According to the Talmud, in order for the Sanhedrin to sentence a man to death, the majority of them must agree to it. However

R. Kahana said: If the Sanhedrin unanimously find [the accused] guilty, he is acquitted. (Babylonian Talmud, Tractate Sanhedrin, Folio 17a)

Scott Alexander has a devious mind and considers how he would respond to this rule as a criminal:

[F]irst I’d invite a bunch of trustworthy people over as eyewitnesses, then I’d cover all available surfaces of the crime scene with fingerprints and bodily fluids, and finally I’d make sure to videotape myself doing the deed and publish the video on YouTube.

So, suppose you were on a panel of judges, all of whom had seen overwhelming evidence of the accused’s guilt, and wanted to make sure that a majority of you would vote to convict, but not all of you. And suppose you cannot communicate. With what probability would you vote to convict?

Test your intuition by guessing an answer now, then click below:

My gut instincts were that (1) we should choose really close to , probably approaching as and (2) there is no way this question would have a precise round answer. As you will see, I was quite wrong.

Tumblr user lambdaphagy is smarter than I was and wrote a program. Here are his or her results:

As you can see, it appears that is not approaching , or even coming close to it, but is somewhere near . Can we explain this?

We want to avoid two events: unanimity, and a majority vote to acquit. The probability of unanimity is .

The probability of a majority vote to acquit is . Assuming that , and it certainly should be, almost all of the contribution to that sum will come from terms where . In that case, . And we’ll roughly care about such terms. So the odds of acquittal are roughly .

So we roughly want to be as small as possible. For large, one of the two terms will be much larger than the other, so it is the same to ask that be as small as possible.

Here is a plot of :

Ignore the part with below ; that’s clearly wrong and our approximation that is dominated by won’t be good there. Over the range , the minimum is where .

Let’s do some algebra: , , (since is clearly wrong), . Holy cow, is actually right!

First of all, actually do some computations.

Secondly, I was wrongly thinking that failing by acquittal would be much more important than failing by unanimity. I think I was mislead because one of them occurs for values of and the other only occurs for one value. I should have realized two things (1) the bell curve is tightly peaked, so it is really only the very close to which matter and (2) exponentials are far more powerful than the ratio between or and anyway.

Finally, for the skeptics, here is an actual proof. Assuming , we have

The main step is to replace each by the largest it can be.

But also,

Here we have lower bounded the sum by one of its terms, and then used the easy bound since it is the largest of the entries in a row of Pascal’s triangle which sums to .

So the odds of failure are bounded between

and . We further use the convenient trick of replacing a with a , up to bounded error to get that the odds of failure are bounded between and .

Now, let be a probability greater than other than . We claim that choosing conviction probability will be better than for large. Indeed, the -strategy will fail with odds at least , and the strategy will fail with odds at most . Since , one of the two exponentials in the first case is larger than , and the -strategy is more likely to fail, as claimed.

Of course, for a Sanhedrin of members, , so our upper bound predicts only a one percent probability of failure. More accurately computations give . So the whole conversation deals with the overly detailed analysis of an unlikely consequence of a bizarre hypothetical event. Fortunately, this is not a problem in the study of Talmud!

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This is the corner of a crystal of salt, as seen under an electron microscope. (I took the image from here, unfortunately I couldn’t find better information about the sourcing.) As you can see, the corner is a bit rounded, where some of the molecules have rubbed away. They ask the question: “What is the shape of that rounded corner?”

The molecules of a salt crystal form a cubical lattice. We can index them as . But it’s not the ones from the interior that are missing – if is missing and , and , then is missing as well.

A finite subset of with the property that implies for , , is what I’ll call a **three dimensional partition**. (Here I am deliberately rebelling against the awful classical terminology, which is a “plane partition“.)

The opening of Kenyon, Okounkov and Sheffield’s paper discusses the question: “What is the shape of a random three dimensional partition of size ?”

It was only after I had worked through this paper, and its sequels 1 2 fairly carefully that I realized they hadn’t actually answered their motivating question. They certainly implied what the answer should be, and they laid out all the necessary tools, but they never came back and said “let’s do the salt crystal example”.

