User moonface - MathOverflow

Answer by moonface for Very strong multiplicity one for Hecke eigenforms

2009-10-29T22:50:06Z

It is not expected that any a>0 would work. If E, E' are quadratic twists of one another they have the same number of points at a set of primes of D. density 1/2.

First sentence added by David Speyer, to explain (hopefully correctly) which part of the question is being addressed.

Answer by moonface for Can the "physical argument" for the existence of a solution to Dirichlet's problem be made into an actual proof?

2009-10-29T05:15:10Z

Well, I don't understand the electrostatics, but here is another physical heuristic:

Impose a temperature distribution at the exterior, and measure (after some time has passed) the temperature in the interior. This gives a harmonic function extending the exterior temperature. [What's the electrostatic analogue? Formerly I had written "charge density", but now I am not sure if that's right.]

I think this strongly suggests a mathematically rigorous argument: We are naturally led to model the time-dependence of temperature in the interior. This satisfies a diffusion (or heat) equation, but in words:

"After a time \delta, the new temperature is obtained by averaging the old temperature along a circle of radius \sqrt{\delta}."

This process converges under reasonable conditions, as time goes to infinity, to the solution of the Dirichlet problem. Anyway, we are led to the Brownian-motion proof of the existence, which I personally find rather satisfying. Another personal comment: I think one should always take "physical heuristics" rather seriously.

[In response to Q.Y.'s comments below, which were responses to previous confused remarks that I made: neither the electric field nor the Columb potential is a multiple of the charge density on the boundary: the former is a vector, and in either case imagine the charge on the boundary to be concentrated in a sub-region; neither the electric field nor the potential will be constant outside that sub-region.]

Answer by moonface for Division Algebras as Algebraic Groups

2009-10-28T11:38:00Z

Choose an F-basis of D. The multiplication is described by certain quadratic functions, with respect to this basis; D* is given by the nonvanishing of a polynomial function (the norm). So the multiplication can be understood as defining an algebraic group structure on the complement of a hypersurface in an affine space.

Answer by moonface for Solving polynomial equations when you know in which number field the solutions live

2009-10-25T03:41:24Z

Practically speaking: I would try to reduce the equations modulo several small primes (or prime powers), solve them by brute force, and then try to lift to solutions in your number field using the chinese remainder theorem. It might be helpful to combine this with solving them to high precision over the real (or complex) numbers.

Answer by moonface for Is there a good way to think of vanishing cycles and nearby cycles?

2009-10-21T13:48:55Z

I also have a lot of difficulty seeing what's going on, and am no expert, so take this with a grain of salt.

Here's one small picture (which you might already know) that I found helpful: Massey's description of "vanishing cycles at angle \theta" (see http://arxiv.org/abs/math/9908107, around page 23) which gives geometric intuition for the passage from X_{\infty} to its universal cover. Namely, restrict your family to the segment where t is in e^{i \theta} [0,\epsilon], i.e., a ray emanating from the origin in the complex plane; then proceed as you did before, except no crazy covers needed, because your base is now contractible. This gives a functor (isomorphic to nearby cycles for any fixed \theta) together with an action of monodromy.

Answer by moonface for Is any representation of a finite group defined over the algebraic integers?

2009-10-17T15:03:06Z

Not a satisfying argument: We can, first of all, find a basis in which the entries lie in some algebraic number field K. Let O be the ring of integers of K. Then there is a locally free O-module M of rank n preserved by G: add up all the translates of O^n under G. Now, M need not itself be free, but it is isomorphic as an O-module to the sum of various ideals of O. Now pass to an extension L/K so that every ideal class of K trivializes in L, e.g. the Hilbert class field; then G preserves a free rank n module for the ring of integers of L. Sorry!

Answer by moonface for Arithmetic progressions without small primes

2009-10-17T14:16:06Z

I don't have a reference for you, but what GRH proves has nothing to do with the truth; the proof is "lossy." The expected answer would be p^{1+\epsilon} for the smallest prime congruent to 1 mod p: an integer around N is prime with probability 1/log(N), and in most regimes (although not all) this makes accurate predictions.

Comment by moonface

2009-10-18T17:51:50Z

(This might show up twice by accident, sorry.) Let e > 0. Mindlessly applying the heuristic, the chance f(p) that "there is no prime less than p^{1+e} in the progression" decreases very fast with p; in fact, the sum of f(p) over all p converges, suggesting this event happens only finitely often. For an example of the limitations of such reasoning, see "Primes in short intervals" by H. Maier.