User jared weinstein - MathOverflow

Answer by Jared Weinstein for Torsion points in Abelian varieties over number fields

2011-02-19T07:22:22Z

Let's take a page from Silverman's book, VII.3. Let $\mathfrak{p}$ be one of the primes of good reduction of $A$. Let $K/k$ be any extension, and let $\mathfrak{P}$ be a prime of $K$ above $\mathfrak{p}$. The reduction map $A(K)\to A(\mathcal{O}_K/\mathfrak{P})$ becomes injective when you restrict to torsion points of order prime to the residue characteristic of $\mathfrak{p}$ -- this is proved using an appeal to formal groups.

Now choose two such primes $\mathfrak{p}$ and $\mathfrak{p}'$ with distinct residue characteristics. Convince yourself that there exists a $K/k$ and primes $\mathfrak{P},\mathfrak{P}'$ of $K$ for which $A(K)\to A(\mathcal{O}_K/\mathfrak{P})\times A(\mathcal{O}_K/\mathfrak{P}')$ has nontrivial kernel. Nontrivial points in the kernel must not be torsion.

Answer by Jared Weinstein for "Purely local" proof of local Langlands

2012-05-12T04:22:02Z

The short answer to the question is that all currently known proofs of the local Langlands correspondence (and I'm just referring to GL(n) here) are "global" in the sense that they involve embedding the local problem into a global one. That is, the local field in question is realized as the completion of a global field at one of its places. Then the theory of automorphic forms over the global field may be applied. In particular, under certain circumstances, we know that Galois representations may be attached to automorphic representations. A purely local proof would not make reference to global fields at all.

Kevin commented that a purely local characterization (he uses the word statement) of the correspondences is a prerequisite for a purely local proof. The established characterization for GL(n) (and indeed, the one used in the proofs of Henniart and Harris-Taylor) is, as Kevin points out, through epsilon factors of pairs, and the existence of these is only defined through global means. (Rob is correct that Langlands has unpublished notes on the subject, but these are so complicated as to be unsatisfactory, and in any case it is truly unclear what the right characterization is for groups other than GL(n).)

Now to Alexander Chervov's important comment: what is the right characterization in the case of $n = 1$? Sure, you can make some quantitative conditions involving ramification. But let's recall that the most elegant path to local CFT is unquestionably through Lubin-Tate theory: the maximal totally ramified abelian extensions of a nonarchimedean local field are obtained by adjoining the torsion of a one-dimensional formal module of height one. Let us declare that Lubin-Tate theory itself provides the correct characterization of the local Langlands correspondence in the $n=1$ case (and to hell with conductors, Gauss sums, etc.).

This point of view suggests that variations on the theme of formal modules ought to provide the right purely local characterization of local Langlands (and also a hope for a purely local proof). Now already by 1990, Carayol conjectured ("Nonabelian Lubin-Tate theory") that certain deformation spaces of formal modules ("Lubin-Tate spaces") exhibit the local Langlands correspondence in their cohomology, at least for some classes of representations of GL(n). Harris and Taylor prove Carayol's conjecture for supercuspidal representations, which is enough to prove the existence of the correspondence in general. Here the characterization is still through epsilon factors of pairs, and therefore still global in nature.

The next big development along these lines is Peter Scholze's new proof of the correspondences for GL(n). While still global in nature, Scholze gives a purely local characterization of the correspondences, which satisfies Kevin's requirements for a "natural bijection", and which is compatible with the global theory. Suppose $\pi$ is a smooth irreducible representation of $\text{GL}_n(F)$ ($F$ a $p$-adic field). Scholze characterizes the corresponding (semisimplified) Weil representation $\sigma$ by giving an actual formula for the trace of $\sigma(\tau)$, for any element $\tau$ in the Weil group of $F$! Alas, the other side of Scholze's formula is too involved to describe here, but it involves deformation spaces of $p$-divisible groups in an ingenious way. When $n=1$, the formula reduces to the statement that local class field theory is realized in the torsion of Lubin-Tate formal modules. In my mind, purely local attacks on the local Langlands correspondence ought to start here.

(Not that any of the preceding is going to be mentioned in my talks tomorrow. My own meager contributions to this story don't yet connect to Scholze's work, but only to the theory of types, which figure prominently in the Bushnell-Henniart book mentioned by Keerthi.)

Galois invariants in a ring of fractional power series over a finite field

2012-04-25T01:55:27Z

Let $\mathbf{F}_q$ be a finite field, and let $A=\mathbf{F}_q [[ x^{1/q^\infty} ]]$ be the completion of $\mathbf{F}_q[x^{1/q^\infty}]$ with respect to the $x$-adic topology. Then the $q$th power Frobenius map $\pi\colon A\to A$ is an automorphism. If $F$ is the local field $\mathbf{F}_q((\pi))$, there is an action $F^\times\to \text{Aut} A$ by continuous automorphisms; an element of $F$ sends $x$ to the corresponding linear fractional power series. Let $\mathcal{O}_F=\mathbf{F}_q[[\pi]]$.

My question is: Can anyone produce an explicit nonconstant element of $A$ which is $\mathcal{O}_F^\times$-invariant?

I assure you that such elements exist! Here's why: Let $f(X)=\pi X+X^q\in F[X]$. Let $x_1,x_2,\dots$ be a compatible family of roots of iterates of $f$, with $x_1\neq 0$. Let $F_n=F(x_n)$. Then by Lubin-Tate theory, $\text{Gal}(F_n/F)\cong (\mathcal{O}_F/\pi^n)^\times$ . If $F_\infty$ is the union of the $F_n$, then $F_\infty/F$ is a maximal totally ramified abelian extension of $F$, with Galois group $\mathcal{O}_F^\times$ . Let $K$ be the $\pi$-adic completion of $F_\infty$.

I claim that $\mathcal{O}_K$ is isomorphic to $A$. Indeed, it isn't hard to see that the sequence $x_1^q,x_2^{q^2},\dots$ converges in $\mathcal{O}_K$ to an element $x$, all of whose $q$th power roots lie in $\mathcal{O}_K$. Observe that $x^{1/q^n}-x_n$ is divisible by $\pi$, hence by $x^{1/q}$, and therefore (since $\mathcal{O}_K$ is topologically generated by the $x_n$) we have $\mathcal{O}_K=\mathbf{F}_q[x^{1/q^\infty}]+x^{1/q}\mathcal{O}_K$. This shows that $\mathbf{F}_q[x^{1/q^\infty}]$ is dense in $\mathcal{O}_K$, which proves the claim.

