Tension in teaching

The following quote is by Matt Emerton (in a comment on MathOverflow)

I think there is a genuine tension between proofs that a professional will like (where professional here may mean professional algebraist!) and ones that are elementary. For professionals, reductions and devissages are easy, natural, and we don’t even think of them as real landmarks in the proof; they are just serve as passages between the key points and ideas. But in writing things out, they can take a lot of words, and seem (as you wrote) mysterious and difficult. I don’t know the best way to deal with this tension.

Interestingly, Matt posted it as part of a discussion about exactly what I wanted to talk about in this post, the teaching of the structure theorem for finitely generated abelian groups, or more generally, of finitely generated modules over a PID.

My personal connection is that I taught this as part of our third-year algebra course this year at the University of Melbourne, and am slated to do so again next year. I think that I did not do a particularly good job of teaching it in 2022, primarily because I got distracted by the reductions and devissages and tried to proceed along those lines as much as possible, when what I have learned is more appropriate for one of these courses is the more prosaic approach involving matrix manipulations. It is with the matrix manipulations (directly proving Smith Normal Form) that I plan to teach this part of the course in 2023 (and beyond, if necessary).

For completeness, allow me to state the professionals’ proof: Split off the quotient by the torsion subgroup to reduce to the torsion case. Then canonically decompose the module into a direct sum of its p-primary components. Then use the fact that R/(p^e) is injective over itself to manually split the remaining short exact sequences needed to complete the classification.

While it may not be reasonable to expect a third-year student to follow this proof, I think it is fair to expect any PhD student of mine to be able to understand and execute this proof.

SMMC 2022 A4

The Simon Marais Mathematics Competition happened last weekend. It is a maths competition for undergraduate students across Europe, Asia, Africa and Oceania. This post is about problem A4, which I submitted. I’ll talk a bit about where the problem came from, a generalisation, a conjecture and also provide a solution. The entire paper is available on the Marais website, and solutions should be put up there at some time in the near future.

Problem (SMMC 2022 A4)
Let n be a positive integer, and let q\geq 3 be an odd integer such that every prime factor of q is larger than n. Prove that

    \[ \frac{1}{n!(q-1)^n}\prod_{i=1}^n (q^i-1) \]

is an integer that has no prime factor in common with \displaystyle{\frac{q-1}{2}}.

Origins

Let G=GL_n(\mathbb{F}_q) and let N be the subgroup of monomial matrices (a matrix is a monomial matrix if and only if it has exactly one nonzero entry in each row and column). I show below in my solution that this question is equivalent to the fact that the integer |G|/|N| is coprime to (q-1)/2. Now why would I ever care about that?

This coprimality fact implies that the cohomology of G with mod (q-1)/2 coefficients is isomorphic to the cohomology of N with mod (q-1)/2 coefficients. And I was interested in these cohomology groups because the second cohomology group classifies central extensions, which is what I used to think about back in my PhD days. The group N feels somewhat more “combinatorial” than G, so it is nice to be able to pass information from N to G for free.

Generalisations (known and conjectural)

Let G be a split reductive group over \mathbb{F}_q, which I conflate with its \mathbb{F}_q-points below in an abuse of notation. Let T be a maximal split torus and N its normaliser in G. Then

    \[ \frac{|G|}{|N|}=q^{|\Phi^+|}\prod_i \frac{q^{d_i}-1}{d_i(q-1)}. \]

Here \Phi^+ is the set of positive roots and the collection of integers \{d_i\} are the exponents of the Weyl group. Then the same argument as in my proof below shows that this fraction is an integer, relatively prime to \frac{q-1}{2}.

If we remove the assumption that G is split, then I suspect the same conclusion is satisfied, but there is an additional argument needed as the formula for the quotient has additional factors. I have not worked out this argument and really don’t want to resort to case by case arguments, so there is your conjecture (I expect we now need to say T is a maximal torus containing a maximal split torus).

