User david loeffler - MathOverflow

Answer by David Loeffler for Is there an algebraic curve over Q which is not modular?

2013-05-11T21:02:50Z

One expects that the majority of algebraic curves over number fields having genus $> 1$ should not be modular in this sense.

For instance, take a sufficiently general genus 2 curve $C$ over $\mathbf{Q}$. Then its $\ell$-adic $H^1$ (which is just the $\ell$-adic Tate module of its Jacobian) will be a 4-dimensional Galois representation whose image lands inside $\mathrm{GSp}_4(\mathbf{Z}_\ell)$ . If $C$ is sufficiently generic, then the image of this Galois representation should be the whole of $\mathrm{GSp}_4(\mathbf{Z}_\ell)$ for all but finitely many $\ell$; in particular, it will be absolutely irreducible. (I don't know if this is known, but certainly one expects it to be the case.) On the other hand, if $C$ admits a non-constant map from $X_0(N)$, then the its $H^1$ would have to be a quotient of the $H^1$ of $X_0(N)$, and this can be calculated in terms of modular forms; in particular all its absolutely irreducible subquotients have dimension 2. So most genus 2 curves $C$ will not be modular in your sense. and if you get one that is, you should regard it as a rather unlikely coincidence.

(A more high-powered interpretation of this is that $H^1(C)$ should be the Galois representation attached to a degree 2 Siegel modular form. In some very special cases this Siegel modular form will be endoscopic, i.e. describable in terms of lifts from elliptic modular forms, but most Siegel mod forms will not be endoscopic and thus will not have anything to do with $X_0(N)$ for any $N$.)

If you're willing to relax your definition of "modular", though, you can get many more possibilities. There's a very striking result of Belyi stating that any algebraic curve defined over a number field can be obtained as the quotient of the upper half-plane by some subgroup of $PSL(2, \mathbf{Z})$, although the corresponding group will usually not be a congruence subgroup.

Elementary proof of algebraicity of Hecke eigenvalues in weight 1

2011-01-28T11:14:19Z

It's "well known" that, for any weight $k$ and level $N$, the space $S_k(\Gamma_1(N))$ of cusp forms of that weight and level has a basis in which all the Hecke operators act by matrices with entries in $\mathbb{Z}$; consequently all the Hecke eigenvalues are algebraic numbers (indeed algebraic integers).

I was reflecting on how to prove this while teaching an undergraduate course on modular forms. For $k \ge 2$ it's not hard: there's the Eichler-Shimura machinery which relates it to a question about cohomology, and the cohomology with $\mathbb{Z}$ coefficients does the job. Alternatively, and more or less equivalently, you use the pairing with modular symbols. Both of these methods break down for $k = 1$; the only argument I know that works in this case is to use the fact that $X_1(N)$ has a model as an algebraic variety, and weight $k$ modular forms correspond to sections of the $k$-th power of a line bundle that has a purely algebraic definition. But that's not really something I can stand up and explain to a class of undergraduate students!

For cusp forms of weight $k = 1$, can the algebraicity of the Hecke eigenvalues be proved without quoting heavy machinery from arithmetic geometry?

Answer by David Loeffler for Concrete examples of noncongruence, arithmetic subgroups of SL(2,R)

2013-05-08T08:18:17Z

I would quibble with your definition of "arithmetic". There are discrete subgroups of $SL(2, \mathbf{R})$ coming from quaternion algebras, which are not commensurable with $SL(2, \mathbf{Z})$, but are commensurable with $G(\mathbf{Z})$ for some twisted form $G$ of $SL(2)$ which becomes isomorphic to the usual form over $\mathbf{R}$. These are arithmetic subgroups of $G$, but not of the usual $SL(2)$. So the notion of "arithmetic subgroup" depends on a choice of model over a number field.

With that quibble out of the way, examples of noncongruence arithmetic subgroups of $SL_2 / \mathbf{Q}$ are not too hard to come by. The easiest construction I know goes like this. We know $G = PSL(2, \mathbf{Z})$ has a rather simple presentation; it's the free product of $C_2$ and $C_3$. So if we have any other group $H$ containing elements $a,b$ of orders 2 and 3, there is a unique homomorphism $G \to H$ sending the generators to these elements.

In particular, $S_7$ is generated by two such elements, so there is a normal subgroup of $G$ with quotient isomorphic to $S_7$. But $S_7$ is not a quotient of $SL(2, \mathbf{Z} / m)$ for any $m$ (easy check once you know that $PSL(2, \mathbf{Z}/p)$ is simple for $p > 2$) hence this subgroup cannot be congruence.

(If you look not at the kernel of this homomorphism but the preimage of the stabilizer of 1, then you get a noncongruence subgroup of index 7, which is, if I remember correctly, the smallest possible.)

The congruence subgroup problem for more general reductive groups is a difficult problem and one that's far from solved. As I mentioned above, it depends on a choice of model over a number field. The group $SL(n) / \mathbf{Q}$ has no non-congruence arithmetic subgroups for $n \ge 3$ (Bass-Milnor-Serre), while $SL(2)$ over an imaginary quadratic field has lots of them (it behaves like $SL(2) / \mathbf{Q}$). There are many more recent results; googling brings up a survey by Raghunathan.

(EDIT: I suddenly realized I wasn't sure if the bit about $S_7$ being a quotient of $PSL(2, \mathbf{Z})$ was true or not. So I wrote a computer program and it is indeed true that the elements (2,3,4)(5,6,7) and (1,2)(3,5)(4,6) generate $S_7$).