In this post, I want to lay out in outline how this question is answered. The previous posts on Legendre transforms in statistical mechanics, and on random partitions, were meant as warm ups, where it is easier to make complete arguments.

Let me acknowledge right out that I am making no attempt at rigor here. What I want to do is sketch the argument, and hope this encourages some of you to read the amazing sequence of papers I have linked to.

In the previous posts, our first step was to analyze partitions of a given slope. Namely, we showed that there are roughly partitions that fit in a box, where . We would now like to similarly analyze three dimensional partitions with a given slope.

We could specify a normal vector and ask for the partition to have slope roughly , but there is an approach which turns out to be equivalent and is a bit easier to formulate. Let be positive real numbers with . The boundary of a plane partition is made up of squares which lie either in the plane, the plane or the plane. Let us suppose there were some function such that the number of partitions with some specified boundary, using roughly squares of the first type, of the second type and of the third type, is roughly . We would like to know what this function is.

I said “some specified boundary”. What boundary shall we use? As is often the case in mathematical physics, the nicest thing to do is to use “periodic boundary conditions” — which is to say, to avoid the issue of boundaries by wrapping the problem on a torus.

A perspective drawing can give us inspiration. The boundary of a three dimensional partition looks like a tiling of the plane by rhombi with angles and . The three planes which squares can lie in turn into the three possible orientations of the rhombi (colored red, black and blue below).

(Image taken from Eventually Almost Everywhere, who has further nice discussion of the relation between rhombus tilings and plane partitions.)

Tile the plane by equilateral triangles in the standard manner and then quotient that plane by some lattice to produce a torus. Let that torus contain upward pointing and downward pointing triangles; so we can hope to tile it with rhombi.

Let be the number of such tilings which use , and rhombi of the three potential types. So our goal is to determine .

As in the case of two dimensional partitions, the best strategy is to form a generating function. Let .

There is an amazing explicit formula for this generating function.

To tantalize you, I will state it without proof, and simply point you to the search term “Kasteleyn’s method” if you want to learn more.

Let us suppose our torus has a fundamental domain which is , so , and let be even. Set

.

We have (up to possible sign errors on my part)

.

Asymtotically, all four terms contributing to are about the same size, so . And how big is ? We can approximate that sum by an integral:

.

Set . This is known as the “Rankin function”. Then, as before, we conclude that is the Legendre transform of the Rankin function. And we conclude that a random three dimensional partition has the shape for some constant .

Something interesting happens. Suppose that . So cannot be zero for any complex numbers and with , . So is a harmonic function when we restrict and to those discs. The average of a harmonic function over the boundary of a disc is the value at the center of the disc. So simply equals when .

Thus, the surface contains, in particular, the planar region , , and similar planar regions in the and planes. This is a major difference between the two dimensional and three dimensional limit shapes — no part of the two dimensional limit curve is contained in the coordinate axes. While I'm not sure how seriously this picture should really be taken, it might do a bit to explain why those salt crystals above look so cubical.

There is another very important difference between the two and three dimensional pictures. In the two dimensional case, the only solutions of the variational problem were of the form . But, in the three dimensional case, is only one of many solutions to the resulting PDE. There is so much more to say about all of this. If you want to read more, I refer you to Chapter One of Kenyon and Okounkov.

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It’s up on mathjobs, but applicants will need to apply through the university website. Here’s the pitch:

The Mathematical Sciences Institute at the Australian National University is seeking to invigorate its research and teaching profile in the areas of statistics, probability, stochastic analysis, mathematical finance and/or biomathematics/biostatistics. We wish to fill several continuing positions at the Academic Level B and/or Level C (which equates to the position of Associate Professor within the United States of America). Up to 3 full time positions may be awarded.

You will be joining an internationally recognised leading team of academics with a focus on achieving excellence in research and teaching. The Institute comprises of approximately fifty academics, within seven mathematical research programs. Applicants are expected to have an outstanding record in research, teaching and administration. All positions will involve some teaching, in the specialised areas advertised and/or standard mathematics undergraduate courses, but this may be at a reduced level for several years.

It’s a great place to work, excellent opportunities for research grant funding, and a really nice place to live. Feel free to contact me if you have any questions about the job or living in Canberra.

