Daleks

Posted on December 5, 2013 by Persiflage

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

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

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

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

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:

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

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!

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What is my Kasparov Number?

Posted on November 27, 2013 by Persiflage

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 number 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!

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Local representations occurring in cohomology

Posted on November 5, 2013 by Persiflage

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

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

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.

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Abelian Varieties

Posted on October 30, 2013 by Persiflage

Jerry Wang gave a nice talk this week on his generalization of Manjul’s work on pointless hyperelliptic curves to hyperelliptic curves with no points over any field of odd degree (equivalently, \(\mathrm{Pic}^1\) is pointless). This work (link here) is joint with Manjul and Dick, so the exposition is predictably of high quality. But I wanted to mention a result that arose during the talk which I found quite intriguing. Namely, given the intersection \(X\) of two quadrics \(P\) and \(Q\) in projective (2n+1)-space, the variety of projective n-spaces passing through \(X\) turns out (over the complex numbers) to be an abelian variety. For \(n = 1\) this is pretty familiar, but, for general \(n\), I hadn’t seen any construction like this before. It gives, for example, explicit constructions of equations for abelian varieties in surprisingly low degree. It brought me back to a lecture I once went to by Beauville as a graduate student when he talked about intermediate Jacobians (wait – perhaps this construction also has to be isomorphic to an intermediate Jacobian…). Is it possible (in some weak sense) to classify all varieties whose variety of maximal linear subspaces is an abelian variety of suitably high dimension? Are there varieties in which this construction gives rise to abelian varieties which are not isogenous to Jacobians? The geometric result is due (independently) to several authors, but, in a solo paper here, Jerry showed that the result is true arithmetically, and, even better, the construction can more precisely be described as giving an explicit torsor for the corresponding Jacobian. This very nicely generalizes the classical picture between pairs of quadrics and 2- and 4-descent.

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The problem with baseball

Posted on October 27, 2013 by Persiflage

Jordan Ellenberg, in a lovely slate article, explains perfectly what I don’t like about baseball.

I think the fundamentals of baseball as a sport are sound. I like the pace of the game, the variation, the statistics, the quirkiness, the history. But my problem is that I only started following baseball after I came to the US. I supported the A’s and the Giants (no doubt already poor form amongst serious baseball fans), and I went to a few games at each park. The A’s were (as they perennially seem to be) thriving on young talent, in this case the trio of Hudson, Mulder, and Zito; the Giants were in the peak of the Bonds era. But seasons passed and players came and went — either to other teams or into disgrace (Giambi, Tejada, Bonds, etc.). Whilst my move to Boston (presumably) inspired the Red Sox to their ’04 world series win, my allegiance to any single team became even more fractured. I’m not sure you can truly love baseball unless you either grew up with a team or have a deep sense of loyalty to a particular city. I’m not going to lament a past lost era when players spent their entire career with one team (Persiflage supports capitalism!), but I know that my top sporting allegiance will never stray from the Australian cricket team.

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Does Harvard discriminate in favour of Jews?

Posted on October 22, 2013 by Persiflage

In The American Conservative, Ron Unz published an essay that was ostensibly about whether the top Ivy schools discriminate against Asian students but, upon closer reading, was mainly concerned with arguing that Harvard/Yale/Princeton discriminate against white Gentiles in favour of Jews. The essay was widely discussed in various blogs, in part because the question of whether Asians are discriminated against is one that has a resonance in the academic community, and also because the essay contained voluminous appendices to back up its claims.

However, the wheels started to come off around February, when Columbia statistician Andrew Gelman reported on some significant inconsistencies in Unz’s methodology, as pointed out by Nurit Baytch, and Unz’s gross underestimates of the percentage of Jews among US recent IMO team members (as pointed out by Prof. Janet Mertz). There’s now an update on Gelman’s blog here, which links to a more in-depth rebuttal of Unz by Nurit (which you can find directly here). Nurit’s critique rips apart Unz’s argument on many levels (it mainly addresses the issues concerning [the lack of] discrimination in favour of Jews, not against Asians).