Are Kato's zeta elements integral?

2010-11-23T19:00:06Z

Let $E$ be an elliptic curve over $\mathbb{Q}$ and $T$ the $p$-adic Tate module of $E$. Kato's Euler system, constructed in the paper "P-adic Hodge theory and values of zeta functions of modular forms" (Asterisque 295, 2004), gives rise to an element $\mathbf{z}_{\rm Kato}$ lying in the Iwasawa cohomology $ H^1_{\mathrm{Iw}}(\mathbb{Q}_p, T)[\frac{1}{p}]$ . In Theorem 12.5(4) of the paper, Kato shows that if the image of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ in $\mathrm{Aut}(T)$ contains $\mathrm{SL}_2(\mathbb{Z}_p)$, then in fact $\mathbf{z}_{\rm Kato} \in H^1_{\mathrm{Iw}}(\mathbb{Q}_p, T)$ .

Is it known if there are weaker conditions that are sufficient to ensure that $\mathbf{z}_{\rm Kato}$ has this integrality property? Are there examples where it is genuinely non-integral, or is it conjectured that it should always be so?

I would be the last person to claim I understand Kato's argument, but it looks to me as if he only actually uses the weaker statement that the mod $p$ Galois representation $T/pT$ is irreducible. I'd be interested to know if this weaker condition is indeed sufficient, and whether anything is known in this direction if the weaker condition doesn't hold (i.e. if $E$ admits a $p$-isogeny).

(EDIT: Added more detail and references.)

Answer by David Loeffler for On the Cartan decomposition of unitary group

2013-05-06T12:38:25Z

Theorem: Let $G$ be a reductive algebraic group over a local field $F$, let $K$ be any maximal compact subgroup of $G(F)$, and let $Z = Z(G)$. Then $K \cap Z(F)$ is the unique maximal compact subgroup of $Z(F)$.

Proof: Let $W$ be the maximal compact subgroup of $Z(F)$. Then the natural multiplication map $K \times W \to G(F)$ has image a compact subgroup of $G(F)$ containing $K$; since $K$ is maximal by assumption, the image is equal to $K$ and thus $W \subseteq K$. QED.

In particular, in your case $Z$ is $U(1)$ and hence $Z(F)$ is compact, and thus $Z(F)$ is contained in $K$ for every maximal compact $K$.

Answer by David Loeffler for j-invariant duplication, triplication and quintuplication formulae... how?

2013-05-02T18:06:44Z

One can explain this conceptually in terms of "modular equations".

It's been known since the late 19th century that for any integer $N$, there is some polynomial $\Phi_n(X, Y)$ with the property that $\Phi_n(J(\tau), J(N\tau)) = 0$ for all $\tau$ in the upper half-plane. This is the "classical modular equation". For a few small values of $N$, the plane curve defined by $\Phi_n(X, Y) = 0$ is a rational curve, so we can find rational functions $a(u), b(u)$ of an auxilliary variable $u$ such that for every $\tau$, there is a $u$ such that $J(\tau) = a(u)$ and $J(N\tau) = bu$. Moreover, the map $\tau \mapsto -1/N\tau$ gives an involution on the curve, and you can choose your parametrization so this corresponds to $u \mapsto 1/u$, which tells you that $a(u) = b(1/u)$. This is what is going on in the formulae quoted on MathWorld.

What MathWorld doesn't tell you is that the curve defined by $\Phi_n(X, Y) = 0$ is actually a rather important object. It's singular in general, but its normalisation is a smooth curve over $\mathbb{C}$ which turns out to be isomorphic to the quotient of the upper half-plane by the group of matrices $$\Gamma_0(N) = \left\{\begin{pmatrix} a& b \\ c & d\end{pmatrix} : N \mid c, ad-bc = 1\right\}.$$

The curve thus obtained is called $X_0(N)$ and is very important in the theory of modular forms. In particular one knows that it is of genus 0 if and only if $N$ is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18, or 25, so a formula of the kind that you quote exists if and only if $N$ is in this list.

Motivic cohomology and cohomology of Milnor K-theory sheaf

2013-02-17T15:40:58Z

Let $X$ be a smooth variety over a field $k$. (Assume $k$ has characteristic 0 if it helps; in fact I'd be happy to assume that $k$ is a finite extension of either $\mathbf{Q}$ or $\mathbf{Q}_p$).

Then there is a sheaf $\mathscr{K}_m^M$ on $X$ (in the Zariski topology), for each $m \ge 1$, which comes from sheafifying the Milnor $K$-theory of function rings of affine opens of $X$. So we can take sheaf cohomology groups $H^*(X, \mathscr{K}_m^M)$. We can also consider Voevodsky's motivic cohomology groups $H^*(X, \mathbf{Z}(i))$, where $\mathbf{Z}(i)$ is Voevodsky's motivic complex.

Is it true that there are isomorphisms $$H^r(X, \mathscr{K}_s^M) = H^{r+s}(X, \mathbf{Z}(s))$$ for all integers $r, s \ge 0$?

(Note that the right-hand side is known to be coincide with Bloch's higher Chow group $CH^s(X, s-r)$.)

