concerning your second remark: I meant that $p'=cv$ for some $c\in\mathbb R$, and that $p'$ wasn't equal to the (orthogonal) projection. So $p'$ was a potential candidate that didn't equal $p$

Well, surge protectors only help if lightning zaps the power supply, not the computer itself. I wasn't home when it happened.

@ShaV, but saying $p'=cv$ is definitely wrong. Just leave all that out completely. $p'$ need not at all be in the plane spanned by $v$ and $p$. In general, it will most definitely NOT be.

Anyways now I can't play my games because no headphones :(

Gee, poor Zach might have to be a productive member of society, instead.

LOL

minus society

Oh, I thought the idea of a projection onto a vector was that we had a scalar times that vector, whose error is perpendicular to this vector

Oh so I mentioned earlier the result about equidecompositions of circles and squares

This is it: arxiv.org/abs/1501.06122

@ShaV, are you drawing 3-D pictures of these things?

Project onto a plane, not onto a line.

oh, I did 2-D

gulps at 3-D pictures

ok I'll try to visualise that

I have pictures all over my linear algebra book, but that doesn't help you ... :D

alright, but I projected onto a vector

That's prob what messed me up in physics, I couldn't visualize motion too well, or 3-D anything

Hey guys... Can someone tell me what time does it take for your reputations to comeback after your suspension ends?

No, project onto a PLANE.

not onto a space spanned by possibly more vectors

well I get it's not a linear combination of the vector if we project onto a plane

then it's just a linear combination of the basis vectors of the plane

Right, so you should be using the plane picture to get intuition.

alright, so I'll be proving a more general case

projecting on subspaces

I guess

Your proof was the right proof (except for that $p'$ nonsense).

it was no nonsense if we projected on the vector right?

$p'$ is just another vector in the subspace $V$.

Or whatever you're calling the subspace onto which you are projecting.

that subspace is $\{cv:c\in\mathbb R\}$

No, no, no.

$v$ is a fixed vector and you're projecting onto a subspace (which I thought was $V$).

Draw $v$ as a vector coming out from the origin at a random angle from $V$.

I don't want to be too stubborn, but I'm not projecting onto a subspace, I'm projecting onto a fixed vector?

I can show you where I'm getting this from

oh wait

you're right indeed :P

okay my bad, I totally misunderstood the lecture then

I will just rematch that entire part of the lecture, trying to understand why we project on a plane

Oh, I'm sorry.

You're right that what you wrote was trying to project $b$ onto the line spanned by $v$. I just didn't pay attention. So do make sure what your lecturer wants you to do.

I usually do the projecting on a line the third day of class, right after I've defined dot product. Then I do projection on subspaces in general much later.

@Daminark i need your help please to find a closed ball, $B_{d'}'(0,1)$ on $\mathbb{R}^2$ where $d'=\ln(1+d_1)$ and $d_1(X,Y)=|x_1-y_1|+|x_2-y_2|$ i found $$B'_{d'}(0,1)=\{X\in\mathbb{R}^2, |x_1|+|x_2|\leq \exp{1}-1\}$$ how to continue please

Heyo @Ted

Heyo @Danu

About that short exact sequence stuff: The thing that helped me (the UCT) turned out to actually be the reversed sequence :D

You know, I hesitated between watching your lectures or those of Gilbert Strang. But when I looked at the title of your videos, it seemed to involve a lot of calculus too, I guess? Or are the lectures you're talking about not online?

The course has calculus and linear algebra both in it. If you're trying to learn something about proofs, my lectures will help you far more than Strang's. He's a good friend and a very smart guy, but his style is not proof-oriented.

I'm happy I managed to resolve the problem. Which brings me to the next :D

yes, I've noticed!

@Danu: Life as a mathematician is always a fight to get to the next.

but I just take the opportunity to prove it for myself as a small exercise

I was just trying to tell you that you needed more information to decide between $\Bbb Z$ and $\Bbb Z\oplus\Bbb Z_2$.

