Or constructing the weird first counterexample whose name I can't remember.

something Lindemann, probably

Huh? @DogAteMy. That sounds like crap.

Oh, I see. I thought you wanted to conclude that $\pi$ was in fact transcendental.

You merely are arguing that there exists some transcendental without having to exhibit one.

Sure.

Easy to prove by diagonalization that transcendentals are a thing, hard to prove what any of them are.

It seems like there should be a clear-cut example, though

Okay @TedShifrin so your argument is the same as Huybrechts. I understand the computations. I just don't get why I'm supposed to go from $\Omega\in C^0(\{U_j\},\mathcal A^2)$ ($\mathcal A^k$ is my notation for $k$-forms) to an element of $C^2(\{U_j\},\mathcal A^0)$. It must be something obvious.

Why am I chasing through this diagram?

It's easy to come up with examples where you can disprove a claim by a nonconstructive argument.

Because you're trying to find what Cech cocycle in $C^2(\Bbb Z)$ corresponds to the curvature $2$-form.

Seems a lot harder to come up with an example where no constructive argument is possible period.

@TedShifrin So what does "corresponds" mean, here? I know I'm sounding real dumb, but I just somehow don't see why I'm doing these calculations.

$c_1$ is represented by the differential form $\frac i{2\pi}\Omega$. I want to chase the coboundary map in the LES to get the $2$-cocycle with coeffs in $\Bbb Z$.

So this diagram is how you do that.

Coboundary map in the LES = the $\delta:C^j\to C^{j+1}$?

@semi "there exist no well-ordering of the reals"

anything relying heavily on AC really

$H^1(\mathscr O^*)\to H^2(\Bbb Z)$ on the cocycle level, yes.

Not what you wrote.

Well, what you wrote could be anything.

Ah, okay. So I guess my problem is that I don't see why that map is the same as that composition

@TedShifrin I'm referring to (for instance) the $\delta_1,\delta_2$ from the diagram I just posted. The Cech coboundary maps

No, I'm talking about the interesting map in the long exact sequence of cohomology. Yours are just the individual coboundary maps for each complex.

The interesting "snake" map is defined by a bunch of lifting back and taking $d$.

Right. What you wrote is not the same as what I wrote. I realize now.

@TedShifrin Sure, the "stairs" in the diagram I posted.

(When I taught this stuff I doubtless drew the diagrams and chased. I didn't bother doing that in my notes. Sorry.)

Yes, the stairs.

Last seven quizzes....

It's funny reading posts on FB by former students of mine who are now TAs in grad school and apoplectic about grading exams and how badly their students are doing. I can only chuckle. I said something like "You're expecting me to impart words of wisdom? I am unwise." :D

Need to get six quizzes graded in 15 minutes. That's not actually hard, I'm just indecisive.

Just do it, man.

Yea.

Okay, so perhaps my problem is the following: I see that we have a way of going from $C^0(\mathcal Z^2)$ (in your notation) to $C^2(\mathcal Z^0)\supset C^2(\Bbb C)$. What I don't see is (i) why this way is particularly natural/right for this situation (ii) why it is unique?

Shaddap and calculate like a good physicist (@Semi)

It's not unique, but everything is unique on the level of cohomology. That's the proof of the LES.

OK.

And it's totally "natural/right" in this situation because you're trying — as I said — to turn the curvature $2$-form into a Cech cohomology class with integer coefficients.

Five.

@TedShifrin But is that a natural thing to do (I mean, before knowing that you can/want to prove this theorem)?

If you have the $c_1$ map and you're trying to understand what it does, yes.

Hmm yeah I guess

I don't know... I'm just missing some kind of philosophical satisfaction :P

I think it's amazing that this thing you define using differential geometry magically measures this algebraic construction.

I... guess? Urgh, I feel like when I was reading books for adults at age 12 and I just didn't really get it.

Or when you took physics the first time

Four.

You just want the X-rated version, I know, @Danu.