The Galois action of $\mathcal{O}_F^\times$ on $F_\infty$ extends to the completion $K$. It's a fun exercise to see that the isomorphism $\mathcal{O}_K\cong A$ respects the action of $\mathcal{O}_F^\times$. Meanwhile, we have the element $\pi\in \mathcal{O}_K\cong \mathbf{F}_q[[x^{1/q^\infty}]]$. This element is invariant under the Galois group $\mathcal{O}_F^\times$. What is its power series development in $x$?

If $p=0$ and $df=0$, is $f$ a $p$th power?

2011-11-18T04:06:19Z

This question is a follow-up to http://mathoverflow.net/questions/75329/when-does-the-relative-differential-df0-imply-that-f-comes-from-the-base. There it was asked, for an $A$-algebra $B$, under what conditions does $df=0$ (in the module of relative differentials $\Omega_{B/A}$) imply that $f$ is ``constant", i.e. lies in $A$. The answer relied on a characteristic 0 assumption. My question is about rings $A$ in which $p=0$ for a prime $p$.

Assume that $p=0$ in $A$. Let's also assume that $A$ is perfect in the sense that $A^p=A$. I don't want to assume, however, that $A$ is integral or even reduced. Let $B$ be an $A$-algebra. Let $f\in B$ be such that $df=0$ in $\Omega_{B/A}$.

Under what conditions on $A$ and $B$ may we deduce that $f\in B^p$?

Notice that the converse is always true, because $d(f^p)=pf^{p-1}df=0$. Also notice that if $f\in A$, then of course $df=0$, but by our hypothesis on $A$ we have $f\in A=A^p\subset B^p$.

The conclusion is true, for instance, when $B$ is a polynomial ring over $A$, and also (I think) when $B$ is etale over $A$.

Answer by Jared Weinstein for Riemann's $\zeta$ function and the uniform distribution on $[-1,0]$

2011-09-18T08:07:24Z

$X$ is the uniform distribution on $[-1,0]$, so the moment generating function of $X$ is $$E(e^{tX})=\lim_{k\to\infty} \frac{1}{k}\left(1+e^{-t/k}+\dots+e^{-(k-1)t/k}\right)=\lim_{k\to\infty} \frac{1}{k}\frac{1-e^{-t}}{1-e^{-t/k}}=\frac{1-e^{-t}}{t}$$ The cumulant generating function is the logarithm of this, call it $g(t)$. It'll be easier to compute the Taylor series of $tg'(t)+1$: $$tg'(t)+1=-\frac{t}{1-e^t}=-\sum_{k=0}^\infty te^{kt}=-t\sum_{k=0}^\infty \sum_{n=0}^\infty \frac{k^nt^{n}}{n!}$$ Rearrange the order of summation: $$tg'(t)+1=-t\sum_{n=0}^\infty \zeta(-n)\frac{t^n}{n!}$$ Now compare coefficients on either side: the $n$th cumulant of $X$ is plainly $-\zeta(1-n)$.

What? Why are you looking at me like that?

When does the normalization have regular special fiber?

2011-08-03T02:27:58Z

Let's say $\mathcal{O}$ is a complete DVR with fraction field $K$ and algebraically closed residue field $k$. (The case I had in mind here was with $\mathcal{O}$ of equicharacteristic $p$, so assume this if you like.)

Let $A$ be a complete local $\mathcal{O}$-algebra of finite Krull dimension. Say we are given a presentation of $A$ as $$ A=\mathcal{O}[[X_1,\dots,X_n]]/(f_1,\dots,f_m)$$

Let's also assume that $A$ is a domain, and that $A$ is $\mathcal{O}$-flat.

My question is:

Let $B$ be the integral closure of $A$ in $A\otimes K$. When is $B\otimes k$ a regular local ring (i.e. when is it a power series ring over $k$)?

This is the same as asking whether $B$ is formally smooth over $\mathcal{O}$.

Let's say $t$ lies in the maximal ideal of $\mathcal{O}$. The ring $\mathcal{O}[[X,Y]]/(Y^2-t^2X)$ has this property, but $\mathcal{O}[[X,Y]]/(XY=t)$ does not.

Is there an "algorithm" to determine whether $A$ has this property, based on the presentation above? I put algorithm in quotes because the problem probably involves knowing $f_1,\dots,f_m$ to infinite accuracy. To get around this point, let's assume that all the $f_i$ lie in a polynomial ring $D[X_1,\dots,X_n]$, where $D\subset\mathcal{O}$ is finitely generated, as in the above examples.

How does Tate verify his own conjecture for the Fermat hypersurface?

2011-05-13T04:36:57Z

This question is about Tate's 1963 paper "Algebraic Cycles and Poles of Zeta Functions". Here he announces a conjecture (now known as "the Tate conjecture") which states that certain classes in the cohomology of a projective variety are always explained by the existence of algebraic cycles. In the case of a variety $X/\mathbf{F}_q$ of dimension $n$, the conjecture predicts that the subspace of classes in $H^{2d}(X\otimes\overline{\mathbf{F}}_q,\mathbf{Q}_\ell)(d)$ which are Frobenius-invariant is spanned by the image of the space of algebraic cycles in $X$ of codimension $d$.

As an example, Tate gives the projective hypersurface $X$ defined by $$ X_0^{q+1}+\dots X_r^{q+1} =0,$$ where $r=2i+1$ is odd. Here $X$ admits a large group of automorphisms $U$, namely those projective transformations in the $X_i$ which are unitary with respect to the semilinear form $\sum_i X_iY_i^q$. Using the Lefschetz theorem, it isn't at all hard to compute $H^{2i}(X)$ as a $U$-module: it decomposes as a trivial $U$-module and an irreducible $U$-module of dimension $q(q^r+1)/(q+1)$. And then when you attempt to compute the $q^2$-power Frobenius eigenvalues on the middle cohomology, you find (once again by Lefschetz) that each one is a Gauss sum which in this case is nothing but $\pm q$. (If there is enough demand, I can supply all these calculations here.) Thus, miraculously, all classes in $H^{2i}(X\otimes\overline{\mathbf{F}}_q,\mathbf{Q}_\ell)(i)$ are fixed by some power of Frobenius.

My question is: How did Tate confirm the existence of the necessary cycles? Surely the hyperplane section of codimension $i$ lands in the part of $H^{2i}$ which has trivial $U$-action (for $U$ translates hyperplanes to other hyperplanes, and these all cohomologically equivalent). In order to verify Tate's conjecture, all you need to do is produce a cycle in $X$ whose projection into the big $U$-irreducible part of $H^{2i}$ is nonzero. How did Tate produce this cycle? Did he lift to characteristic zero and appeal to the Hodge conjecture, or what?