Solution

First we show that the fraction in the question is an integer. Since q-1 divides q^d-1 as a polynomial for all d, the statement only depends on the residue class of q modulo n!. Since every prime factor of q is greater than n, q is relatively prime to n!. So by Dirichlet’s theorem on primes in arithmetic progressions, we may assume without loss of generality that q is prime.

Let G=GL_n(\mathbb{F}_q) and let N be the subgroup of monomial matrices. Then

    \[|G|=q^{\frac{n(n-1)}{2}}\prod_{d=1}^n q^d-1 \qquad \mbox{and}\qquad |N|=n!(q-1)^n.\]

By Lagrange’s theorem |G|/|N| is an integer. Since q is relatively prime to |N|, we can further divide by the largest power of q in |G| and deduce that

    \[ \frac{|G|}{q^{\frac{n(n-1)}{2}}|N|}=\frac{1}{n!(q-1)^n}\prod_{d=1}^n (q^d-1) \]

is an integer.

Now let p be a prime dividing \frac{q-1}{2} and let d be a positive integer. To conclude, it suffices to show that the fraction

    \[ \frac{q^d-1}{d(q-1)} \]

has zero p-adic valuation. Write q=1+2m, then by the binomial theorem,

    \[ \frac{q^d-1}{d(q-1)}=\frac{1}{2dm}\sum_{i=1}^d {d \choose i}(2n)^i. \]

Let a=v_p(d) and b=v_p(m). Since v_p(i!)=\lfloor\frac{i}{p}\rfloor+\lfloor\frac{i}{p^2}\rfloor+\lfloor\frac{i}{p^3}\rfloor+\cdots <\frac{i}{p-1}, we get

    \[ v_p\left( {d \choose i}(2n)^i \right)\geq v_p\left(\frac{d(2n)^i}{i!}\right)>a-\frac{i}{p-1}+i(b+v_p(2)). \]

We have the inequality v_p(2)-\frac{1}{p-1}\geq \frac{v_p(2)-1}{2}, so

    \[ v_p\left( {d \choose i}(2n)^i \right)>a+i\left(b+\frac{v_p(2)-1}{2}\right). \]

For i\geq 2, we therefore get

    \[ v_p\left( {d \choose i}(2n)^i \right)>a+2(b-\frac{1}{2})=a+2b+v_p(2)-1\geq a+b+v_p(2)=v_p(2dm) \]

as b\geq 1 from our assumption that p divides m.
Thus in our sum, the term with i=1 has a strictly smaller p-adic valuation than every other term, so determines the p-adic valuation of the sum, and we get

    \[ v_p \left( \frac{q^d-1}{d(q-1)} \right)=v_p\left(\frac{2md}{2md}\right)=0, \]

completing the proof.

Infinite dimensional vector spaces

This is a quick note to prove that two bases of an infinite dimensional vector space have the same cardinality. We freely use the axiom of choice and other standard facts about cardinalities of infinite sets. We will in fact prove the following:

Theorem: Let V be a vector space with basis \{v_i\}_{i\in I} with I an infinite set. Let \{w_j\}_{j\in J} be a linearly independent subset of V. (e.g. a basis of a subspace). Then |J|\leq |I|.

To prove this, WLOG J is a basis of V (by extending \{w_j\}_{j\in J} to a basis of V if necessary). For all i\in I, write

    \[v_i=\sum_j c_{ij}w_j.\]

Let E\subset I\times J be the set of pairs (i,j) with c_{ij}\neq 0. Then E\to I has finite fibres, since the sum above is finite, and E\to J is surjective, since the v_i lie in the span of the w_j with j in the image of V, but also the v_i generate V. Since I is assumed infinite, this is enough to prove that |I|\geq |J|, as required.

