@BalarkaSen Did you see that Peter Woit has compiled a lightly edited version of the mathematically substantive comments on his blog post about the IUT papers? It is freaking 51 pages.
@AlessandroCodenotti it was all sorts of stuff. We started with interpolation of L^p spaces, did some Hahn-Banach stuff, Hilbert spaces, Fourier analysis, distributions, and the Sobolev inequality.
I did an essay on Banach algebras, the Gel'fand transform, and the connection to the Fourier transform (abstract harmonic analysis), which is why you heard my exclamations the other day about that.
They are strictly related, the functor from unital commutative C*-algebras to compact hausdorff spaces is given by taking the spectrum $\hat{A}$, while in the opposite direction you take the algebra $C_0(X)$ of functions vanishing at infinity
I do remember my physics teacher in high school getting mad at me because he couldn't understand what was going on and I tried to help the class by explaining things to them. And I ended up teaching the last month or two of the calculus class (the teacher didn't know it at all) after I took the AP test. And I remember having a history teacher — senior year, finally — whom I really liked; he also taught me drivers ed and I remember teasing him because he was so great.
@MikeMiller or @BalarkaSen from our discurssion earlier today: I fail to show that we can always find a chart that is divided into two opens (that are connected). would you mind helping me through it?
@MikeMiller we had our jordan curve right, which divided the plane into two opens, say $U$ van $V$ (along with itself as a boundary). now if I take a point $p$ on the curve, I would like to find a neighbourhood around $p$ that is divided again into two connected opens
@anakhro Yes, but the nature of test is quite weird. Actually, I don’t like any kind of tests, because the other man is checking me by his way of thinking
I didn't necessarily want to find a different chart, but I thought I needed to make our original slice chart possibly smaller to achieve this connectedness
dear @loch, thanks for your reply; it looks reasonable, but may I ask you how do you think is defined outside the $(m_i)_i$'s? thanks again, if you want to spend some minutes editing it as an answer I'll accept it ;)
@loch oh maybe I see: so you pick $(m_i)_i\in \oplus_i M/M(\sigma_i)$ and map to $(m_i-m_j)_{i<j})$. I thought this was defined only for a cartier data, but $(m_i)_i$ is any element in that direct sum , sorry
@CaptainAmerica16 I mean I am going to start a math company called MathCity. We outsource to undergrads and grads with a b.s./m.s./phd in mathematics/physics and they prove managements theorems. Then management compiles the information and writes papers with it
no problem ! i'm too lazy to do that but feel free to ping me if you have related questions (although like i said i can't say i really know toric geometry :p )
@Mike ok I can argue using compactness that there exists a positive $\delta$ such that for each point on a component of the curve that is not the components where $p$ is an element of, the distance from that point to $p$ is at least $\delta$. so we can now shrink the ball so that there is only one component of the curve left. I still feel like I don't have it clear enough in my head why we would only have two components left now
so there were two options in the case our U intersected with the ball wasn't connected; either there was a strand that was touching the chosen strand, or it wasn't touching it
Choose a point $p \in S^1$ and let $q = \gamma(p)$. Find chart neighborhoods $U, V$ around $p$ and $q$ such that $\gamma(t) = (t, 0)$ in the corresponding local coordinates.
WLOG I can assume $U, V$ are ball neighborhoods, i.e., $U$ is a small arc around $p$ in $S^1$ and $V$ is a small 2-disk around $q$ in $\Bbb R^2$
So now $V$ is a ball neighborhood. Consider $\gamma^{-1}(V) \subset U$, which is an open subset of $S^1$ containing $p$. Open subsets of $S^1$ are union of open arcs, choose an open arc containing $p$
@ShaVukila Now, your issue: $V$ is a ball around $q$ in $\Bbb R^2$ and $\gamma(U)$ is a properly embedded curve in $V$, which is part of the curve $\gamma$.
But there are many other strands of $\gamma$ which exit and enter $V$
Let $B \subset V$ be an even smaller open ball, arbitrarily chosen, such that $\overline{B} \subset V$. Then $\overline{B} \cap \gamma$ consists of various strands of $\gamma$ which exit and enter.
Let $z \to \frac{az+b}{cz+d}$ $(a,b,c,d \in \mathbb{C}$, and $ z \not = \frac{-d}{c}$) be the Mobius transformotation. A point $z_{0}$ if fixed if and only if $\frac{az_{0}+b}{cz_{0}+d}=z_{0}$,so $cz_{0}^{2}+(d-a)z_{0}-b=0$. This is an equation of degree at most $2$, hence has at most two soluti...
How to extend this answer to the extended complex plane?
Oh ok, so I wanted to show that $\pi_1(S^2)$ is trivial I think by stereographically projecting to $\Bbb R^2$ but you threw surjective loops on $S^2$ at me so I had to argue why every loop is homotopic to a nonsurjective one by homotoping the finitely many pieces in a compact ball to its boundary or something like that @Balarka
You could have crazy loops that enter a ball and cover it entirely without ever leaving the ball. There's no reason why the image of a small interval should have empty interior
Screw this topology nonsense, I'm going to think about groups again, bye
@Alessandro I remembered the argument (much thanks to Mike's patience because I told him at least 5 wrong arguments privately elsewhere before coming up with the right one): If $f : S^1 \to S^2$ is a map, choose a point $p \in S^2$ and consider $f^{-1}(p)$. This is a compact subset of $S^1$, cover it with finitely many arcs $U_1, \cdots, U_n$. Then $\gamma|U_i$ are these finitely many arcs which contain $p$.
Each of these arcs can actually touch $p$ infinitely often Cantor set's worth of times
I am trying to calculate the value of $\ln j$ where $j^2=1, j\ne\pm1$($j$ is split complex).
This is how I did it:
given $e^{j\theta}=\cosh\theta+j\sinh\theta$ I can set $\cosh\theta=0\implies \theta = i\pi n - \frac{i \pi}2, n \in \Bbb Z,i$ is the imaginary number, for convenience sake i'll...
@ShaVukila Yeah this seems annoying to argue. Take a disk around $q$, then basically what you want to say is that there is some $\epsilon > 0$ such that $\epsilon$-arcs on each of the strands are contained in the disk
It's most certainly true but I can't seem to find a quick and dirty argument
I am trying to calculate the value of $\ln j$ where $j^2=1, j\ne\pm1$($j$ is split complex).
This is how I did it:
given $e^{j\theta}=\cosh\theta+j\sinh\theta$ I can set $\cosh\theta=0\implies \theta = i\pi n - \frac{i \pi}2, n \in \Bbb Z,i$ is the imaginary number, for convenience sake i'll...
