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.

Simon Marais Problem Competition 2017

A couple of weeks ago, there was the inaugural Simon Marais Mathematics Competition, a new maths competition for undergraduate students living in the time zones between New Zealand and India inclusive.

The students’ scripts have not yet been marked. As a member of the problem committee, I will be especially interested to see how they did. The questions and solutions can be found on the Simon Marais website. There will be a greater variety of solutions posted some time in November after the scripts have been marked. I will currently permit myself a few brief comments on the problems.

I actually came up with B1, which surprised me! The idea here was to specifically come up with an easy problem. One day I was leafing through some old Olympiad material, and on a sheet of preparatory problems for a maths camp from when I was a student, I saw a problem about classifying configurations of 4 points in the plane such that the sum of the distances from each point to the other points was the same. I wondered to myself what happened in three dimensions and the problem was born.

Problem A4 is expected to be very hard and has an interesting solution which I found and wrote up here. This is similar to an approach to the fundamental theorem of algebra which I believe goes back to Gauss and which I may discuss here if I have time.

As a member of the problem committee, I am always looking out for problem submissions for future competitions and welcome anybody who has what they think might be a suitable problem to submit it (which can be directly to me, or the chair of the committee, via email).

The medians of a triangle are concurrent

In case anyone reading this does not know, a median is a line connecting a vertex of a triangle to the midpoint of the opposite edge. The theorem is that the three medians of a triangle are concurrent (i.e. they meet in a single point). Here are four proofs.

Proof 1: (Transformation geometry)

Let E and F be the midpoints as shown and let BE and CF intersect at G. Consider the dilation about A with factor 2. It sends E to C, F to B and G to Q (this is the definition of Q). Then EG and CQ are parallel, as are FG and BQ. Thus BGCQ is a parallellogram and the diagonals of a parallelogram bisect each other QED.

Proof 2: (Vectors. Efficient and boring). Write {\bf a}, {\bf b} and {\bf c} for A, B and C respectively. Let G be ({\bf a}+{\bf b}+{\bf c})/3. It is easy then to check that the midpoint D of AB is ({\bf a}+{\bf b})/2 and that A, D and G are concurrent.

Proof 3: (Why not just prove Ceva’s Theorem)

Ceva’s Theorem states that in the situation shown, AD, BE and CF are concurrent if and only if

\displaystyle \frac{BD}{DC}\frac{CE}{EA}\frac{AF}{FB}=1.

To prove this, note that in the case of concurrence

\displaystyle \frac{BD}{DC}=\frac{|ABG|}{|ACG|}.

The rest of the proof is routine.

Proof 4: (my favourite) WLOG the triangle is equilateral. Now the statement is obvious (e.g. by symmetry).

Perhaps some elaboration should be made to the WLOG. An affine transformation of \mathbb{R}^2 is a map of the form {\bf{v}}\mapsto A{\bf{v}}+{\bf{b}} where A is an invertible matrix and {\bf{b}} is a vector. The affine transformations act transitively on the set of (nondegenerate) triangles and the property of having concurrent medians is clearly invariant under these transformations.

Acknowledgements: Thanks to Inna Lukyanenko for the first proof and tikz files, and to pdftoppm for converting the .pdf output to .png.

A different proof of the fundamental theorem of arithmetic

The fundamental theorem of arithmetic states that every integer can be uniquely written as a product of primes (i.e. \mathbb{Z} is a unique factorisation domain).

The usual proof proceeds through the Euclidean algorithm. Yesterday at lunch I was surprised to learn (thanks to Ole Warnaar) of a different proof bypassing the Euclidean algorithm which I reproduce below. Its primary attraction is its cuteness, as it provides a weaker result than the usual proof (i.e. doesn’t prove that \mathbb{Z} is a Euclidean domain, or even a principal ideal domain).

Suppose that n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k} where p_1,p_2,\ldots,p_k are distinct primes. Consider the finite cyclic group C_n. It has a composition series where the group C_{p_i} appears a_i times as a simple subquotient (and no other simple factors appear). Therefore by the Jordan-Holder theorem, the primes p_i together with their multiplicities a_i are unique. QED.

Pi is trancendental

First the rabbit. The introduction of this function is the part which I don’t know how to motivate. Let f be a polynomial and define

\displaystyle I(z,f)=\int_0^z e^{z-t}f(t)\,dt.

Integration by parts gives the recursion

\displaystyle I(z,f)=e^z f(0)-f(z)+I(z,f')

and therefore we have the formula

\displaystyle I(z,f)=e^z\sum_{j\geq 0} f^{(j)}(0) - \sum_{j\geq 0}f^{(j)}(z).

Now suppose (for want of a contradition) that \pi i \in \overline{\mathbb{Q}}. Let the set of Galois conjugates of \pi i be \{a_1,\ldots,a_k\}. Then we have \prod_{i=1}^k (1+e^{a_i})=0, expand this as \sum_{J\subset[k]}e^{\sum_{j\in J}a_j}=0 and rewrite as

\displaystyle (2^k-d)+\sum_{i=1}^d e^{\theta_i}=0

where the \theta_i are the nonzero exponents.

