## The Thick Diagonal

Suppose that $F$ is an imaginary quadratic field. Suppose that $\pi$ is a cuspidal automorphic form for $\mathrm{GL}(2)/F$ of cohomological type, and let us suppose that it contributes to the cohomology group $H^1(\Gamma,\mathbf{C})$ for some congruence subgroup $\Gamma$ of $\mathrm{GL}_2(\mathcal{O}_F)$. Choose a prime $p$ which splits in $F$ so that $\pi$ is ordinary at $v|p$. Hida proves that the corresponding cohomology class lives in a Hida family $\mathcal{H}$ over the appropriate weight space, which in this case is (up to connected components) just $\Lambda = \mathbf{Z}_p[[X,Y]]$. However, unlike the classical situation, this Hida family will not be flat, because the specialization to any local system which is not invariant under complex conjugation is necessarily finite. Thus the support $D$ of $\mathcal{H}$ has co-dimension at least one over $\Lambda$. Hida proves that it does indeed have co-dimension one.

What does the support $D$ of $\mathcal{H}$ look like? Let us suppose that we are normalizing $\Lambda$ so that the point $X = Y = 0$ corresponds to $\pi$. One can imagine two possibilities:

1. $D$ contains the diagonal $\Delta: X = Y$.
2. the components of $D$ passing through $X = Y = 0$ only contains finitely many classical points.

It seems as though these are the only possibilities. Certainly, by a Zariski closure argument, $D$ either contains the diagonal $\Delta$ or intersects it in finitely many points. Hence, it is true that if the first condition does not hold, then the components passing through $[0,0]$ contain only finitely many crystalline automorphic forms. However, there could be more classical points on $D$, namely, those of parallel weight but non-parallel finite order nebentypus character. To be concrete, the possible points of $\mathrm{Spec}(\Lambda)$ which may give rise to automorphic forms have (with some normalization) the following shape:

$1+X \mapsto (1+p)^k \zeta, \qquad 1+Y \mapsto (1+p)^k \xi,$

where $\zeta$ and $\xi$ are $p$-power roots of unity, and $k$ is a non-negative integer. So one is really considering not simply the intersection of $D$ with the diagonal $\Delta$, but with the thick diagonal ${\Delta\kern-0.6em{\Delta}}$, which is the union of the infinitely many translates of $\Delta$ by $p$-power roots of unity. In particular, the Zariski closure of ${\Delta\kern-0.6em{\Delta}}$ is all of weight space.

I wrote a paper with Barry Mazur where, as an illustrative example, we found an explicit Hida family which did not satisfy the first condition and claimed that it therefore satisfied the second, whereas we should only have made the weaker claim that $D$ (which was irreducible in this particular case) contains only finitely many crystalline points. (The main point of the paper was, by studying infinitessimal deformations of Artin representations, to give evidence that $D$ should only ever contain the diagonal when $\pi$ is either a base change form or CM.) The error was pointed out to me by David Loeffler.

I am pleased to say, however, that my student Vlad Serban has overcome this error! Namely, suppose one has a non-trivial power series $\Phi(X,Y) \in \mathbf{Z}_p[[X,Y]]$, and suppose that

$\Phi((1+p)^k \zeta - 1,(1+p)^k \xi - 1) = 0$

for infinitely many triples $(k,\zeta,\xi)$ with $k$ a non-negative integer, and $\zeta$, $\xi$, $p$-power roots of unity. Let $D$ be a component of the zero set $\Phi(X,Y) = 0$ passing through $(0,0)$. Then, after possibly replacing the roles of $X$ and $Y$, Vlad proves the following. Either:

1. $D$ contains the diagonal $\Delta$,
2. $\Phi(\zeta - 1, \zeta^N - 1) = 0$ for all $p$-power roots of unity $\zeta$, for a fixed $N \in \mathbf{Z}_p$.

Certainly the latter is possible, because one could have $\Phi(X,Y) = (1+X)^N - (1+Y)$. In fact, he proves a more general theorem than this for all the components (not necessarily passing through $(0,0)$. After translation, this amounts to working over ramified extensions of $\mathbf{Z}_p$.

This theorem allows one to prove (with finite computation) that any particular $D$ only contains finitely many points (when that is true). It also shows, without any computation at all, that $D$ either contains $\Delta$, or it only contains finitely many classical points of weight different from $\pi$. A nice way to think about this theorem is that it is of the flavour as the multiplicative Manin-Mumford conjecture. That is, one is intersecting a sub-variety with a particular arithmetically defined discrete set (inside ${\Delta\kern-0.6em{\Delta}}$), and one wants to deduce that this can only happen for a well defined geometric reason. In fact, if one replaced $\Phi(X,Y)$ by a polynomial with coefficients over $\mathbf{C}$ and specialized to the case when $k$ is always zero, then this would exactly be the Multiplicative Manin-Mumford conjecture in two dimensions.

As a special case, letting $k = 0$, one ends up with the following pretty result. Suppose that $\Phi(X,Y) \in \mathbf{Z}_p[[X,Y]]$ is a power series, and suppose that

$\Phi(\zeta_1 - 1,\zeta_2 - 1) = 0$

for infinitely many pairs of $p$-power roots of unity. Then the zero set of $\Phi$ contains a translate of $\mathbb{G}_m$. This exactly answers the puzzle asked by Jordan here. Explicitly, it says that the only quotients of $\mathbf{Z}_p[[\mathbf{Z}^2_p]]$ of co-dimension one which have lots of “arithmetic” points really do come from a one-dimensional subgroup!

I think that this special case (with $k = 0$) is probably easier than the general case, because one has other methods available. The argument was, however, inspired by a result of Hida which came up during his last number theory seminar at Northwestern. Translated into the language of this post, Hida’s rigidity lemma corresponds to the puzzle of Jordan above in the case when $\Phi(X,Y) = Y - F(X)$ for some function $F(X) \in \mathbf{Z}_p[[X]]$.

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### 5 Responses to The Thick Diagonal

1. In the case k=0, I think I have a simple argument. Switching x and y if necessary, write infinitely many of the solutions as $\zeta_2 = \zeta_1^a$, with $a$ depending on the solution, of course. Now take a subsequence of the a’s that converge p-adically to N. Then $\Phi(t,(1+t)^N-1)$ is a power series with infinitely many zeros, so vanishes identically.

• Deer Felipe, I think you have to be slightly careful; even if the $a$‘s converge to $N = 1$, that doesn’t mean that $\Phi(t,t) = 0$ has to have any zeros. For example, the nth root could be

$(\zeta_{p^{2n}},\zeta^{1 + p^n}_{p^{2n}})$

On the other hand, Vlad’s argument does proceed (in this case) by first showing that $\Phi(t,(1+t)^N - 1) = 0 \mod p$; he uses the fact that the valuation of $\zeta - 1$ becomes arbitrarily small.

2. Good point. Thanks for the clarification. Is there a preprint?

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