bio | website | people.ucsc.edu/~weissman |
---|---|---|
location | Around and about | |
age | 38 | |
visits | member for | 5 years, 7 months |
seen | 6 hours ago | |
stats | profile views | 5,774 |
Associate Professor, Yale-NUS College, Singapore.
Associate Professor of Mathematics, UC Santa Cruz. (On leave)
Research interests: Automorphic representations and representations of p-adic groups, especially exceptional groups and "metaplectic" groups lately. Theta correspondences (exceptional ones). Geometric methods in representation theory. Periods and Hodge theory. Model theory applied to number theory and geometry.
Book blog: Illustrated Theory of Numbers
Jul
26 |
comment |
Is the twisted symmetric fifth power $L$-function holomorphic?
@7-adic: Nope -- $Sym^5$ and higher are out of range of Langlands-Shahidi. Basically getting up to $Sym^4$ requires some special cases of Levi subgroups in exceptional groups, and sadly the special cases run out. Garland has a long-term program to try to extend these methods to infinite-dimensional groups -- if one is allowed to use Kac-Moody groups, one could go further. But alas, the results in that direction are not nearly strong enough for $Sym^5$ L-functions as far as I know. |
Jul
7 |
comment |
Why is there a $\sqrt{5}$ in Hurwitz's Theorem?
My favorite proof is in Ford's aptly titled article "Fractions" (Amer. Math. Monthly, Vol 45, No 9 (Nov 1938)). He gives the "Ford circle" proof of Dirichlet's approximation theorem, and the $\sqrt{5}$ comes straight out of the geometry he uses. So, if "visual" suffices for "intuitive," this might suffice for your needs. |
May
21 |
awarded | Nice Answer |
Feb
24 |
comment |
$ n $-Cats-in-a-Bed Problem: Picking $ n $ points in a given planar domain to maximize the sum of their pairwise distances
In the case n=2, a Google image search indicates that pairs of cats in a bed tend not to maximize their pairwise distance. This may be a case of sampling bias however. |
Feb
23 |
comment |
$ n $-Cats-in-a-Bed Problem: Picking $ n $ points in a given planar domain to maximize the sum of their pairwise distances
When $n=1$, the solution can be found at sleepingcatsw.com/images/cats/luther_bed.jpg |
Jan
26 |
awarded | Yearling |
Nov
16 |
comment |
Weil index computation, p-adic integral
These computations should be straightforward, using the explicit formulas of Ranga Rao. See the appendix of "On some explicit formulas in the theory of the Weil representation" in Pacific J. of Math., Vol. 157, No. 2, 1993. |
Nov
3 |
comment |
What are the higher homotopy groups of a K3 suface?
A new entry in the encyclopedia of integer sequences, perhaps? |
Oct
28 |
comment |
Visibility interpretation of Riemann zeta zeros on the critical line?
You forgot to mention, the fraction of $\mathbb Z$ lattice points visible from the origin is $1 / \zeta(1) = 0$. |
Oct
26 |
comment |
Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section?
@Jason: But I guess that by removing a bunch of codim 1 subvarieties from Y, we'd get something $U$ which is "anabelian" or "hyperbolic" in some sense of the word. Then a rational section (existing by anabelian section conjecture) from $P^1 C \rightarrow U$ would extend to a regular section $P^1 C \rightarrow Y$. Again, all contingent on some anabelian conjectures that I don't fully understand. |
Oct
25 |
comment |
Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section?
Does this follow from some anabelian conjecture? From such a conjecture, we might expect rational sections $P^1 C \rightarrow Y$ to arise from sections of $Gal(Y) \rightarrow Gal(P^1 C)$ (abs. Galois group of function fields). It's known (Harbater, Pop, Haran, I think) that $Gal(P^1 C)$ is profinite free, and so sections exist. A rational section $P^1 C \rightarrow Y$ gives a regular section by the valuative criterion, right? Or perhaps a counterexample to this type of anabelian conjecture provides a counterexample here? |
Sep
22 |
awarded | Enlightened |
Sep
15 |
comment |
Results true in a dimension and false for higher dimensions
Every cubic polynomial map from $\mathbb C^n$ to $\mathbb C^n$ with nowhere vanishing Jacobian is a bijection. This is true for $n \leq 17$ but fails for larger $n$. Just kidding... but perhaps someday I'll edit the answer, change 17 appropriately and will gain some upvotes. |
Sep
5 |
comment |
Recognize this strange expression from linear algebra?
Ahh - or arrgh.. This gets more interesting though. I'll think about how to phrase things in terms of a cocycle. And please don't take offense for the removal of acceptance. I'm also hoping that someone might recognize it from another context and contribute another answer. |
Sep
3 |
comment |
Recognize this strange expression from linear algebra?
Wonderful, thank you! (And you did say "cocycle".) |
Sep
2 |
comment |
Recognize this strange expression from linear algebra?
Nick - I can only take credit for the pasta-related one. But I'll keep the other to maximize the tag-contribution. |
Sep
2 |
awarded | Nice Question |
Sep
2 |
asked | Recognize this strange expression from linear algebra? |
Aug
29 |
answered | Are the quaternions not uncountably categorical? |
Aug
16 |
awarded | Enlightened |