I think it takes a little while to be happy with that. :)

^

My topology professor drew this once

I omitted it in my notes because I didn't see how it worked :P

Algebraic geometers draw it always.

But you become more comfortable with these things by thinking about and with them n

Using that picture you can see (/$\Bbb R$) why a parabola, a hyperbola, and an ellipse are all projectively equivalent.

That's also why I'm picking on you for not doing exercises, regardless of how little time you have.

But you're picking on me, too, @MikeM, as usual.

You can pick on me all you want---I can't help ya. I'd love to have 4-5 years of secured position to work on exercises, like you ;D

(poke back)

It's a matter of helping you, @Danu. We're not being this way to be mean or arbitrary. But anyhow ...

@Danu: Okay, I'll upgrade from picking. If you don't stop to think about the stuff you're thinking beyond surface level, it will be gone in three months.

So are we done with your list, @Danu?

@TedShifrin No, but it's okay. I have to get to doing some otehr stuff now :(

OK.

Glad I've improved your insight about $L^k$ at least.

@MikeMiller If I don't end up doing my thesis on this, it's okay. If I do, I'm obviously not going to work like this for the rest of the year.

Danu, I took a course in Riemann surfaces and a course in several complex variables and a course in complex manifolds while I was an undergraduate. I then did more complex manifolds and algebraic geometry in grad school. So I obviously was in the trenches for a pretty long time.

@MikeMiller: Was the thing I pinged you above right?

And some of it I really only learned when I taught it. Math takes time and repetition.

@Balarka I don't understand what you're saying.

SW classes of the tangent bundle don't work like that.

My point is this: Under current constraints I cannot afford to spend more time on this than I'm already am. And if that means I'll just get the bare minimum, then so be it. I'm already letting my supervisor down by being so slow to get through the book.

The SW class of a pullback bundle is the pullback of the SW class. But the pullback of the tangent bundle under some random map is rarely the tangent bundle!

Obviously I agree with you guys that it's not the right way to do things, and I hate doing it. But shit happens. I switched to math (too) late and now I have to live with this.

I am not criticizing, but I think you do need to get more big picture intuition and spend less effort on all the technical details ... for now.

@MikeMiller Hm, yes, I misunderstood the pullback condition.

@TedShifrin What can I say? I'm studying on my own, with you as basically my main "mentor"---I can't get the big picture without some guidance, and I'm already stretching what I can ask from you too far as is.

Aren't there other grad students you can talk to about this stuff?

I don't know if anyone is even working on complex geometry at all in the department.

Ah. :(

Maybe inquire, though?

I don't know any of the mathematics students---I switched (too) late!

Ask your adviser if he can recommend some grad students. There might be some who've thought about this stuff even if they don't work in it.

My supervisor probably won't be able to even give me a work place in the building, so I won't be "around" the mathematicians either.

Oh :(

So I work from home, alone.

That sucks.

Anyhow, I don't mind helping when I can.

I will warn you that Nov 1 I'll be traveling for 2 weeks.

Any place exciting?

hey @TedShifrin

Northern California: Stanford, SF, wine country, Berkeley, then home via Yosemite.

I was wondering when

@TedShifrin You've already been very helpful.

when we have the identity permutation what would be the sgn of that thing ?

What's $(-1)^0$, Karim?

I see

so it is by definition like that

Someone asked me last night what I had enjoyed most about teaching ... I had trouble answering.

an advisor s only half of the value of grad school. the other half is your fellow students

For me, fellow students were more than half, @MikeM, no question.

But Danu's in an awkward position. Hopefully he'll improve it in a year.

I'm not in the same position as you guys were as grad students.

fine.

For one thing, I only spoke with my adviser(s) for less than an hour a week, often less. Or by letter occasionally.

Letter?

Yeah, Griffiths was half my adviser and he was at Harvard. I visited for a few weeks in the summers.

@TedShifrin Nice.

I simply don't personally know anyone seriously working on math---all my friends are doing physics and so they can't really help me.

Anyways, enough of my self-pity

I gotta run, bye guys.

@Danu All my friends are engineers

I think you should make an effort to go to teas or anything in math to get to know some of the grad students.

Bubye.

Which is worse

Actually I might appear again on phone

once I'm in the metro

@TedShifrin Teas?

In the US most departments have a weekly or daily tea/coffee hour (for grad students and faculty, although occasionally undergrads crashed 'em at UGA).

Guess undergrads aren't invited if my dept has that

@TedShifrin Don't exist here.

Sigh. Europe is so antisocial.

Some of the professors are actually sworn enemies here :P

One more analysis problem, then I'm done with this hell

The two geometry guys hate each other

@Ted I only meet with him once every week or so. Nonetheless, he's had a big influence. But fine - drop him to 30%.

