@TedShifrin This isn't the standard proof, which uses Schoenflies. But I don't like Schoenflies.
@TedShifrin I have no idea what I'm turning into. Whatever discipline I end up doing will have "topology" or "geometry" in the title, and that's all I can say at present.
@TedShifrin One of the main reasons I was interested in number theory was because of its interplay with geometry. There's a paragraph or so in the introduction of one of Silverman's books that I wasn't able to read about how geometry determines arithmetic - and that paragraph hooked me.
@TedShifrin I'm not sure why I'm bringing this up, but someone the other day said that $\pi$ was a roughly arbitrary choice of "angle" for the circle. I'm still offended.
@TedShifrin, I suspect it was in a discussion regarding why $e^{i\pi} = -1$ is beautiful. Because it seems that one could just as easily write $e^{i*180^\circ} = -1$.
@TedShifrin The context was an $e^{i\pi}$ question, and they quoted the formula $e^{ix} = \sin x + i\cos x$, and then said that $\pi$ was an essentially arbitrary definition of the angle. But that's so far from true, unless one says it's arbitrary vs. $2\pi$ which would be more natural.
@KajHansen wiki is one of the best places for a math diorreah. i had one once upon a long time ago. the symptoms includes babbling a lot of names without knowing the math behind them.
No, in terms of multilinearity properties, @Hippa, which is what I'll need for differential forms in a bit. I did the permutation formula in the last lecture.
@DanielFischer I was wondering, have you heard of a "Cauchy transform"? I'm reading a paper where they call something like $$\tilde f(z) = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{f(\zeta)}{\zeta - z}\,d\zeta$$ one and say such and such relation follows from "standard properties of the Cauchy transform".
I am getting too addicted by the chat. I think I'll need everyone's opinion about deletion/suspension of my account for sometimes to get rid of this habit.
@TedShifrin So, I think the point has to be that, under the injection $T' \hookrightarrow T$, the copy of $S^1 \times \mathbb R$ plus the deleted point is $D^2$. If that's the case, we can easily smooth it.
But I don't see why the deleted point must be a "point at infinity" of the copy of $S^1 \times \mathbb R$. Surely this follows from compactness of its complement.
@TedShifrin That set is compact, so we can cover it with finitely many charts. We can also put a single smooth chart in a neighborhood of the punctured point. Since we have finitely many of these we can smooth "near the edges" appropriately to make them compatible with the smooth structure on $S^1 \times \mathbb R$.
I have to admit I get annoyed when people send me emails informing me I have mistakes in my book when they are totally NOT mistakes and they're just being sloppy.
You might have a point. The videos are a good compliment to the textbooks though. Having the 3500 videos would make up for the book's price, considering its binding :P
@Ted You may be pleased ot know that a number of people made to look really dreadful by the memes that carry their faces have parlayed that attention into modest sums of money
I want to rush off to the meta and propose a system where all votes must be signed. I don't like the idea that someone can express their opinion but not put their name to it. Im just weird like that
@AntonioVargas I didnt consider that people would engage in downvote retaliation. I guess I think 2 things about that 1) If something needs ot be downvoted then its worth being downvoted in return; 2) Im fine with kicking people out who just downvote in response to being downvoted
@PedroTamaroff, I don't think the conjecture you gave me is true. Consider, for example, an arbitrarily large $n \times n$ matrix such that odd-numbered columns are all $0$ and even-numbered columns are all $1$.
