@Knight No. He was at NeXT for most of the time I worked at Apple. When he came back to Apple, there were massive layoffs and the project I was working on was slashed.
@Ted: I have a naive question. When do complex things admit holomorphic tubular neighborhoods? That is to say, a tubular neighborhood along with a holomorphic retraction to the base
@TedShifrin I wrote a program for teaching Logic at UCLA, and although the professor in charge of the project said that the way the program presented derivation proofs was a huge improvement (and similar comments on the symbolization of English sentences into symbolic logic, and other facets of the program), when he got an award from UCLA for the program, he did not share the spotlight.
@TedShifrin That's better than what happened to a friend when I was at Princeton. His advisor said that one of my friend's papers was garbage, and then quoted it, word for word, without attribution, in one of his papers.
My only analog of that is I once told Phil Griffiths (who was de facto half my thesis adviser) some easy thing I'd figured out on the way to something more interesting. A year later he asked how it went. It ended up in a paper of someone else he was advising (or maybe a postdoc, I forget), no attribution to me.
What is the statement you want? Given any complex submanifold, there's a neighborhood which is biholomorphic to an nbhd of the 0-section of a complex bundle on it?
@TedShifrin But this cannot possibly be right. In particular this implies you have found some nbhd which admits holomorphic retraction to the submanifold -- take some torus in $\Bbb P^2$, then arbitrarily close to it you can find torii which are not biholomorphic to the original torus
@Billy: I had notes on limits (giving less crazy $\delta$-$\epsilon$ proofs) and a bunch of stuff on Taylor polynomials and applications which he incorporated. So some problems got added, as well.
I won't ask who it was, @robjohn. I might have known this person.
@TedShifrin Do you still have the notes and/or could share them with me? I'll eventually buy the fourth edition but at the moment it's kinda impossible because the dollar is too expensive 1USD ~ 5.36BRL and we earn much less money on average.
I can send you the pdfs of the short notes I handed out in class. Spivak embellished some on his own. There were also some other exercises added/changed, which I cannot remember at all. But email me and I'll send you the two pdfs, sure.
Nothing hugely novel, but things I thought were important to do.
@Balarka: To be honest, I've never thought about the point you raise. But does the fact that $\Bbb P^1\subset\Bbb P^2$ has self-intersection $1$ tell you we can't do it there? I'm being addled.
So I have to rethink the usual proof I used to give of the tubular nbhd theorem.
I think the proof in the compact case for the tubular neighborhood theorem should go over just fine. You just use the holomorphic version of the inverse function theorem. Riddle me that, Batman.
Good question. Is there a holomorphic tubular neighborhood theorem?
Interestingly, @Balarka, that Overflow question has an answer which says that $\Bbb P^1$'s in a surface with negative self-intersection have holomorphic nbhds.
I read through these in grad school, but I'll have to look more closely after lunch. Good thing I kept Griffiths's Selected Works when I gave everything away.
So I'm getting hundreds of emails because UGA is about to open up to in-person classes (thanks to a wonderful Tromp-loving governor) and the faculty are freaking out, @MikeM.
What I know is true is that totally real submanifolds are more flexible than complex ones; if L is a totally real submanifold of a complex manifold M then it always admits Stein neighborhoods
Unsurprising/uninspiring because being totally real is an open condition
@Tobias: Just read your earlier exchange that was starred. I have corrected so many people on MSE who think that you solve $A=0$, $B=0$ by setting $A=B$. Even people who've studied linear algebra.
@TedShifrin And if the person had just done an example, they should have immediately (hopefully) recognized that they would get the equation of a plane again
I also commented on an answer a few weeks ago where the answerer had done the elimination of a variable and said that the resulting projection was the intersection. Got upset at me for saying this was incorrect. "Oh, anyone would know what I mean."
I'm thinking my teaching days are getting close to the end. :D
So why is $V=\bar V$ an open condition, @Balarka? I'm confuzled.
@TedShifrin No shock. In addition to GA's ... general temperament, let's say ... I think you'll see on the balance sheets that they're not doing very healthy. Probably desperate to stay open for fear of lost tuition.
For less-than-half dimensional subspaces $V$, you can find a maximal totally real subspace $W$ containing $V$ which is half-dimensional, and then argue.
Suggestion to the Reader. Browse through all theorems/inequalities in the previous as well as in the following sections, formulate their possible homotopy parametric versions and try to prove some of them.
@TedShifrin I did some more investigation and evidently the line of housewares was not his advisor, but someone with a very close name. I did however find a very scathing article about this guy from somewhere else.
@DarkRunner looks to be $\vec a=(\overbrace{\delta\cos(\phi)}^{\rho\text{-component}},\overbrace{-\delta\sin(\phi)}^{\phi\text{-component}},\overbrace{\,\,\quad\gamma\quad\,\,\vphantom{\phi}}^{z\text{-component}})$
I'm trying to follow this proof for "If f(z) satisfies CR at $z_0$, then so must f(z)^n for any positive integer n"
But I don't quite understand how the partial derivative of $f(z)^n$ w.r.t $x$ turns out to be $f(z)^{n-1}$ multiplied by the partial derivative of $f(z)$ w.r.t x
Nothing to do with partials. It's just the usual chain rule for a single variable. But if you're doing complex analysis, you do need to know your multivariable calculus.
Am I right in saying that since partial derivatives are normal derivatives where the variables not concerned are taken constant, then all the algebraic properties of derivatives apply to partials as well?
Oh, so they aren't making all classes remote, which means undergrads at least have to risk their lives on campus?
UGA seems to be way behind ... the faculty are revolting now, but it's intended that most classes will be in person — until tragedy strikes and they're up the creek.
I don't know how much of an option professors (who tbh are prob at highest risk) are given regarding teaching in person vs online, the grad classes here are all online anyway as well as the huge lectures
Yeah testing is gonna be tough for sure. Last semester there was def some grade inflation in my calc class on the second midterm, so on the final things were made tougher
There were 4 versions of the test, and people didn't have as long of a window
I think we're partnering up with some online proctoring thing for next fall?
But yeah the online proctoring program I imagine would video people, supposedly an AI detects suspicious behavior (and it can tell if you open a new tab or program, so that's flagged), presumably students would have to make very subtle movements to check on a cell phone or smth
Ted: No I didn't actually! It's been a little while since I was last here
Can someone explain to me how $Qc\zeta_5 + \zeta_4) = Q(\sqrt{5})$? It's said I can solve for this in Dummit and Foote. I just know that $x = \zeta_5 + \zeta_4$ satisfies $x^ 2 + x - 1$ and that both fields are degree $2$ over $Q$, but I don't see how that shows $x = \sqrt{5}$
This question is related to the computation suggested here: Finding the subfields of the cyclotomic field of order $5$
Let $K = \Bbb{Q(\zeta)}/\Bbb{Q}$ where $\zeta$ is a primitive 5th root of unity.
We can identify the Galois group with the following 4 automorphisms:
$$
\mathrm{id}: \zeta \map...