Here is why I think this. Both sides are zero if $r > s$ or $r > \dim X$. For $r \le s$, one has a candidate for the isomorphism using Kerz's Gersten complex for Milnor K-theory and a construction due to Landsburg; and for $r = s$ or $r = s-1$ this map is indeed known to be an isomorphism. On the other hand, for $r = 0$ and $X = \operatorname{Spec} k$, this is just the isomorphism of Nesterenko--Suslin--Totaro, $$H^s(\operatorname{Spec} k, \mathbf{Z}(s)) = K_s^M(k).$$

(This is related to my earlier question http://mathoverflow.net/questions/106021, where I asked essentially the same thing for the Quillen $K$-theory sheaf instead of the Milnor one. It was pointed out there that for $r = 0$ and $X = \operatorname{Spec} k$ the motivic cohomology is just the Milnor K-theory, which leads me to wonder if one does get an isomorphism using the Milnor $K$-theory sheaf; for $r = s$ or $r = s-1$ the cohomology of the Milnor and Quillen $K$-theory sheaves agrees.)

Answer by David Loeffler for Can we find a set of elliptic curves over rationals associated with $f$?.

2013-04-27T14:54:29Z

This has already had some votes to close, but I'll see if I can answer it anyway...

The answer is "no". There are lots of motivic L-functions that are not elliptic curve L-functions, just because there are lots of motives that are not $H^1$ of an elliptic curve! For instance, the L-function attached to a modular form of weight $k > 2$ (which is motivic, by a theorem of Scholl) does not have anything to do with the L-function of any elliptic curve, because the form of the $\Gamma$-factors is different.

A nontrivial relevant statement that might interest you is perhaps this one: if $L(s) = \sum_{n \ge 1} a_n n^{-s}$ is a Dirichlet series with coefficients $a_n \in \mathbf{Q}$, and $L$ and all of its twists by Dirichlet characters have analytic continuation to all $s \in \mathbf{C}$ and satisfy a functional equation of the same kind as the $L$-function of an elliptic curve (in particular, with the same $\Gamma$-factors), then $L(s)$ is indeed the $L$-function of an elliptic curve. This follows from Weil's converse theorem (which is essentially the same statement with "modular form" in place of "elliptic curve") together with the fact that one can attach an elliptic curve to any weight 2 modular form with coefficients in $\mathbf{Q}$.

Answer by David Loeffler for Sum of two random variables following K0 (modified 2nd kind Bessel) distributions

2013-04-19T09:53:31Z

This an easy exercise using moment generating functions. By scaling we can assume WLOG that $a = 1$. Then the MGF of the modified Bessel distribution is $$ E[e^{tX}] = \frac{1}{\sqrt{1-t^2}}, $$ and hence the MGF of the sum of two independent such variables is $\frac{1}{1 - t^2}$, which is the MGF of the Laplace distribution.

Answer by David Loeffler for central/critical/special values of L-functions terminology

2013-04-17T15:24:53Z

How are you normalizing your $L$-function? I ask because in some fields, it's conventional to normalize everything by shifting so the functional equation always takes $s$ to $1-s$. If the original motivic weight is odd, this means that what I would call critical values may sometimes get moved to half-integers rather than integers. (This is why I personally dislike that normalization.)

If your $L$-function is the $L$-function of a motive and you don't do any strange shiftings (so your usage is consistent with the motives literature), then (2) and (3) are unambiguously correct, (4) is a consequence of (1), and (1) accords with how I use the term but I've never seen it written down as a formal definition.

E.g. if the $L$-function is $L(f, s) = \sum_{n \ge 1} a_n(f) n^{-s}$ for $f = \sum_{n \ge 1} a_n q^n$ a modular cusp form of weight $k$, then the functional equation sends $s$ to $k - s$, the special values are at $s \in \mathbb{Z}$, the critical values are at ${1, \dots, k-1}$, and if $k$ is odd then the central value at $s = k/2$ is not a critical or special value.

Is the smallest primitive root modulo p a primitive root modulo p^2?

2010-06-09T13:43:44Z

Let $p \ne 2$ be a prime and $a$ the smallest positive integer that is a primitive root modulo $p$. Is $a$ necessarily a primitive root modulo $p^2$ (and hence modulo all powers of $p$)? I checked this for all $p < 3 \times 10^5$ and it seems to work, but I can't see any sound theoretical reason why it should be the case. What is there to stop the Teichmuller lifts of the elements of $\mathbb{F}_p^\times$ being really small?

Answer by David Loeffler for Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficients

2013-04-09T12:57:15Z

I typed a comment but the formatting wouldn't come out right, so here it is as an answer!

I cannot work out why you expect the "Plancherel or Parseval type" formula to work. Does it not bother you a little that $a_n(f)$ and $a_n(g)$ are perfectly capable of being integers for all $n$, so your series is obviously divergent?

Much better is to consider the series $$ L(f, g, s) = \sum_{n \ge 1} a_n(f) \overline{a_n(g)} n^{-s},$$ which converges for $Re(s) > 2$ (this is not so easy to see, but it is easy to show that it converges for $Re(s) \gg 0$). This has meromorphic continuation to all $s \in \mathbb{C}$ with a pole at $s = 2$ at which the residue is (maybe up to a normalizing constant depending on $N$ that I have forgotten) the Petersson product $\langle f, g \rangle$.