The entire episode points to a few disturbing facts. First, if you present a lot of data, people will trust your arguments even if your statistical analysis is completely flawed (the sociological “power of math”!). Second, if you couch your crazy argument (“The Jews” control Harvard — look, even President Drew Faust’s husband is Jewish [seriously, Unz uses this argument]) in something which is a genuine concern (discrimination against Asians), people will take you seriously. Finally, the essay was promoted by David Brooks; that should be a warning to anyone that it should not be taken seriously.

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Who is D.H.J. Polymath?

Posted on October 20, 2013 by Persiflage

D.H.J. Polymath is the assumed collective pseudonym for the authors of a number of papers which have arisen as a result of the polymath project initated by Gowers. Presumably, since it is a matter of open record, one can go through and identify the participants and their various individual contributions.

But what is it that mathematicians on the street think when they answer this question? The answer is that D.H.J. Polymath is equivalent to T.Gowers, T.Tao, et. al. I don’t claim that this is an accurate reflection of the contribution of the participants, but simply the perception in the community (as far as I have gauged through a number of conversations). Credit for joint publications can be a tricky issue at the best of times. In this case, the people responsible for making the decision to choose a pseudonym are presumably those for whom receiving credit is the least relevant. I find myself sympathetic to the remarks made by T.Brown in the mathscinet review of the first polymath paper:

This reviewer would have preferred to see, rather than the pseudonym “Polymath”, a list of authors. In other fields there are papers with a hundred co-authors. Why not in mathematics a paper with twenty or thirty co-authors, with extra credit for the person(s) who wrote the exposition?

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Why is my paper taking so long to review?

Posted on October 1, 2013 by Persiflage

The question in the title does not refer to any of my own papers; rather, I want to *answer* the question from the perspective of an editor. Here, roughly, is how the sausage is made (this is a medium case scenario, your mileage may vary). Keep in mind that this is a journal which has relatively good standards (for number theorists, we are talking somewhere between JNT and Duke).

  • Day 0: After carefully selecting a suitable journal and performing a final check on your paper for typographical errors, you submit your precious baby to the whims of fate.
  • Day 20(?): The paper works its way though the editorial system and is assigned to me as an editor.
  • Day 40: I have had a chance to take a look at the paper and determine whether it is obviously rubbish or not. Moreover, I have identified someone (usually at the level of professor) whom I trust to give an honest opinion of both how interesting the paper is and whether it is suitable for the journal in question. I email that person asking for a quick opinion and any suggestions they may have for possible reviewers.
  • Day 60: I email the expert again because they have not yet responded to my original request. Often, at this point, the expert will say that they are not qualified to give an opinion, and I return to the previous step.
  • Day 80: The expert has usually found time to respond, often to suggest another expert to consult (go back two spaces).
  • Day 100: I have a response from the expert. If they are only lukewarm, I reject the paper. So far, 80% of papers have now been rejected. Measured by the “standards of the industry,” I think that rejecting papers within about 3 months is acceptable to good. If the expert is enthusiastic, they either agree to referee the paper themselves or suggest someone else (often someone younger) to to the job. I then send out a detailed review request, either to the person suggested by the expert or to someone else.
  • Day 120: I email a different reviewer, because the first review declines for one of the standard excuses (busy/not qualified/lazy and so makes up something about not liking commercial publishers). I email someone else.
  • Day 130: They agree to review! I give them three months.
  • Day 230: I email the reviewer to follow up on my previous email. They start reviewing the paper.
  • Day 250: The paper is accepted. 25% of the time, the comments consist of minor typographical remarks. 50% of the time, there are a few requests for clarification, references, and corrections of minor inaccuracies. 25% of the time, there are substantial comments and corrections. In the majority of cases, the referees do a conscientious job (some papers don’t need many corrections!)

    Some General Remarks:.