I did not mean porn. I meant famous World War II novels :P

suuure

Three

Heya @Faraad.

Hey @TedShifrin: I was just reading through previous messages. How are you?

One(!)

@Danu: The truly beautiful computations are essentially Chern's original ones, showing that symmetric polynomials in $\frac i{2\pi}\Omega$ (the curvature matrix of the universal bundle on the Grassmannian) are Poincaré dual to the appropriate Schubert cycles. That's just so gorgeous to me.

Doing OK, thanks, @Faraad. Learn anything cool lately?

So... From MS's book I really dislike these universal bundles. They're really terrible and ugly-sounding, to me. Is it MS, or is it me?

@Danu: That computation is at the end of those notes. In the recent course I taught, I also did the Euler class (with the Pfaffian) and proved the general Gauss-Bonnet-Chern theorem.

It's MS. I told you I don't like that book.

So where the heck can I find a less terrible and ugly treatment of this stuff? Ain't many books on characteristic classes...

What's to dislike about the tautological bundle? It's just $\mathscr O_{\Bbb P^n}(-1)$ generalized to Grassmannians.

Reminds me, one thing I always wanted to understand was KP hierarchy stuff

^the infinite-dimensionality in MS is already a pretty big yuck factor for me :P

which is related to Grassmannians.

I keep coming back to G/H, I suppose. I taught a whole course on bundles and characteristic classes. Did I not send you those notes?

somehow

@TedShifrin I don't think so, no.

@Danu: There's only infinite-dimensionality when they go to the classifying space.

@TedShifrin Yeah but that's all the book is about.

Again, they'll probably be a bit too cryptic, but I'll send 'em to you. Just a minute.

Also, the f@(!&R)(F-ing Steenrod squares were probably the biggest letdown I've ever experienced in my mathematical life. Spend 100 pages developing ugly theory just to end up quoting some other stuff that is equally mysterious/unproven as the final "proof"?!

Not really. I've just been reading ODE's by V.Arnold. I've just been working through the proofs of Inverse and Implicit function theorem (he uses them a lot). I've used them so much (well the former at least), and now I finally know enough to understand the proofs.

Yeah, I hate the Steenrod square treatment.

More stuff from the blue book, Faraad :P

@TedShifrin Cryptic? How so?

Written as lecture notes for me, not as a text for the world.

You mean not-at-my-level because they're just personal notes?

Yeah. okay.

I did a bit of stuff on symmetric spaces and invariant forms/cohomology, too, in that class.

Sent.

There were exercises, too, but I know you don't want those. :D

Pfft. I love exercises!

@TedShifrin: I know they are in there. I wanted something that wasn't too clever for me. Someone posted some proofs on the archives a few years ago that are really good and quick.

I ended up doing a few from Huybrechts because he ends up relying on them near the end...

Got it. THanks!

Nothing clever in my proof, @Faraad. It's the standard proof (admittedly the one that works in Banach spaces, which is far better than the one Spivak and Munkres do).

@TedShifrin Seriously though, I'd like to have the exercises.

Yeah, I'm looking for them. BTW, the end of those notes is the proof of one of my theorems :P

@TedShifrin: I stay away from almost anything Spivak writes. I don't have enough years left in my life to read through extremely terse arguments.

@TedShifrin D'aww ;D Which one?

Using Chern class stuff to count "cusps of the Gauss map" on a surface in $\Bbb P^3$. (I mentioned this to PVAL a few months ago.) I can send you the paper if you're curious.

Got it. Thanks!

Ehh, sure!

@Faraad: His diff geo 5 volume-text is anything but terse, but Calculus on Manifolds sucks, yeah.

Sent, @Danu.

(Three emails was hardly efficient.)

Can I use the ratio test on sequences ?

Or only series ?

Wouldn't want it any other way :3

NO, only series.

On sequences only limit and comparison ?

Why have I never heard of this "ratio test" ?

@TedShifrin: The latter is the one I am really referring to. All the fan boys however recommend it to people wanting to learn calculus. Drives me insane, but that's just my opinion.

