> By the end of this year, Martinis says, his team will build a device that achieves “quantum supremacy,” meaning it can perform a particular calculation that’s beyond the reach of any conventional computer.
@vzn What more than the word of someone who works with him do you want? What about scientists sticking to standard terminology they dislike for better communication do you not understand? You are essentially accusing DanielSank of lying there, is that what you want to say?
@0ßelö7 it should come with it's own dictionary, or at least mine did on mac and pc. did you do what I said? because I did it myself right now and it worked.
@0ßelö7 I don't know what "shadow banned" means. If you are found to have sockpuppets you'll be suspended with a corresponding message as usual in such cases, but I have literally zero reason to believe that.
for me, when I click "import" and choose the open office dictionary, it brings up a window prompt and if you click "yes" it links it automatically and applies the spellcheck.
@ACuriousMind There are three possibilities. 1) The user is trolling. 2) In a drunken rage I created a sock and got banned. 3) He somehow got my IP and someone else from my school got banned. (Not sure how IP works on school wifi.) I take it 3) will not be an issue?
Damn, out of milk. Do I: a) walk to the corner shop (in the rain) b) have a black coffee - I don't like black coffee c) put up the potentially fatal caffeine withdrawal symptoms
Some decent, actual, non-dust (technical term for the tea that goes in standard tea bags) tea is the solution :) Although yes, it's possible that I'm slightly mad
So, uh, this: Suppose $X$ is a Riemann surface and $P(T) = T^n + c_1 T^{n-1} + \cdots + c_n$ be a polynomial in the ring $\mathscr{M}(X)[T]$ over the field of meromorphic functions $\mathscr{M}(X)$ over $X$.
Forster says there's a branched cover $p : Y \to X$ with a function $f \in \mathscr{M}(Y)$ such that $(p^* P)(f) = 0$
I want to say this is an analytic continuation result but I can't seem to parse myself
I guess I mean that $P$ can be thought as a meromorphic function on a holomorphic chart $U \subset X$ by feeding $P(T)$ points in $U \cong \Bbb C$ in place of $T$, and if I take a germ of that at some point $p \in U \subset X$, $f$ is really the analytic continuation of that germ?
@BalarkaSen I'm not so sure what you mean by "analytic continuation" there. The way I know to construct that $f$ is to reduce to the case where the zeroes of $P(T)$ are simple, take the space of local zeroes of $P(T)$ (ie. $\hat{Y} = \{ (x,f_x)\in X\times \mathcal{O}_x\mid P(f_x,x) = 0\}$), observe that $\hat{Y}$ covers $X$ with constant degree except at the poles of the coefficients of $P(T)$, which are isolated, so we have extend that to a covering $Y \to X$ with these points added in.
Now $f$ is the function that sends $(x,f_x)\in \hat{Y}$ simply to the germ's value $f_x(x)$, and is extended meromorphically to the rest of $Y$.
Ah, I see what you mean
Yes, that $f$ is the analytic continuation of all germs that fulfill $P(f_x) = 0$.
@ACuriousMind Ah, right, that's probably what I meant
So, if I take $X = \Bbb C$ and look at $P(T) = T^2 - z$ over $\mathscr{M}(X)[T]$
I guess the branched cover we get is the two-fold covering $p : Y = \Bbb C \to \Bbb C$, $z \mapsto z^2$ with the coordinate function $f(z) = z$ over $Y$?
Because away from the origin, a germ $g \in \mathscr{O}_p, p \neq 0$ which satisfies $P(g) = 0$ would be germ of the holomorphic function $\sqrt{z}$ in a neighborhood of $p$, and we'd be trying to analytically continue that
@ValterMoretti Do you know what "multiple characteristics" means? The best I can find is some old paper by Lax that has a complicated limit of symbols in the pseudodifferential sense.