    • Of all the papers I have edited, a small number (at most 2 or 3) have ultimately been rejected because of a fatal mathematical error (i.e., the paper would have been accepted if it had turned out to be correct). In all of those cases, I was the one who found the error.
    • I end up rejecting quite a few papers because there is a fixed number of pages I can accept per year. I would anticipate doubling the number of acceptances if there were no such constraint.
    • Sometimes papers do fall through the cracks. It can be very hard to find a reviewer for a very technical paper, especially one that builds off previous technical work of the author. Can one reject a paper on the basis that you couldn’t find anyone to review it? I honestly think we may be heading in that direction.
    • The main task of the editor is not summary judgement, but administration. It’s not enough to email someone (say, a reviewer) and then consider one’s job done; you have to keep track of when you emailed them, so you know when to email them again (or someone else) if (or frequently when) they don’t respond. (I admit, I’m by no means perfect as a reviewer, either.)
    • Any online system set up to coordinate and facilitate communication with authors/editors is more annoying than useful; I work off the grid as much as possible.
    • Posted in Mathematics, Politics | Leave a comment

    Virtual Congruence Betti Numbers

    Posted on September 27, 2013 by Persiflage

    Suppose that \(G\) is a real semisimple group and that \(X = \Gamma \backslash G/K\) is a compact arithmetic locally symmetric space. Let us call a cohomology class tautological if it is invariant under the group \(G\). For example, if \(X\) is a 3-manifold, then the tautological classes are all multiples of either the trivial class in \(H^0\) or the fundamental class in \(H^3\). We say that \(X\) has positive Betti number if there exist any non-tautologial classes in the cohomology of \(X\). One can pose the following question:

    Problem: Show that there exists a finite congruence cover \(\widetilde{X} \rightarrow X\) such that \(\widetilde{X}\) has positive Betti number.

    An automorphic way of phrasing this question is as follows: do there exist any automorphic forms besides the trivial representation for the \(\mathbf{Q}\)-group \(\mathbb{G}\) associated to \(\Gamma\). If \(G\) admits discrete series, then the result is obvious for automorphic reasons (from the trace formula, by de George-Wallach). If \(X\) has non-zero Betti number Euler characteristic, then the result is obvious for topological reasons. In fact, as I leant from Gross one day at tea, these two situations coincide (this certainly follows from Borel-Wallach, even in the stronger form that the contribution from each \(\pi\) via Matsushima’s formula has zero Euler characteristic if it is not a discrete series; I’m not sure if there’s a slicker argument).

    The problem is obviously related to the virtual positive Betti number theorem of Agol, but there are a few important subtle differences. The first is that we insist that the cover \(\widetilde{X}\) is congruence. Hence, the problem remains open for a general arithmetic 3-manifold. Second, we also allow (as we must) cohomology in any degree. Another example to consider is \(G = U(2,1)\). In this case, \(X\) is a compact complex hyperbolic manifold. It is an open problem whether such manifolds have virtual positive first Betti number. In contrast, by a theorem of Rogawski, they certainly don’t have virtual positive first Betti number in congruence covers, although they clearly do have virtual positive Betti number in congruence covers for the two equivalent reasons given above.

    What I want to do in this post is discuss a related problem, namely, can one find arbitrarily large congruence covers \(\widetilde{X}\) which all fail to have positive Betti number? Specific examples of this kind (for a compact arithmetic 3-manifold \(X\)) were given in my paper with Dunfield (conditional on local-global compatibility of certain Galois representations, now known), and Boston-Ellenberg shortly thereafter found a different (unconditional) argument using group theory (which applied to the same example). I want to explain how to generalize these results to higher dimension, contingent on computations which might be hard to carry out explicitly.

    Choose:

    • An imaginary quadratic field \(F\).
    • A prime \(p\) which splits as \(\mathfrak{p} \overline{\mathfrak{p}}\) in \(F\).
    • A central simple algebra \(D/F\) with local invariants \(1/N\) and \(-1/N\) at the primes dividing \(p\).

    Associated to \(D\) is a maximal lattice \(\Gamma\) in \(G/K = \mathrm{SL}_N(\mathbf{C})/\mathrm{SU}_N(\mathbf{C})\) whose quotient is a compact finite volume orbifold of real dimension \(N^2 – 1\). For sufficiently large \(n\), the congruence covers \(X(\mathfrak{p}^n)\) are manifolds which are \(K(\pi,1)\) spaces with fundamental group \(\Gamma(\mathfrak{p}^n)\). When \(F = \mathbf{Q}(\sqrt{-2})\), \(N = 2\), and \(p = 3\), one recovers the manifolds considered in my paper with Nathan.

    Let me now make another definition. Let \(F_S\) be the maximal pro-p extension of \(F(\zeta_p)\) unramified outside the primes dividing \(p\).