But this does not help you to get lower bounds on $\langle f, g \rangle$ as far as I can see. Some quite grotty things can happen, e.g. if $f$ is an eigenform and there is another newform $f'$ with $f = f'$ modulo some integer $N$, then one can take $g = (f - f')/N$, and this will be integral but its Petersson product with $f$ will be $\langle f, f \rangle / N$. So the issue of bounding Petersson products below is quite closely related to the issue of congruences between eigenforms.

Twists of CM modular forms

2013-04-08T12:44:38Z

Let $K$ be an imaginary quadratic field of class number 1 and $E$ an elliptic curve over $\mathbf{Q}$ with CM by $\mathcal{O}_K$ . Let $\psi$ be the Groessencharacter of $K$ attached to $E$, and $$ g_{\psi} = \sum_{\mathfrak{a}} \psi(\mathfrak{a}) q^{N(\mathfrak{a})} $$ the corresponding CM-type modular form, where the sum is over the integral ideals prime to the conductor $\mathfrak{f}$ of $\psi$. So the level of $g_{\psi}$ = the conductor of $E$ = $|\Delta_K| N(\mathfrak{f})$; let's call this integer $M$.

Up to some fudge-factor (the Manin constant), the $\Gamma_1(M)$-invariant differential $g_{\psi}(z) \mathrm{d}z$ on the upper half-plane is the pullback of the invariant differential $\omega_E$ on $E$ under a modular parametrization $\pi: X_1(M) \to E$.

Suppose I choose an auxilliary ideal $\mathfrak{b}$ of $K$ (coprime to everything in sight, i.e. to $\mathfrak{f}$ and the discriminant of $K / \mathbf{Q}$). Then I can consider, for each character $\eta$ of the ray class group $G_{\mathfrak{b}}$ of $K$ modulo $\mathfrak{b}$, the Groessencharacter $\psi\eta$ and hence the modular form $g_{\psi\eta}$. These all have level $M B$, where $B = N(\mathfrak{b})$.

Does there exist a morphism of algebraic varieties $\pi_\mathfrak{b}: X_1(M B) \to E$, defined over the ray class field $K(\mathfrak{b})$, such that the span of the conjugates of $\pi_\mathfrak{b}^*(\omega_E)$ under $\operatorname{Gal}(K(\mathfrak{b}) / K) \cong G_{\mathfrak{b}}$ is equal to the span of the forms $g_{\psi \eta}$ for all characters $\eta$ of $G_{\mathfrak{b}}$?

(By way of motivation: if one takes a general (not necessarily CM) modular form $g$ of level $N$ and a rational integer $m$, then the space of modular forms at level $m^2 N$ spanned by the twists of $g$ by Dirichlet characters mod $m$ is also the span of the Galois conjugates of the pullback of $g$ under a map $X_1(m^2 N) \to X_1(N)$ defined over $\mathbf{Q}(\mu_m)$. So I'm hoping for a sort of "CM version" of this.)

Answer by David Loeffler for critical values of motives

2013-03-21T15:57:09Z

In your example (with Hodge structure of weight 3 concentrated in bidegrees (2, 1) and (1, 2)) there is only one critical value, at s = 2. At all other integer values of $s$, either $L_\infty(M, s)$ or $L_\infty(M^\vee, 1-s)$ will have a pole.

Does the Manin-Drinfeld theorem hold over number fields?

2013-03-14T11:35:24Z

The Manin-Drinfeld theorem has various equivalent statements. Let $\Gamma$ be a congruence subgroup of $\mathrm{SL}_2(\mathbb{Z})$. Then:

for any congruence subgroup $\Gamma$ of $\mathrm{SL}_2(\mathbb{Z})$, the subgroup of $\operatorname{Jac}(X(\Gamma))$ generated by the cusps is finite;
if $c_1, c_2$ are cusps and $f \in S_2(\Gamma)$ is a Hecke eigenform, the integral $\int_{c_1}^{c_2} f(z) \mathrm{d}z$ is a linear combination of the periods $\Omega_+(f)$ and $\Omega_-(f)$ with coefficients in the field generated by the Hecke eigenvalues of $f$;
the natural map $H^1_c(Y(\Gamma), \mathbb{Q}) \to H^1(Y(\Gamma), \mathbb{Q})$ has a unique Hecke-equivariant section.

I'm interested in extensions of this to the context of arithmetic quotients of $\mathrm{GL}_2(\mathbb{A}_K)$, where $K$ is a number field. The first statement only makes sense if $K$ is totally real, but the second and third can be formulated for any $K$. Are they true in this generality?

(I have a translation of a 1978 paper by Kurcanov which gives the proof for $K$ an imaginary quadratic field; and I believe there is a more recent paper of Kurcanov that covers CM fields, but I can't read Russian and there doesn't seem to be an English translation.)

Answer by David Loeffler for Unramified Galois representations not from smooth and proper stacks

2013-02-17T15:02:19Z

Here's a class of examples which isn't in your list as far as I know. Take an even unimodular lattice, e.g. the $E_8$ lattice. The corresponding orthogonal group is a reductive group over $\mathbf{Q}$ which is split at every finite place and compact at $\infty$ (see Gross, "Groups over $\mathbf{Z}$", Inventiones 124 (1996)). There will be lots of algebraic automorphic representations of this group $G$ if you take the weight large enough, and these will give you automorphic representations of $GL(n)$ which are self-dual and unramified at all finite places. It's known that these have Galois representations attached, which will be crystalline at $\ell$ and unramified everywhere else; but I don't think these are known to come from smooth proper stacks over $\mathbf{Z}$.

