## Short thoughts on my visit to Berkeley

The marine biologists at Monterey Bay Aquarium give their octopuses hand massages. So do the fishmongers at Eataly.

It’s quite an experience to come face to face with this antediluvian monster:

Are you allowed to turn right on red in California? I hope so, because I did so at every opportunity.

Almost mathematics is an endless source of almost humor.

Matt complained that I was not live blogging the hot topics conference from MSRI. But given that they record all the lectures, it seemed somewhat unnecessary.

The winter olympics are still on; don’t forget the greatest ever moment in winter olympic history (video in Dutch, but then Dutch sounds pretty much like English):

Does one ever tire of watching the sunsets in Berkeley?

Wine tasting with Ken in Arlington. Why do I not yet have a wine subscription from Kermit Lynch?

Heavyweight matchup: userxxx versus answer_bot live!

The pork burrito from Gordo Taqueria was simply the best burrito I have had for a very long time.

The initial letters of “perfectoid space” and “Peter Scholze” are both PS. Coincidence? You decide.

What foods do you miss from your native land (those of you who are immigrants)? Answers included sour cherries and instant custard.

A tip to anyone planning to visit wineries in Napa valley: don’t casually turn up around 5:00 only to find that many tasting rooms close before then.

The Maître d’ at Chez Panisse Cafe asked me to give my best wishes to Hendrik Lenstra.

Popular consensus was that the only way the conference could have been better was if Peter Scholze had given every talk.

Cheeseboard Pizza versus Sliver: Cheeseboard wins with a tapenade pizza.

Going to Berkeley and having lunch at MSRI is like going to the grand canyon and never leaving the visitor’s center.

Babette is the place to have lunch and hipster coffee near campus; hat-tip to Tony Wirth.

Kiran is happy that his research is now a hot topic.

Is “perfectoid” a portmanteau of “perfect” and “affinoid” or does it mean “somewhat perfect”?

And men of mathematics now-a-bed
Shall think themselves accurs’d they were not here,
And hold their manhoods cheap whiles any speaks
That thought with us upon this Pres’dents’ day.

## Local crystalline deformation rings

I just returned from a very pleasant conference in Puerto Rico courtesy of the Simons Foundation (general advice: if you live in Chicago, always accept invitations to conferences in January). One thing I learnt from Toby Gee was the following nice observation. Suppose that

$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p)$

is a modular Galois representation, which for convenience we shall assume is unramified outside $p$. Consider deformations of this representation which are crystalline with fixed Hodge-Tate weights $[0,k-1]$ where $k$ is even. According to Kisin, the global minimal crystalline deformation ring contains a point on every component of the corresponding local crystalline deformation ring. (All discussions of components refer to the generic fibres.) One natural question is how many components the local deformation rings actually have (when the weight is very small, it’s usually the case that there is only one such component and it is smooth — this was crucial in the original Taylor-Wiles method before Kisin). For higher weight, one can distinguish between components which are “ordinary” and “not ordinary”, but it is not clear what else there is. (Indeed, Kisin seemed to think some years ago that this would be it, using the meta-argument that amongst any finite set one should be able to distinguish different points by some naturally available property.)

Now suppose we also now assume that $\overline{\rho}$ is locally reducible. According to Buzzard’s conjectures, all the slopes of the global crystalline lifts of $\overline{\rho}$ will be integral. Suppose one wants to prove this by local methods. Then one is ultimately led to conjecturing that each component of the local crystalline deformation ring has a fixed integral slope (recall we are in the locally reducible case, this is certainly false for locally irreducible representations in general). As a first consequence, one sees that in very high weights there will be many different components. Moreover, if one takes a different
global representation $\overline{\varrho}$ which is the same locally as $\overline{\rho}$, then the set of slopes arising from lifts of $\overline{\varrho}$ will be the same as for $\overline{\rho}$. These ideas do not quite give a complete conjectural explanation of why Buzzard’s slope conjectures are true, but it is a good start.

Something that is a little disturbing in this picture, however, is the case when $\overline{\rho}$ is reducible. It becomes clear that, in high weight, there will be many crystalline representations with reducible residual representations, but the set of components of local crystalline deformation space which have a global point will be a proper subset of the set of components (assuming that components can be distinguished by slope).
For example, all the slopes at level one when $p = 2$ are (besides the Eisenstein series) $\ge 3,$ but there certainly exist modular forms of higher tame level with the same local residual representation of slope one. So is there any way to predict when a reducible representation will have a global lift on any component of local deformation space?