Lovely.

And I'm not sure if there are more than 2 or 3 PhD students in geometry

I'm not minimizing the importance, @MikeM. And I was more than flattered that Chern worked with me. But they have lots of students and lots of things to do, and it's much easier to spend hours talking/learning with fellow students.

@MikeMiller @TedShifrin Not in grad school but my advisor is 90% of the reason I even like math.

Ok

I did not enjoy mathematics when I first came to school

My horrid teaching is what impelled a lot of students to major in math and, in some cases, go on to geometry type stuff in grad school. So I suck :P

I don't think I have yet turned a student into a topologist.

Give it time.

@MikeMiller sounds like something a vampire would say

The only thing I have tricked students to do is to take algebra a year early.

Isn't second year early enough?

It's always too early.

I actually had one student who took it as a first-year at UGA, concurrently with my Multivariable Math class. He's now a postdoc at Cambridge in England. He's done OK.

@TedShifrin Is it common for profs to give hints that make the problem unreasonably hard to punish them for not thinking about it themselves

I believe it should be taken before real analysis.

I actually agree for most students, @AndrewT. The logical structure is several layers easier.

Exactly. In addition the course tends to be more computational, allowing for a more smooth transition into proof-based mathematics.

No, @0celo. But in some cases one has to think about the hints the right way or they're no help. It's not necessarily designed to be opaque.

@AndrewT: I don't know that a Herstein algebra course would be more accessible. More formal, yes. But certainly the algebra that I taught was more concrete.

I have not read Herstein. We followed Fraleigh (which I now have mixed feelings about, but enjoyed as a student.)

I don't like Fraleigh, but, yes, he's more down-to-earth.

The books that I truly understood things from were rarely the ones I first enjoyed.

(Or papers.)

My internet's being annoying.

Fraleigh once had a bad error in his answers to his problems, and a student from somewhere wrote me to tell me that my algebra book was wrong because it contradicted Fraleigh. He essentially said that $\deg_{\Bbb Q}\alpha = 2^k$ implies $\alpha$ constructible. Oops.

I currently have 6-7 papers I want to jumble together ideas from. I think I should TeX notes from the relevant parts of each of them so that my brain doesn't fry. Ugh.

your multivariable math is one of the best @TedShifrin

learned many things

Thanks for the kind words, Karim.

I like that book a lot.

@AndrewThompson I like to write notes on the papers themselves and file them away appropriately.

@TedShifrin I just checked out the postdoc research in cambridge

really cool

algebraic geometry (higher-dimensional algebraic geometry, algebraic cycles, abelian varieties, mirror symmetry, geometric aspects of representation theory);

very cool

@MikeMiller i.e. print them out, write handwritten notes on them?

You mean my former student specifically, or things in general, Karim?

no things in general @TedShifrin

dpmms.cam.ac.uk/vacancy/postdoc_opportunities.html

@AndrewT I always print out papers. Do you do all your reading on a laptop?

they have many cool research topics

@MikeMiller Most of it. Sometimes I bother PhD-students to print things for me, but I tend to lose physical papers.

(For non-PhD students printing is not free.)

Heya mr inactive

Hiya @Ted

I'm nowhere near Haiti, Jasper, but thanks.

Don't think GA got hit

or did it?

I'm not in GA :)

The coast got some ...

In the metro now. To recap that hypersurface thing @TedShifrin...

cool @WillHunting

I remember when I was taking mechanics as undergrad

I remember reading some notes from cambridge

was quite cool

So if I take a linear P1 it PD's to a top-2 degree form and wedging with w and then integrating over X will give me 1 because of the intersection thing

oh yeah

cache

No, integrating over general X gives you the degree.

Similarly if I take a linear P2 and wedge with w^2

@TedShifrin X being the entire space

No, codim, not dim.

Codim of what

For a $k$-dimensional complex submanifold $X$ of $\Bbb P^n$, $\int_X \omega^k = \#(X\cdot \Bbb P^{n-k}) = \deg(X)$.

Okay, I think you are saying the same as me

I'm saying $\int_{\Bbb P^n} (PD(X)\wedge \omega)$

Where X is now the submnf

There are other ways to write Poincaré duality, but that's what I was talking about. You can of course do $\int_{\Bbb P^n}\psi\wedge\omega^k = \int_{\Bbb P^{n-k}}\psi$ for any closed $(n-k,n-k)$-form $\psi$, etc.

This is the same as what you're saying I think

Jasper, what two algebra books?

Not doing any more publishing.

Nah.