There is much more on level 1 automorphic representations in the recent preprint of Chenevier and Renard, http://www.math.polytechnique.fr/~chenevier/articles/dimform.pdf.

Answer by David Loeffler for What is a(n algebro-geometric) family of modular forms?

2013-02-06T09:28:03Z

Perhaps it is easier to explain what's going on here in the context of Galois representations. I know of two almost wholly disjoint ideas that are, confusingly, both described using phrases like "families of Galois representations".

One can consider "families of representations" of any group, which are just homomorphisms from G to GL_n(A) for some (usually commutative) ring A, or more generally GL(V) for some locally free sheaf V on a base scheme S, etc. These are "families" in the sense that the image of a group element is a matrix whose entries are functions on Spec(A) (resp. on S, etc). Note that, intuitively, the group is fixed and the coefficients are varying.
One can also consider the kind of "geometric" family you mentioned in your question and Michigan J Frog enlarged upon: given a family of geometric objects over a base S, you can do various kinds of relative cohomology to give sheaves on S whose fibres have an action of some kind of Galois group depending on the fibre, and in particular the generic fibre has an action of something like the fundamental group of S. So here the group is, so to speak, varying in the family as well.

The first kind of family, over a p-adic base and with G being a Galois group, comes up a lot in the context of modular forms (Hida, Coleman-Mazur, etc). These can be viewed as sections of a family of sheaves on a subvariety of the rigid-analytic space you get by analytifying the modular curve. Note that we are varying the coefficients and not the group, again.

The second kind of family doesn't come up so much in modular form theory, although it makes a notable appearance in Kato's work on Iwasawa theory for modular forms.

Variant of Leopoldt's conjecture

2013-01-17T17:52:28Z

Let $K$ be a number field with $[K : \mathbf{Q}] = d$, and let $p$ be a prime. Let $\sigma_1, \dots, \sigma_d$ be the embeddings of $K$ into $\mathbf{C}_p$. Let $u_1, \dots, u_k$ be a basis of $\mathcal{O}_K^\times$ modulo torsion, so $k = r_1 + r_2 - 1$ where $r_1$ and $r_2$ are the number of real and complex infinite places.

Consider the $d \times k$ matrix $M$ whose $i, j$ entry is $\log \sigma_i(u_j)$. Leopoldt's conjecture claims that $M$ has full rank, i.e. it has some $k \times k$ minor which is non-singular.

I'm wondering about the following conjecture:

If $K$ does not contain a CM field, then all $k \times k$ minors of $M$ are nonsingular.

Obviously this is stronger than Leopoldt's conjecture, so there is no hope of proving it; but does anyone know if such questions have been studied? Are counterexamples known?

(The statement is definitely false if $K$ contains a CM field $E$ with $[E : \mathbf{Q}] > 4$.)

Answer by David Loeffler for How to solve n! mod p?(p is a prime)

2013-01-17T16:17:58Z

I posted this as a comment but the formatting got screwed up, so here it is as an answer :-)

This looks like rubbish to me. As the original poster clearly realizes, $n! \pmod p$ is zero if $n \ge p$.

The genuine question which this algorithm almost answers is this. For any nonzero integer $m$, there are unique $a \in \mathbb{Z}_{\ge 0}$ and $b \in (\mathbb{Z} / p \mathbb{Z})^\times$ such that $m = p^a b \pmod{p^{a + 1}}$. This algorithm more or less calculates these for $m = n!$, but if $n \ge p^2$ it would need to be applied recursively, to deal with the $(n / p)!$ bit.

Incidentally, $(p - 1)!$ is $-1$ modulo $p$, by Wilson's theorem, so the last line can be simplified a great deal.

Answer by David Loeffler for Diferent abelian varieties over local field with the same p-adic representation?

2013-01-15T17:18:41Z

Yes, this can happen. Here is a counterexample (which is probably not the simplest possible, but it's the one that first came to mind).

There are not very many 2-dimensional representations of the Galois group of $\mathbf{Q}_p$ which are "crystalline" in Fontaine's sense. Fontaine's functor $\mathbf{D}_{cris}$ classifies them by linear data: 2-dimensional vector spaces over $\mathbf{Q}_p$ with a filtration and a linear operator $\varphi$ (the Frobenius) satisfying some compatibility properties. If $V$ is the $p$-adic Galois representation coming from an elliptic curve over $\mathbf{Q}_p$ with good reduction, then $\varphi$ has characteristic polynomial $X^2 - a_p(E) X + p$, where as usual $a_p(E) = p + 1 - \# \overline{E}(\mathbf{F}_p)$.

If $a_p = 0$, then this uniquely determines $\mathbf{D}_{cris}(V)$ as a $\varphi$-module, and the conditions on the filtation ("weak admissiblity") mean that if $a_p(E) = 0$ then there is (up to isomorphism) a unique possibility for the filtration. So, in other words, all elliptic curves over $E$ with good supersingular reduction have isomorphic $p$-adic Galois representations [edit: if $p>3$, at least]. But they certainly aren't all isomorphic (or even isogenous) to each other, so that gives a counterexample.