Jucys-Murphy elements and induction

This post concerns the representation theory of the symmetric group over the complex numbers. Recall that the irreducible representations of the symmetric group S_n are indexed by partitions of n. Let S^\lambda be the irreducible representation indexed by \lambda. I want to say some words about the theorem that the decomposition of the induced module \operatorname{Ind}_{S_{n}}^{S_{n+1}}S^\lambda is given by the decomposition into eigenspaces under the action of the Jucys-Murphy element.

First, the relevant Jucys-Murphy element is

    \[X:=(1,n+1)+(2,n+1)+\cdots+(n,n+1)\in \mathbb{C}[S_{n+1}].\]

The way it acts on \operatorname{Ind}_{S_{n}}^{S_{n+1}}S^\lambda=\mathbb{C}[S_{n+1}]\otimes_{\mathbb{C}[S_n]}S^\la is not as an element of \mathbb{C}[S_{n+1}] but by X\cdot (a\otimes v)=aX\otimes v. This is well-defined since X commutes with \mathbb{C}[S_n].

What this action defines is a natural transformation from the functor \operatorname{Ind}_{S_{n}}^{S_{n+1}} to itself. The induction functor is (bi)-adjoint to the restriction functor and this natural transformation is even simpler to construct on the adjoint side. Recall that if \mathcal{F} and \mathcal{G} are adjoint functors, then there is an isomorphism

    \[\operatorname{End}\mathcal{F}\cong\operatorname{End}\mathcal{G}.\]

Here \operatorname{End}\mathcal{F} refers to the natural transformations from \mathcal{F} to itself, and the map in this isomorphism is given by pre- and post-composition by the unit and counit of the adjunction.

And the way that X yields a natural transformation from \operatorname{Res}_{S_n}^{S_{n+1}} to itself is very simple, it’s just by its usual action as an element of \mathbb{C}[S_{n+1}]. If you transport this natural transformation to a natural transformation of the induction functor via the method I just mentioned, then you get the formula mentioned above.

Now given a pair of adjoint functors \mathcal{F} and \mathcal{G}, a natural transformation X from \mathcal{F} to \mathcal{F} (and hence from \mathcal{G} to \mathcal{G}) and a complex number a, we can define a functor \mathcal{F}_a by

    \[\mathcal{F}_a(V)=\{w\in \mathcal{F}(V)\mid Xw=aw\}.\]

and similarly for \mathcal{G} (this requires some linearity assumptions, but they’re satisfied here. Also you could take generalised eigenspaces if you wanted to, but in our application there is no difference).

When you do this, the functors \mathcal{F}_a and \mathcal{G}_a are adjoint:

Proof: Both \operatorname{Hom}(\mathcal{F}_aV,W) and \operatorname{Hom}(V,\mathcal{G}_aW) are the a-eigenspace of the action of X on \operatorname{Hom}(\mathcal{F}V,W)\cong\operatorname{Hom}(V,\mathcal{G}W).

Now apply this to our situation. We also use the following standard fact about the action of the Jucys-Murphy element (as developed e.g. in the Vershik-Okounkov approach):

Consider the decomposition

    \[\operatorname{Res}_{S_n}^{S_{n+1}}S^\mu=\bigoplus_{\lambda\to\mu}S^\lambda.\]

Then the Jucys-Murphy element X acts by the scalar c(\alpha) on S^\lambda, where c(\alpha) is the content of the box \alpha added to \lambda to get \mu.

Now translating this statement via the above yoga onto the adjoint side, we get

In the decomposition

    \[\operatorname{Ind}_{S_{n}}^{S_{n+1}}S^\lambda=\bigoplus_{\lambda\to\mu}S^\mu,\]

the Jucys-Murphy element X acts by the scalar c(\alpha) on S^\mu, where c(\alpha) is the content of the box \alpha added to \lambda to get \mu.

Singularities of Schubert varieties within a right cell

Martina Lanini and I recently posted our preprint Singularities of Schubert varieties within a right cell to the arXiv. In it, we show that every singularity which appears in a type A Schubert variety appears between two permutations lying in the same right cell. This shows that any behaviour controlled by the singularities of Schubert varieties manifests itself within a Specht module. Some exmples are discussed.