In fact, the failure of lifts in the reducible case is an old problem. In the most naive sense, one can find reducible representations at levels where there are no cusp forms, but to play the game honestly we should also allow (globally) reducible lifts. Perhaps the first genuine example corresponds to extensions:

$1 \rightarrow \mathbf{Z}/p\mathbf{Z} \rightarrow V \rightarrow \mu_p \rightarrow 1$

where the extension is completely split at $p$ but ramified at an auxiliary prime $N$. These representations are locally split and so certainly admit local lifts (namely, $\mathbf{Z}_p \oplus \mathbf{Z}_p(1)$). If $p > 3,$ then such extensions exist whenever $N \equiv \pm 1 \mod p$, but (by Mazur) one knows that there exist weight two level $\Gamma_0(N)$ lifts only when $N \equiv +1 \mod p$ (in fact, one can prove the analogous claim that there only exist global crystalline lifts with the appropriate conductors under the same congruence condition). This is related to the general problem of understanding when certain reducible representations can be lifted to cusp forms, which seems to be a tricky problem (Ken Ribet’s student Hwajong Yoo has thought about this).

This also reminds me of a fact I learnt from Kevin Buzzard. Take the representation

$\rho: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{Q}_2)$

associated to the cusp form $\Delta$. Then there exist lattices for this representation such that the corresponding residual representation is any one of the four (three non-split) extensions of $\mathbf{Z}/2\mathbf{Z}$ by itself which are unramified outside $2$. (Question: does this immediately imply the same is true for all $2$-adic representations coming from level one modular forms?)

## Gross Fugue

Here are some variations on the theme of the last post, which is also related to a problem of Dick Gross.

In this post, I want to discuss weight one modular forms where the level varies in the “vertical” aspect (that is, $N$ is a growing power of a fixed prime, rather than simply an increasing integer). First of all, consider the spaces

$S_1(\Gamma(M \cdot \ell^n),\mathbf{C})$

for fixed $M$ and growing $n$. For example, if $M = 1$, the corresponding Galois representations are associated to number fields unramified outside a single prime $\ell$. Given a cusp form $f$, the twists $f \otimes \chi$ by a finite order character of $\ell$-power order will also be modular (possibly with larger $n$), so all the finiteness statements below should be interpreted “up to twist.”

The first observation is that there exist only finitely many exceptional cusp forms (with projective image $A_4, S_4, A_5$) because, by a theorem of Hermite, there are only finitely many fields with a fixed Galois group unramified outside a fixed set of primes. (This echos a very general conjecture which says [very loosely] that if one fixes an infinitesimal character and varies the level in a $\ell$-adic tower, one should only see finitely many automorphic forms which do not arise via functoriality from constructions using discrete series.)

The second observation is that all the other cusp forms are easy to describe: they are induced from finite order characters of a fixed number of easily determined quadratic fields $K$.

So far so good. But what happens if one replaces $\mathbf{C}$ by $\mathbf{F}_p$, or more generally $\mathbf{Q}_p/\mathbf{Z}_p$? Here is the following optimistic guess:

Question: For a fixed prime $p \ne \ell$, are there only a finite number of non-liftable forms in the $p$-power tower?

Here we have to take the usual caveats — not only do we have to take into account twisting, but also the $\mathbf{GL}_2(\mathbf{Q}_{\ell})$-action (old forms).

This question is supposed to be a $\mathrm{GL}(2)$-analogue of Washington’s famous theorem on the $p$-part of the class group in the $\ell$-adic cyclotomic tower. We shall see that it is more than an analogy.

What will the source of torsion classes be?