**Please pass this on to friends with relevant interests!**

(Oh, and don’t forget those two postdocs I’m hiring in quantum algebra, higher category theory, subfactors, representation theory, etc.)

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By a partition of , we mean positive integers with . We draw a partition as a collection of boxes: For example, this is :

Suppose we let , select partitions of uniformly at random and rescale the size of the boxes by , so that the diagram of the partition always has area . What is the shape of the most likely diagram?

We want to describe, given a particular shape, how many partitions have roughly that shape. To that end, let , , …, be a sequence of points with and . Let's count partitions which fit inside the box and whose boundary passes through the points .

To describe such a partition, we simply must describe the portion between and , and then describe the portion between and , and so forth; each of these portions is independent. Set and (so ). Set .

So the number of such partitions is

.

From our computations in an earlier post, this is roughly

where . Moreover, if we rescale all of the and by , this approximation becomes better.

So, if we want there to be a lot of partitions, we should make large.

We have now described what happens if we require the boundary of the partition to pass through finitely many fixed points. Suppose instead we insist that the partition lie near a fixed curve? Parametrize that curve as , with the normalization that . I’m not even going to say what I mean precisely, but we can extrapolate from the preceding formula that the number of partitions whose boundary is near is roughly

We thus obtain a problem in the calculus of variations: Consider curves , with the normalizations that and the area enclosed by and the coordinate axes is . Subject to these conditions, maximize . The next several paragraphs are solving this variational problem.

Rotating the coordinates , we see that the condition on the enclosed area is simply that .

As always in problems involving the calculus of variations, the first step is to perturb our proposed solution and see how it changes. Replace by . If was optimal, than should be a maximum, so the first order variation in should vanish.

This is only valid if our perturbation preserves the conditions on . We want the perturbed functions to obey , so we take , and we want to preserve the area, so we take .

For such a perturbation, the first order variation in is supposed to vanish. We compute:

.

Integrating by parts:

.

We want this to hold for any obeying . This will happen if and only if is a constant. So we want for some constants and . So .

The formula , in some sense, solves the variational problem. Our next task is to simplify that formula.

From the previous post, we know that . So , where is another constant of integration.

As , we want . This shows that .

As , we want . This shows that . So we have and .

To evaluate the final constant , we need to use that the area under the curve is . It is convenient to eliminate and relate and directly as , or, . Now, the area under the curve is . We deduce that .

In other words, our final answer is that the curve is .

How good is this approximation? See for yourself! Here is a random partition of , and the approximating curve.

The thing which I think is really cool is that we never needed the explicit formula ; we only needed that it was Legendre dual to . Suppose that we looked at some other sort of local probability distribution on partitions: For example, maybe we would weight the likehood of a partition by or something like that. We could still build a generating function for partitions fitting in a box with this weighting. We could then try to compute . If we succeeded, the exact same logic as before would show that was the shape of a random partition of area chosen from this distribution. Which brings me to the thing I really want to talk about … next time.

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We are going to be considering systems with parts, and asking how many states they can be in. The answers will be exponential in , and all that we care about is the base of that exponent. For example, the number of ways to partition an element set into two sets of size (if is even) is which Stirling’s formula shows to be . All we will care about is that .

How many ways can we partition an element set into a set of size roughly and a set of size roughly ? More mucking around with Stirling’s formula will get us the answer

I’ll show you a non-rigorous way to get that answer without getting into the details of Stirlings formula.

Let’s suppose there is a function so that the number of ways to partition an element set into a piece of size and a piece of size is roughly . Let’s look at the generating function

Then we expect

For fixed , there is presumably some which maximizes . The terms coming from near that will overwhelm the others, so we should have

or

Set . So we should have

In other words, we expect and to be Legendre dual. In particular, we expect to have all of the following formulas:

To spell the last one out in words, is a function of and is a function of ; they should be inverse functions. (Keeping the signs straight is one of the real nuisances in this subject.)

In this case, we can compute explicitly. We have , so so . We use the relation that the derivatives of Legendre dual functions are inverse. Inverting gives so

and we can compute the integral to get

I’m trying to learn to write shorter posts, so I’ll stop here for now. Next time, we compute the shape of a random partition of .