His Calculus is beautiful, @Faraad, but Calculus on Manifolds is not.

The single-variable Calculus is very expansive and has superb exercises (some of which I wrote, I admit).

@Astyx: I'm sure you have. $\sum a_n$ converges if $\limsup |a_{n+1}/a_n|<1$.

You wrote Spivak's exercises?!

@Astyx D'Alembert's ?

A few hundred, @Danu, in the later editions, yes.

Damn

Ah right

@TedShifrin: Okay, I lied. I do think highly of his regular calculus text. I guess I should distinguish between his calculus, calculus on manifolds text.

Yup, @Faraad, and I agree with you re the latter.

I should do calculus sometime.

I would be very upset if you said my book was just as unreadable.

Thanks

@TedShifrin How can I prove $ a_n = \dfrac {2^{2n} n!^2} {(2n)!} $ diverges ?

The sequence or the series, @Maks?

Isn't this a series problem?

I'm trying to find another sequence to compare it to because I cant find its limit

sequence

You're sure?

Let me recheck

Uh huh

@TedShifrin: I really think those who wish to one day write textbooks, should start immediately when they have the thought. The reason I say this is because we easily forget the things that "tripped" us up when we were learning the material and instead of shedding light on those particular instances, we unconsciously submerge them because you're so far past that stage by the time you write the book, you forgot.

Yes ted

Well, @Faraad, I come at this from the perspective of having taught it and decided how best to teach it.

Is Striling's equivalent well known ?

Why did you get the thought it was a series?

Because this sort of problem shows up with the ratio test for series.

@FaraadArmwood I'm not sure I agree---it's best to start taking notes early, maybe. But polishing is valuable.

@Astyx asked the right question: Has your instructor shown you Stirling's formula for estimating $n!$?

I almost always change a lot of my first draft when presenting something

@Faraad: But you'll note that I assailed you several times for not knowing where your Calc III students were, intellectually or in terms of what they actually needed. Having that awareness is crucial.

Me too, @Danu. And tenth draft.

@TedShifrin And how would you solve that sequence ?

@Maks: Astyx's question is appropriate. Do you know about Stirling's formula for estimating $n!$?

@TedShifrin Nope, let me look for it

Without that, this is a very hard question.

My question still holds :)

Even with it, it is very subtle.

I know, @Astyx. But you need the refined version, not the sloppy one, for this problem.

?

@Danu: I think we are saying the same thing. I'm just encouraging authors to write earlier before they become a seasoned vets.

@TedShifrin when computing flux , does any normal work ?

I usually just tell people $n! \sim (n/e)^n$ (without the $\sqrt{2\pi n}$).

This is wrong though, is it not ?

What do you mean "any normal," @Kasmir? It has to be the unit normal pointing the right way.

Usually it's good enough, @Astyx. Not in this problem, though.

$ ln(n!) = n ln (n) - n $ ?

@TedShifrin i meant does it have to be unit normal ?

i.sstatic.net/T9HJG.png @TedShifrin You are one of the few scholars left. I know you really take the time to drive ideas home for your students. If at any time you felt like I was bashing your texts, just look at this that I wrote about you the other day.

@Maks, approximately equal, yeah. But it turns out that won't be good enough for this problem.

But I mean writing $n! \sim (n/e)^n$ is not right since ${n!\over(n/e)^n} \rightarrow 0$

@TedShifrin doesnt that area element dS take care of the length ?

or $ \dfrac {n!} {\sqrt{2\pi n} (\dfrac {n} {e})^n} $

How can that help me ?

I know what you mean, @Astyx.

Oh, yeah, well that's exactly what you need to use, @Maks.

Huh

But that series converges...

@Faraad: That guy is going to have very miserable students.

@Maks: You should have $$\lim_{n\to\infty} \frac {n!}{\sqrt{2\pi n}\left(\frac ne\right)^n} = 1.$$

Exactly

How would I use that sequence to solve mine ?