1. One source are Galois representations: $\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_q)$ with big image that are unramified outside $\ell$ (with $q$ a power of $p$). Of course there are only finitely many such representations for any fixed $q$, but some heuristics I learnt from Akshay convince me that there should only be finitely many even if one varies $q$ over all powers of $p$ (taking into account twisting, of course).
2. Another source of torsion comes from deformations of big image Galois representations $\overline{\rho}$ as above, or from representations with projective image one of the exceptional groups. Since each unramified deformation ring will be finite, each $\overline{\rho}$ should only give rise to finitely many extra torsion classes.
3. A third source of torsion classes comes from reducible indecomposable representations. The residual representations $\overline{\rho}$ which arise in this way occur when $L(0,\chi)$ is divisible by $p$ for an odd character $\chi$ of finite order. In particular, there are only finitely many such representations which occur exactly if all but finitely many $L$-values $L(0,\chi)$ are prime to $p$, where $\chi$ is an odd character of conductor $M$ times a power of $\ell$. But this exactly the content from Washington’s Theorem (the oddness assumption is not, however, necessary).
4. The final class come from deformations of dihedral representations. If $\overline{\rho}$ is the induction of a character $\psi$ of $K/\mathbf{Q}$, then the tangent space to the unramified deformation ring of $\overline{\rho}$ gives rise to torsion classes when there are no everywhere unramified classes in $H^1(\mathbf{Q},\mathrm{Ind}(\psi/\psi^c))$ — the unramified dihedral representations in $H^1(\mathbf{Q},\eta_K)$ are seen globally. By inflation-restriction, this is equal to a certain invariant part of the class groups of the anti-cyclotomic tower. There are non-vanishing results concerning L-values of Hida that are relevant here, although I haven’t checked to see if they imply the finiteness statement or not.

The only way to start thinking about answering this question is to think in terms of the torsion in the cohomology of modular curves. But, I confess, I do not really have any ideas on how to prove it. (To be honest, I still find Washington’s proof very mysterious.)

On related matters, it would be nice if one could prove — say by analytic means — that $H^1(X_H(N),\omega)$ has torsion (prime to $N$) for all sufficiently large $N$. Taking $N$ to be a power of a prime, this would give a different construction of non-solvable Galois representations unramified outside a single prime (for all $\ell$) from the one suggested by Dick and carried out in for $\ell \in \{2,3,5,7\}$ by Dembélé and others. Moreover, although (as in those examples) it would involve the group $\mathrm{PSL}_2(\mathbf{F})$ as a simple factor, the residue characteristic would be different from $\ell$ rather than equal to $\ell$ in the previous constructions. (George Schaeffer told me he tried computing torsion coming from $X_H(343)$ but didn’t find any.) It might also (for suitable $H$) give a lower bound for $\pi_1(\mathcal{O}_{K})$ where $K = \mathbf{Q}(\sqrt{-D})$ which is better (at least for some primes) than one gets from class numbers.

## The mystery of the primes

No, this is not the sequel to Marcus du Sautoy’s book, but rather a curious observation regarding George Schaeffer’s tables of “ethereal” weight one Katz modular eigenforms (which you can find starting on p.64 here). Let $N$ be a positive integer, let $\chi$ be an odd quadratic character of conductor dividing $N$, and let $p$ be an a prime not dividing $N$. Recall that the reduction map between spaces of Katz modular forms:

$M_1(\Gamma_1(N), \chi, \mathbf{Z}_p) \rightarrow M_1(\Gamma_1(N), \chi, \mathbf{F}_p)$

is not surjective in general, although it will be surjective for all but finitely many $p$. For what pairs $(N,p)$ is the map not surjective? As originally observed by Mestre (and predicted by Serre), such pairs do exist. One way to think of the primes which arise in this way are as the primes dividing the torsion subgroup of $H^1(X_H(N),\omega)$, where $H \subset (\mathbf{Z}/N \mathbf{Z})^{\times}$ is the subgroup of squares, and $X_H(N)$ is the corresponding modular curve (as a stack, if necessary) over $\mathbf{Z}[1/N]$. The reduction mod-p map is Hecke equivariant; let $\mathfrak{m}$ denote a maximal ideal of $\mathbf{T}$ in the support of the cokernel. Associated to $\mathfrak{m}$ is a Galois representation:

$\overline{\rho} = \overline{\rho}_{\mathfrak{m}}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_p)$

which is unramified at primes not dividing $N$ (including $p$). It is not necessarily the case that $\overline{\rho}$ does not lift to characteristic zero, although this is typically the case for the examples arising in the tables (and is always the case if the image of $\overline{\rho}$ contains $\mathrm{SL}_2(\mathbf{F}_q)$ for some $q > 5$). Not surprisingly, it turns out there are no such forms for small $N$. The reason is that the fixed field $K$ of the kernel of $\overline{\rho}$ would be a high degree field with a root discriminant which (for very small $N$) would violate the GRH discriminant bounds of Odlyzko, and for smallish $N$ would still give fields of unusually low root discriminant.