I have this complex vector bundle, which as a real vector bundle splits as $V=A\oplus B$. Can I equate the formula for the Pontryagin class in terms of the Chern classes to the product $p(A)p(B)$?! (there is no 2-torsion to worry about)

@TedShifrin And you were right :-) Luckily the UCT gave me just that.

@ShaV: Watch what you find most helpful. I've tried to watch a few of Strang's lectures and his sloppy boardwork drives me nuts :P

hahahaha :P yea, I'll see what works best. I might watch partly his, partly your stuff, whatever works best

That's an excellent question, @Danu, and if it were some years ago I'd know the answer instantaneously.

(I want to hear a NO to that question, because doing so I'm getting some total crap)

So you mean it splits as oriented real bundles ...

Ehhh... Probably :D Yes

It's the tangent bundle of the total space of this fibration I've been working with. And the base is oriented (and the fiber is $\Bbb CP^1$, each fiber being a holomorphic submanifold)

Hmm, but the usual formula you're talking about for $p_i$ in terms of $c_j$ applies only to complex bundles, so I don't think your question even makes sense.

@Daminark are you there please

@Ted You don't need orientations to define pontryagin numbers, I don't think, @Ted. Just tb sir away.

@TedShifrin But I know the total space has a complex structure.

Oh, that was a mental slip. Only for Euler class. Yup.

This is the quesition I was on about earlier @Mike

Just tensor away*

(right; the Pontryagin class "orients it for you" by complexifying)

But I still don't know what you mean by the usual formula in terms of Chern classes when you don't have a complex bundle.

@Danu If you have the Pontryagin classes, does that give you the Chern classes?

So I ahve

@Mike I didn't buy my tickets yet, but I had to look to estimate the cost for funding.

There's some formula, yes?

$TZ=\pi^*TM\oplus T\pi$

whre I know $Z$ is Kaehler but $M$ is not considered as an (almost) complex manifold

So $TZ$ is complex on its own

@PVAL Yah, got it. I'm waiting for funding info.

@MikeMiller Exactly

@Danu: Of course, all we need is a SES, not a splitting. But you don't even have that in the complex category, so I'm dubious.

But can I equate it to the formula obtained from the real splitting

@Danu If you can calculate the chern classes of E from Pontryagin, and you have a splitting of E as real line bundles, then you can calculate the pontryagin classes from this splitting and use those tond tube the Chern classes.

Don't do formulas. Just calculate p and use that to get x

c

Mike's confusing when autocomplete is in play :P

Yeah, definitely confusing

Then you tell him what I just said

Also I don't think you understnad what exactly I'm trying to do Mike. I want to figure out $p_2(Z)$, then use that to determine $c_3(Z)$

Yes, and you have a formula for pontryagin numbers of a direct sum

I know $p(M)$ and $p(T\pi)$

@MikeMiller Yes, but I don't think I can equate it to the formula for the Chern numbers since the splitting is not one of complex vector bundles.

So you know p(Z) up to 2-torsion

So, right, I guess I haven't thought about a Whitney product formula for $p_k$, but it does follow by just tensoring with $\Bbb C$, as Mike said, I guess.

Who cares? You now know the pontryagin classes. P is P is P is P, whether or not you're complex.

Tensoring what with $\Bbb C$?

The real split bundle.

Now since you're complex and know the pontryagin classes you know the Chern classes.

@Vrouvrou I've got a manifolds pset I'm hoping to finish tonight, so I won't be able to help out much, since this seems like a somewhat heavier problem

@TedShifrin But that will give me $TZ\oplus \overline{TZ}$, not $TZ$ itself

Sorry

But so what, @Danu? That's where the usual formulas in terms of Chern classes come from.

Demonark: I'm interested to see some of those manifolds exercises down the road.

Salut, DODO !

In other news, I know that if I do do that then I get bogus.