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We’ve just put up an ad for a new 2 year postdoctoral position at the ANU, to work with myself and Tony Licata. We’re looking for someone who’s interested in operator algebras, quantum topology, and/or representation theory, to collaborate with us on Australian Research Council funded projects.

The ad hasn’t yet been crossposted to MathJobs, but hopefully it will eventually appear there! In any case, applications need to be made through the ANU website. You need to submit a CV, 3 references, and a document addressing the selection criteria. Let me know if you have any questions about the application process, the job, or Canberra!

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Recently, Gavin Larose put up a lengthy footnoted post on the effort that goes into running our “Gateway testing” center, and the benefits we get from it. This is a room designed for proctoring computerized tests of basic skills, and we use it for things like routine differentiation or putting matrices into reduced row echelon form, which we want every student to know but which are a waste of class time. Check it out!

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Even if you are not requesting support, I’d appreciate knowing that you are coming.

Our speakers are Jonah Blasiak (Drexel), Laura Escobar (UIUC), Joel Kamnitzer (Toronto) and Tri Lai (IMA and Minnesota). Please see our website for more information.

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Let be a finite group, and let be a positive integer dividing . Then the number of solutions to in is divisible by .

This is a 1907 theorem of Frobenius. Along with the Sylow theorems, it is one of the few nontrivial elementary results about a completely general finite group. And it has some nice applications, which you can read about on Mathoverflow. But it has never made it into the standard basic group theory syllabus the way the Sylow theorems have. I wanted to give it as a challenging problem last time I taught group theory, but I didn’t find a proof that I liked enough.

The last few days, I’ve been thinking about the problem again, and I found what I think is a decent counting proof. I have the feeling there is a really slick proof in here waiting to get out. Let me know if you can find it!

As with almost all counting proofs in group theory, we’ll be using Burnside’s Lemma (due to Cauchy): If a finite group acts on a finite set , then . Actually, we only need the following corollary:

**Corollary of Burnside’s theorem** If a finite group acts on a finite set , then .

**Notation** Fix a finite group and a positive integer dividing . For any dividing , let denote the number of -torsion elements in , and let denote the number of elements of order precisely . So

Let act on the set of element subsets of . We will apply the Corollary to this action. So, we each , we must determine how many element sets are taken to themselves under multiplication by .

A set is fixed by if and only if is a union of cosets for . If the order of does not divide , this is impossible. If the order of is , a divisor of , then there are cosets of , and we must choose of them to make up a set of size . So there are sets fixed by , if is the order of . We obtain

Our desired result is that . If all of the binomial coefficients were equal to , we’d be done. Our strategy is to show that the binomial coefficients act enough like that we win.

Let’s start with a special case. Suppose that is a prime. Equation says . We have (exercise!). We deduce that , so as desired. The argument that follows is just this argument for a more general .

Substituting the Mobius inversion formula from into , we get

Interchanging summation and putting , we get

We now need a lemma to evaluate the inner sum.

**Lemma:** Let . Then .

**Proof:** We give a combinatorial interpretation. Consider necklaces with beads, of which are black and the rest are white. Then is the number of necklaces which are taken to themselves under a rotation of period . So the Mobius sum is the number of such necklaces with no nontrivial periodicity. Rotating such a necklace will produce distinct necklaces, so the number of such aperiodic necklaces is divisible by . (Here are some similar examples.)

We remark that, if we had in place of , the sum would be zero and the result obvious. So this Lemma is an example of our strategy of showing that binomial coefficients are -adically close to fractions.

Putting and , the inner sum of is divisible by . Also, inductively, for every , we know that . So every summand in except for possibly the term is divisible by . We deduce that the term is also divisible by .

In other words, . So . **QED**

The reason I am unsatisfied is that we started with direct counting to get to . We then messed around with a bit of Mobius inversion (which, sadly, is not considered standard material) and get to another sum which we evaluated combinatorially (modulo ) in order to win. Let denote the number of aperiodic necklaces on beads, of which are black. I feel like, if I were really clever, I could give a counting argument to jump directly to the equation . Then I’d never have to mention Mobius inversion. At that point I’d be completely happy giving this proof in class, or as a exercise with a series of hints.

Can you be that clever?

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