@TedShifrin: Well I think my words (your advice) made him think a bit. But yes, as soon as I read that question I went, "Oh dear".

So use that for the stuff in your limit. You have to figure out what goes with $(2n)!$ by substituting $2n$ for $n$. Then you have to use this thing squared for the numerator.

I'm going, bon après-midi à tous

Bonne nuit, @Astyx.

Merci

ted you're french ? :D

@KasmirKhaan: No, but he is a grammarian so you should probably edit and write you're.

@Faraad: When I taught multivariable calc and some linear algebra to a class of mechanical engineers at Berkeley, I did it two quarters in a row. The first quarter, one of the postdocs chose the text. It was way too hard for the students, but we survived. The second quarter, I just used Thomas (even though I hated it) because I knew the level was right for them, along with some linear algebra notes I'd written for MIT classes. Those students did better.

ROFL @Faraad

@Kasmir: Je n'suis pas français du tout, mais je parle français, évidemment :P

haha :D

Each day you surprise me =p

J'ai même donné une conférence de maths en français il y a longtemps.

started with those lectures you had on Youtube , never thought id speak to you =p

Hmm where in france?

or canada?

Ecole Polytechnique

(even a publication in French, basically the lecture — but then I gave another 5 lectures or so to graduate students).

@TedShifrin Hahaha I'd love to see that :D I've never heard an American speak proper French (though I imagine the percentages (of good speakers) are comparatively sky-high among mathematicians, particularly ones who know something about algebraic geometry).

@NaCl hellooo :D

that is very good , i wish i could do something like you , i want to be a teacher as well

@TedShifrin: The math community is doing a good job in making us think that things should be full of extreme rigor and terse arguments. And haha, I thought you might find that funny.

hi null :)

My French is a bit rusty now, @Danu. My tongue gets more tangled-up, but I did OK in France for 2 weeks.

@TedShifrin :)

Well, @Faraad, I was trying to stop you from doing the same thing!! :)

There are a lot of bad lecturers/teachers.

Regarding students being overly ambitious in teaching: I also suffer from this a little bit. I wonder how many will be able to completely follow my proof that the Euler class is PD to generic zero locus...

But you're doing a seminar for graduate students, not teaching a first-year undergraduate, @Danu.

Many of them are not that solid with their cohomology 'n' stuff.

Nevertheless, try to give intuition sometimes and don't give all the technical details.

@TedShifrin I know, I know.

Choose your battles.

Why compute unit normal in flux integral ? when the length will be taking care of with the area element dS ?

I also teach 2nd year undergrads though. The quantum mechanics.

Oh yeah, right.

I was outraged at what they got to work with when it came to tensor products

Look at this:

I saw that on mit lecture

Yeah, that's right, if you do it with the parametrization, @Kasmir. I often taught my students shortcuts for spheres and cylinders, though.

@TedShifrin oh the only way we taught is with parametrization =p

If you're going to just write down (as MIT often does, or at least when I taught it there) $dS = a^2\sin\phi d\phi d\theta$, then you do need the actual unit normal to dot with $F$.

@TedShifrin: I listened and my reviews were great because of it. I pass the advice along to incoming graduate students.

got it ! :) thanks! :D

Ah, good, @Faraad, I'm glad you decided I wasn't in fact the enemy. :P

I wish faculty would do a better job of mentoring grad student (and postdoc) teachers. That's something UGA apparently takes way more seriously than most everybody else.

@Danu: benützen wir :D

That's all they got to go on for tensor products. It turns $\psi\otimes \phi+\psi\otimes \xi=\psi\otimes (\phi+\xi)$ into a result.

is the adjective "fair" before dice a tautology?

@TedShifrin Gotta love it... :P

No, @Null. Some dice aren't fair.

^

@TedShifrin but are then they dice by definition?

Null must be easy to scam...

Of course.

Real-world dice are rarely perfectly fair!!

are ok, in german we have only cube for both, the perf cube and the game cube

(the game cube being the dice)

Oh, @Danu, I thought you'd written that paragraph and I was correcting your German :D

@Null: Still, the physical act of constructing the cubes is rarely 100% perfect.