Of course, as $N$ increases, there do exist many such forms, sometimes in quite large characteristic. However, something peculiar happens in the range of the tables, namely, there is not a single example with $N$ prime. This leads to the (incredibly) vague question: can this be predicted in advance? If there is going to exist a $\mathrm{PGL}_2(\mathbf{F}_{199})$ representation unramified outside $N$ for small $N$, is it more likely that $N = 82$ (see here) rather than $N = 83$? One can try to use heuristics predicting the number of fields with certain ramification behavior, but these heuristics are much better behaved for fixed Galois groups $G = \mathrm{PGL}_2(\mathbf{F}_p)$ or $G = \mathrm{PSL}_2(\mathbf{F}_p)$ and increasing discriminant, not in the regime of fixed root discriminant and Galois group $G$ as above for varying $p$. Is there any conspiracy ruling out certain kinds of number fields with small root discriminant ramified at a single prime? For example, if you fix some arbitrary constant, say $M = 1000$, do there exist infinitely many primes $p$ such that there is a number field $K$ different from $\mathbf{Q}$ which is unramified away from $p$ and has root discriminant less that $M$?

These questions are hard to pin down, because they are really questions concerning the law of small numbers. Namely, they ask about the behavior/distribution of various quantities in the range before asymptotic behavior begins. Since the asymptotic behavior is (in these contexts) already mostly conjectural, it’s probably hard to say anything intelligent about these even more delicate questions. (Idle question: are there similar problems for which one does understand what happens before the asymptotic regime begins, even heuristically?)

Here’s one reason to consider these questions. Suppose one wants to compute “ethereal” Siegel modular forms. At what level does one first expect to find such forms? The numerics above suggest that it might be easier to find such forms at small composite levels rather than prime levels. Is that a reasonable inference?

Posted in Mathematics | | 4 Comments

## Daleks

I’ve wanted to write a post about the new Doctor Who series for a while, but this is not that post. Instead, this post is about a Macintosh game called Daleks, which I first played on a Mac 512 (running OS 3) in the mid-80′s. Research indicates that this game was based on a Unix game called robots, and that some wag came up with the idea of rebranding it under the name of the classic Doctor Who monster. The first version I played had a very peculiar high score table: all the high scores were attributed to a fellow named “fingers,” and the high scores were not in any sort of numerical order. Moreover, no matter what one scored, it was impossible to permanently make it onto the high score list. My second encounter with the game was during a summer research program with Alf van der Poorten in 94/95, where I was impressed to find that he had broken 10000. Later, I had a copy on an ancient laptop given to me by my brother, and still later, I played classic Daleks in classic mode under OS X. I am not ashamed to say that I am proud of my high score, 15670, a feat which is probably meaningless to almost everyone. Anyway, today’s post is about some mathematical problems related to this game. If you have a mac computer, I recommend playing around with some current incarnations of the game, for example super Daleks (presumably robot is available on Gnome games as well):

Consider the following game: the Doctor is positioned on the lattice $\mathbf{Z}^2$ at the origin $(0,0)$, and Daleks are distributed on the rest of the lattice with uniform density $\rho \in (0,1)$. It turns out that it is more convenient to work with the parameter $q = 1 - \rho$, although all the graphs below are drawn with respect to $\rho$. On each move, the Dalek at point P moves to the unique neighbouring square (out of 8) which is closest to the origin in the taxicab metric. In particular, Daleks always move diagonally towards the origin unless they lie on one of the axes. If two or more Daleks occupy the same square, then they crash and are destroyed, leaving a pile of debris which remains at that square forever. Moreover, any other Dalek which later moves on to the same square now occupied by the debris is also destroyed. If a Dalek reaches the origin unscathed, the Doctor is exterminated. However, if the debris resulting from Dalek collisions prevents all other Daleks from reaching the origin, then the Doctor survives. What is the probability that the Doctor survives? (In the computer game the Doctor can also move about, but not in our simplified version.)

If a Dalek starts on either the diagonal or the anti-diagonal then it will never crash with another Dalek (in general, Daleks can only crash on the axes). Hence, we modify the game by forbidding Daleks from either of these diagonals. This effectively separates the playing area into four quadrants which do not interact, and so we may as well confine ourselves to a single quadrant, and assume that all Daleks lie in the quadrant $(1,0) + Q$ where $Q = (x,y)$ with $x \ge |y|$. A sample game is as follows, with the positions at time $t =$ 0, 1, and 2. The Doctor will win this game, because the debris at $(1,0)$ will prevent all other Daleks in the quadrant from reaching the origin (they will crash into the debris and be destroyed):

Let $Q_d$ be the truncated quadrant consisting of $(x,y)$ with $|y| \le x \le d$. Let $w_d$ denote the probability of surviving the game where Daleks only exist with density $\rho$ in this quadrant. It is clear that

$1 - w_d = \displaystyle{\sum_{P \in (1,0) + Q_d} E(P)},$

where $E(P)$ is the expectation of being exterminated by a Dalek which originates at point P. (If there is a Dalek which kills the Doctor, it is unique.) Note that $E(P)$ is independent of $d$, providing that $P \in (1,0) + Q_d$.