Then you've made a calculation error on the way

Naw, I triple-checked and had others look at it too

Yeah, I'm on Mike's side now.

But I've made my share of stooopid errors on this kind of thing in my career (thankfully over).

Remember I told you that my coauthor and I did a Chern class computation something like 20 times before we got two answers that agreed. :P

I would just explain the whole calculation to you with all details but I guess nobody is waiting for that.

Do you have it typed up?

@Ted I do that a lot

Sure

Haha, yeah so far it's more basic stuff, showing that tangent spaces as defined by charts is equivalent to when defined by curves

OK, email it to me. I'll try to read it.

But it's long, and references a paper a bunch of times

There's this one neat problem

I'll back off unless I'm needed

Ah, Demonark, that's stuff I typically saved for the grad course and didn't belabor in the undergrad course.

All I want is this particular calculation, @Danu.

I no longer have Milnor's characteristic classes. Doesn't he explicitly state this fact about Pontryagin classes?

Depending on where I start, it's a few lines or a few pages.

Or Hirzebruch? I don't have him either.

You should start by verifying this in one or both of those books.

Yeah, the splitting formula is in Milnor (modulo 2-torsion, which I don't have any of)

@TedShifrin can you help me please

I already did that stuff

Yeah, $2$-torsion can show up because of the $E\oplus\bar E$ stuff.

Gosh this Thurston guy must have been pretty smart.

i have to find a closed ball, $B_{d'}'(0,1)$ on $\mathbb{R}^2$ where $d'=\ln(1+d_1)$ and $d_1(X,Y)=|x_1-y_1|+|x_2-y_2|$ i found $$B'_{d'}(0,1)=\{X\in\mathbb{R}^2, |x_1|+|x_2|\leq \exp{1}-1\}$$ how to continue please

@TedShifrin None in this case.

So assuming most of the background calculations as facts I can make it real short.

@PVAL he's a cool dude

You haven't stated what the problem is, @Vrouvrou. You have to find a closed ball such that what?

Basically, you define an equivalence relation that gives 2 points as being equivalent if they differ by a lattice point. First, we show the set of equivalence classes is homeomorphic to the torus, then we define $F:\mathbb{R}\to S^1\times S^1$ by $F(t) = (e^{it}, e^{i\sqrt{2}t})$, and we should show that it's injective, along with its derivative. Finally, show that $F(\mathbb{R})$ is dense in $S^1\times S^1$

That does sound fun

@PVAL, he was. Although sometimes his lecturing style made me nuts. But it was a long time ago.

(I haven't started yet so don't say anything :P)

That's a classic, Demonark. There are lots of ways of arguing that density. I used to assign this at the beginning of the grad course and I got 3 or 4 different proofs of the density. That was cool.

You know me better than that. I'm not the one who goes around ruining problems for people. :P

@Ted has nobody to look at suggestively because Akiva is not online.

It was more clarification that I haven't started, as opposed to started and got stuck

LOL, Zach.

You mean look at accusingly. :P

So, Zach, you working on anything these days?

I don't think Ted looks at anyone in this room suggestively.

Hope not, at least.

@TedShifrin i don't know what i must find just the closed ball $B_{d'}(0,1)$

I certainly don't mean to, @MikeM. I got accused of such things by you-know-who.

I know you're not me, also k e k

I don't know who I-know-who is

Which is kind of contradictory I guess

No, Zach, no one here other than Mike and Semiclassic probably does, thank goodness.

@Vrouvrou: Closed ball with what property?

I dunno either

I can't help if your question makes no sense.

Yeah, @Balarka, you might.

@Ted Have you ever gotten an e-mail from that John Gabriel or whatever his name is?

Huh, who?

I haven't, I'm just wondering

@Ted Wild guess, Is it it Trump?

He's the pseudo-mathematician guy

@TedShifrin what do you mean by "what property"? please

I have no idea to whom you refer, Zach.