Answer by Jared Weinstein for Geometric construction of depth zero local Langlands correspondence

2011-05-07T07:49:09Z

Yoshida considers the Lubin-Tate tower in his geometric realization of the depth zero supercuspidals for $GL(n)$. For unitary groups, I'm sure that the answer to your question will be found in a similar analysis for the corresponding Rapoport-Zink spaces. I don't believe this has been done yet, but it ought to be done soon -- I'm an interested party, as is (at least) Tasho Kaletha at Princeton. In the meantime, I have a reinterpretation of Yoshida's thesis that uses the $p$-adic period mappings out of the Lubin-Tate tower in an essential way. I've just posted some notes yesterday (!) here. The bit about DL varieties appears at the very end. I resort to using coordinates and deriving (as Yoshida does) an explicit equation for the DL variety, but with a little more work I bet a more abstract approach can be found. Since the theory of $p$-adic period maps for more general RZ spaces has all been worked out, I bet this approach can be used for unitary groups as well.

Answer by Jared Weinstein for p-adic representations of a quaternion algebra over a local field

2011-03-01T17:17:23Z

If $E$ is an algebraic closure of $F$, then $D\otimes_F E\simeq M_2(E)$. (In fact this is also true if $E$ is taken to be, say, the unramified quadratic extension field of $F$.) We get an algebraic representation $$\phi\colon D^\times\hookrightarrow (D\otimes E)^\times=\text{GL}_2(E).$$ And then for each $a\geq 0$ and $b\in \mathbf{Z}$ we get the representation $\text{Sym}^{a}\phi\otimes (\det\phi)^b$ . My feeling is that these exhaust the irreducible algebraic representations of $D^\times$, but I'm afraid I don't have a proof at the ready.

As the other answerers show, the question of classifying the admissible representations of $D^\times$ (with complex coefficients) is a far more subtle issue!

What are the endomorphisms of Drinfeld's "special formal O_D-modules"?

2011-02-23T18:53:43Z

Let $F$ be a nonarchimedean local field, and let $D/F$ be the central division algebra of invariant $1/d$. Let $k$ be the algebraic closure of the residue field of $F$ and let $\pi$ be a uniformizer.

Let $X$ be a $\pi$-divisible group over $k$. Following Drinfeld, $X$ is called a formal $\mathcal{O}_D$ -module if it is connected and if it is endowed with an action of $\mathcal{O}_D$ for which $\text{Lie}(X)$ becomes a locally free $\mathcal{O}_{F_d}\otimes_{\mathcal{O}_F} k$-module of rank 1. Here $F_d\subset D$ is an unramified extension of $F$ of degree $d$.

Such an $X$ has $F$-height divisible by $d^2$; let us assume that the height is exactly $d^2$. Then $X$ is unique up to isogeny. It is easy to concoct such an $X$ in the case that $F=\mathbf{F}_{q}((\pi))$ has positive characteristic. Here $\mathcal{O}_D$ is generated over $\mathcal{O}_F=\mathbf{F}_q[[\pi]]$ by $\mathbf{F}_{q^d}$ and by an element $\Pi$ which has $\Pi^d=\pi$ and $\Pi\zeta=\zeta^q\Pi$, $\zeta\in\mathbf{F}_{q^d}$ . We can let $X$ be the formal $\mathcal{O}_D$ -module whose law is determined by

$$ [\Pi]_X(T_0,\dots,T_{d-1})=(T_1,\dots,T_{d-1},T_0^{q^d}) $$

$$ [\zeta]_X(T_0,\dots,T_{d-1})=(\zeta T_0,\dots,\zeta^{q^{d-1}}T_{d-1}) $$

My question is:

What is the algebra of $\mathcal{O}_D$ -endomorphisms of this particular $X$?

Through an examination of the Dieudonnè module of $X$, one shows that its endomorphism algebra becomes $M_d(F)$ when tensored with $F$. So $\text{End} X$ is an order in $M_d(F)$ , but which one? Can one exhibit the endomorphisms explicitly on the level of coordinates?

Answer by Jared Weinstein for Deligne's letter to Piatetskii-Shapiro from 1973

2011-02-18T03:57:35Z

I have typeset Deligne's letter, and placed the result here:

http://www.math.ias.edu/~jaredw/DeligneLetterToPiatetskiShapiro.pdf

I have made some minor edits so that the text reads more naturally to a native speaker of English. Also I made a few annotations where I truly believe there is an error in the original. Any other errors are mine.

The letter struck me with how much it accomplishes in such a short space. Actually, I would say that the meat of the argument is confined to the final three pages, with the rest there only to establish notation. This letter ought to be required reading for anyone studying automorphic forms in arithmetic!

Answer by Jared Weinstein for How to get explicit unramified covers of an elliptic curve?

2011-01-14T23:27:50Z

The covers you seek are exactly the elliptic curves which admit cyclic $n$-isogenies into $E$ (or out of $E$, it doesn't matter). Proof: if $X\to E$ is unramified, then $X$ is a genus one curve (by the Riemann-Hurwitz formula). Choose a base point on $X$ lying above the origin of $E$; then $X$ is an elliptic curve and $X\to E$ is an isogeny (standard theorem about morphisms between abelian varieties).

The curves $X$ you want are what you get when you quotient $E$ by each of its cyclic subgroups of order $n$. Finding these $X$ is exactly the purpose of the modular polynomial $\Phi_n(x,y)$. The way that this works is that if $j(E)$ is the $j$-invariant of $E$, then the roots of $\Phi_n(x,j(E))$ are exactly the quantities $j(E')$, where $E'$ runs over the elliptic curves related to $E$ by cyclic $n$-isogeny. The modular polynomial $\Phi_n(x,y)$ is difficult to calculate, and its coefficients grow very quickly with $n$; a Google search comes up with this reference on calculating them. Once you have the $j$-invariants, it's easy to find the corresponding elliptic curves (this is one nice thing about working over $\mathbf{C}$).

