300 reputation
19
bio website
location Minsk, Belarus
age 20
visits member for 1 year, 11 months
seen Jul 23 at 8:01

I like math and computer science.


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.