User dror speiser - MathOverflow

What is the algebraic closure of the field with one element?

2009-11-23T02:03:27Z

If doing geometry over $\mathbb F_p$ means also using its algebraic closure, it must be interesting to talk about the algebraic closure of $\mathbb F_1$ - the field with one element.

I saw that the finite extensions of $\mathbb F_1$ are considered as $\mu_n$, but an article by Connes et al says that it is unjustified to think of the direct limit of these. In their paper, the group ring $\mathbb Q[\mathbb Q/\mathbb Z]$ appears a lot. Maybe it's one of $\mathbb Q/\mathbb Z$, $\mathbb Q[\mathbb Q/\mathbb Z]$, $\mathbb Z[\mathbb Q/\mathbb Z]$ ?

What is the algebraic closure of the field with one element?

And then, what is $\overline{\mathbb F_1} \otimes_{\mathbb F_1}\mathbb Z$? This seems like a very interesting question...

When does a modular form satisfy a differential equation with rational coefficients?

2013-03-27T02:22:45Z

Given a modular form $f$ of weight $k$ for a congruence subgroup $\Gamma$, and a modular function $t$ with $t(i\infty)=0$, we can form a function $F$ such that $F(t(z))=f(z)$ (at least locally), and we know that this $F$ must now satisfy a linear ordinary differential equation $$P_{k+1}(T)F^{(k+1)} + P_{k}(T)F^{(k)} + ... + P_{0}(T)F = 0$$

Where $F^{(i)}$ is the i-th derivative, and the $P_i$ are algebraic functions of $T$, and are rational functions of $T$ if $t$ is a Hauptmodul for $X(\Gamma)$.

My question is the following:

given a modular form $f$, what are necessary and sufficient conditions for the existence of a modular function $t$ as above such that the $P_i(T)$ are rational functions?

For example, the easiest sufficient condition is that $X(\Gamma)$ has genus 0, by letting $t$ be a Hauptmodul. But, this is not necessary, as the next condition will show.

Another sufficient condition is that $f$ is a rational weight 2 eigenform. I can show this using Shimura's construction* of an associated elliptic curve, and a computation of a logarithm for the formal group in some coordinates (*any choice in the isogeny class will work).

Trying to generalise, I have thought of the following: if $f$ is associated to a motive $h^i(V)$ of a variety $V$, with a pro-representable Artin-Mazur formal group $\Phi^i(V)$ of dimension 1, then we can construct formal group law a-la Stienstra style, and get a logarithm using the coefficients of powers of a certain polynomial. This makes the logarithm satisfy a differential equation with rational functions as coefficients. Since the dimension is 1, the isomorphism back to "modular coordinates" will be a single modular function $t$, and this answers the question positively.

Now, some people, without naming names, believe that rational eigenforms should correspond to the middle cohomology of certain rational Calabi-Yai varieties. I'm not entirely certain that such people exist. Probably. If this is true, then this should answer my question for rational eigenforms.

Putting non-eigenforms aside, since I'm not interested as much in them, we are left with non-rational eigenforms. We can try to perform the same Stienstra construction, but this time we get that the galois orbit of $f$ is associated to a "formal group law" of a motive with dimension greater than one. This will make for an interesting recurrence for the vector of the galois orbit, but not necessarily for each form individually, as the isomorphism of formal groups laws (between Stienstra's and those with the modular forms as logarithm) might scramble them together. Maybe not, and this solves might the question. I realise this last paragraph might be difficult to understand, for the wording is clumsy, and the mathematical notions are even worse. If you're really interested in this, I'd be happy to elaborate.

Class number measuring the failure of unique factorization

2010-01-06T17:05:32Z

The statement that the class number measures the failure of the ring of integers to be a ufd is very common in books. ufd iff class number is 1. This inspires the following question:

Is there a quantitative statement relating the class number of a number field to the failure of unique factorization in the maximal order - other than $h = 1$ iff $R$ is a ufd?

In what sense does a maximal order of class number 3 "fail more" to be a ufd than a maximal order of class number 2?

Is it true that an integer in a field of greater class number will have more distinct representations as the product of irreducible elements than an integer in a field with smaller class number?

Points on Deligne-Lusztig varieties: Interpreting Borels in relative position as flags with conditions

2012-07-09T22:43:43Z

Background

I am studying the paper "On the Green polynomials of classical groups" by Lusztig, in which he computes the values of the Deligne-Lusztig representation, corresponding to a Coxeter element of minimal length in a classical group, on unipotent elements. I am interested in computing the values of representations not corresponding to a Coxeter element of minimal length. (Note that this is done in a vast generalisation in later work by Lusztig, and in the work of Shoji. But I am not in a place to be able to use their methods)

Question

Let $G$ is a classical group defined over a finite field with frobenius morphism $F$, $w$ an element of the Weyl group, and let $X(w)$ be the Deligne-Lusztig variety - all Borel subgroups $B$ of $G$ such that $B$ and $F(B)$ are in relative position $w$.

My specific question is: how do I translate this definition into the language of flags? I.e. I would like an alternative definition for $X(w)$ as the variety of flags satisfying some conditions involving to $F$.

In the original Deligne-Lusztig paper, my question is answered, for the case $w$ a Coxeter element of minimal length, in a short section.

Is the Gelfand-Graev character isomorphic to a cohomology group for some sheaf on a Deligne-Lusztig variety?

2011-11-02T12:20:13Z

Deligne-Lusztig theory

is awesome. You take a maximal torus $T$, you take a character $\theta$, construct a variety $X_T$$^*$, take etale cohomology, get a virtual character $R_T^\theta$, maybe it's reducible, so you try to decompose it.

Gelfand-Graev character

is awesome. You take a maximal unipotent subgroup in some maximal split Borel subgroup, take a generic character, induce the character to the whole group, and you get many interesting subrepresentations.

My question

Is the Gelfand-Graev character equal to the character of the cohomology of some sheaf on some nice variety, similar to a Deligne-Lusztig character?

Why is this interesting?

Say you have an $R_T^\theta$ that is reducible. Before trying to find explicitly all constituents, let's try to decompose it first into constituents of the Gelfand-Graev character (generic), and the rest (not generic). If $R_T^\theta$ has exactly two subrepresentations, one generic and one not generic, then we needn't look further.

What am I looking for?

The best thing would be if there was a sheaf $F_{GG}$ on $X_T$ with cohomology, in the $\ell(w)$-degree, with character equal to the Gelfand-Graev character. Then we might have a sequece of sheaves $$0\rightarrow F' \rightarrow F_\theta \rightarrow F_{GG} \rightarrow F^{''} \rightarrow 0$$

and we might get that the cohomologies of $F'$ and $F^{''}$ will break our $R_T^\theta$ into two parts.

So, in essence, what I'm looking for, is a geometric way to break a Deligne-Lusztig character into its genereic and non-generic parts.

This might not be possible, at least not in the way I described, which is very naive and wishful. The sentence before last should be regarded as the real question.

(*) Non-standard notation, I know. Fix some maximal $F$-stable torus, let $w$ be the Weyl element that twists the torus to desired $T$, and let $X_T=X(w)$, where $X(w)$ is the standard notation Deligne-Lusztig variety.

Galois representations attached to Maass form

2012-02-29T11:27:54Z

So, how does one construct a galois representation from a Maass form?

For a modular cusp eigenform, I am familiar with the work of Eichler-Shimura, Deligne, Deligne-Serre, and realize these are different situations because of the involved geometry. I am also familiar with Maass's construction of Maass forms of weight zero from Hecke characters on real quadratic fields, so I can reverse this to answer a tiny bit of my question. There is also Langlands-Tunnell, which I am not familiar with. Finally, I realize that most Maass forms are not conjectured to be associated to galois representations.