Incidentally, the degree of $\Phi_n(x,j)$, which is (generically) the number of $X$ that you want, is $$ n\prod_{p\vert n}\left(1+\frac{1}{p}\right).$$

A family of hypersurfaces with many points

2010-11-08T23:46:01Z

This question is a sequel to an earlier question, which asked about the zeta function of a certain affine variety over a finite field $k$. The unusual thing about this variety is that it had the maximum number of rational points over every extension of $k$, relative to the constraints imposed by the topology. Here we're going to construct a family of varieties which we suspect behaves the same way. If my hunches are correct, the result will be a means to construct irreducible characters of certain unipotent algebraic groups over finite fields, akin to the way Deligne-Lusztig varieties are used to construct irreducible characters of Chevalley groups.

Let $n\geq 3$ be prime. We define a group variety $U/\mathbf{F}_q$ as follows: for an $\mathbf{F}_q$-algebra $R$, we set

$$U(R)=\{1+x_1\tau+\dots+x_n\tau^n\biggm\vert x_i\in R\}$$

Here $\tau$ is an indeterminate satisfying $\tau x_i=x_i^q \tau$ and $\tau^{n+1}=0$. Thus $U$ is a unipotent group, abstractly isomorphic to $\mathbf{A}^n$. Let $p\colon U\to \mathbf{A}^1$ be the projection onto the coefficient of $\tau^n$. Also let $g\mapsto g^{(q)}$ be the automorphism of $U$ which raises each coordinate to the $q$th power.

Let $X\subset U$ be the hypersurface defined by $p\left(g^{(q^n)}g^{-1}\right)=0$. Then $X$ is a nonsingular affine variety of dimension $n-1$.

My question is:

Prove that the zeta function of $X$ over $\mathbf{F}_{q^n}$ is $$Z(X/\mathbf{F}_{q^n},T)=\left(1-q^{n(n-1)/2}T\right)^{-q^{n(n-1)/2}(q^n-q)} \left(1-q^{n(n-1)}T\right)^{-q}$$

The form of this zeta function is going to make more sense when we consider that $X$ has a large group of $\mathbf{F}_{q^n}$ -linear automorphisms, namely $U(\mathbf{F}_{q^n})$ , which acts on $X$ on the right.

Let $Z\subset U$ be the center: $Z=\{1+a_n\tau^n\}$ . Call a character $\psi\colon Z(\mathbf{F}_{q^n})= \mathbf{F}_{q^n}\to\overline{\mathbf{Q}}_\ell^\times$ generic if it does not factor through the trace map $\mathbf{F}_{q^n}\to\mathbf{F}_{q}$ . It can be shown that if $\psi$ is generic, then there is a unique irreducible representation $V_\psi$ of $U(\mathbf{F}_{q^n})$ whose central character is $\psi$. The dimension of $V_\psi$ is $q^{n(n-1)/2}$ . (Quick construction: the subgroup $U^{(n-1)/2}\subset U$ defined by $x_1=\dots=x_{(n-1)/2}=0$ is abelian. Extend $\psi$ to a character $\tilde{\psi}$ of $U^{(n-1)/2}(\mathbf{F}_{q^n})$ any way you like, and let $V_\psi$ be the induction of $\tilde{\psi}$ to $U(\mathbf{F}_{q^n})$ .)

Each non-generic character $\psi$ extends to a one-dimensional character of $U(\mathbf{F}_{q^n})$ , which we also call $V_\psi$. Indeed, there is a "reduced norm" map $N\colon U(\mathbf{F}_{q^n})\to Z(\mathbf{F}_q)$ extending the trace map $T\colon Z(\mathbf{F}_{q^n})\to Z(\mathbf{F}_q)$ , and if $\psi=\psi_0\circ T$ then let $V_\psi=\psi_0\circ N$ . (In fact this $N$ is really a morphism $U\to Z$, albeit not a homomorphism, and an alternate definition of $X$ is $\{g\vert N(g)\in Z(\mathbf{F}_q)\}$ . Thus $X$ has at least $q$ connected components.)

Our question now becomes:

Show there is an isomorphism of $U(\mathbf{F}_{q^n})$-modules $$H^*_c(X\otimes\overline{\mathbf{F}}_q,\overline{\mathbf{Q}}_\ell)\cong \bigoplus_\psi V_\psi,$$ where the sum is over all characters of $\mathbf{F}_{q^n}$ and each $V_\psi$ appears exactly once. The eigenvalue of the $q^n$-power Frobenius on $V_\psi$ equals $q^{n(n-1)/2}$ if $\psi$ is generic and $q^{n(n-1)}$ otherwise.

There will be a generalization of these statements to the case of $n$ composite, but they are more complicated; there will be contributions to $H^*_c$ of degrees strictly between $n-1$ and $2(n-1)$.

Answer by Jared Weinstein for What is the L-function version of quadratic reciprocity?

2010-10-22T22:35:47Z

Say $K=\mathbf{Q}(\sqrt{p^{*}})$ is the unique quadratic field in which $p$ is the only ramified finite prime. That is, $p^*=(-1)^{(p-1)/2}p$. There are two L-functions one can concoct out of $K$, and quadratic reciprocity manifests in the fact that the L-functions coincide.

First, there is the L-function arising from the Galois character $\chi$ which cuts out $K$. This is a one-dimensional Artin L-function. Its Euler factor at a prime $q$ is $(1\pm p^{-s})^{-1}$, where you place $-$ whenever $q$ splits in $K$, and $+$ whenever $q$ is inert in $K$.

The other L-function is a Dirichlet L-function. Let $\varepsilon$ be the unique character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order exactly two. This extends to a Dirichlet character $\varepsilon$ on $\mathbf{Z}$, and the Dirichlet L-function is $\sum_{n\geq 1} \varepsilon(n)n^{-s}$.

Let us see that if the two L-functions are equal, then quadratic reciprocity holds. Indeed, look at the coefficient of $q^{-s}$. The coefficient in the first L-function is $\left(\frac{p^*}{q}\right)$. The coefficient in the second L-function is $\left(\frac{q}{p}\right)$. (This requires some explanation: $x\mapsto \left(\frac{x}{p}\right)$ is a character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order two, so it must equal $\varepsilon$ by uniqueness.) We find

$$\left(\frac{q}{p}\right)=\left(\frac{p^{*}}{q}\right)=\left(\frac{(-1)^{(p-1)/2}}{q}\right)\left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right).$$

Of course you didn't need L-functions to state any of this, but they are nonetheless very important. For instance, you can't prove that the Artin L-function has an analytic continuation and functional equation without knowing that it equals a Dirichlet L-function. The generalization of this coincidence to higher-degree Artin L-functions is still quite conjectural!

Answer by Jared Weinstein for How do you explicitly compute the p-torsion points on a general elliptic curve in Weierstrass form?

