Sergey Finsky
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 Mar 21 comment Trying to prove that operator is compact Do you have any problems in understanding locally bounding criteria? Feb 16 comment Boolean ring. Representation as direct product? Thank you! Yes, the reason I tangled up is that I haven't notice that the proof of the theorem in the paper, I provided, was for general(infinite) case. Feb 16 comment Boolean ring. Representation as direct product? Sorry guys, I forgot to say that I am only interested in finite case. I will edit it now. Feb 11 comment Asymptotics with prime of form 4k+3 Sorry for not responding for such a long time, I kept forgetting to respond that your answer solves original question, which can be easily deduced from standart $\delta - \epsilon$ arguments. Jan 30 comment Induced representation, Ind(Res(U)) I will have a clear look on the 1st problem in tomorrow morning. For the second - sorry, I meant not $W$, but $Ind(W)$ and now we can form such sum, since we are trying to reconstruct $U$ from $\underline{G}$-module $U \otimes Ind(W)$ Jan 30 comment Induced representation, Ind(Res(U)) For the 1st problem: (at first, just to be sure that I understood everything correctly, we need $H$-modules $\alpha$ and $\beta$, don't we?) and the problem is: what would happen if we wanted to send some element of the form $(x \otimes_H w) \otimes yu$, where $x^{-1}y \ne 1 mod H$, we would send it to $x \otimes (w \otimes_H x^{-1}yu$, but $H$ acts differently on $U$ and $x^{-1}yU$ in general case... That's seems to be a problem here. For the second part: let we consider $U \otimes (\sum_{g \in G} gx)$ for some $x \in W$, it seems that we can recover action of $G$ on $U$. Jan 30 comment Induced representation, Ind(Res(U)) There are 2 issues: first is dealt with module map $\alpha$ : why it sends elements to $kG \otimes H(W \otimes U_{H})$, but not to $kG \otimes H(W \otimes U)$?And the second is a bit psychological: it doesn't seems that restriction serves the whole information about the original representation. (apart from the $U_{H}$ I do understand everything from the proof, but psychologically I can't admit it, because there are many representations, restrictions of which collides (there are trivial and alternating representations of $Sym($n$)$, restrictions to $A_n$ of which equal),despite the formula. Dec 23 comment Number of representable as sum of 2 squares Yes, it'd be nice if somebody posted whole proof, because it's know quite straightforward. But thank you very much, I would never expect to see asymptotics for this problem, thought, there'll be only some bounds. Dec 23 comment Number of representable as sum of 2 squares Sure, I meant $\Omega$. Thank you. Dec 23 comment Asymptotics with prime of form 4k+3 Yes, but in the paper we consider $L$-function, which is but it's definition a series (not finite), so we have to somehow jump from the infinite series to finite. (It's easy to do when the convergence in series is uniform, but it's obviously not the case) I mean, your previous comment is obviously true, but why $\sum_{p \equiv a \mod d \le n} \frac{1}{p} \sim \frac{\log(\log(n))}{\phi (d)}$. Do you see what I mean? Dec 23 comment Asymptotics with prime of form 4k+3 Thank you very much for your answer, but I have one question left: in papers it's proven that $\sum_{p \equiv a \pmod d} \frac{1}{p^{s}} \sim \dfrac1{\phi(d)}\sum_{p} \frac{1}{p^{s}}$ when $s \to +1$. But how does original problem follows from this? (It is not obvious) Dec 22 comment Asymptotics with prime of form 4k+3 Thank you, sure I did. =) Nov 23 comment Lebesgue measure is invariant under isometry Thank you guys very much. I just want to supply your beautiful answers with some links. proof that translation invariant measure is Lebesgue measure up to a constant factor Nov 17 comment Power sums, fast algorithm Sorry, really, there is a misunderstanding in terminology. Nov 17 comment Closed surjective map It turns out that wiki says : "surjective closed map is not necessarily an open map". So yours explanation is wrong. I assume, there is just a typo in a book. They don't really use it a lot, it just matters in one theorem. (They also give in the beginning that map is continuous (but don't use it in sub-statement), maybe with this restriction it's true, or you have counterexample?) Nov 17 comment Closed surjective map This is exactly what I did, but I don't see how it can be trivially deduced from this. I see that the image of $U$ in union with image of it's complement must give the whole $Y$. But they can intersect, because we are not given that map is injective. Could you, please, give a more detailed proof.