Searching the web did not yield much. But I do want to ask an interesting precise question, so here it is:

Is there an infinite family of Maass eigenforms, such that an irreducible galois representation of infinite image is constructed to each form, and these do not somehow arise from Maass's original construction or Langlands-Tunnell?

If not, is there a conjectural association that has been checked (without proof) computationally?

When is a product of elliptic curves isogenous to the Jacobian of a hyperelliptic curve?

2010-08-09T23:28:50Z

David's question Families of genus 2 curves with positive rank jacobians reminded me of a question that once very much interested me: when is a product of elliptic curves isogenous to the jacobian of a hyperelliptic curve?

How is this related to David's question? Well, if we can multiply two elliptic curves over $\mathbb{Q}(t)$ with large rank, and the result is isogenous to the jacobian of a hyperelliptic curve, then this will probably produce record families answering David's question, i.e. genus two curves with very large rank. It is also interesting for all genera, so don't restrict answers to 2. On the other hand, answers containing arithmetic information, for example on elliptic curves over the rationals, are more than welcome.

Answer by Dror Speiser for Ring of algebraic integers in a quadratic extension of a cyclotomic field

2011-12-17T14:35:48Z

As requested by Franz, here is the short Magma code looking for solutions in cyclotomic units:

test_n := function(n)
    K<z> := CyclotomicField(n);
    O := MaximalOrder(K);
    I := ideal<O|4>;
    R := quo<O|I>;
    G,p := MultiplicativeGroup(R);
    p1 := p^-1;
    H := sub<G|[p1(z)] cat [p1(c) : c in CyclotomicUnits(K)]>;
    return [p : p in PrimeDivisors(n) | p ne 2 and p1(1-z^(n div p)) in H];
end function;

I have run this for $n$ up to $171$, always returning an empty set. This is naive code and can be speeded up like Franz says.

Given David's answer, it is now interesting to look at the even case. Here there are in fact many solutions. For $n\le 100$ there are solutions when $n$ is 28, 56, 60, 92, with $p$ being 7, 7, 3, 23, respectively.

Class number of non-maximal order in imaginary quadratic function field?

2010-07-15T18:37:21Z

It is well known that for $K=\mathbb{Q}(\sqrt{D})$, $D < 0$, the non-maximal order of squarefree conductor $f$, relatively prime to $D$, has class number $$h_K \prod_{p|f} (p-(\frac{D}{p}))$$

What is the class number of a non-maximal order in an imaginary quadratic extension of $\mathbb{F}_p[t]$? Is it proven using ideas involving zeta function, as appears in many books for the case above?

Finite field analogue of representations in same packet have equal central character

2011-09-07T15:39:38Z

In Kevin Buzzard's recent question, a warm up question was: if two automorphic representations are nearly equivalent, then are the central characters of their local components equal?

Working my way up to local and later global automorphic representations, I am currently studying the situation over finite fields.

What is the analogue of Buzzard's question in the finite field case?
Is it true?

Here are some of my own thoughts on this.

If $$(T_1,\theta_1)\sim (T_2,\theta_2)$$ are two geometrically conjugate pairs of torus+character, then I think the characters have to agree on the center of $G^F$ (elements in the center are norms [?], so the geometric conjugacy equation shows equality). Using 7.2 in [DL] we see that the value of $R_T^\theta$ on the center is $\theta$ (maybe up to sign), i.e. $$\frac{R_{T_1}^{\theta_1}(z)}{R_{T_1}^{\theta_1}(1)} = \frac{R_{T_2}^{\theta_2}(z)}{R_{T_2}^{\theta_2}(1)}=\theta(z)$$ for $z\in Z(G^F)$.

This might be considered an analogue as requested, but a naive one at that. A deeper analogue should consider the irreducible representations of $G^F$, and not the Deligne-Lusztig virtual characters, which can be reducible.

Consider the two cross sections $\rho_x$, $\rho_x^'$ from the set of geometric conjugacy classes to irreducible representations (10.7.1/2 in [DL]): $$\rho_x=\sum_{[(T,\theta)]=x} \frac{(-1)^{\sigma(G)-\sigma(T)}}{\lt R_T^\theta,R_T^\theta\gt }R_T^\theta$$ $$\rho_x^'=(-1)^{\sigma(G)-\delta_x} \sum_{[(T,\theta)]=x} \frac{1}{\lt R_T^\theta,R_T^\theta\gt }R_T^\theta$$

Do these have the same central character?

If the $\theta$'s are trivial on the center, then the answer is yes. Computations that I have done before show that this is the case for the $\theta$'s that are not in general position in $Sp_4$.

We can divide in an obvious way the $[(T,\theta)]\in x$ into two sets, of "positive" and "negative", such that $\rho_x^'$ is a sum and $\rho_x$ is a difference. We see that for the two representations to have equal central character, either the sum of "negative" terms, or "positive" terms (depending on $\delta_x$), must be zero on the center (minus the identity).

Note that $\rho_x$ appears in the Gelfand-Graev representation of $G^F$, whose character restricted to the center (minus the identity) is zero, so this supports in spirit the paragraph above. I'm not sure if it can be extended from spirit to an actual proof.

[DL] - this is, of course, the original Deligne-Lusztig paper from 1976.

Why is there a weight 2 modular form congruent to any modular form

2011-07-08T11:47:16Z

I got my copy of Computational Aspects of Modular Forms and Galois Representations in the mail yesterday. The goal of the book is "How one can compute in polynomial time the value of Ramanujan's tau at a prime", well, or any other modular form of level 1. It's all very thrilling!

The following fact is essential: for any modular form $f$ of level 1, and any prime $l$, the mod $l$ reduction of the semisimplification of the galois representation attached to $f$ by Deligne is a 2-dimensional subrepresentation of the galois representation of the $l$-torsion of the jacobian of the modular curve of level $l$.

Why?

From what I understand, this is somewhat equivalent (after Shimura and Deligne) to there being a modular form of weight 2 and level $l$ that is congruent to $f$ mod $l$, or something similar. Is this the right statement? Why is it true then?

Searching far and wide for an introduction to this topic yields very little.

Answer by Dror Speiser for Why does the definition of modularity demand weight 2?

2011-08-15T00:15:03Z

$\newcommand\Q{\mathbf{Q}}$ $\newcommand\Qbar{\overline{\Q}}$ $\newcommand\Gal{\mathrm{Gal}}$ $\newcommand\C{\mathbf{C}}$ $\newcommand\Sym{\mathrm{Sym}}$ $\newcommand\E{\mathcal{E}}$ $\newcommand\Betti{\mathrm{Betti}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Hom{\mathrm{Hom}}$ $\newcommand\T{\mathbf{T}}$ To answer this question, it might be best to start with the following:

Q. What do the Galois representations attached to a variety know about the variety?

In order make this more precise, let us introduce some notation. Fix a prime $p$ and a non-negative integer $n$. Let $X$ be a proper smooth scheme over $\Q$, and let $V = H^n_{et}(X/\Qbar,\Q_p)$ denote the $n$th etale cohomology group of $X$. The basic and fundamental properties of etale cohomology tell us that:

$V$ is a vector space of dimension $H^n_{\Betti}(X(\C))$, where $H_{\Betti}$ denotes Betti (or singular) cohomology, and $X(\C)$ denotes the complex points of $X$ thought of as a topological manifold.
$V$ (with the $p$-adic topology) has a continuous action of $G_{\Q}:=\Gal(\Qbar/\Q)$.

Grothendieck and Serre further conjecture that the $G_{\Q}$-representation $V$ is semi-simple. The strongest possible conjecture one might make is to ask whether the functor from smooth projective varieties over $\Q$ to semi-simple $G_{\Q}$-representations (or the collection of all such representations for $n \le 2 \cdot \mathrm{dim}(X)$) is fully faithful. However, this is too much to ask, for the following reasons.