2010-03-04T23:13:24Z

Since you asked about software, I'd just like to point out (if you don't know already) that SAGE (available at sagemath.org) can compute division polynomials easily. The commands

R.<A,B> = PolynomialRing(GF(5))

E = EllipticCurve([A,B])

f = E.division_polynomial(5)

f

return the result

2*A*x^10 - A^2*B*x^5 + A^6 - 2*A^3*B^2 - B^4.

The warnings of the commenters apply; you must be cautious about interpreting the division polynomials. It is true that the roots of this polynomial give you the "physical" 5-torsion points of the elliptic curve in characteristic 5, but that's about all it says. The polynomial does not tell you, eg, the structure of the group scheme E[5] over the supersingular locus.

A little more helpful might be the formal group associated to this family:

G = E.formal_group()

G.mult_by_n(5,30)

which returns

2*A*t^5 + (2*A^6 - A^3*B^2 - B^4)*t^25 + O(t^30)

There's also a command G.group_law() whose output I'm not going to give here. Together these data give you the structure of E[5] over an infinitesimal neighborhood of the zero section of your elliptic curve over $\text{Spec}\mathbf{F}_5[A,B,\Delta^{-1}]$.

I'm a firm believer in explicit calculations as a means of developing intuition about algebro-geometric concepts. But by all means read Katz-Mazur :-).

Answer by Jared Weinstein for Locally profinite fields ?

2010-03-04T09:06:58Z

I was going to leave this as a comment, but I have a firm conviction about this, so here's my answer. The fields in your (1) are called (a) locally compact non-archimedean fields, or (b) non-archimedean local fields, and I fear that attempts to use other terminology (however logical) might lead to confusion. (A "locally profinite field" sounds like it should admit an open profinite subfield, which doesn't make sense.) Which is not to say that authors vary in how they refer to such objects, but always there is complete precision. Eg:

Lubin and Tate, Formal Complex Multiplication in Local Fields: "Let $k$ be a field complete with respect to a discrete valuation, with finite residue field..."

Harris, On the Local Langlands Correspondence: "The local Langlands correspondence for GL(n) of a non-Archimedean local field $F$ parametrizes irreducible admissible representations..."

Bushnell and Henniart, Calculs De Facteurs Epsilon De Paires Pour GL(n) Sur Un Corps Local, I: "Soit $F$ un corps commutatif localement compact non archimédien; notons $p$ sa caractéristique résiduelle et $q$ le cardinal de son corps résiduel."

Whether you want to capitalize "archimedean" is another story. But the moral is: defer to tradition and leave no doubt in the reader's mind as to what you mean.

Does Con(ZF) imply Con(ZF + Aut C = Z/2Z)?

2010-02-28T09:29:20Z

How many field automorphisms does $\mathbf{C}$ have? If you assume the axiom of choice, there are tons of them -- $2^{2^{\aleph_0}}$ I believe. And what if you don't -- how essential is the axiom of choice to constructing "wild" automorphisms of $\mathbf{C}$? Specifically, if you assume that ZF admits a model, does that imply that ZF admits a model where $\mathbf{C}$ has no wild automorphisms: $\mathop{Aut}\mathbf{C}=\mathbf{Z}/2\mathbf{Z}$?

I suppose if that's true, then the next logical question is to construct models of ZF where $\mathop{Aut}\mathbf{C}$ has cardinality strictly between 2 and $2^{2^{\aleph_0}}$--pretty disturbing if you ask me. Which finite groups can you hit?

Answer by Jared Weinstein for etale fundamental group and etale cohomology of curves

2010-02-27T00:44:41Z

The two groups you want to compare are canonically isomorphic, so long as C is connected. See Example 11.3 of Milne's notes:

http://www.jmilne.org/math/CourseNotes/lec.html

Convergence of a sequence of continuable Dirichlet series

2010-02-20T01:37:16Z

Let's say $f$ is a Dirichlet series which converges on the half-plane $\text{Re }s>\sigma$ to a function $f(s)$. Suppose further that $f(s)$ admits an analytic continuation to an entire function, together with the standard sort of functional equation. Let $g_n$ be a sequence of Dirichlet series, also convergent on $\text{Re }s>\sigma$, which each admit an analytic continuation and functional equation, though their precise FEs may vary. We assume that $g_n$ converges to $f$ in the following sense: for every $m>0$ there exists an $N$ for which the series $g_n$ and $f$ match on every term up to the $m$th, for all $n>N$. Note this implies that $g_n(s)$ converges to $f(s)$ for every $\text{Re }s>\sigma$.

Can it be said that $g_n(s)$ converges to $f(s)$ for any $s$ outside the domain of convergence?

Perhaps that's too much to hope for, and you can't even expect that $g_n(s)$ converges to $f(s)$ even for the point $s=\sigma$. I'd certainly be interested in a counterexample which does this!

Cohomology of rigid-analytic spaces

2010-01-13T19:58:15Z

Let $R$ be a complete discrete valuation ring and let $K$ be its field of fractions. Suppose $X$ is a smooth rigid-anaytic space over $K$. Often it is convenient to have a model of $X$ whose reduction has singularities which are as mild as possible--a semistable model. This amounts to having an admissible covering of $X$ by open affinoids $X_i$ , each of which has good reduction, such that the reductions of any pair $X_i$ and $X_j$ meet transversally, if at all. (See the paper of Bosch/Lütkebohmert for definitions.) Let us assume such a semi-stable model of $X$ exists. Then the étale cohomology of $X$ can be computed from the combinatorics of the covering $X_i$, together with the étale cohomology of each $X_i$, via the weight spectral sequence of Rapaport-Zink.

Now suppose I have an open affinoid $Z\subset X$ which happens to have good reduction. My question is: Does there admit a semi-stable model of $X$ for which $Z$ belongs to the covering? Failing this, is there some sense one can make of my intuition that the cohomology of the reduction of $Z$ ought to contribute to the cohomology of $X$?

Feel free to edit/criticize my question to smithereens if you like.

Answer by Jared Weinstein for modular eigenforms with integral coefficients [Maeda's Conjecture]

2010-01-13T19:04:30Z

The statement that the Hecke algebra acts irreducibly on $S_k(\Gamma(1))$ is known as Maeda's conjecture, and it is still open. So an affirmative answer to your question about eigenforms with integer coefficients would provide a negative answer to Maeda's conjecture. The negation of your question--that there are only finitely many integral eigenforms of level 1--is a weaker form of Maeda's conjecture, but one which nevertheless seems very hard to me.