Definition: The occludation $O(P)$ of P consists of the squares R different from P where a Dalek at square R will reach the origin before or at the same time as P, and which reach the $x$-axis at least as near to the origin as P reaches the $x$-axis.

The Daleks $R \in O(P)$ are those for which R is in the shadow of R, namely, those R which eventually occlude P from the origin (thus the name). It is not a great name, but I couldn’t think of anything better. Explicitly, if $P = (x,y)$, then

$O(P) \cup P = \{(a,b) \in (1,0) + Q \ \text{such that} \ a \le x, \ \text{and} \ a - |b| \le x - |y|\}$

The Doctor can only be killed by Dalek at point P if the occludation $O(P)$ is empty of Daleks. The reason is that any Dalek R in the occludation can only crash at points on the $x$-axis where P must eventually travel, and R will reach this point either at or before P does. We have

$|O(P)| = |O((1,0) + (x,y))| = x^2 - y^2 - 1.$

On the other hand, conditional on the assumption that the occludation contains no daleks, then the probability that P exterminates the Doctor only depends on $y$; namely, it is equal to the probability of surviving the game with $Q_d$ and $d = |y|$. (This the the quadrant not in the occludation of $P$.) It follows that

$E(P) = q^{|O(P)|}(1-q) w_{|y|} = q^{x^2 - y^2-1}(1-q) w_{|y|},$

and hence

$1 - w_d = \sum_{n=1}^{d} \sum_{|m| < n} E((n,m)) = \sum_{n=1}^{d} \sum_{|m| < n} q^{n^2 - m^2 - 1}(1-q) w_{|m|}.$

We may simplify this slightly by writing

$w_{d-1} - w_{d} = (1-w_{d}) - (1-w_{d-1}) = \sum_{|m| < d} q^{d^2 - m^2 - 1}(1-q) w_{|m|},$

This simplifies even further to

\begin{aligned} & (w_{d} - w_{d+1}) - q^{2d+1} (w_{d-1} - w_{d}) \\ = & \sum_{|m| < d+1} q^{(d+1)^2 - m^2 - 1}(1-q) w_{|m|} - q^{2d+1} \sum_{|m| < d} q^{d^2 - m^2 - 1}(1-q) w_{|m|} \\ = & \ 2 q^{2d}(1-q) w_{d}, \end{aligned}

and hence, subject to $w_0 = 1$ and $w_1 = q$,

$w_{d+1} = (3 q^{2d+1} - 2 q^{2d} + 1) w_d - q^{2d+1} w_{d-1}.$

The resulting recurrence relation for $w_d$ gives a decreasing convergent sequence (for each fixed $q$ and also in $\mathbf{Z}[[q]]$) with limit

$w_{\infty} = q - 3q^3 + 3q^4 - 2q^5 + 2q^6 + 4q^7 - 13q^8 + 13q^9 + \ldots$

Here is a graph of this function (with respect to $\rho$, remember that $\rho = 1 - q$):

Although it appears from the graph that the maximum occurs at $\rho = q = 1/2$, closer inspection reveals that the optimal density is $\rho = 0.517208\ldots$ The maximum value is approximately $\sim 0.28116\ldots$, which means that, on an entire plane with all four quadrants, the largest possible chance of winning (with daleks on the diagonal and anti-diagonal removed) is approximately 1 in 160.