(i). The target category is semi-simple, but the category of varieties is far from semi-simple. (In particular, the existence of a map $X \rightarrow Y$ does not imply the existence of a non-trivial map $Y \rightarrow X$.)

(ii). Varieties built in a combinatorial way from projective spaces (think toric varieties) tend to have etale cohomology groups indistinguishable from products of projective spaces. This is because their cohomology groups are generated by geometric cycles, on which Galois acts in a well understood way (essentially by some power of the cyclotomic character).

These are - in some sense - manifestations of the same reason: A correspondence in $X \times Y$ gives rise to a cohomology class in $H^*(X \times Y)$; then by the Künneth formula, this leads to a relation between the cohomology of $X$ and $Y$ even when there is not necessarily any non-trivial map from $X$ to $Y$ (or vice versa). In order to account for this, one can try to take the quotient category of the category of algebraic varieties in which one is allowed to "break up" smooth proper varieties into pieces given the existence of certain correspondences on $X$. There are a variety of ways in which one might do this. Conjecturally, these constructions are all essentially the same, and the corresponding category is the category of pure motives. The Tate conjecture now says that etale cohomology is a fully faithful functor from pure motives to semi-simple $G_{\Q}$-representations.

Example If $E$ is an elliptic curve over $\Q$, and $n = 1$, then the etale cohomology group $V$ is the (dual) of the usual representation attached to the $p$-adic Tate module of $E$. Suppose that $E'$ is another elliptic curve over $\Q$ with first etale cohomology group $V'$. For curves, the theory of "motives" is essentially the theory of abelian varieties. (More generally, the theory of $H^1$ is essentially the theory of abelian varieties, since, for any proper variety $X$, there is an isomorphism $H^1(X) \simeq H^1(A(X))$, where $A(X)$ is the Albanese of $X$.) Tate's conjecture in this case says that $$\Hom(E,E') \otimes \Q_p \rightarrow \Hom_{G_{\Q}}(V,V')$$ is an isomorphism. This is how you will see the Tate conjecture stated for elliptic curves, for example, in AOEC. The Tate conjecture for abelian varieties is a theorem of Faltings. (Suggestion: to understand what the Tate conjecture really is about, and why it is hard, you should really think about the special case of Elliptic curves.)

If we now return to our question, we can (tautologically) say the following: assuming the Tate conjecture, the etale cohomology knows about the motive corresponding to the original variety. What does that really mean? One way of thinking about motives is as a ``universal cohomology theory''. In particular, we can recover from the motive not only the etale cohomology groups, but also the algebraic de Rham cohomology groups. Recall that de Rham cohomology is another cohomology theory that gives vector spaces of the "correct" dimension for a smooth proper variety $X/\Q$. The de Rham cohomology groups do not have associated Galois representations, but they do have a Hodge filtration. Over $\C$, if one takes the associated graded of the Hodge filtration, one recovers the Hodge decomposition: $$H^n_{dR}(X,\C) = \bigoplus_{p+q=n} H^{p}(\Omega^q_X).$$ The dimensions of the latter space are called the Hodge numbers $h^{pq}$. So, assuming the Tate conjecture, from $V$ we can recover the underlying motive, from which we may reconstruct the de Rham cohomology, and then the Hodge numbers. The Tate conjecture seems to be very hard. However, Grothendieck asked the following: given $V$, can we directly recover the (algebraic $p$-adic) de Rham cohomology along with its filtration without first constructing the motive? This was a great question, and the answer (yes!) constitutes one of the major achievements of $p$-adic Hodge Theory. I can do no more than give a cartoon description here. In order to do so, first recall the much more classical story connecting de Rham cohomology to Betti (singular) cohomology. These groups can both naturally be defined as vector spaces over $\Q$ (one has to define de Rham cohomology in the correct way), but the isomorphism relating these spaces comes from integrating forms over cycles. Yet these integrals are typically transcendental numbers, so to pass from Betti to de Rham cohomology one first has to tensor with a field bigger than $\Q$ which contains all these periods (usually, one simply tensors with $\C$). In order to pass from etale cohomology to algebraic de Rham cohomology, one might ask for a period ring in which we can compare both groups. In this refined setting, the period ring should both have a Galois action and a filtration. The most basic verion of a period ring is $B_{HT}$, specifically, $$B_{HT} := \bigoplus_{\Z} \C_p(n),$$ where $\C_p$ is the completion of $\Qbar_p$, and $\C_p(n)$ is $\C_p$ twisted (as a local Galois module) by the $n$th power of the cyclotomic character. The ring $B_{HT}$ has a natural filtration (indeed, it is even graded). Now we can consider $$D_{HT}(V) = (V \otimes B_{HT})^{\Gal(\Qbar_p/\Q_p)}.$$ The Galois group acts on both $V$ and $B_{HT}$. The result is a graded (and so filtered) module. On the other hand, one can also consider the ring $B_{HT} \otimes H^n_{dR}(X/\Q_p)$, where there is a natural way to make sense of the corresponding filtration. An important theorem of Faltings then says that $$H^{n}(X/\Qbar_p,\Q_p) \otimes B_{HT} = H^n_{dR}(X/\Q_p) \otimes B_{HT},$$ and $D_{HT}(V) = H^n_{dR}(X/\Q_p)$. In particular, from a geometric Galois representation, we can recover the Hodge filtration and the Hodge numbers.

Modular Forms. The Eichler-Shimura isomophism relates modular forms of weight $k \ge 2$ to $H^1(X_0(N),\Sym^{k-2}\Q)$. If $k = 2$, this is just $H^1(X_0(N),\Q)$. The Hecke algebra $\T$ acts on $H^1_{\Betti}(X_0(N),\Q)$, and (since it is constructed functorially) also on the etale cohomology $H^1(X_0(N),\Q_p)$. Now the Hodge decomposition of $H^1$ is $H^1 = H^{0,1} \oplus H^{1,0}$, where $h^{0,1} = h^{1,0}$ is the genus of $X_0(N)$. The Hecke algebra breaks up the cohomology into two dimensional pieces corresponding to the Galois representations associated to eigenforms; it turns out that each two dimensional piece contains one dimension from $H^{0,1}$ and one dimension from $H^{1,0}$. The result of Faltings above tells us that we can read off that $h^{0,1} = h^{1,0} = 1$ directly from the Galois representation.

For $k > 2$, recall that (technical issues aside) there is a universal elliptic curve $\E \rightarrow X_0(N)$. The Kuga-Sato variety is (again, roughly) The $k-1$ dimensional variety $K = \E \times_X \E \ldots \times_X \E$ where $X = X_0(N)$. There is a natural map $\pi: K \rightarrow X$. The local system $\Sym^{k-2}(\Q^2_p)$ is trivialized over $K$, and so, using the proper base change theorem, Deligne shows that $H^1(X_0(N),\Sym^{k-2}\Q_p)$ is a sub-quotient of the cohomology group $H^{k-1}(K,\Q_p)$. (Warning: this requires more than simply a formal cohomological argument, it also requires some trickiness with weights to show that terms on different diagonals the Leray spectral sequence don't "mix", and hence the sequence degenerates.) The Galois representation associated to a modular form is now a two-dimensional piece of $H^{k-1}(K,\Q_p)$. Faltings proves that the corresponding "piece" of de Rham cohomology seen by this representation is $H^{0,k-1} \oplus H^{k-1,0}$. In particular, the representation has Hodge numbers $h^{0,k-1} = 1$ and $h^{k-1,0} = 1$.