There's lots of computational evidence in support of Maeda's conjecture, as Google will reveal. For instance, I don't think there is a weight $k$ known for which the operator $T_2$ has reducible characteristic polynomial (let alone has a linear factor).

Answer by Jared Weinstein for Explicit computations of small Deligne-Lusztig varieties (e.g. Drinfeld curve)

2010-01-02T07:16:16Z

I found Teruyoshi Yoshida's exposition of the subject very helpful:

http://www.dpmms.cam.ac.uk/~ty245/Yoshida_2003_introDL.pdf

As JT commented, the curve you wrote down is really the Deligne-Lusztig variety for SL_2, not GL_2. Ben is also right about the curve being $\mathbf{P}^1 - \mathbf{P}^1(\mathbf{F}_q)$, only he is using a different definition of DL variety from you I would presume. The way Ben has it, the DL variety is a subvariety of G/B, but the curve you want is a subvariety of G/U, where U is the unipotent radical. One formulation is a cover of the other with galois group equal to the rational points on a twist of the torus T. We'll take the G/U point of view here.

So let's start with $G=\text{SL}_2$ over the field $\mathbf{F}_q$. We'll let $B$ be the usual Borel and $U$ its unipotent radical. We can then identify $G/B$ with $\mathbf{P}^1$ and $G/U$ with $\mathbf{A}^2$. The latter identification sends $(a,b,c,d)$ to $(a,c)$.

Let $w=(0,1,1,0)$ be the nontrivial Weyl element. We let $X_w$ be the subvariety of $G/B$ consisting of elements $x$ for which $x$ and $F(x)$ are in relative position $w$, where $F$ is the Frobenius map. This is $\mathbf{P}^1 - \mathbf{P}^1(\mathbf{F}_q)$ as Ben says.

For cosets $x,y\in G/B$ in relative position $w$, and a coset $gU\in G/U$ for which $gB=x$, we are going to define a new coset $w_{x,y}(gU)\in G/U$ as follows. First find a $g'\in G$ for which $g'B=x$ and $g'wB = y$. We may further take $g'$ so that $g'U=gU$. (This can be done because of the Bruhat decomposition of $G/B \times G/B$--wait a moment to see how this plays out for $\text{SL}_2$ .) Then define $w_{x,y}(gU) = g'wU$. (Pardon the abuse of notation of the symbol $w$.) The Deligne-Lusztig variety $Y_w$ is defined as the set of $gU\in G/U$ for which $F(gU)=w_{gB,F(gB)}(gU)$.

When does a point $(x,y)\in\mathbf{A}^2=G/U$ lie in $Y_w$? We need to calculate $w_{gB,F(gB)}(gU)$, where $g=(x,*,y,*)\in G$ . We have $gB=g\cdot\infty=x/y$ and $F(gB)=(x/y)^q$. So we must now find $g'\in G$ with $g'U=gU$ and $g'wB=F(g)wB$. The first condition means that $g'=(x,*,y,*)$ and the second means that $g'\cdot 0=(x/y)^q$ . Thus $g'=(x,ux^q,y,uy^q)$, where $u$ must satisfy $u(xy^q-x^qy)=1$. We find that $w_{gB,F(gB)}(gU)=g'wU=(ux^q,uy^q)$. The condition that $(x,y)\in Y_w$ is exactly that $(x^q,y^q)=(ux^q,uy^q)$, which implies that $u^{-1}=x^qy-xy^q=1$. So that's the equation for the Deligne-Lusztig variety.

The equation for the DL variety for the longest cyclic permutation in the Weyl group of $\text{SL}_n$ is $\det(x_i^{q^j})=1$, where $0\leq i,j\leq n-1$.

I believe Lusztig calculated the zeta functions of his varieties in a very general setting, but I was never able to trudge through it all. There must be a simple answer for the behavior of the zeta functions for the $\text{GL}_n$ varieties--if you ever write it up I'd certainly love to read it! I can start you off: for $\text{SL}_2$ over $\mathbf{F}_q$ , the DL curve has a compactly supported $H^1$ of dimension $q(q-1)$, and the $q^2$-power Frobenius acts as the constant $-q$. (The behavior of the $q$-power Frobenius might be a little subtle--I suspect it has to do with Gauss sums.)

Good luck!

Answer by Jared Weinstein for Etale cohomology and l-adic Tate modules

2009-12-01T03:37:57Z

IMO, the scenario is closer to your (a). I'll sketch an explanation of the duality between $H^1(E,\mathbf{Z}_l)$ and the dual to the Tate module. We have $H^1(E,\mathbf{Z}_l)=\text{Hom}(\pi_1(E),\mathbf{Z}_l)$ ,

where that $\pi_1$ means etale fundamental group with base point the origin $O$ of $E$. Thus the isomorphism we really want is between $\pi_1(E)\otimes\mathbf{Z}_l$ and $T_\ell(E)$.

What is $\pi_1(E)$? In the topology world, we'd consider the universal cover $f\colon E'\rightarrow E$ and take $\pi_1(E)$ to be its group of deck transformations. Then $\pi_1(E)$ has an obvious action on $f^{-1}(O)$. If $E$ is the complex manifold $\mathbf{C}/L$ for a lattice $L$, this is just the natural isomorphism $\pi_1(E)\cong L$.

But in the algebraic geometry world, there is no universal cover in the category of varieties, so the notion of universal cover is replaced with the projective system $E_i\to E$ of etale covers of $E$. Then $\pi_1(E)$ is the projective limit of the automorphism groups of $E_i$ over $E$.

One nice thing about $E$ being an elliptic curve is that any etale cover $E'\rightarrow E$ must also be an elliptic curve (once you choose an origin on it, anyway); if $E\rightarrow E'$ is the dual map then the composition $E\rightarrow E'\rightarrow E$ is multiplication by an integer. So it's sufficient to only consider those covers of $E$ which are just multiplication by an integer. Since it's $\pi_1(E)\otimes\mathbf{Z}_l$ we're interested in, it's enough to consider the isogenies of $E$ given by multiplication by $l^n$.

What are the deck transformations of the maps $l^n\colon E\rightarrow E$? Up to an automorphism of $E$, they're simply translations by $l^n$-division points. And now we see the relationship to the Tate module: A compatible system of deck transformations of these covers is the exact same thing as a compatible system of $l^n$-division points. Thus we get the desired isomorphism. Naturally, it's Galois compatible!

In the end, we see that torsion points were tucked away in the construction of the etale cohomology groups, so it wasn't exactly a coincidence. Hope this helps.