Reverse the Polarity: Here is a different way to estimate $w_{\infty}$, this time from below. In order for the Doctor to survive, two or three Daleks must eventually coincide at $(1,0)$. Call such Daleks savior Daleks. All savior Daleks must be in the same row, and at least one such Dalek must lie on the edge of the quadrant. Let us now consider the probability $s_n$ that one will be “saved” by a Dalek in the $n$th row. If $P = (n,n-1)$ is a savior Dalek, then the Dalek $P$ creates the first crash at the point $(0,1)$, and no Dalek exterminates the Doctor before this point. It follows that no Daleks may occlude $P$, and hence $O(P)$ must be free of Daleks, with the possible exception of $-P$. Note that $|O(P)| = n^2 - (n-1)^2 - 1 = 2n$. Suppose that $-P$ is not occupied. Then (assuming that $O(P)$ is empty) $P$ will be a savior Dalek if and only if the remaining restricted quadrant of size $n-1$ would otherwise result on the Doctor being exterminated at the final term, equivalently, the probability that, from a quadrant of size $n-1$, the Doctor would be exterminated by a Dalek in the last row. Yet the probability of this is

$(1 - w_{n-1}) - (1 - w_{n-2}) = w_{n-2} - w_{n-1},$

and hence the contribution to $s_n$ is

$2q^{2n-2}(1-q)(w_{n-2} - w_{n-1}).$

On the other hand, if both $P$ and $-P$ are to be savior Daleks, then one simply requires that, in addition to the rest of occlusion $O(P)$ being empty, that in the remaining quadrant of size $n-2$ (removing the final row and the occlusion) no Doctor is exterminated, and this has probability $w_{n-2}$. Hence

$s_{n} = 2q^{2n-2}(1-q)(w_{n-2} - w_{n-1}) + q^{2n-3} (1-q)^2 w_{n-2}.$

Let $t_n$ be the probability that there exists a savior Dalek at a row at most $n$. Then clearly

$t_{n} = \sum_{m=1}^{n} s_m.$

Moreover, we naturally have inequalities $w_n \ge t_n$, and

$w_{\infty} = \lim_{n \rightarrow \infty} w_n = \lim_{n \rightarrow \infty} t_n,$

where the limit is pointwise and $q \ne 1$. However, the behavior of $w_n$ and $t_n$ is quite different in the regime $\rho \rightarrow 0$ or $q \rightarrow 1$, as the following graph of $w_5 \ge t_5$ shows:

Behavior as $\rho \rightarrow 0$:

In order to estimate the behavior of $w_{\infty}$ as $q \rightarrow 1$, we consider the following problem:
What is the probability that the first row with any Dalek contains exactly two Daleks, and that at least one of these Daleks lies at the edge of the quadrant? In such a situation, the Daleks necessarily annihilate one another at $(1,0)$, and the Doctor is saved. Call the resulting function $A(q)$, so $w_{\infty} > A(q)$. Since there are $(4n-1)$ pairs of elements in $1,\ldots,2n+1$ which contain at least one of the end points, we have

\begin{aligned} A(q) = & \ \sum_{n=1}^{\infty} (4n-1) q^{n^2}(q^{2n-1} (1-q)^2) = (q-1)^2 q^{-2} \sum_{n=2}^{\infty} (4n-5) q^{n^2} \\ = & \ (q-1)^2 q^{-2} \left(5 + q + \sum_{n=0}^{\infty} 4n q^{n^2} - 5 \sum_{n=0}^{\infty} q^{n^2}\right) \\ = & \ (q-1)^2 q^{-2} \left(\frac{5}{2} + q +2 \sum_{n=-\infty}^{\infty} |n| q^{n^2} - \frac{5}{2} \sum_{n=-\infty}^{\infty} q^{n^2}\right) \\ = & \ (q-1)^2 \left(2 \sum_{n=-\infty}^{\infty} |n| q^{n^2} - \frac{5}{2} \sum_{n=-\infty}^{\infty} q^{n^2}\right)(1 + O(1)) + O((q-1)^3).\end{aligned}

Let $q = e^{-\tau}$. As $q \rightarrow 1$, we have $\tau \rightarrow 0$, and so $1-q \sim \tau$. On the other hand, by Poisson summation, we have

$\displaystyle{\sum_{n=-\infty}^{\infty} e^{-n^2 \tau} \sim \sqrt{\frac{\pi}{\tau}} + O(\tau^N),}$

$\displaystyle{\sum_{n=0}^{\infty} |n| e^{-n^2/\tau} \sim \frac{1}{\tau} - \frac{1}{6} - \frac{\tau}{60} - \frac{\tau^2}{252} + O(\tau^3),}$

from which it follows easily that $A(q) \sim 2 \tau \sim 2(1-q)$, and thus

$\limsup_{q \rightarrow 1} w_{\infty} \ge 2(1-q).$ In fact, we can actually prove that

$w_{\infty}(e^{-\tau}) = \displaystyle{\frac{2}{\tau} + O \left(\frac{1}{\sqrt{\tau}}\right)}.$

In other words, the simple model above is very accurate in the limit $q \rightarrow 1$. However, the combinatorics required to prove this are actually somewhat involved and annoying, and this is a blog, so I will omit it here. (The arguments are somewhat timey–wimey.)