Given a Galois representation $V$, one can twist $V$ by the cyclotomic character. How does this effect the Hodge decomposition? One can compute this on the Hodge side by seeing what happens to the cohomology of $X \times \mathbf{G}^1_m$ and comparing with the Künneth formula. It turns out that $h^{p,q}(V(n)) = h^{p-n,q-n}$. Thus, if only know $V$ up to twist, we still recover some information about the Hodge numbers.

Returning to modular forms. The coefficients $a_p$ determine the Galois representation, by Cebotarev. A modular form of weight $k$ has Hodge numbers $h^{0,k-1} = h^{k-1,0} = 1$. The determinant of the representation is the $k-1$th power of the cyclotomic character (up to a finite character) which can be read off from the "degree". By twisting, we can easily change the determinant, and change the Hodge numbers to $h^{-d,k-d-1} = h^{k-d-1,-d} = 1$. Yet, it is clear that we cannot twist so that $h^{1,0} = h^{0,1} = 1$ unless $k = 2$. Thus, given a modular form of weight $k > 2$, it cannot be associated to an elliptic curve even after twisting. This is Kevin's answer.

Secondly, any motive has (conjecturally) an $L$-function. The recipe of building this $L$-function breaks up into two parts. The first involves the factors at finite primes, which give rise to the Euler product. The second involves the infinite primes, which give rise to Gamma factors. The information at $\infty$, however, (by Tate's conjecture) can be read off from the Galois representation, and the recipe of Deligne shows that it will exactly depend on the Hodge numbers of the motive, and visa versa. Moreover, twisting by $\epsilon^k$ some power of the cyclotomic factor has the effect of replacing $L(s)$ by $L(s+k)$ (and shifting the corresponding central value) In particular, given an elliptic curve, one knows the Gamma factors (because one knows the Hodge decomposition of $E$), and one sees that even after twisting one cannot get Gamma factors that "look like" the Gamma factors associated to a modular form of weight different from $2$. This is GH's answer.

More generally, arithmetic conjectures of Langlands type imply that all motives should be "automorphic", and that the Hodge structure of the motive determines the infinity type of the automorphic form, which in turn determines the Gamma factors. So, at least morally, given a pure irreducible motive $V$, we know that if it is automorphic, it must be automorphic of a particular weight determined by the underlying geometry of $V$.

Of course, even before the result of Faltings, one had enough faith in terms of how these things were connected to be very confident that Elliptic curves over $\Q$ should correspond exactly to weight two forms - GH's remark that "It was an experimental fact that the gamma factors are always the same, hence the precise form of the modularity conjecture was formulated, which then turned out to be right, namely it was proved by great efforts of great mathematicians" seems spot on.

Related problems. Given a modular eigenform $f = \sum a_n q^n$ of weight four (in the arithmetic normalization), one can ask: is it possible to show that there does not exist a weight two modular form $g = \sum b_n q^n$ where $a_p = p b_p$ for all primes $p$ without using $p$-adic Hodge theory? I think this is not so easy. For example:

(i) The arithmetic approach: The weight $4$ form $f$ would have the property that it is not ordinary at every prime, since clearly $a_p = p b_p \equiv 0 \mod p$. One conjectures that a set of primes of density one are modular (or $1/2$ if $f$ has CM). Yet it is still unknown whether any form of weight $\ge 4$ has even a single ordinary prime.

(ii) The analytic approach: What do the distributions of coefficients of weight $2$ and $4$ forms look like? Sato-Tate says that the normalized coefficients satisfy a precise distribution (now a theorem!). Yet the "normalized" coefficients of $f$ and $g$ are by construction exactly the same, so Sato-Tate says nothing. In particular, it is hard to see any analytic estimates of functions involving the $a_p$ being able to distinguish two classes of numbers with the same underlying distribution. A related argument: The Hasse bound in weight four is satisfied by $p b_p$ if and only if the weight two Hasse bound is satisfied by $b_p$.

Summary. Conjectures arise organically from heuristics and computation. I was "known" that Elliptic curves should be associated to weight two forms long before one could actually formally prove that they weren't associated to twists of weight $4$ forms. To prove the latter fact, one has to use $p$-adic Hodge theory.

(* I am not sure about a lot here, so I'm putting this community wiki. Also, there is no mention of weight 1 or half integers. Change whatever needs changing, or in the extreme cases, peacefully leave a comment to delete...)

When is induction to Siegel parabolic of cuspial representation reducible

2011-08-01T22:13:52Z

Hello,

I'm currently looking at the symplectic group $Sp_4(\mathbb{F}_q)$ ($q$ odd), trying to understand some of its comlpex representation theory. I note that it has been completely calculated by Srinivasan (1968). A partial description appears in "Seminar on Algebraic Groups and Related Finite Groups", Springer et al (1970). My question regards what is written in the second to last section of the description: my calculations turn out a tiny bit different from theirs.

There are two parabolics, $P_1$, $P_2$, with Levi decompositions $P_i=M_iU_i$, and $$M_1 \simeq SL_2\times GL_1\,\ M_2 \simeq GL_2$$

Let $\tau$ be an irreducible cuspidal representation of $GL_2$, and $\hat\tau$ be its lift to $P_2$ (by way of $M_2$). Let $\theta$ be the corresponding character of the anisotropic torus $T$ in $GL_2$ (well, one of them, doesn't matter which). Let $\rho$ be the representation $Ind_P^G \hat{\tau}$

Enough notation:

If I did my first exercise correctly, a simple application of Mackey theory shows that $\rho$ is reducible if and only if the central character of $\tau$ is trivial.

If I did my second exercise correctly, the central character of $\tau$ is trivial if and only if $\theta^{q+1}\equiv 1$ (the center being exactly the $q+1$ powers), if and only if $\theta^{-1}=\theta^q$.

This is not what it is written in "Seminar on". Instead, the condition $\theta=\theta^q$ is claimed. In fact, if I did a third exercise correctly, if $\theta=\theta^q$ then the virtual character $R_T^\theta$ is reducible and $\theta$ doesn't actually give rise to an irreducible cuspidal representation of $GL_2$.

What is the correct condition on $\theta$ for $\rho$ to be reducible?

Is the Brauer group of a surface an elliptic curve?

2011-06-14T22:46:08Z

Of course not.

But after reading a bit, some points make me believe it should be:

Let $S$ be a nice$^{*}$ surface defined over $Spec\ \mathbb{Z}$.

The Brauer group $Br(S\otimes \bar{\mathbb{Q}})$ is an abelian divisible group,
It is also a $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ module,
For good primes there are reductions $Br(S)\rightarrow Br(S\otimes \mathbb{F}_q)$,
These $Br(S\otimes \mathbb{F}_q)$ are finite,
There is a formal Brauer group $\hat{Br}(S)$ of dimension 1,
The coefficients of $\hat{Br}$, in suitable natural coordinates, relate to $|Br(S\otimes \mathbb{F}_q)|$.
There are some examples where the associated L-function comes from a modular form (of weight 3). I'm not sure if this is conjectured (let alone known) in general.

Since the Brauer group observes many characteristics of an abelian variety (all properties) of dimension 1 (properties 5 and 7 [weight isn't two, but it's the right space]), my vague question is: how far is it from actually being a variety?

There are some easy examples of $S$ with $|Br(S\otimes \mathbb{F}_q)|$ varying between $1$ and $4(q-4)$, as $q$ varies over the primes. This is a clear point of departure from elliptic curves and varieties in general.

Maybe there's a family of natural galois-module homomorphisms into certain abelian varieties defined over $\mathbb{Q}$, commuting with the reduction maps and restriction (or some other appropriate term) to formal groups?

What's going on with these Brauer groups?

$^*$ say a K3 surface. Something that (1) is true for (so not a rational surface) and (4) is proven for.

Computing the dimensions of representations in a reducible induced representation

2011-06-12T22:07:40Z

This is a question on math.se that got no answers.