So what we say agrees because of $c\cap (\alpha\cup\beta)=(c\cap\alpha)\cap\beta$

Sounds right, @Danu.

just a quick question

I had an analogous argument yesterday :)

the symmetrizing operator always gives the same value for a linear function right ?

Huh? Karim.

Now I'd still like to know what you mean by the

Now I'd still like to know what you mean by the degree

Let us say we have a k linear function f we define the symmetrizing operator $Af(v_1,...,v_n) = \Sigma_{\sigma \in S_k} f(v_{\sigma(v_1)},...,v_{\sigma(v_k)})$

It has to be degree of the defining eqn

Algebraic subvarieties have a degree. For a hypersurface, it's the degree of the homogeneous polynomial that defines it.

Yeah

@TedShifrin it is a way to force a k-linear function to be symmetric.

For higher codimension, if you have the right number of equations, and the hypersurfaces intersect transversely, then it's the product of the degrees. But the most interesting easy example is the twisted cubic $(t,t^2,t^3)$ in $\Bbb P^3$. You need three equations to cut it out; you cannot do it with just two.

Related: what is the degree of a line bundle? Is it the intersection no. of a generic section with tge zero section?

What do you mean by "always gives the same value," Karim?

On a curve (dim 1 cx manifold), @Danu, it's $c_1(L)$.

It's the degree of the associated divisor.

nvm I understand where I was confused about now @TedShifrin this will just collect all permutations in $S_k$ and force any permutation of k-variables to be same value that is why it symmetric.

Yeah I thought about that but I only learned how to associate line bindles to effective divisors

Well, if your manifold is higher-dimensional, you won't get a number that way, @Danu. You need to do more intersections.

So if its not effective...

Right, Karim, you have to relabel the permutations.

yeah

Degree can be negative, @Danu, for a divisor or line bundle.

Yeah

But I never learned how to connect the two in that case

You need to get to Kodaira embedding, @Danu. Then you can use a positive line bundle to define a projective embedding for which the hyperplane class is the line bundle.

Eff div gives line bundle and holo section gives eff div

Any divisor gives a line bundle, remember?

But nothing if it's negative

You work with meromorphic sections of the line bundle.

Sure sorry

But the other way doesnt work because no holo sections

Anyhow, I'm running away now ... Bye for now, all.

Yeah thats kind of clear that you need mero. But i wouldnt be able to write down how to make a divisor out of that

Bye

cya

@BalarkaSen My QM problem sets are typed up in Hatcher's font.

Interesting.

Group theory terms and theorem progress:

(Currently span 2 A4 papers)Tomorrow need to finish the other half

Are you writing an exam?

@Danu: BTW, in line with our discussion, you can interpret primitive cohomology geometrically. Sticking with $\Bbb P^n$ (but you can make sense in any compact Kähler guy), a closed $(p,p)$-form $\phi$ corresponds to a submanifold $X$ of dimension $n-p$. Saying $\phi$ is primitive means that $X$ doesn't intersect generic $\Bbb P^{p-1}$'s. (It has to intersect any $\Bbb P^p$, but it's in enough general position that it misses most things of one dimension less.)

Of course, in $\Bbb P^n$ this is all automatic, but it needn't be in more general Kähler manifolds. (Everything is primitive in $\Bbb P^n$.)

--- I think. You should double-check this.

@ItachíUchiha should that be 2013^3 ?

you'll need to use some kind of formula to factor it of course

@arctictern Thanks for the correction :)

@TedShifrin Thanks a lot. The geometric stuff is the most important to me.

@ItachíUchiha I mean work on a general formula for n^3-(n-1)^3+...-3^3+2^3-1^3

It might be useful to note that $2014^3+2^3=(2014+2)(2014^2-2014+1)$

and similarly $2013^3+3^3=(2013+3)(2013^2-2013+1)$

i.e. they've got that same 2016 factor in common

not sure that'll be decisive, but it seems handy

@ItachíUchiha if you post on main I can answer how I'd do it there

there is a straightforward systematic way to find closed forms for these sorts of things

If i have a+4= sqrt(b/(2pi)) can i clear the fraction by multiplying 2pi to get sqrt(b) or do i have to remove the sqrt first?