A conditional game:

Consider the game which is pre-conditioned on the first square $(1,0)$ being empty. Since that square containing a Dalek is not consistent with survival, the new game results in a win with probability:

$\displaystyle{c_{\infty} = \frac{w_{\infty}}{1 - \rho} = \frac{w_{\infty}}{q}.}$

Apropos of nothing, here’s Davros enjoying a cuppa:

The TARDIS:

Suppose that the Doctor has a TARDIS. This allows him, at any point, to dematerialize and the materialize somewhere else. In the context of the classic Daleks game, the player appears at a random point in the plane with uniform distribution. Although this doesn’t quite make sense on an infinite plane, we can take it to mean that we have moved sufficiently far away from the axes that it is as if the game has started again. Hence this will be the context in which we shall consider rematerialization, namely, as if the game has started again. The catch with using the TARDIS is that the Doctor may materialize next to a Dalek, in which case he is immediately exterminated. The optimal strategy is to continue to continue rematerializing until one has a winning game. The chance of surviving a rematerization is $(1 - \rho)$; the resulting game is the same, but now conditional on not being annihilated by the initial Dalek, hence is equivalent to the conditional game descibed above. It follows that the chances of survival are:

$d_{\infty}:=w_{\infty} + (1 - w_{\infty})(1 - \rho)(c_{\infty} + (1 - c_{\infty})(1-\rho)(c_{\infty} + \ldots = \displaystyle{\frac{(2-q) w_{\infty}}{1 - q + w_{\infty}}}.$

The asymptotic behavior of this function as $\rho \rightarrow 1$ (or $q \rightarrow 1$) requires the correct asymptotic $w_{\infty} \simeq 2(1-q)$, and from this we can deduce that

$d_{\infty} \rightarrow 2/3 \ \text{as} \ \rho \rightarrow 0.$

In this case, we see that the optimal probability is that the density $\rho$ tends to zero. Here is a graph of $d_{\infty}$:

The Sonic Screwdriver: Like John Nathan-Turner, I find the sonic screwdriver to be somewhat ridiculous. Although it does exist in some versions of the game, I will only mention a minor modification here. The “sonic” in the game allows the Doctor to survive for one round when he would otherwise be exterminated; it has only one use. We shall additionally assume that the sonic can only be used on the very first round. This essentially changes the game (at the beginning) into the conditional game described above. If one is allowed to use the TARDIS as above, the resulting probability of winning is

$(c_{\infty} + (1 - c_{\infty})(1-\rho)(c_{\infty} + (1 - c_{\infty})(1-\rho)(c_{\infty} + \ldots = \displaystyle{\frac{w_{\infty}}{q(1-q + w_{\infty})}}.$

As $\rho \rightarrow 1$, this function tends to 1, and as $\rho \rightarrow 0$, it tends to $2/3$. The behavior of this function in a neighbourhood of $0$ appears to be of the form

$2/3 - A \rho^{1/2} + \ldots$

for some constant $A$, possibly around $0.3$. Note that this function is not monotone; the most dangerous density of Daleks is approximately $\rho = 0.127$, where the resulting probability of surviving dips below $3/5$.

Here’s an example of the vanilla game at the optimal $\rho \sim 0.517$: the Doctor lives!

Posted in Mathematics, Waffle | | 1 Comment

## What is my Kasparov Number?

This has been a fun week in sport, what with England slaughtered at the Gabbatoir and Anand sliced up by Carlsen’s endgame magic. The latter games were fascinating if not necessarily exciting per se; consisting more of slow grinds rather than Kasparov style flourishes. Speaking of Kasparov, following Andrew Gelman, one defines the Kasparov number as the length of the shortest (ordered) chain of people (starting at you and ending at Kasparov) such that each person has beaten his or her successor at a game of chess. Let me also define the weaker “Draw Kasparov” number where one now allows either wins or draws. Being a little light on official tournament play myself, I have felt free to suitably relax the requirement of where the games take place.

The best upper bounds I could come up with for my Kasparov are around 6, which is probably pretty close to the right answer. However, my “draw” number against Kasparov is 2: I drew* with the British GM Tony Miles in 1991, and Miles’ best result against Kasparov was a draw (he was crushed by Kasparov 5.5-0.5 in 1986, but that 0.5 point counts!)