1) Is there a relatively general method of computing the dimensions of representations in a reducible induced representation?

An explicit specific example is: $G=Sp_4(\mathbb{F}_q)$, $J$ the antidiagonal $(1,1,-1,-1)$, $P$ the parabolic corresponding to $4=2+2$, $\tau$ a representation of $P$ that comes from a representation of $GL_2(\mathbb{F}_q)$, which will also be denoted $\tau_0$ (the matrix $[[g,*],[0,*]]$ acts as $\tau_0(g)$). Mackey's irreducibility criterion shows that $Ind_P^G \tau$ is reducible iff the central character $\omega_{\tau_0} =1$. How does one compute the dimensions of the (two) subrepresentations in this case? Note that this specific case is solved: "The characters of the finite symplectic group $Sp(4,q)$", B. Srinivasan, but the computations there are quite elaborate and not in the spirit of the question.

In the above example a short attempt to use Deligne-Lusztig theory seems to fail since the irreducible subrepresentations seem to be geometrically conjugate. Maybe the short attempt is ridiculous.

2) Is there a way to compute the dimensions, or even the characters themselves, using Deligne-Lusztig theory? (i.e. same question only restricted to reductive groups over finite fields)

How is this surface related to the square of that CM elliptic curve?

2011-05-15T15:51:35Z

I have come across the following surface: let $X$ be the double covering of $\mathbb{P}_\mathbb{Z}^2$ defined by the equation $$y^2=x_0^6+x_1^6+x_2^6$$ where $y$ is a variable of degree 3.

There is an obvious action of $\mu_6 \times \mu_6$.

Less obvious, I have made some short calculations that show a great link between this surface over $\mathbb{F}_p$ and $E^2$ over $\mathbb{F}_p$, for any prime $p$, where $E$ is the well-known CM elliptic curve $$y^2=x^3+1$$

Without getting too much into specifics, the surface's Brauer group seems to give rise to a modular representation coming from a weight $3$ cusp form, that seems to be a modular form also attached to $E^2$ (in the same way), maybe up to a character (of degree 6). This is just some very simple mod $p$ calculations and strong multiplicity one. But the question isn't about the validity of these calculations, so I don't want you to dwell on this.

The described link makes me guess that the surface is geometrically (and not only arithmetically) related to the squared elliptic curve. Are they birationally equivalent? Maybe after dividing $E^2$ by some subgroup of $\mu_6 \times \mu_6$? How does one begin to check this in sage?

Answer by Dror Speiser for Odd powers of the theta function as eigenforms

2011-04-25T19:20:12Z

The case of even powers was (asked and) solved by P. T. Bateman in "Problem E 2051", Amer. Math. Monthly 76 (1969), the solution being that the powers 2, 4, 8 are the only even powers of $\Theta$ that are eigenforms. A similar idea shows that indeed the only odd powers of theta that are eigenforms are 1, 3, 5 and 7:

Proposition. If $k$ is an odd integer greater than $7$, then $\Theta^k$ is not an eigenform for $T_{3^2}$.

Proof. The idea is to compute explicit expressions for the first two fourier coefficients of $\Theta^k$ and $T_{3^2} \Theta^k$, depending only on $k$. For a fixed $n$, in order to compute the $n$-th coefficient of $\Theta^k$ we consider "square-partitions" of $n$ - all the ways to write $n$ as a sum of squares of positive numbers, regardless of order.

Each such partition contributes to all large enough powers of $\Theta$, specifically, as soon as the power is greater than the number of summands in the partition. We can calculate exactly how much a square-partition contributes to a coefficient. Namely, the partition: $$n = \alpha_1 \cdot n_1^2 + \alpha_2 \cdot n_2^2 + \ldots + \alpha_r \cdot n_r^2$$

where the $n_i$ are distinct, constributes to the $n$-th coefficient of $\Theta^k$: $$2^{\alpha_1+\ldots+\alpha_r} \binom{k}{\alpha_1} \binom{k-\alpha_1}{\alpha_2}\cdot \ldots\cdot \binom{k-\alpha_1-\ldots-\alpha_{r-1}}{\alpha_r}$$

The power of 2 accounts for different choices of sign, and each binomial coefficient accounts for the number of ways left to choose the positions of $n_i$'s.

If $n$ is fixed, each such expression is a polynomial in $k$. So, we can sum over all square-partitions of $n$ to get the $n$-th coefficient of $\Theta^k$.

Now we need to get the coefficients of $T_{3^2}\Theta^k$. This is well known. We use theorem 1.7 of "On Modular Forms of Half Integral Weight", G. Shimura (1973), which can also be found as proposition 13 of chapter 4 in "Introduction to Elliptic Curves and Modular Forms", N. Koblitz (1984) (I am slightly weakening the theorem to exclude $p=2$):

Theorem (Shimura). Let $p$ be an odd prime number, and $f \in G_k(N, \chi)$ (an integral modular form of half-integer weight $k/2$, level $N$, and character $\chi$).
Put: $$f=\sum_{n=0}^{\infty} a_n q^n$$ $$T_{p^2}f=\sum_{n=0}^{\infty} b_n q^n$$ Then: $$b_n = a_{p^2n} + \chi (p) \bigg(\frac{(-1)^\lambda n}{p}\bigg) p^{\lambda -1}a_n + \chi(p^2) p^{k-2}a_{n/p^2}$$ where $\lambda = \frac{k-1}{2}$, and $a_{n/p^2}=0$ if $p^2 \not| n$.

For $\Theta^k$ we have $N = 4$ and $\chi = 1$. Using the above with $p=3$: $$b_1 = a_9 - (-3)^{\lambda-1}a_1$$ $$b_2 = a_{18} + (-3)^{\lambda-1}a_2$$

And $a_1$, $a_2$, $a_9$, $a_{18}$ are polynomials in $k$, or $\lambda$ (which is more convenient). If $\Theta^k$ is an eigenform for $T_{3^2}$, we must have $\frac{b1}{a1}=\frac{b2}{a2}$. We will show this is doesn't happen for $k\gt 7$.

The idea is simple: writing $b_1a_2-b_2a_1$ out we see a polynomial of $\lambda$ of degree 19, and $(-3)^{\lambda-1}$ times a quadratic polynomial in $\lambda$. As $\lambda \rightarrow \infty$ (so $k\rightarrow \infty$ as well), the exponential factor dominates, and the expression cannot be zero. We compute a lower bound for this, and check remaining cases manually.

The last part isn't very enlightening, but since you expressed keenness for an algorithm (and someone might want to check that I'm not rambling nonsense), I will include the sage computation below.

def square_partitions(n, min=1):
    """ Returns a list of partitions of n into squares, the least of which can be min^2. """
    if n == 0:
        return [[(-1,1)]]
    else:
        l = []
        for i in [min..sqrt(n)]:
            for p in square_partitions(n-i^2, i):
                if p[0][0] == i:
                    p[0] = (i, p[0][1]+1)
                else:
                    p = [(i,1)] + p
                l += [p]
        return l

def theta_coeff(k, n):
    """ Computes the nth coefficient of $\theta^k$. """
    f = 0
    for p in square_partitions(n):
        g = 1
        j = k
        for a in p:
            if a[0] != -1:
                g *= 2^a[1] * binomial(j, a[1])
                j -= a[1]
        if (not k in ZZ) or j >= 0:
            f += g
    return f

R.<l,s> = QQ['l','s']
# s = (-3)^(l-1)

k = 2*l+1
a1, a2, a9, a18 = theta_coeff(k,1), theta_coeff(k,2), theta_coeff(k,9), theta_coeff(k,18)
b1, b2 = a9 - s*a1, a18 + s*a2

g = b1*a2-b2*a1
lc = LCM([denominator(c) for c in g.coefficients()])
g *= lc

# Check the quadratic factor increases in absolute value when $x\ge7$
x = QQ['x'].0
print factor(g.coefficient(s)(l=x)) # (-6252318072000) * x * (x + 1/2)^2

# Lower bound where exponential factor dominates. Can do better since we're not using the quadratic factor.
m1 = sum(abs(c) for c in g(s=0).coefficients())
print m1*116^19 < 3^115 # True

# Check remaining cases manually.
print [2*i+1 for i in [0..120] if g(i, (-3)^(i-1)) == 0] # [1, 3, 5, 7]

Answer by Dror Speiser for Examples of Galois-invariant central simple algebras which aren't base change?