The work was conducted during our recent visit to the thematic trimester program on representation theory at the Institut Henri Poincaré in Paris. I spent an enjoyable first month there before returning to Australia. Originally I was scheduled to be on a plane right now to return to Paris for the end of the program, but alas this is no longer possible. Oh well.

We’re hiring

The University of Melbourne maths department is hiring. All our ads can be found on mathjobs.

In pure maths, we have a continuing position in analysis. Applications should be submitted by the earlier deadline of 25 October to ensure full consideration.

There are also other positions available in applied and statistics, as well as a temporary position open to all areas.

Sydney Mathematical Research Institute International Visitor Program

There is a new (established in the last couple of years) mathematical institute based at the University of Sydney. I want to call attention to a visitor program they run, for mathematical visits to Australia.

They fund research visits of at least one month’s duration. At least two weeks must be spent at the University of Sydney, but the rest can be taken anywhere in Australia.

Funding rounds for these research visits happen periodically. Currently there is one open for visits within the period April-December 2020, with applications closing on 28 July.

Anyone interested should check out the research intstitute’s webpage.

Latex in WordPress

To activate Latex support in this blog, I have now activated the QuickLatex plugin. This has solved the problems with vertical alignment of mathematics that previously plagued this blog (only in new posts using QuickLatex, the back catalogue has not been converted).

The plugin webpage linked above shows its usage (for both posts and comments), allowing Latex to be typed natively once the word “latexpage” in square parentheses appears (so no inserting of the word “latex” after a dollar symbol anymore).

The binary tetrahedral group as a p-adic Galois group.

Let G be the binary tetrahedral group. This group appears as the double cover of the group of rotations of the tetrahedron (under SU(2)\to SO(3)), as a group of units in an appropriate \mathbb{Z}-form of the quaternion group, or as SL_2(\mathbb{F}_3).

Consider a local field of residue characteristic p. Now consider the Galois group of a finite Galois extension. It has a large pro-p part, together with two cyclic parts corresponding to the tamely ramified part and the unramified part. This structure alone shows that G cannot be such a group unless p=2.

Alternatively, every 2-dimensional irreducible representation of the Galois group of a local field of odd residue characteristic is induced, and G has no index two subgroups, so again cannot occur as a Galois group when p is odd.

So what about when p is even? By considering the fixed field of a Sylow-2-subgroup, we see that if G appears as a Galois group, then it is the Galois closure of a degree 8 extension.

Now the degree 8 extensions of \mathbb{Q}_2 are finite and number and have all been computed, together with their Galois groups. They can be found at this online database. A quick examination shows that our binary tetrahedral group does not appear.

But it does appear as an inertia group. So G is not a Galois group over \mathbb{Q}_2, but is a Galois group over the unramified quadratic extension of \mathbb{Q}_2.

An exceptional isomorphism

We will construct the exceptional isomorphism S_6\cong Sp_4(\mathbb{F}_2).

The group S_6 acts on \mathbb{F}_2^6 preserving the usual pairing \langle e_i,e_j\rangle=\delta_{ij} where the e_i are the usual basis vectors.

There is an invariant line L, the span of \sum_i e_i and an invariant hyperplane H=\{\sum_i a_ie_i|\sum_i a_i=0\}. Let V=H/L. S_6 acts on V.

Since L is the radical of the pairing \langle \cdot,\cdot\rangle on H, the pairing \langle \cdot,\cdot\rangle descends to a non-degenerate bilinear pairing on V. As it is symmetric and we are in characteristic 2, it is automatically skew-symmetric.

The S_6-action preserves this pairing, hence we get our desired homomorphism from S_6 to Sp_4(\mathbb{F}_2).

To check injectivity, it suffices to show that (12) is not in the kernel, since we know all normal subgroups of S_6. Surjectivity then follows by a counting argument, so we get our desired isomorphism.