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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I briefly discussed vertex operator algebras in my earlier post on generalized moonshine. While an ordinary commutative ring has a multiplication structure , a vertex operator algebra (or VOA) has a “meromorphic” version , and there is an integer grading on the underlying vector space that is compatible with the powers of z in a straightforward way.

I won’t say much about VOAs in general, but rather, I will consider those that satisfy some of the following nice properties:

Rational: Any V-module is a direct sum of irreducibles.

Holomorphic: Any V-module is a direct sum of copies of V.

cofinite: This is a rather technical-sounding condition that ends up being equivalent to a lot of natural representation-theoretic finiteness properties, like “every representation is a direct sum of generalized eigenspaces for the energy operator L(0)”.

It is conjectured that every rational VOA is cofinite.

As usual, when we have a collection of nice objects, we may want to classify them, or at least find ways of building new ones and discovering invariants and constraints.

Some basic invariants are the central charge c (a complex number), and the character of a module , given by the graded dimension , where the grading is given by the energy operator L(0), and we view the power series as a function on the complex upper half plane using . One of the first general results for “nice” VOAs is Zhu’s 1996 proof that if V is rational and cofinite, then the characters of irreducible V-modules form a vector-valued modular form for a finite dimensional representation of . Furthermore, he showed that in this case, the central charge c is a rational number, and if V is holomorphic, then c is a nonnegative integer divisible by 8.

Dong and Mason classified the holomorphic cofinite VOAs of central charge 8 and 16 – there is one isomorphism class for central charge 8, and 2 isomorphism classes for central charge 16. All three are given by a lattice VOA construction. In general, if you are given an even unimodular positive definite lattice (which only exists in dimension divisible by 8), you get a a holomorphic cofinite VOA from it, so the central charge 8 object comes from the lattice, and the central charge 16 objects come from the and lattices. Central charge 24 is at a sweet spot of difficulty, where Schellekens did a long calculation in 1993 and conjectured the existence of 71 isomorphism types. Central charge 32 is more or less impossible, since lattices alone give over types.

For central charge 24, because the L(0) eigenspace with eigenvalue 1 is naturally a Lie algebra, the proposed isomorphism types are labeled by finite dimensional Lie algebras. Schellekens’s list is basically

1. The monster VOA, with .

2. The Leech lattice VOA, with commutative of dimension 24.

3. 69 extensions of rational Kac-Moody VOAs by suitable modules (here the Lie algebras are products of simple Lie algebras and in particular noncommutative).

As far as existence is concerned, 23 of the 69 come from lattices, known as the Niemeier lattices. An additional 14 come from Z/2 orbifolds of lattices. Another 18 come from a “framed VOA” construction, given by adjoining modules to a tensor product of Ising models according to some codes (Lam, Shimakura, and Yamauchi are the main names here). The remaining 12 are more difficult, and after this recent paper, there are 2 that have not been constructed. There are only a few cases where uniqueness is known, such as the Leech lattice VOA. The case is wide open, and perhaps the worst for uniqueness, since there isn’t any Lie algebra structure to work with.

One of the results of van Ekeren, Möller, and Scheithauer was a reconstruction of Schellekens list, i.e., eliminating other choices of Lie algebras from possibility. This was desirable, since the original paper was quite sketchy in places and didn’t have proofs. A second result was a collection of new examples, in particular nearly filling out this list of 69. They did this by solving an old problem, namely the construction of holomorphic orbifolds. The idea is the following: Given a holomorphic cofinite VOA V, and a finite order automorphism g, take the fixed-point subalgebra , and take a direct sum with some -modules not in V to get something new. In fact, the desired -modules were more or less known – there is a notion of g-twisted V-module V(g), and one takes the submodules of all fixed by a suitable lift of g. To show that this even makes sense requires substantial development of the theory.

First, the existence and uniqueness of irreducible g-twisted V-modules V(g) was a nonconstructive theorem of Dong, Li, and Mason in 2000. Then, to get a multiplication operation on the component -modules, one first shows that irreducible -modules have a nice tensor structure (in particular, are simple currents), so that the space of suitable multiplication maps is highly constrained. This requires recent major theorems of Miyamoto ( is rational and cofinite – 2013), and Huang (if V is rational and cofinite, then Rep(V) is a modular tensor category and the Verlinde formula holds – 2008). By some clever applications of the Verlinde formula, van Ekeren, Möller, and Scheithauer showed that once we have simple currents with suitable L(0)-eigenvalues, the homological obstruction to a well-behaved multiplication vanishes, and one gets a holomorphic VOA.

The intermediate results that I found most useful for my own purposes were:

1. assembly of an abelian intertwining algebra (a generalization of VOA where the commutativity of multiplication is allowed some monodromy) from all irreducible -modules.