2011-04-09T14:32:22Z

[big edit]

(1) Let $L/K$ be as above. Take any non-identity element $\sigma \in G_{L/K}$, and let $F=L^{<\sigma>}$ be the fixed field of the cyclic subgroup generated by $\sigma$. By your comment above about the vanishing of $H^3$, every galois invariant CSA of $L$ is a base change of a CSA of $F$. Hence, there are no galois invariant CSA that are not a base change from any proper subfield.

But, if you fix $L$ and $K$, then there can be galois invariant CSA's of $L$ that are not a base change from $K$. Since the smallest non-cyclic group is $C_2\times C_2$, we would like to search for examples with such a galois group.

For $K=\mathbb{Q}$, probably, the simplest example is: $$L=\mathbb{Q}(\sqrt{-3},\sqrt{13}),\ (43)=P_1\cdot P_2\cdot P_3\cdot P_4,$$ $$u=\frac{1}{4}P_1+\frac{1}{4}P_2+\frac{1}{4}P_3+\frac{1}{4}P_4 \in \bigoplus_v Br(L_v)$$

I will prove below that this element of the Brauer group is not a base change, and that in fact a similar construction can be given for any extension.

(2) Near the end of Tate's "Global Class Field Theory" section in Cassels-Frohlich, there is a small paragraph devoted to $H^3(G_{L/K},L^\times)$:

"$H^3(G,L^\times)$ is cyclic of order $n/n_0$, the global degree divided by the lowest common multiple of local degrees, generated by $\delta u_{L/K}$ ($\delta:H^2(C_L)\rightarrow H^3(L^{\times})$), the "Teichmuller 3-class." ..."

Therefore, assume $n_0 < n$ for our galois extension $L/K$. Let $v_0$ be an unramified finite place of $K$ that splits completely (exists by Chebotarev's theorem). We can construct an element of the Brauer group similar to the one above: $$u := \sum_{w|v_0} \frac{1}{n} w$$

This is in fact an element of $Br(L)$ since the number of $w|v_0$ is $n$, so that $n\frac{1}{n} = 1 \in \mathbb{Z}$.

Proposition. For any prime $p$ that divides $\frac{n}{n_0}$, $\frac{n}{pn_0}\cdot u$ is not a base change.

From which we immediately get:

Corollary. The map $Br(L)^{G_{L/K}}\rightarrow H^3(L^\times)$ is onto.

Proof of proposition. Assume that $\frac{n}{pn_0}\cdot u$ is the base change of some $u'$, i.e. $$u' = \sum_v n_v v \mapsto \sum_v \sum_{w|v} [L_w : K_v] n_v w = \frac{n}{pn_0}\cdot u$$

Since for any $w|v_0$: $[L_w : K_{v_0}] = 1$, we must have $n_{v_0} = \frac{n}{pn_0}\cdot \frac{1}{n} = \frac{1}{pn_0}$. And since $\sum_v n_v \in \mathbb{Z}$, at least one other place $v_1$ has $$v_p(n_{v_1}) \le v_p(n_{v_0}) = v_p(\frac{1}{pn_0}) < 0$$

Where $v_p$ is the usual $p$-adic valuation. So, using that $n_0$ is the lcm of local degrees: $$v_p([L_{v_1}:K_{v_1}] n_{v_1}) \le v_p(n_0) + v_p(\frac{1}{pn_0}) = -1$$

Contradicting the zero coefficient of any $w|v_1$ in $u$.

A small computation shows that for the extension $\mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}$, $n_0=2$, and that indeed this is the smallest (discriminant-wise) $C_2\times C_2$ such extension.

Can an even degree galois extension complete p-adically to an even galois extension

2011-04-10T15:48:54Z

Given a galois extension of number fields $L/K$ of even degree, set $n_0=\text{lcm} (\{[L_v:K_v] : v \in M_K \})$ ($L_v$ is any completion corresponding to a place dividing $v$).

Does $2$ divide $n_0$?

This comes up in this question.

Answer by Dror Speiser for Are there any rational solutions to this equation?

2011-03-02T00:28:13Z

[Complete revamp of answer. It is based on the one before, but is better!]

In Tim's hyperelliptic equation, make the change of variables $y$ to $y/(-7)^2$, and $x$ to $x/(-7)$, to get: $$y^2=x(x+7)(x^3+56x^2+245x-343)$$

For every prime $p$ with $v_p(x)<0$, the valuation must in fact be even, thus appears to an even power in the factorisation of each of the factors on the right (as non-zero rational numbers).

We need only consider the squarefree parts of the factors. Assume $p$ divides the numerator of at least two of the factors, to an odd power. By a small gcd calculation (3 in fact), we see that $p$ must be $7$.

Thus, we must have that each factor is a square times a number in $\{ -7, -1, 1, 7 \}$. For each triple $(a,b,c)$ of numbers in the set, with $abc$ a square, there is a possible element of the 2-Selmer group defined by the curve $C_{a,b,c}$ (in $\mathbb{P}^4$): $$x = au^2,$$ $$\ x+7=bv^2,$$ $$x^3+56x^2+245x-343=cw^2$$

Some of these might not have points locally and will not define an element of the 2-Selmer. But we will not make any explicit local computations.

For each such triple, the curve $C_{a,b,c}$ has a morphism into each of the curves: $$C_{a,b,c}^1:\ y^2=c(x^3+56x^2+245x-343)$$ $$C_{a,b,c}^2:\ y^2=acx(x^3+56x^2+245x-343)$$ $$C_{a,b,c}^3:\ y^2=bc(x+7)(x^3+56x^2+245x-343)$$

A sage computation shows that for each such triple $(a,b,c)$, at least one of these three curves has no rational points (other than ones at infinity, or points that don't correspond to solutions of the original equation, i.e. $x=1$ or $x=7$).

Therefore, the original equation has no rational solutions.

Here is the sage code of the computation:

def cubic_to_ellipticcurve(f):
    a, l = f.coeffs()[-1], f.coeffs()
    return EllipticCurve([0,l[2],0,l[1]*a,l[0]*a^2])

def quartic_to_ellipticcurve(f):
    for fac in factor(f):
        if fac[0].degree() == 1:
            r = fac[0].roots()[0][0]
            v = f.variables()[0]
            f = f(v+r)
            f = sum([f.coeffs()[4-i]*v^i for i in [0..3]])
            return cubic_to_ellipticcurve(f)
    return None

R.<x> = QQ[]

possible_sels = []
for a in [-7,-1,1,7]:
    for b in [-7,-1,1,7]:
        c = a*b
        E1 = cubic_to_ellipticcurve(c*(x^3+56*x^2+245*x-343))
        if E1.rank() == 0 and E1.torsion_order() == 1:
            continue

        E2 = quartic_to_ellipticcurve(a*c*x*(x^3+56*x^2+245*x-343))
        if E2.rank() == 0 and E2.torsion_order() == 1:
            continue

        E3 = quartic_to_ellipticcurve(b*c*(x+7)*(x^3+56*x^2+245*x-343))
        if E3.rank() == 0 and E3.torsion_order() == 1:
            continue

        possible_sels += [(a,b,c)]

print possible_sels # prints []

Tunnel like thereom: is there an interesting function with fourier coefficients related to $L'(E_n,1)$ instead of $L(E_n,1)$?

