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Sep
24
awarded  Autobiographer
Sep
1
comment Natural surjection from complex upper half plane into modular curve
Approach $1)$ unfortunately doesn't work, what you want to prove is that $\pi^{-1} (\pi (U))$ is open for anu $U$ open, and not the other way around. And $2)$ is using too much structure, you actually don't need to consider the complex structure on $Y(\Gamma)$. For the answer you could check this previous question: math.stackexchange.com/questions/61173/… .
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
13
awarded  Popular Question
Jun
11
accepted Example of a meromorphic function with no analytic continuation outside the unit disc
Jun
11
comment Example of a meromorphic function with no analytic continuation outside the unit disc
Thanks, this does exactly what I was looking for! I'm going to do the exercise then!
Jun
11
comment Example of a meromorphic function with no analytic continuation outside the unit disc
Thank you very much! I actually did miss it...
Jun
11
comment Are the two statements about continuous functions equivalent?
I think that the implication arrow in your first statement has to be reversed to get the definition of continuity.
Jun
11
asked Example of a meromorphic function with no analytic continuation outside the unit disc
Apr
20
awarded  Yearling
Jan
9
comment How important are automorphic representations among admissible ones?
Thanks for the answer!
Jan
2
comment How prove this $XY=YX$
If $X$ and $Y$ commute with $A$ then they also commute with $A^{-1}$...
Oct
6
revised Help solving this Linear First order ODE
There were no "{}"-brackets for exponentials, making the formulae unreadable. I added them where it seemed the case, the OP should check.
Oct
6
suggested approved edit on Help solving this Linear First order ODE
Sep
17
asked Analytic Continuation of a nowhere existing Mellin Transform
Sep
16
comment How to calculate the following double integral
Do you have a proof for the $a=c=0$ case? I think that the knowledge of the right integral representation of $K_\nu$ could unlock the problem...
Sep
16
comment How to calculate the following double integral
There is a small difference between your expression and the one of the Normal Product Distribution you cite, i.e. in the denominators of the exponents you should take $2b^2$ and $2d^2$ instead of $2b$ and $2d$. Is it just a typo or you want to compute exactly what you wrote?
Jul
10
revised Turning a Line Integral into a Contour one
added 136 characters in body
Jul
10
comment Turning a Line Integral into a Contour one
@DanielFischer Yes, this agrees with my computation. I think I could try to compute the additional pole (and relative residue) by brute force, but it's gonna be long and not elegant. While the observation of the authors of the paper on the possibility of turning it into a contour integral with only one pole is very tantalizing. Do you think it is possible to do that? On the other hand, according to the suggestions of Avitus I was trying to play a bit with different contours to exclude this additional pole. Thanks for your interest!