    Definition: The prime \(p\) is very regular in \(F\) if the map:

    \(\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p) \rightarrow D_v \subset \mathrm{Gal}(F_S/F)\)

    is surjective for either \(v|p\).

    The notion of very regular primes arose in my latest paper on \(K\)-theory and completed cohomology in the stable range, but more on that later. One last definition: say that an ideal \(\mathfrak{m}\) of a Hecke algebra \(\mathbf{T}\) is Eisenstein if the image of any Hecke operator \(T\) in \(\mathbf{T}/\mathfrak{m}\) coincides with multiplication by the degree \(\deg(T)\). This is how \(\mathbf{T}\) acts on the trivial representation. We then have the following:

    Conditional Theorem: Suppose that \(p\) is very regular, and that \(\mathfrak{m}\) is an Eisenstein maximal ideal. Then for all \(n\) there is an equality:

    \(H^*(X(\mathfrak{p}^n),\mathbf{Z}_p)_{\mathfrak{m}} \otimes \mathbf{Q}_p = H^*(\mathrm{SU}(N),\mathbf{Q}_p)\)

    In particular, if the only maximal ideals of \(\mathbf{T}\) on \(H^*(X(\mathfrak{p}),\mathbf{Z}_p)\) are Eisenstein, then all the \(X(\mathfrak{p}^n)\) are rational \(SU(N)\)-homology spaces.

    Example: The prime \(p = 3\) is strongly regular for \(F = \mathbf{Q}(\sqrt{-2})\), and — by a computation — the only maximal ideals of \(\mathbf{T}\) on \(H^*(X(\mathfrak{p}),\mathbf{Z}_p)\) are Eisenstein. Of course, a rational \(SU(2)\)-homology space is a homology 3-sphere.

    Proof: Suppose that there is exists a non-trivial class in the cohomology of \(X(\mathfrak{p}^n)\). It will give rise to an automorphic representation \(\pi\) which is tempered, because \(X\) are Shimura varieties manifolds for which we can show (reference?) have no endoscopic forms. Hence, by HLTT or Scholze, there exists a corresponding Galois representation

    \(r(\pi): G_{F} \rightarrow \mathrm{GL}_n(\mathbf{Q}_p)\)

    that is unramified away from \(p\). We now assume (this may be proved soon, but this is the reason for the “conditional” in the statement) that we know enough about local-global compatibility to deduce that this representation is also ordinary at the prime \(\mathfrak{p}’\). Note that the reason it should be ordinary is that the level is prime to \(\mathfrak{p}’\), and since the quaternion algebra is ramified at this prime we know that \(\pi_{\mathfrak{p}’}\) is Steinberg. We deduce that \(r(\pi)\) is completely reducible after restriction to \(D_v\) for \(v = \mathfrak{p}’\). The Eisenstein assumption and the ramification assumption imply that \(\overline{r(\pi)}\) and hence \(r(\pi)\) factor through \(\mathrm{Gal}(F_S/F)\). Hence, using the fact that \(p\) is very regular, we immediately deduce that \(r(\pi)\) itself is reducible and ordinary. It follows that, after semisimplification, \(r(\pi)\) is a direct sum of characters, which leads to an easy contradiction.

    Experts will recognize this argument as a generalized and more streamlined version of what appears in my paper with Nathan. One may naturally ask whether there is a generalization of the Boston-Ellenberg argument as well. Emerton and I already explained that the correct way to view that argument was as follows. What one really wants to prove is that the partially completed cohomology groups:

    \(\displaystyle{\widetilde{H}^*(\mathfrak{p}) = \lim_{\rightarrow} H^*(X(\mathfrak{p}^n),\mathbf{F}_p)}\)

    all vanish identically outside degree zero. For 3-manifolds, it suffices to prove this for \(\widetilde{H}^1\). For what \(X\) might one be able to prove such vanishing? As Matt and I explained in our paper on 3-manifolds, for all these groups to vanish there has to be a delicate balancing act between the dimension of the group acting on completed cohomology and the dimension of the manifold. For example, it is crucial that there is an equality

    \(\dim(G/K) = \dim(\prod_{S} G(F_v))\)

    where one partially completes at primes \(S\) above \(p\). (Otherwise one obtains an immediate contradition by Hochschild-Serre.) In the case at hand, this inequality is satisfied, since:

    \(\dim(G/K) = \dim(\mathrm{SL}_N(\mathbf{C})) – \dim(\mathrm{SU}_N(\mathbf{C})) = N^2 – 1 = \dim(\mathrm{SL}_N(\mathbf{Z}_p))\)

    Hence, it is really possible that all the completed cohomology groups may vanish in this case. In fact, if one instead considers the split group \(\mathrm{GL}(N)/F\), then the partially completed cohomology groups do vanish in the stable range exactly for very regular primes. (This is where the definition of very regular primes comes from.) By Nakayama’s Lemma, one can explicitly compute at some finite level to determine whether the \(\widetilde{H}^*(\mathfrak{p})\) vanish or not. In fact, it suffices to compute that the maps:

    \(H^*(G(p),\mathbf{F}_p) \rightarrow H^*(X(\mathfrak{p}),\mathbf{F}_p)\)

    are isomorphisms, where \(G(p)\) is the congruence subgroup of \(\mathrm{SL}_n(\mathbf{Z}_p)\).
    If one wanted to find an explicit example where these theorems applied for \(N \ge 3\), the first place to look would probably be to take \(F = \mathbf{Q}(\sqrt{-2})\), \(p = 3\), and \(N = 3\). One would then have to compute the cohomology of a certain \(8\)-dimensional manifold! (The resulting manifolds would potentially all be rational \(SU(3)\)-homology space = rational \(S^5 \times S^3\)-homology space). This computation is within the realms of plausibility. To rule out characteristic zero representations, we can pass by functoriality to the split side. So, if there is a characteristic zero class which is not Eisenstein mod-p, that residual representation also has to occur at low(ish) level inside the cohomology of \(\mathrm{GL}_3(\mathbf{Z}[\sqrt{-2}])\). This is the sort of cohomology that people like Gunnells might almost be able to compute!

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    Life on the modular curve

    Posted on September 24, 2013 by Persiflage

    Alice and Bob live on the modular curve \(X_0(1) = \mathbf{H}/\mathrm{PSL}_2(\mathbb{Z})\). What does the world look like to them, assuming that they view the world in hyperbolic perspective?

    To those who are not used to hyperbolic geometry, there may be a few mild surprises. Suppose that Alice is at the point x=i and Bob is at y = 10i. Let us also imagine that Alice is looking in the direction of the cusp along the projection of the geodesic given by the y-axis. What does she see? Take a moment to think about it if you like; we will give the answer in the next paragraph.

    Lifting Bob to the universal cover, there are infinitely many Bobs spaced equally along the horosphere (10i + t). A naive guess is that all of these Bobs would fill out Alice’s field of vision. But this can’t be true; since geodesics in \(\mathbf{H}\) are given by semi-circles perpendicular to the \(x\)-axis, most geodesics through x=i don’t cross Bob’s horosphere. In fact, Bob only takes up about \(10^{\circ}\) of Alice’s vision, and those Bobs who are at (10i + n) for large integers n appear almost to be directly in front of Alice (although a long way away). Of course, Alice also sees copies of herself receding similarly into the distance directly in front of her.

    All this and more can be seen in the 80’s inspired video game of my undergraduate summer students Jasmine Powell and Justin Ahn (funded by the NSF!). The basic setup is as follows: you are a cube wondering around on \(X_0(1)\) and you need to shoot the monsters, which are in the shape of a pill. Occasionally, some bonus feature will appear (extra shields, freeze, extra life, etc.) which you can collect. Some mathematics that is hiding in the background but is only partially relevant for game play: the monsters travel along closed geodesics, and the goodies appear at CM points. The game was also partly inspired by the video not knot. Here’s a link to a video capture from the game:

     

    Video Capture

    (The transition to video has made it look a little wonky.) If you notice carefully, you will see that at one point in the video you crash into yourself by passing through the cone point \(i\), losing a life.

    The alpha-release of the game itself can also be downloaded here (sorry, macintosh only). Please play around with it and offer suggestions and improvements! Various possibilities include upgrading to a 3-manifold (probably a Bianchi manifold), and also the ability to pass to congruence covers \(X_0(p)\) of \(X_0(1)\).

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