2011-02-14T22:23:32Z

Tunnel's result on the congruent number problem hinges on the fact that there are modular forms with fourier coefficients related to the values $L(E_n,1)$.

Is there an interesting function that has coefficients related to $L'(E_n,1)$ instead? (for a reasonable definition of "interesting" and "related")

This is interesting since it can potentially strengthen the solution to the congruent number problem - it might give an effective algorithm to decide if rank$(E_n)>1$ (which is a step before actually constructing the points, as asked in "constructing non-torsion points ...").

(of course BSD is assumed, along with any other interesting and related conjectures)

When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation?

2011-02-11T16:13:40Z

Given a cubic number field and a basis $\{\gamma_1,\gamma_2,\gamma_3\}$ for it over the rationals, we can write down the norm equation $N(x_1\gamma_1+x_2\gamma_2+x_3\gamma_3)=1$. For almost all substitutions, say $x_1=c$, the resulting affine cubic curve is an affine part of an elliptic curve.

I was wandering what can be said of the converse. If we are given an elliptic curve over the rationals, is there a cubic number field such that for some substitution (any kind) in the norm equation, we get an affine curve isomorphic to an affine part of an elliptic curve?

I've started reading Serre's Algebraic Groups and Class Fields, which seems relevant, since its main results concern rational maps $C\rightarrow G$ from a curve to a commutative algebraic group, which is the case above.

Are affine groups over rings of integers finitely generated?

2011-02-01T22:15:53Z

I'll begin by saying that I'm not sure what I want to ask specifically, but pretty sure what in general, so please don't hold my misunderstandings against me, but do comment on them.

I know that the unit group of a number field is finitely generated, and so is $SL_2(\mathbb{Z})$. I understand that so are $SL_n(\mathbb{Z})$ (or was it $GL$?).

1) What is a known positive generalisation?

I also know that the subgroup of an abelian variety of points over a number field is finitely generated. I noticed that this is relevant after reading Franz Lemmermeyer's "Higher Descent on Pell Conics III. The First 2-Descent" (arxiv). The paper contains a proof that the unit group of a quadratic number field is finitely generated - using heights.

The way I think about it is this: the norm equation isn't a projective variety, so we make up for that by considering it over the integers. So we have heights and parallelogram laws and a proof of finitely generated.

2) Is there a single proof for Mordell-Weil, Dirichlet's Unit Theorem, and any to answer to (1), at the same time, that uses some kind of underlying concept to projective-ness and integral-ness?

I think (2) is more far fetched than (1), so feel free to ignore it :)

Can finitely many values of a polynomial determine it?

2011-01-20T19:19:38Z

Let $d$ be a positive integer greater than 2. Define an equivalence relation on monic integer polynomials of degree $d$: $f\sim g \iff f(k_1 x+k_2)=g(k_3 x+k_4)$ for some integers $k_1,...,k_4$.

Is there a number $m$ such that for any $m$ distinct integers, there is at most one equivalence class that attains these at integer coordinates?

I ask for $d>2$ since it is vacuous for $d=1$ (only one class) and it fails for $d=2$: $x^2-1$ and $2y^2$ are not in the same class and having infinitely many common values.

From some short calculations I think it is true that for $d+1$ (maybe a bit more) distinct values there are at most a finite number of equivalence classes possible, but I don't see how to bound the number of classes uniformly, let alone by 1.

Answer by Dror Speiser for What is the shortest proof of the existence of a prime between $p$ and $p^2$ ? other examples?

2011-01-14T16:37:36Z

It is possible to shorten Bertrand's Postulate's proof so it proves only the above. We can throw away the usually-proven upper bound on the primoral. Explicitly, following Wikipedia's "Proof of Bertrand's postulate":

Lemma 1: $$\frac{4^{\lfloor n^2/2 \rfloor}}{2\lfloor n^2/2 \rfloor+1} < \binom{n^2}{\lfloor n^2/2 \rfloor}$$ For a fixed prime $p$, define $R(p,n)$ to be the highest natural number $x$, such that $p^x$ divides $\binom{n}{\lfloor n/2 \rfloor}$.

Lemma 2: $$p^{R(p,n)} \le n+1$$

If there are no primes between $n$ and $n^2$, then: $$\binom{n^2}{\lfloor n^2/2 \rfloor } = \prod_{p\le n} p^{R(p,n^2)} < (n^2+1)^n$$

This violates lemma 1 as soon as $n \ge 7$.

(* the floors where put in a bit hastily)

Answer by Dror Speiser for Why Are Weber Polynomial Coefficients Smaller than Hilbert Polynomial Coefficients?

2010-12-22T22:50:01Z

Simply because they satisfy an equation of the form $P(f)-fj$ for some polynomial $P$. This immediately implies that the height of $f(z)$ will be around $1/deg(P)$ of the height of $j(z)$, or more precisely, asymptotic to it as the discriminant of $z$ goes to infinity.

See A. Enge and F. Morain's "Comparing invariants for class ﬁelds of imaginary quadratic ﬁelds", ANTS-V 2002.

Families of number fields of prime discriminant

2010-01-04T21:33:34Z

When I am testing conjectures I have about number fields, I usually want to control the ramification, especially minimize to a single prime with tame ramification. Hence, I usually look for fields of prime discriminant (sometimes positive, sometimes negative).

I get the feeling that I cannot be the only one who does this...

And so, are there families of number fields of prime discriminant for each degree? Or at least degree 3 and 4? (They are the coolest. Except quadratics. Of course.) What about: given a prime - can I find a polynomial of degree d with the prime as its discriminant?

Answer by Dror Speiser for Galois Groups of a family of polynomials

2010-11-04T22:42:44Z

The following proves David's statement on the discriminant.

$f_n(x) := x^{n-1}\ +\ 2x^{n-2}\ +\ ...\ +\ n$

$f_n(x) = x\frac{x^n-1}{(x-1)^2}-\frac{n}{x-1}$

Say $f_n(\alpha)=0$,

$f'_n(x) = \frac{x^n-1}{(x-1)^2} + x\frac{nx^{n-1}(x-1)^2-2(x-1)(x^n-1)}{(x-1)^4}+\frac{n}{(x-1)^2}$

$f'_n(\alpha) = \frac{n}{\alpha(\alpha-1)}+(n(\frac{n}{\alpha(\alpha-1)}+\frac{1}{(\alpha-1)^2})-\frac{2\alpha}{\alpha-1}\frac{n}{\alpha(\alpha-1)})+\frac{n}{(\alpha-1)^2}$

$=\frac{n}{\alpha(\alpha-1)^2}((\alpha-1)+(n(\alpha-1)+\alpha)-2\alpha+\alpha)$

$=\frac{n(n+1)(\alpha-1)}{\alpha(\alpha-1)^2}=\frac{n(n+1)}{\alpha(\alpha-1)}$

$\Delta_{f_n} = (-1)^{\frac{(n-1)(n-2)}{2}} \text{Nm}(f'_n(\alpha)) = (-1)^{\frac{(n-1)(n-2)}{2}}\frac{(n(n+1))^{n-1}}{nf(1)} = (-1)^{\frac{(n-1)(n-2)}{2}}2n^{n-3}(n+1)^{n-2}$

This shows that whenever $n+1$ is twice an odd square the discriminant is a square.

Is the direct limit of Weil restriction of an elliptic curve a scheme?