2. the explicit description of the action on the characters of irreducible -modules. This also solves a conjecture of Dong, Li, and Mason concerning the graded dimension of twisted modules.

In particular, if g has order n, then the simple currents are arranged into a central extension , where the kernel is given by an action of g, and the image is the twisting on modules. The group A is also equipped with a canonical -valued quadratic form. One obtains an A-graded abelian intertwining algebra with monodromy determined by the quadratic form (up to a certain coboundary), and the action is by the corresponding Weil representation (up to the c/24 correction).

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ALGECOM, the twice annual midwestern conference on algebra, geometry

and combinatorics on Saturday, October 24. We will feature four

speakers, namely,

Jonah Blasiak (Drexel University)

Laura Escobar (University of Illinois at Urbana-Champaign)

Joel Kamnitzer (University of Toronto)

Tri Lai (IMA and University of Minnesota)

as well as a poster session. If you would like to submit a poster, please e-mail (David Speyer) with a quick summary of your work by September 15.

A block of rooms has been reserved at the (Lamp Post Inn) under the name of ALGECOM.

This conference is supported by a conference grant form the NSF. Limited funds are available for graduate student travel to the conference. Please contact (David Speyer) to request support, and include a note from your adviser.

More information will be added to our website as it becomes available.

We hope to see you there!

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Suppose that a drug company wishes to create evidence that a drug is beneficial, when in fact its effect is completely random. To be concrete, we’ll say that the drug has either positive or negative effect for each patient, each with probability . The drug company commits in advance that they will state exactly what their procedure will be, including their procedure for when to stop tasks, and that they will release all of their data. Nonetheless, they can guarantee that a Bayesian analyst with a somewhat reasonable prior will come to hold a strong belief that the drug does some good. Below the fold, I’ll explain how they do this, and think about whether I care.

To be concrete, let’s suppose that the drug company knows that the analyst begins with a uniform prior on the drug’s efficacy: she thinks it is equally likely to be any real number between and . And the drug company’s goal is to get her to hold a greater than percent belief that the drug’s benefit is greater than .

The drug company chooses (and announces!) the following procedure: They will continue to run patients, one at a time, until a point where they have run patients and at least have benefited. This will eventually happen with probability . At this point, they stop the study and release all the data. If the analyst updates on this, she will believe that the drug has effectiveness with a probability that is roughly a bell curve around and standard deviation . (I didn’t check the constants here, but this is definitely the right form for the answer and, if the constants are wrong then just change to .) In particular, the analyst would be willing to bet at 19 to 1 odds that the drug does some good.

If we think that the key to this error is that the length of the experiment is allowed to be infinite, perversesheaf gives some practical numbers based on simulation, which I have also checked in my own simulations. If the experiment is cut off after patients, or when are helped, which ever comes first, then it is the latter situation about 30% of the time.

I mostly want to open this up for discussion, but here are some quick points I noticed:

The uniform prior isn’t important here. As long as the analyst starts out with some positive probability assigned to the whole interval for some , you get similar results.

As Reginald Reagan points out, the analyst rarely thinks the drug is very good.

To state the last point in a different manner, if the drug was even mildly harmful (say it helped 45% of patients and harmed 55%), this problem doesn’t occur. With those numbers, I ran a simulation and found that only 6 out of 100 analysts were fooled. Moreover, in the limit as the simulation goes to , the fraction of analysts who are fooled will stay finite: If a random walk is biased towards , the odds that it will be greater than , let alone greater than , drop off exponentially.

Normally, I’d like to think a bit more about the question before saying something, but I am getting tired and I want to put up this post for one very key reason: Tumblr is an absurd awful interface for conversations. So, I am hoping that if I get a conversation started here, maybe we will be able to actually talk about it usefully.

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These are excellent positions — available for up to 3 years, with no teaching requirements, and salaries in the AUD81-89k range.

Applications close at the end of January, and I hear Amnon is keen to hire as soon as possible.

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Dear Colleagues,We the undersigned announce that, as of today 15 September 2014, we’re starting an indefinite strike. We will decline all papers submitted to us at the Journal of K-Theory.Our demand is that, as promised in 2007-08, Bak’s family company (ISOPP) hand over the ownership of the journal to the K-Theory Foundation (KTF). The handover must be unconditional, free of charge and cover all the back issues.The remaining editors are cordially invited to join us.Yours Sincerely,Paul Balmer, Spencer Bloch, Gunnar Carlsson, Guillermo Cortinas, Eric Friedlander, Max Karoubi, Gennadi Kasparov, Alexander Merkurjev, Amnon Neeman, Jonathan Rosenberg, Marco Schlichting, Andrei Suslin, Vladimir Voevodsky, Charles Weibel, Guoliang Yu

More details to follow!

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