Re the edit: I believe your best bet is to work locally. First of all, you didn't mention which Galois representation you wanted exactly; let's say you want the representation on $H^i$ of your variety for a given $i$. Let's assume this space has dimension $d$.

Step 1. For each prime $p$ at which your variety $V$ has good reduction, you can compute the local zeta function of $V/\mathbf{F}_p$ by counting points on $V(\mathbf{F}_{p^n})$ for $n\geq 0$. In this way you can compute the action of the $p^n$th power Frobenius on $H^i(V\otimes\overline{\mathbf{F}}_p,\mathbf{F}_5)$ for various primes $p$.

Step 2. Do this enough so that you can gather up information on the statistics of how often the Frobenius at $p$ lands in each conjugacy class in the group $\text{GL}_d(\mathbf{F}_5)$ . In this way you could guess the conjugacy class of the image of Galois inside $\text{GL}_d(\mathbf{F}_5)$ .

Step 3. Now your job is to find a table of number fields $F$ whose splitting field has Galois group equal to the group you found in the previous step. I found a table here: http://hobbes.la.asu.edu/NFDB/. You already know which primes ramify in $K$ -- these are at worst the primes of bad reduction of $V$ together with 5 -- and you can distinguish your $F$ from the other number fields by the splitting behavior your found in Step 1. Then $K$ is the splitting field of $F$.

A caveat: Step 1 may well take you a very long time, because unless your variety has some special structure or symmetry to it, counting points on $V$ is Hard.

Another caveat: Step 3 might be impossible if $d$ is large. If $d$ is 2 then perhaps you're ok, because there might be a degree 8 number field $F$ whose splitting field has Galois group $\text{GL}_2(\mathbf{F}_5)$ . If $d$ is large you might be out of luck here.

You are free not to accept this answer because of the above caveats but I really do think you've asked a hell of a tough question here!

Answer by Jared Weinstein for examples of admissible representations of $GL_{n}(K)$ over p-adic field

2009-12-08T05:17:02Z

I second L Spice's recommendation of the book by Bushnell and Henniart, called "The local Langlands conjectures for GL(2)."

After you master the principal series representations, it's not too hard to tinker with some supercuspidals. Easiest among these are the tamely ramified supercuspidals. To construct these, let's start with the unramified quadratic extension $L/K$, with corresponding residue fields $\ell/k$. Choose a character $\theta$ of $L^\times$ which has these properties:

(a) The character $\theta$ is trivial on $1+\mathfrak{p}_L$ , so that $\theta\vert_{\mathcal{O}_L^\times}$ factors through a character $\chi$ of $\ell^\times$. (b) $\chi$ is distinct from its $k$-conjugate. (In other words, $\chi$ does not factor through the norm map to $k^\times$.)

It's a standard fact that there's a corresponding representation $\tau_\chi$ of $\text{GL}_2(k)$, characterized by the identity $\text{tr}\tau_\chi(g)=-(\chi(\alpha)+\chi(\beta))$ whenever $g\in\text{GL}_2(k)$ has eigenvalues $\alpha,\beta\in\ell\backslash k$. (This is somewhere in Fulton and Harris, for instance.)

Inflate $\tau_\chi$ to a representation of $\text{GL}_2(\mathcal{O}_K)$ , and extend this to a representation $\tau_\theta$ of $K^\times\text{GL}_2(\mathcal{O}_K)$ which agrees with $\theta$ on the center. Finally, let $\pi_\theta$ be the induced representation of $\tau_\theta$ up to $\text{GL}_2(K)$ ; then $\pi_\theta$ is an irreducible supercuspidal representation.

By local class field theory, our original character $\theta$ can be viewed as a character of the Weil group of $L$. In the local Langlands correspondence, $\pi_\theta$ lines up with the representation of the Weil group of $K$ induced from $\theta$. All the supercuspidals of $\text{GL}_2(K)$ arise by induction from an open compact-mod-center subgroup, but the precise construction of these is a little more subtle than the above example.

A hypersurface with many points

2009-12-01T21:31:04Z

Ok, it's time for me to ask my first question on MO.

Consider the affine curve $Y+Y^q=X^{q+1}$ over the finite field $\mathbf{F}_q$ . It's interesting because it has the largest number of points over $\mathbf{F}_{q^2}$ possible relative to its genus, which is $q(q-1)/2$. In other words, this curve realizes the Weil bound over $\mathbf{F}_{q^2}$ . This seems to be well-known in the literature.

Now consider the following hypersurface $\mathcal{X}$ over the finite field $\mathbf{F}_q$ :

$$Z+Z^q+Z^{q^2}=\det\left(\begin{matrix} 0 & X & Y \\ Y^q & 1 & X^q \\ X^{q^2} & 0 & 1\end{matrix}\right)$$

Empirical observation seems to point to the following: The compactly supported cohomology of $\mathcal{X}$ is only nonzero in degrees 2 and 4. In degree 2, the dimension of $H^2(\mathcal{X})$ is $q^2-1$ and the $q^3$-power frobenius acts as the scalar $q^3$. Which is all a fancy way of saying that for all $n$,

$$\#\mathcal{X}(\mathbf{F}_{q^{3n}})=q^{6n}+q^{3n}(q^2-1).$$

Thus $\mathcal{X}$ has the largest number of $\mathbf{F}_{q^3}$-points among any hypersurface with the same compactly supported Betti numbers.

Can anyone help me prove the above formula? (I can do $n=1$ alright...) Bonus points if you can also compute the automorphism group of $\mathcal{X}$. Many more bonus points if you can formulate the generalization to hypersurfaces of higher dimension!

The above hypersurface arises in the study of the bad reduction of Shimura varieties, if anyone cares to know.

EDIT: Admittedly, this is a narrow problem about a very particular surface. Therefore I'm going to accept an answer to the following question: Is there an algorithm to compute the zeta function of a hypersurface of this sort, that's quicker than counting points?

(I've already noticed that it's enough to recur over X and Y, and to test for each pair that the expression on the right lies in the image of the linear map defined by the expression on the left. But I can't think of anything faster than this.)

Answer by Jared Weinstein for SL(2,Z/N)-decomposition of space of cusp forms for Gamma(N)

2009-11-10T06:54:49Z

As usual, once I spot a question on here I have anything useful to say about, somebody has already answered it.

I can sum up that part of my thesis this way: let M be the induced representation of the character (-I) --> (-1)^k of the center up to all of SL(2,Z/NZ). Then S_k(Gamma(N)) is roughly k/12 copies of M, plus some error term which can be given precisely, with some effort.