2010-11-03T20:33:51Z

In a discussion today on the Shafarevich-Tate group of an elliptic curve, the following structure and question came up. I will abuse many notations and be very vague about some things, but am very open to suggestions for clarification. Not to mention that some, if not all, of the following is incorrect.

A key ingredient in the rich structure of number fields is Hilbert 90, asserting that the first galois cohomology of the multiplicative group is trivial. Elliptic curves, on the other hand, have no analogy (as far as I know). The discussion was about looking for a natural object to inject an elliptic curve into, that has trivial first galois cohomology.

Given an elliptic curve, an interesting looking (well, maybe not) object is the direct limit of Weil restriction of the curve, going over all the galois extensions of the base field and the morphisms are the natural inclusions. As a $G_K$ module the Weil restriction is the induced module $Ind_{G_L}^{G_K} E$, so by Shapiro's lemma the cohomology is isomorphic to $H^1(G_L,E)$. Applying the direct limit, we indeed see that the first cohomology of the direct limit of the Weil restriction is trivial (since $G_L$ keeps getting smaller).

And so many questions are raised regarding this direct limit. The first one is in the title, and is my question.

Is the direct limit of Weil restrictions, going over all galois extensions of the base field, of an elliptic curve a scheme?

What is the Shafarevich-Tate group of GL(2)?

2010-10-26T17:33:30Z

Let $K$ be a number field, $\bar{K}$ a fixed algebraic closure, $G_K$ the absolute galois group, $O_\bar{K}$ the ring of integers of the algebraic closure, $\mathfrak{p}$ runs over prime ideals and subscripts mean p-adic completions.

Define the "Integral Shafarevich-Tate group" of $GL_2(O_\bar{K})$, a $G_K$-module, to be (normal galois cohomology follows): $$\mathrm{III}^{int}(GL_2) = Ker\ \lgroup\ H^1(G_K, GL_2(O_\bar{K}) \rightarrow \prod_{\mathfrak{p}} H^1(G_{K_{\mathfrak{p}}},GL_2(O_\bar{K_\mathfrak{p}})) \rgroup$$

For $GL_1(O_\bar{K})$ it is known^* to be isomorphic to the class group of $K$. My question is: $$\text{What is }\mathrm{III}^{int}(GL_2)?$$

It is not clear that the above defines a group since $GL_2$ is not abelian. The article "Abelianization of the First Galois Cohomology of Reductive Groups", M. Borovoi, shows that reductive groups over a number field have abelian Sha (over the field - not "integral" Sha). It might not be related, but I suspect something similar would work for the above.

The same question for $GL_n$, quaternion units, etc. is also very interesting.

(*) The case of $GL_1$ is proved in "Visibility of Ideal Classes" by Schoof and Washington (arxiv:0809.5209).

Comment by Dror Speiser

2013-06-17T01:59:14Z

Note Carmon and Rudnick's recent paper on a function field analogue: The autocorrelation of the Mobius function and Chowla's conjecture for the rational function field. The bound they get is quite far from the analogue of $x^{1/2+\epsilon}$, and by analysing their proof, you can see it is actually wrong. But it is only an specific analogue ($\lim q\rightarrow\infty$, "$x$" a fixed degree).

Comment by Dror Speiser

2013-05-26T15:43:03Z

Looking at the mod p representation attached by Shimura, Deligne, Deligne-Serre, corresponding to the inclusion $K_f\rightarrow\bar{\mathbb{Q}}_p$, we see that for any $q\not | pN$, we have, from the Eichler-Shimura relation, that $a_q\ \text{mod p}$ depends only on the conjugacy class of Frobenius of $q$. By Chebotarev's theorem, there are infinitely many $q$'s in every such conjugacy class. Hence, if you delete such $q$'s, the residue field doesn't change. I'm not sure what happens at the ramified primes of the finite representation.

Comment by Dror Speiser

2013-05-24T08:34:57Z

You can use the Weil pairing to show that same for all integers $n$. This is corollary 8.1.1 in Silverman's The Arithmetic of Elliptic Curves, in the section titled The Weil Pairing.

Comment by Dror Speiser

2013-04-17T00:02:26Z

... explaining why Digne-Michel (well, really Lusztig in his work) turn to the problem of describing $\mathcal{E}(C_\hat{G}(s),1)$. Essentially, your question would be answered if you could describe these Lusztig series with a part of an L-parameter, whether into $\hat{G}(\mathbb{C})$ or $\hat{G}(\mathbb{F}_q)$.

Comment by Dror Speiser

2013-04-17T00:01:29Z

@Will: first of all, not exactly, as the characters I mentioned are the Deligne-Lusztig characters, and not the $R_{ss}$. Second, if you mention 8.4.6, then I'm sure you're also aware of 8.4.7, that also give the regular characters as a sum over Deligne-Lusztig characters. Together, these give all irreducible characters for a few low-dimensional groups, which should be the starting point of an answer to your question. Now, Lusztig's result is that you can parametrize irreducible characters by $\{\chi_{s,\lambda}\ |\ s\in \hat{G}\ \text{semisimple},\ \lambda \in \mathcal{E}(C_\hat{G}(s),1)\}$..

Comment by Dror Speiser

2013-04-16T16:35:26Z

... This is remark 5.21.5 in the original Deligne-Lusztig paper, and also appears in Carter's Finite Groups of Lie Type in the chapter on geometric conjugacy.

Comment by Dror Speiser

2013-04-16T16:23:59Z

@Will: I think you are missing something, though I'm not sure it gives an answer to the big Langlands-analogy question: namely $\hat{F}$-stable semisimple conjugacy classes of $\hat{G}(\mathbb{F}_q)$ are in bijection with geometric conjugacy classes of $(T,\theta)$ on $G$. These in turn give virtual representation - the Deligne-Lusztig characters. For the most part, these are in fact irreducible characters, and to understand their decomposition it is enough to understand the Lusztig series. This is the part I am less familiar with, so can't say much more.

Comment by Dror Speiser

2013-04-12T00:37:25Z

@Joel: Yeah, I think so, isn't this a part of the Langlands-Tunnel theorem?

Comment by Dror Speiser

2013-04-11T14:05:10Z

I hope I don't sound too ignorant, but doesn't this answer ignore the construction from (2) to (1c) for even solvable galois groups?

Comment by Dror Speiser

2013-04-08T07:57:19Z

I'm guessing the words "polynomial time" above are meant in the logarithm of $n$. To answer that, it would be interesting to know how well the Lagarias-Odlyzko algorithm approximates $\pi(x)$ if one is only willing to put it in $log^k(x)$ time.

Comment by Dror Speiser

2013-03-31T00:06:04Z

@robot: hey, thanks for the link. I've seen this paper before, and unfortunately (for me) it suffers from the same problem every paper I've seen suffers from: it begins with any given modular function $t$, instead of constructing an interesting one.

Comment by Dror Speiser

2013-03-27T08:43:54Z

@Francois: thanks for the reference. My french is a bit a rusty (non-existent), but I think the remark says what I wrote above: that the coefficients are in general algebraic.

Comment by Dror Speiser

2013-03-27T02:23:32Z

I also asked this in math.stackexchange: math.stackexchange.com/questions/338453/…

Comment by Dror Speiser

2013-03-05T00:14:19Z

I guess if you want to cover 4th and 6th roots then you need the group to contain elements corresponding to $\mu_4$ and $\mu_6$, i.e. conjugate to $\left(\begin{array}{cc}0 & -1 \\ 1 & 0 \end{array}\right)$ and $\left(\begin{array}{cc}-1 & -1 \\ 1 & 0 \end{array}\right)$

Comment by Dror Speiser

2013-03-03T09:14:20Z

The easiest way to realize that the values of $\phi$ are not random, is to compute $\phi'$ - the polynomial that does the same thing but from $E'$ to $E$. It will look similar, and will have the same property.