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visits | member for | 2 years, 4 months |
seen | Jun 9 at 14:48 | |
stats | profile views | 668 |
Apr 13 |
comment |
Is there a “right” proof of Riemann's Theta Relation?
@J. Martel: I have to confess I started to get a bit out of my algebraic depth with the later (adelic) stuff in Tata III, so haven't yet managed to get a working understanding of Thm 7.4, but will persevere. Possibly I should just try to figure out what that result is saying in the Euclidean case. @Jeff: That would be great! @Everyone: the equation for $x_1$ in Mumford looks like it must be a typo; corrected in the above. |
Apr 13 |
revised |
Is there a “right” proof of Riemann's Theta Relation?
added 12 characters in body |
Apr 12 |
asked | Is there a “right” proof of Riemann's Theta Relation? |
Apr 9 |
comment |
Minimal size of subsets $A,B$ in a finite group $G$ such that $AB=G$
So, the strategy when A = H is a subgroup generalizes to A being your favorite set with $|A A^{-1}| = O(|A|)$, by a similar packing argument (if you accept a constant factor at the end). The aim is then to find such a set with $|A| \approx \sqrt{N}$. This gives rise to the cyclic example above, and should do what Colin Reid wants for nilpotent groups (see above). I don't know Whether you could use something like this to avoid the use of CFSG in the general case. |
Mar 22 |
awarded | Citizen Patrol |
Mar 22 |
awarded | Critic |
Mar 22 |
awarded | Commentator |
Mar 22 |
comment |
More expanders?
My guess is that (1) and (2) are doomed not to expand because they are "too abelian", or "too nilpotent". So for (1), a set like $\{\pm g^{n} : 0 \le n \le N \}$ for $N$ large (but small compared to $p$) is likely to contradict expansion. Something similar should work in (2), considering a set of bounded length words on both your generators. I think this is closely related to this MO question. (3) has a very different flavour. It might be related to expansion in $SL_2$. |
Mar 21 |
comment |
Joint (close to uniform) distribution in finite fields
I think that's right -- third edit lucky. I think the $L^2$ result should give $\varepsilon$ up to a (basically) constant factor like $k / (k - 1)$, meaning the $L^1$ result should go as $k \varepsilon$ times some constant factor; i.e. this hasn't made too much difference to the conclusion. |
Mar 21 |
revised |
Joint (close to uniform) distribution in finite fields
deleted 48 characters in body |
Mar 21 |
revised |
Joint (close to uniform) distribution in finite fields
added 192 characters in body |
Mar 21 |
comment |
Joint (close to uniform) distribution in finite fields
Kevin -- thanks! You're absolutely right, and that means there's also an $\varepsilon^2$ that needs tracing through the argument. I'll fix this now and see what I get. |
Mar 21 |
revised |
Joint (close to uniform) distribution in finite fields
typo |
Mar 21 |
answered | Joint (close to uniform) distribution in finite fields |
Mar 20 |
awarded | Yearling |
Jan 26 |
awarded | Teacher |
Jan 26 |
revised |
Question about the Hardy-Littlewood method (quite basic)
added 7 characters in body; added 2 characters in body; edited body |
Jan 26 |
answered | Question about the Hardy-Littlewood method (quite basic) |
Jun 20 |
comment |
The space of lattices and modular forms of weight 1/2
@David Loeffler: fixed (I think), thanks. |
Jun 20 |
revised |
The space of lattices and modular forms of weight 1/2
added 5 characters in body |