Answer by David Loeffler for Applications of Iwasawa Theory

2012-12-28T15:38:32Z

Aha, an excuse to quote chunks of my most recent grant proposal :-)

Iwasawa theory is heavily used in work on the BSD conjecture. For instance, the first positive result to be proved in the direction of BSD -- the Coates--Wiles theorem that analytic rank 0 implies algebraic rank 0 for elliptic curves over $\mathbf{Q}$ with complex multiplication -- was shown by using Iwasawa theory. This is a statement which has nothing obviously to do with Zp-extensions, on the face of it, although to be sure they are lurking in the proof. More generally, pretty much all the results on BSD that we now have (thanks to Kolyvagin, Rubin, Kato, Perrin-Riou, Kobayashi, etc...) use Iwasawa theory heavily.

This fits into a general philosophy which states that if $M$ is a motive over a number field $K$, the $L$-values $L(M, j)$ for integer values of $j$ encode information about the cohomology of $M$. Iwasawa theory provides a tool to attack such conjectures, by interposing a third object -- a p-adic L-function -- which one can (sometimes) relate both to the cohomology and to the L-values. This is bound up with the idea that the behaviour of $M$ over your original ground field $K$ might be quite complex, but making a tower of extensions $K = K_0 \subset K_1 \subset K_2 \subset \dots \subset K_\infty$ and taking a limit of objects defined over the $K_n$'s can serve to "smooth out" the behaviour, and then you prove things over $K_\infty$ and see what you can recover over $K$ by some kind of descent argument, splitting your problem into two hopefully easier chunks. (For more philosophy along these lines, see Colmez's Bourbaki expose on p-adic L-functions, http://www.math.jussieu.fr/~colmez/851bourbaki.pdf)

Answer by David Loeffler for Algebraic maximal extension and algebraic closure

2012-12-04T10:32:52Z

I think the answer is "hardly ever", because pretty much everything is algebraic maximal in your sense. For any complete discretely-valued field $K$, and any finite extension $L / K$, we have $[L : K] = e(L / K) f(L / K)$, where $f(L/K)$ is the degree of the extension of residue fields and $e(L / K)$ is the index of the value group of $K$ in that of $L$. So any complete discretely valued field is algebraic maximal, and such fields are very far from being algebraically closed!

I can't actually think offhand of an example of a valued field which is not algebraic maximal.

Answer by David Loeffler for Numerical evaluation of the Petersson product of elliptic modular forms

2012-12-01T21:47:06Z

It's easy to reduce to the case of computing the Petersson product of a normalised new eigenform with itself. Here you can use the fact that the product is equal to the value at s=k of the symmetric square L-function of f, and this you can compute using e.g. Tim Dokchitser's algorithms. Here is a thread from the Sage developers mailing list with example code by Martin Raum: https://groups.google.com/forum/m/#!topic/sage-nt/EkBWOogY8yw

For elliptic curves there is also Mark Watkins' Sympow program, which will compute all the symmetric power L-functions.

Answer by David Loeffler for Do L-functions exist for Half-integral weight modular forms?

2012-11-19T11:10:23Z

You can certainly attach $L$-functions to half-integer weight eigenforms, but you don't get anything really new by doing so: they turn out be versions of $L$-functions of integer weight modular forms. More specifically, there is the "Shimura lifting" map from weight $k + 1/2$ to weight $2k$, which sends eigenforms to eigenforms; and the L-function of a half-integer weight eigenform will be closely related to that of its image under the Shimura lift. See e.g. here:

www.math.wisc.edu/~ono/reu06shimura.pdf

In fact this turns out to be a very powerful way of studying the L-functions of integer weight forms (used, for instance, in Tunnell's work on the congruent number problem, which uses modular forms of weight 3/2 to understand the values at $s=1$ of the $L$-functions of twists of elliptic curves.

Answer by David Loeffler for Explicit examples of algebraic Hecke characters with infinite image?

2012-11-08T21:59:39Z

The "most obvious" algebraic Hecke characters of a field $K$ are the characters of the ideal class group of $K$, which have trivial infinity-type and trivial conductor. There might be no non-trivial examples (as in the case of Q). But you can get more examples by:

beefing up the conductor (which gets you Dirichlet characters, or more generally characters of ray class groups, but never anything of infinite order)
changing the infinity-type (the restriction of $\chi$ to the connected component of the identity in $(K \otimes \mathbb{R})^\times \subseteq \mathbb{A}_K^\times$).

Over $K=\mathbb{Q}$, the infinity-types are pretty restricted: the only algebraic characters of the group of positive reals are the maps $x \mapsto x^k$, so you just get powers of the "norm" character (the character of the ideles whose restriction to $\mathbb{R}_+$ is the identity, and which sends a uniformizer at a prime $p$ to $1/p$).

Over a number field the game is more subtle. Let's first suppose $K$ is totally real of degree $n$. The infinity-type of a character looks like $z \mapsto z_1^{k_1} \dots z_n^{k_n}$ for integers $k_1, ..., k_n$, where $z_1, ..., z_n$ are the embeddings into $\mathbb{R}$, but there is a constraint that the infinity-type needs to vanish on a finite-index subgroup of the global units, and this forces the vector $k_1, ..., k_n$ to be orthogonal to the lattice in $\mathbb{R}^n$ generated by the vectors

$$ \{ (\log |u_1|, ..., \log |u_n|) : u \in \mathcal{O}_K^\times\}. $$

By Dirichlet's unit theorem this lattice has rank $r_1 + r_2 - 1 = n-1$, so its orthogonal complement has rank at most 1 -- it's spanned by $(1, ..., 1)$. This tells us that every algebraic Hecke character is just a finite-order character times a power of the norm character, which is a bit boring.