@WDUK sqrt(b/(2pi)) is not a fraction per se, it is a square root of a fraction. you cannot clear a denominator when the whole fraction is stuck inside the square root. you can use sqrt(b/(2pi))=sqrt(b)/sqrt(2pi) though and then clear denominators. what are you trying to do anyway? solve for b?

yeah solve for b

just square both sides first

Tern !! :)

yes?

alrighty , thanks for the tip :)

Just saying hi :)

oh. seemed like an urgent hi. hello

LOL ... nah ... it was crazy nuts in here several hours ago. I was being attacked by 3 or 4 different subjects.

ah

BTW, @Danu, the easiest example of a nontrivial Lefschetz decomposition (and this should only take you a few minutes) is for $X=\Bbb P^1\times\Bbb P^1$ (e.g., the standard nonsingular quadric surface in $\Bbb P^3$). The Kähler form is the sum of the pullbacks of the usual Kähler form on $\Bbb P^1$. Geometrically, this is $L_1+L_2$, where $L_1$ is a line $\Bbb P^1\times \{y_0\}$ and $L_2$ is a line $\{x_0\}\times \Bbb P^1$. Show that $H^2 = P^2 + LP^0$, and tell me what $P^2$ is.

(By $H^2$ I mean $H^2(X,\Bbb C)$, of course.) From your topology course you should know that this is $\Bbb C\oplus\Bbb C$.

hey @TedShifrin I was wondering could you explain me the intuition behind the anti-commutativity of wedge product ?

@Danu: So give me primitive cohomology in terms of $L_1$ and $L_2$ (which generate $H^2$).

Read Chapter 8 of my book, Karim. It's all about oriented area/volume.

cool

So oriented volume changes sign when you switch two vectors.

I see

Ted, how do you know so much about mathematics in all of its realms? Are you a multipurpose professor or something?

LOL @multipurpose.

I'm an old professor :D

There are plenty of parts of math in which I'm woefully ignorant, but I took pride in being pretty broad in my knowledge.

Lol that's awesome. Where do you/have you taught?

that is good @TedShifrin I don't like people who just stick to just one area

Balarka: Go to bed!

that is how I want to be

and understand things from intuition point not just abstract nonesense.

@user3925758: Actually, I had a lot of fun teaching an applied math course in 1986 or so and probability just before I retired ... I had neither taken nor taught either before. It was a blast. ... I taught at MIT for 2 years and at UGA for 34 years.

There's a lot to master, Karim!

Where are you at this point in mathematics, @user3925758?

yeah I agree @TedShifrin

Danu's actually making me rethink stuff I haven't taught or really used in 35 years.

Most of it I know well, but some of it ...

Lol this might be embarassing, but I'm currently in my junior year at high school. I've been doubling up AP Calc BC and Trigonometry H, but I am often bored by the pace of both classes. I enjoy writing up my own ideas in my free time

Sanity-check: if I have an $n$-plane bundle $E/X$ and it's classifying map $X \to \text{Gr}_n^\infty$, saying this map lifts to the universal cover of the Grassmannian and that $E/X$ is oriented are equivalent, right?

Ah, @user3925758, were you one of the people I advised to go get Spivak's Calculus? BC Calculus is just too formula-oriented and doesn't prepare you to do serious math.

@TedShifrin I plan to. But will I?

@BalarkaSen Knowing what a bad boy you are, I'll answer no.

I don't recall anybody advising me to do anything of the sort, but what is Spivak's Calculus?

So the universal cover is the oriented Grassmannian?

I think so.

@user3925758: When people are named userxxxxxxxx, I can't tell one from another.

Lol I'll look into changing my name xD

Look on Amazon for Michael Spivak, Calculus. Super super book. Lots of wonderful calculus, analysis, and challenging exercises.

Does that answer your question, @Balarka?

@Balarka Yes, of course.

Try to understand that fact from at least two different perspectives.

Oh, why not say five?

The 1-skeleton shouldn't be too tough :P

My internet is horrible, dang it.

BTW, @Balarka, I'm gonna whine about your wrong apostrophe, yet again.

@TedShifrin Yeah, since that sounds precisely like the orientation double cover.

Projection given by forgetting the orientation of the subspace.

So now you need to see five ways why $\tilde G_n$ is simply-connected.

@TedShifrin: your avatar looks like the animation in this answer.

Eh, that's BGL_n^+(R), and GL_n^+(R) is path-connected. :)

That's where it came from, @robjohn. Someone asked me that 20 years ago, and I put the problem in two of my books.

And hi :)

OK, @Balarka. Now give me 4 ways I can understand (that one was for Mike).

That'll take me 4 hours, unless I break my wifi before that.

LOL ... Can't you see the $1$-skeleton from $\tilde G(n,n+1)$?

Sure, that's a sphere, ain't it?

It be, last time I checked.

I also believe there should be a hands-on proof, by looking at the oriented n-dim subspaces of R^\infty, looking at a "loop" of them, and explicitly writing down a contraction. It should be a manifestation of the GL_n^+(R) thing I said.

T/F? (I actually want to know.) By compactness, any such loop lives in $\tilde G(n,N)$ for some fixed $N$.