*OK, this game took place as part of a 40 player simultaneous exhibition, but that still counts!

Posted in Chess, Cricket, Waffle | Tagged , , , , | 2 Comments

## Local representations occurring in cohomology

Michael Harris was in town for a few days, and we chatted about the relationship between my conjectures on completed cohomology groups with Emerton and the recent work of Scholze. The brief summary is that Scholze’s results are not naively strong enough to prove our conjectures in full, even for PEL Shimura varieties. Motivated by this discussion, I want to give two quite explicit challenges concerning the mod-p cohomology of arithmetic locally symmetric spaces. The first I imagine will be very hard — it should already imply a certain vanishing conjecture of Geraghty and myself which has strong consequences. However, the formulation is somewhat different and so might be helpful.

Fix an arithmetic locally symmetric space $X$ corresponding to a reductive group $G$ over $\mathbf{Q}.$ Let $\ell$ and $p$ be distinct prime numbers. Consider the completed cohomology groups

$\widehat{H}^d(\overline{\mathbf{F}}_{\ell}) = \displaystyle{\lim_{\rightarrow}} H^d(X(K),\overline{\mathbf{F}}_{\ell}), \qquad \widehat{H}^d(\mathbf{C}) = \displaystyle{\lim_{\rightarrow}} H^d(X(K),\mathbf{C}),$

where we take the completion over all compact open subgroups. The limit has an action of $G(\mathbf{A})$ for the finite adeles $\mathbf{A}$, and so, in particular, has an action of $G(\mathbf{Q}_p)$. What irreducible $G(\mathbf{Q}_p)$ representations can occur in $\widehat{H}^d(\overline{\mathbf{F}}_{\ell})$? Here is a guess:

Conjecture: If the smooth admissible representation $\pi$ of $G(\mathbf{Q}_p)$ occurs as an irreducible sub-representation of $\widehat{H}^i(\overline{\mathbf{F}}_{\ell})$, then there exists an irreducible representation $\Pi$ of $G(\mathbf{Q}_p)$ in characteristic zero such that:

1. The Gelfand-Kirillov dimension of $\Pi$ is at least that of $\pi$. Equivalently,
$\mathrm{dim} \ \Pi^{K(p^n)} \gg \mathrm{dim} \ \pi^{K(p^n)}.$
2. Let $\mathrm{rec}(\Pi)$ and $\mathrm{rec}(\pi)$ be the Weil-Deligne representations associated to $\Pi$ and $\pi$ respectively by the classical local Langlands conjecture and the mod-$\ell$ local Langlands conjecture of Vigneras. Then

$(\overline{\mathrm{rec}(\Pi)})^{\mathrm{ss}} \simeq (\mathrm{rec}(\pi))^{\mathrm{ss}}.$

3. The representation $\Pi$ occurs in $\widehat{H}^j(\mathbf{C})$ for some $j \le i$.

Roughly speaking, this conjecture says that the irreducible representations occurring in characteristic $p$ are no more complicated than those which occur in characteristic zero. One naive way to try prove this conjecture would be to show that any torsion class lifts to characteristic zero, at least virtually. This conjecture is too strong, however, as can be seen by considering K-theoretic torsion classes in stable cohomology — the mod $3$ torsion class in $H^3(\mathrm{GL}_N(\mathbf{Z}),\mathbf{F}_3)$ can never lift to characteristic zero for sufficiently large N because the cohomology over $\mathbf{Q}$ is zero for all congruence sugroups by a theorem of Borel. The conjecture as stated seems very hard.

In a different direction, here is the following challenge to those trying to understand completed cohomology through perfectoid spaces. (I expect one can prove this by other means, but I would like to see a proof using algebraic geometry.)

Problem: Fix an integer $d$, and let $X_g$ be the Shimura variety corresponding to the moduli space of polarized abelian varieties of genus $g$. Prove that, for $g$ sufficiently large, the completed cohomology group $\widetilde{H}^{d}(X_g,\mathbf{F}_p)$ is finite over $\mathbf{F}_p$.

An equivalent formulation of this problem is to show that the only smooth admissible $\mathrm{GSp}_{2g}(\mathbf{Q}_p)$-representations $\pi$ which occur inside $\widetilde{H}^{d}(X_g,\mathbf{F}_p)$ are one dimensional.