For non-totally-real fields the game gets more interesting because there are not so many units. If $K$ is a CM field of even degree $2d$, then the unit group has rank $d-1$, and you can show that the weights of algebraic Hecke characters span a lattice of rank $d + 1$ (spanned by the norm character and characters of the form $x \mapsto \sigma_i(x) / \overline{\sigma_i(x)}$ for each embedding $\sigma_i: K \hookrightarrow \mathbb{C}$).

If $K$ is not either totally real or CM, things are more interesting still: the lattice spanned by the logs of the units has rank $r_1 + r_2 - 1$, so its orthogonal complement has dimension $1 + r_2$ over $\mathbb{R}$, but you can't find enough integer vectors in the orthogonal complement. For instance, if $K = \mathbb{Q}(\sqrt[3]{2})$ then the only possibility is the norm character, again (it is a fun exercise to check this by hand in this case). This is an instance of a theorem of [edit: Emil Artin and] Andre Weil: for any number field $K$, and any algebraic Groessencharacter of $K$, the infinity-type of the character must factor through the norm map to the maximal CM subfield of $K$ [edit: or to $\mathbb{Q}$ if there is no such subfield].

Answer by David Loeffler for Blueprint of L-functions and need for introducing them ( Hasse-Weil L-functions )

2012-11-06T19:28:41Z

There is an excellent reason why the exponential term and the division by $n$ are there, although they look a bit mysterious at first.

Firstly, a correction to your formula: it should be $|C(\mathbb{F}_{q^n})|$ , the number of solutions over the field with q elements, not $|C(\mathbb{F}_{q})|$ . (Notice that this means that the $n$th term really depends on $n$ and $C$ in a subtle way, because it "knows" how many points C has over every extension of the original field.)

With this correction made, a miracle occurs: the quantity $\zeta_{C / \mathbb{F}_q}$ -- a priori just some formal power series -- is a rational function.

To get some idea of the magic that's going on here, let's consider some simple examples. Firstly, you can take $C$ to be $\mathbb{P}^1$. That's not a very interesting curve, I know, but it's a curve. Then $C$ has exactly $q^n + 1$ points over $\mathbb{F}_{q^n}$ -- one for each element of the field, together with the point at $\infty$ -- and we get

$$ \zeta_{C / \mathbb{F}_q} = \exp( \sum_{n \ge 1} \frac{q^n + 1}{n} u^n) = \exp(-\log (1-u) - \log (1 - qu)) = \frac{1}{(1 - u)(1 - qu)}.$$

As promised, this is a rational function! If the funny exponential term and the $1/n$ factor hadn't been there in the definition, we wouldn't have got anything so nice.

It turns out that this happens for any curve $C$ (this was proved by Andre Weil) and in fact for higher-dimensional varieties too (this was proved by Dwork).

PS. If $C$ is an elliptic curve, then one can show (e.g. this is in Silverman's book "The Arithmetic of Elliptic Curves", in section V.5) that $\zeta_{C/\mathbb{F}_q}$ is given by $$ \zeta_{C/\mathbb{F}_q}(u) = \frac{1 - a u + q u^2}{(1 - u)(1 - qu)}$$ where $a = q + 1 - |C(\mathbb{F}_q)|$. So this quadratic term appearing here really appears for a reason; it wasn't just plucked out of midair. I hope that answers another part of your question, which is where this quadratic comes from.

Why does Tate's conjecture imply semisimplicity of crystalline Frobenius?

2011-02-20T19:41:27Z

I'm trying to find a reference for the following fact:

If Tate's conjecture is true for all smooth projective varieties over $\mathbb{F}_p$, then the Frobenius endomorphism on the crystalline cohomology of any such variety is semisimple.

This is stated in the Coleman-Edixhoven paper on the semisimplicity of the $U_p$-operator on modular forms. They reference Milne's paper "Motives over finite fields" (in the 1991 Motives conference proceedings, ed. Janssen/Kleiman/Serre).

I found Milne's paper on the web, and it gives two references for the corresponding statement for $\ell$-adic cohomology ($\ell \ne p$) and then says "There is an analogous statement ... for the crystalline cohomology" without giving a reference (or a precise statement) for this. Moreover, one of the references for the $\ell \ne p$ case is to Milne's book "Arithmetic Duality Theorems" but points to an apparently non-existent section 8.6; while the other reference Milne gives is to Tate's article in the same proceedings, which does not seem to prove anything about semisimplicity as far as I can see.

Can anyone tell me where I can find a proof of the above implication written down?

Answer by David Loeffler for The existence of an elliptic curve with a specific Galois representation induced by a character

2012-10-28T09:46:46Z

In this context, if $\rho$ is a mod $\ell$ representation of $Gal(\overline{F} / F)$, and $A$ is an elliptic curve over an extension $F' / F$, then the statement "$A[\ell] \cong \rho$" needs a little bit of interpretation, because the two sides are representations of different things: $A[\ell]$ is a mod $\ell$ representation of the subgroup $Gal(\overline{F} / F') \subset Gal(\overline{F} / F)$. So the statement is to be read as "$A[\ell]$ is isomorphic as a $Gal(\overline{F} / F')$-representation to the restriction of $\rho$". Now, the bigger $F'$ is, the weaker this condition becomes: in particular, if we take any elliptic curve $A$ over $F$ and define $F'$ to be the extension of $F$ generated by the $\ell$-torsion points of $A$ and the splitting field of $\rho$, then the statement is automatic (both sides are the trivial representation).