@TedShifrin No---that's what my students got fora definition.

@TedShifrin Also you're wrong about that umlaut :P

The professor left out his umlaut. Shocking. I didn't read it for math.

Am I really?

@TedShifrin: I never thought you were the enemy. At UGA, I didn't really know you until the year before graduate school. In response to the mentoring, yes! I also think that faculty should listen more to what graduate students say about their courses.

Yes

...I think :D

Hahahha

Who cares

I never even read that

I was taught it with the umlaut.

there's no umlaut in benutzen (in any tense)

My dictionary says both.

Ah...

Now I'm looking at the Wahrig.

I was taught without I think.

Wahrig has both, too, but makes it sound like without umlaut is more common.

I don't have to hang my head completely in shame ... only partially in shame.

@TedShifrin i see. But then a fair dice is theorethical. I don't even know why I come up with this.

duden looks like without umlaut is the most common variant

@TedShifrin :D

Anyways

Yeah, weird that I remembered it with.

I wrote a 5 page exposition on the tensor product as a result :P

Back to work.

I almost went "Free vector space on $V_1\times V_2$ quotiented by ..."

Physicists don't need it to be pedantic, @Danu. Just the operations. Which is how most mathematicians do it, anyhow.

But decided against it.

Good. I would have smacked you.

I just told them the basis plus the bilinearity :D

Yes, the whole point is bilinearity.

Also Riesz representation

Note how bilinearity is not mentioned in what I just showed you

Riesz representation is heavily used in the bra-ket notation.

It is implicitly with the summation formulas at the bottom.

meh

It's a result

which is totally wrong IMO

Well, it incorporates the bilinearity. I don't necessarily know what's best for physicists.

In my opinion: To have bilinearity as a direct feature of the construction

Well, he didn't construct. Nor did you :P

He defined symbols.

how to differentiate g(t)*x with respect to x , if t= x^2+y^2+z^2 ?

Why are you differentiating, @Kasmir?

But the answer is: Use chain rule and product rule.

You can't differentiate, anyhow. Remember they just told you $g$ is continuous, not differentiable.

@TedShifrin the question ask me for F = g(t) ( x,y,z) to make the div F = 0 and g(1) =1

Oh, then you need $g$ differentiable.

its part c) of the question ><

and they said g is of class C^1

:)

Aha. And div = 0 except at the origin?

exactly

for t>0

Then you should know an example that does this (or your book should have discussed it). It arises all the time from physics.

thats the thing ! we just study math not physcis

not sure why they give us these kind of questions and we didnt do any of this type =p

I solved part a) and b)

Well, I guess they want you to write down the formula in terms of $g'$ and figure out when it will be $0$. I'm fine with that.

So use product rule and chain rule and compute div.

That's a fine question.

@TedShifrin But at least I told them that bilinearity is a direct feature of the construction :P

thanks :)

I'm fine with that, @Danu :P

OK, I'm gone.

Bye Ted!

Bye sir Ted!

Ted is the best man alive :D

if i want to show an equality about spans. I have to show an equality about sets or? (sets of vectors to be precise)

@Semiclassical Actually, Cantor's diagonalization argument actually does construct a transcendental for us.

@TedShifrin Yes

@AkivaWeinberger how?

sorry i had to take a nap

ive had a long day :/

np meow

@Sophie Enumerate the algebraics. Let the $n$ digit of the transcendental be different from that of the $n$th algebraic (and not $9$ to avoid $0.\bar9$ stuff). That gives us our transcendental.

@meow-mix be happy that it was a day. i have long nights :D

@KajHansen hi =)

Bonsoir @Null

@Brody Guten Tag :D

and @KajHansen

Hey all

@KajHansen why is the limit $\sqrt[n]{n}$ easier than $\sqrt[n]{b}$? b is some constant. Our prof mentioned that to explain the useful of the sandwhich method