When you ask instead about Hilbert modular forms over a totally real field K, the "1/12" becomes the absolute value of zeta_K(-1).

Answer by Jared Weinstein for What's the "best" proof of quadratic reciprocity?

2009-10-20T16:46:52Z

The proof involving Gauss sums always seemed the best to me. I'm going to run my own undergrad number theory students through that proof, right after we develop some experience with roots of unity.

If you remove the constraint of accessibility to students of a first number theory course, then you can avoid computations with roots of unity altogether: By Galois theory (and some knowledge of discriminants), a square root of p or -p lives inside of the cyclotomic field of index p. An examination of the action of a Frobenius element at q on this square root relates Legendre(p,q) to Legendre(q,p).

Answer by Jared Weinstein for What is the base change in number theory?

2009-10-12T19:13:02Z

In number theory, base change can also refer to an operation on automorphic representations. If L/K is an extension of number fields, and pi is an automorphic representation of a reductive group G over K, then pi should "lift" to a new automorphic representation of G over L. This is the sense of the phrase used in, e.g., Langlands' book "Base Change for GL(2)". The existence of a certain kind of base change for GL(2) was used to prove the modularity of some mod 3 Galois representations, which in turn played a role in proving Fermat's last theorem.

Comment by Jared Weinstein

2012-05-14T03:27:48Z

@Jef: Yes, the existence of $f$ is proved by purely local means in Chapter 1 of the Arthur-Clozel book on base change. That result is used to characterize automorphic base change. On the other hand, [AC] uses global means to establish the existence of automorphic base change.

Comment by Jared Weinstein

2012-04-25T16:33:29Z

Dror: This is top-rate work (and certainly answer-worthy). I suspected there wouldn't be a clean formula for $\pi$, but at least now I know there's an algorithm.

Comment by Jared Weinstein

2011-12-09T18:34:28Z

Q1 is the same question as mathoverflow.net/questions/3871/…, where the answerer noted that the residue fields of maximal ideals not arising from points are necessarily more exotic than $\mathbf{C}$ (or $\mathbf{R}$ in the context of the current post).

Comment by Jared Weinstein

2011-10-19T04:37:43Z

@Joel: Please help us poor number-theorists understand your comment. If ZFC is consistent, is it not the case that one could construct a model of it whose N (defined as the minimal set which contains {} and which is closed under successor) contains non-standard elements? If that's the case, why are you calling any N in a model of ZFC a standard model of arithmetic?

Comment by Jared Weinstein

2011-09-19T01:14:08Z

I had in my mind a scene where I am presenting this bogus argument on a blackboard to a skeptical, wincing audience.

Comment by Jared Weinstein

2011-05-13T07:51:38Z

Re your addendum: Yes, now it is much clearer, thanks. This is very natural; it is probably what Tate had in mind.

Comment by Jared Weinstein

2011-05-13T07:20:47Z

Ah! I should read more carefully. I retract my comment. I like how your construction requires that $r$ be odd, which is what we expect. I also like that it requires passing to the degree 2 extension, which is also what we expect! You're probably correct about this, but I'm still having trouble understanding your argument. Isn't $L$ the intersection of $X$ with the linear subspace defined by $x_{2j+1}=\zeta x_{2j}$ $(j=0,\dots,i)$, and therefore isn't $L$ necessarily a power of the hyperplane section?

Comment by Jared Weinstein

2011-05-13T06:23:13Z

Hi Torsten, I'm a bit confused. That linear span will contain points such as $(a:a\zeta+b:b\zeta:0:\cdots)$ and these do not lie on $X$ in general. I agree that Shioda et al should have all the answers; it seems that they work primarily in characteristic 0.

Comment by Jared Weinstein

2011-02-09T20:45:26Z

I would like to typeset Deligne's letter if no one else has begun doing so. I agree we should seek his permission again, but this will be easy because he is probably having tea downstairs at the moment.

Comment by Jared Weinstein

2011-02-06T20:55:05Z

I have wanted to see this letter also, so thank you. Now it just remains for KConrad to post a translation of the text...

Comment by Jared Weinstein

2011-01-15T00:12:10Z

Yes, I should have addressed the CM issue in my answer. These other curves are going to have CM by $\mathbf{Z}[ni]$. In fact, the OP is essentially just asking for the minimal polynomial of $j(ni)$, which generates a ray class field of $\mathbf{Q}(i)$. I don't know a quick way to do this, but maybe it's addressed in Cox' book, "Primes of the form $x^2+ny^2$."

Comment by Jared Weinstein

2010-12-01T04:41:15Z

No no, I really meant the basis part of the question! Given a finite set B of algebraic numbers (described, let's say, using their minimal polynomials, together with sufficiently many decimal digits), and another algebraic number x, I thought it could be decided in finite time whether x belongs to the span of B. Certainly it can be decided whether x belongs to the field generated by B (by factoring the minpoly of x over that field), and the rest is linear algebra. What am I missing?

Comment by Jared Weinstein

2010-12-01T03:43:33Z

"At each stage of his construction, he is asking a negated existential question 'is this number not in the span of the earlier numbers?', which we cannot expect to answer computably in finite time." Joel, can you comment on why this is so? Let's suppose that after every iteration of Kevin's construction, you throw in enough new elements so that what you have is the basis for a field, call it K. Then you look up the next polynomial on your list; you have to decide whether each of its roots belong to K or not. But this can be accomplished using the LLL algorithm.

Comment by Jared Weinstein

2010-11-23T19:18:29Z

Thank you for this, Torsten. The problem is still open, but your comments have helped me generalize this construction to the situation where $U$ is replaced by a more general type of unipotent group. (I knew about the action of the group of roots of unity.) I have yet to absorb the significance of Deligne-Fourier transforms to this situation. Can you see any way to at least compute the dimension of the cohomology?

Comment by Jared Weinstein

2010-11-11T16:20:22Z

@François, it is indeed true that $X(\mathbf{F}_{q^n})=U(\mathbf{F}_{q^n})$, so that $\#X(\mathbf{F}_{q^n})=q^{n^2}$. The equation determining $X$ can be written explicitly as one equation in the $n$ variables. When $n=3$, this equation is $\det M=0$, where $M=\begin{pmatrix} x_1^{q^3}-x_1 & x_2^{q^3}-x_2 & x_3^{q^3}-x_3 \\ 1 & x_1^q & x_2^q \\ 0 & 1 & x_1^{q^2} \end{pmatrix}$.