(This is kind of a stupid example, but maybe you can believe now that there exist non-stupid examples as well!)

Answer by David Loeffler for Extending arithmetic functions (and associated Dirichlet series) to arbitrary rings of integers

2012-10-04T11:24:23Z

The answer to your Question 1 is "yes". It's clear that the number of ideals of $\mathcal{O}$ of norm $\le M$ is bounded above by a polynomial in $M$, so one can manipulate Dirichlet series term-by-term for $Re(s) \gg 0$ and argue that

$$ \zeta_K(s)^2 = \sum_{A, B} N(A)^{-s} N(B)^{-s} = \sum_{C} \#\{ (A, B) : AB = C\} N(C)^{-s} = \sum_C d(C) N(C)^{-s}.$$

As for Question 2 it's not entirely clear to me what your precise question is, but philosophically at least the answer is "yes" -- any arithmetical function $a$ definable purely in terms of ideals of $\mathbb{Z}$ will have a natural generalization to ideals of a number field, and if you can express $\sum_n a(n) n^{-s}$ in terms of the Riemann zeta then the same argument should give you an expression for $\sum_{I \triangleleft \mathcal{O}} a(I) N(I)^{-s}$ in terms of $\zeta_K(s)$.

Answer by David Loeffler for Maximal tamely ramified extension of $\mathbf Q_p$

2012-09-18T16:12:40Z

Yes, there is. The maximal unramified extension is obtained by adding all roots of unity of order prime to $p$. The maximal tame extension is obtained by adding on top of that all $n$-th roots of $p$, for $n$ prime to $p$; so its Galois group is isomorphic to $\prod_{\ell \ne p} \mathbf{Z}_\ell$, and conjugation by $\operatorname{Gal}(\overline{\mathbf{Q}}_p^{nr} / \mathbf{Q}_p)\cong \widehat{\mathbf{Z}}$ acts on each $\mathbf{Z}_\ell$ factor via the $\ell$-adic cyclotomic character.

(EDIT: For a reference for this see Pete Clark's notes at http://math.uga.edu/~pete/8410Chapter4.pdf, in particular Theorem 11 and Corollary 12.)

Comment by David Loeffler

2013-05-14T10:45:55Z

It's still not clear why you expect there to be a meaningful definition of "period matrix" in this wider context.

Comment by David Loeffler

2013-05-12T20:46:49Z

(Sorry, that should say $GSpin(2g + 1)$, of course.)

Comment by David Loeffler

2013-05-12T20:46:08Z

@Keerthi: it is true that irreducible 2g-dimensional symplectic Galois representations should correspond to automorphic forms on $GSpin(2n+1)$, but there is an exceptional isomorphism between $GSpin(5)$ and $GSp(4)$.

Comment by David Loeffler

2013-05-12T09:19:31Z

The H^1 is (up to twist) the Tate module of the Jacobian, and the representation on the Tate module of the Jacobian has to preserve the Weil pairing into roots of unity, which is a symplectic pairing.

Comment by David Loeffler

2013-05-10T09:15:24Z

I fixed the formula. Sometimes putting backticks around formulae helps.

Comment by David Loeffler

2013-05-10T06:28:59Z

I am no expert here, but I thought the CSP was still open for lots of real rank 1 groups. I guess it all depends wht kind of groups you consider "natural".

Comment by David Loeffler

2013-05-09T09:01:55Z

This has nothing to do with modular forms, and is too elementary for this site. Voting to close. You might have better luck at math.stackexchange.com.

Comment by David Loeffler

2013-05-09T06:12:20Z

Of course! Very pretty. This one could certainly use in an undergrad course.

Comment by David Loeffler

2013-05-09T06:09:04Z

The title of this question is nonsense -- "modularity" means showing that a given Galois representation is associated to an automorphic form, but you're going the other way here. Also, "modular forms of weight one modular forms"?

Comment by David Loeffler

2013-05-08T20:06:53Z

I added an explicit example for index 7.

Comment by David Loeffler

2013-05-08T10:58:48Z

I think filling in all the steps here is actually quite a lot harder than the argument I mentioned above. Nonetheless, if you assume Deligne--Serre it's certainly a nice short argument!

Comment by David Loeffler

2013-05-07T06:42:07Z

The "exact form of K" depends rather on whether $F$ is archimedean or nonarchimedean. For $F = \mathbb{R}$, you can use the fact that the indefinite special unitary group in 2 variables, $SU(1,1)$, is isomorphic to $SL(2)$ and the max compact subgroup of the latter is well known.

Comment by David Loeffler

2013-05-03T08:14:23Z

You probably want to read a book on modular forms; but modern books on this subject (Diamond+Shurman is the standard one) tend to avoid stressing these sorts of explicit special-functions aspects in favour of more abstract treatments based on Riemann surfaces. You could perhaps try the section on modular forms + modular functions in Knapp's book on elliptic curves, or McKean + Moll's book on elliptic functions + elliptic curves

Comment by David Loeffler

2013-04-29T16:35:10Z

I'm voting to close this, sorry. Please formulate your questions a bit more carefully.

Comment by David Loeffler

2013-04-27T20:32:46Z

@RH: I don't understand, what exactly are you asking?