Now consider

\displaystyle \sum_{i=1}^d I(\theta_i,f)=(d-2^k)\sum_{j\geq 0}f^{(j)}(0)-\sum_{j\geq 0}\sum_{i=1}^df^{(j)}(\theta_i)

Let N be an integer such that N\pi i is an algebraic integer. Let p be a (large) prime and we choose to take

\displaystyle f(x)=N^{dp}x^{p-1}\prod_{i=1}^d(x-\theta_i)^p\in\mathbb{Z}[x].

There are absolute constants A and B (independent of p) such that

\displaystyle |\sum_{i=1}^d I(\theta_i,f)|\leq A B^p

(look at the integral definition of I(z,f) and apply the naive estimate).

Now consider \sum_{i=1}^d I(\theta_i,f). It is a Galois-invariant algebraic integer, hence an integer. We have

\displaystyle \sum_{i=1}^d I(\theta_i,f)=(d-2^k)f^{(p-1)}(0)+\text{higher order terms}.

Here higher order means at least p derivatives appearing. Each of these higher order terms is divisible by p!, hence by p. Since p is prime, for large enough p, \sum_{i=1}^d I(\theta_i,f) is not divisible by p, hence nonzero.

Now every term is divisible by (p-1)!, so we get the lower bound

\displaystyle |\sum_{i=1}^d I(\theta_i,f)|\geq (p-1)!

As there are infinitely many primes, we can send choose p large enough to get a contradiction, QED.

Single variable polynomial division in sage

As far as I can tell, this isn’t done purely within sage, and requires a call to maxima.

If you want to divide the polynomial a by b to produce a quotient and remainder, the code is.


The first element of the output is the quotient and the second is the remainder.

So for example

sage: R=QQ['x']
sage: a=x^210-1
sage: b=R.cyclotomic_polynomial(210)*(x-1)
sage: q,r=a.maxima_methods().divide(b)
sage: q
x^161 + 2*x^160 + 2*x^159 + x^158 - x^156 - x^155 - x^154 - x^153 - x^152 + x^150 + 2*x^149 + 2*x^148 + 2*x^147 + 2*x^146 + 2*x^145 + x^144 - x^142 - x^141 + x^139 + x^138 + x^137 + x^136 + x^135 + x^134 + x^133 + x^132 + x^131 + x^130 + x^129 + x^128 + x^127 + 2*x^126 + 3*x^125 + 3*x^124 + 2*x^123 + x^122 + x^116 + 2*x^115 + 3*x^114 + 3*x^113 + 3*x^112 + 3*x^111 + 3*x^110 + 2*x^109 + x^108 + x^105 + 2*x^104 + 2*x^103 + 2*x^102 + 2*x^101 + 2*x^100 + 2*x^99 + 2*x^98 + 2*x^97 + 2*x^96 + 2*x^95 + 2*x^94 + 2*x^93 + 2*x^92 + 2*x^91 + 2*x^90 + 2*x^89 + 2*x^88 + 2*x^87 + 2*x^86 + 2*x^85 + 2*x^84 + 2*x^83 + 2*x^82 + 2*x^81 + 2*x^80 + 2*x^79 + 2*x^78 + 2*x^77 + 2*x^76 + 2*x^75 + 2*x^74 + 2*x^73 + 2*x^72 + 2*x^71 + 2*x^70 + 2*x^69 + 2*x^68 + 2*x^67 + 2*x^66 + 2*x^65 + 2*x^64 + 2*x^63 + 2*x^62 + 2*x^61 + 2*x^60 + 2*x^59 + 2*x^58 + 2*x^57 + x^56 + x^53 + 2*x^52 + 3*x^51 + 3*x^50 + 3*x^49 + 3*x^48 + 3*x^47 + 2*x^46 + x^45 + x^39 + 2*x^38 + 3*x^37 + 3*x^36 + 2*x^35 + x^34 + x^33 + x^32 + x^31 + x^30 + x^29 + x^28 + x^27 + x^26 + x^25 + x^24 + x^23 + x^22 - x^20 - x^19 + x^17 + 2*x^16 + 2*x^15 + 2*x^14 + 2*x^13 + 2*x^12 + x^11 - x^9 - x^8 - x^7 - x^6 - x^5 + x^3 + 2*x^2 + 2*x + 1
sage: r

hat tip: this question

A free semigroup

In the appendix here, we make the claim that the semigroup

\Gamma_A=\langle \begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}b & 1 \\ 1 & 0 \end{pmatrix}\mid 1\leq a,b\leq A\rangle

is free on the given generators.

Here, I will give a proof. It is a simpler version of the ping-pong argument commonly used to prove that a group is free on a given set of generators.

First a simple observation. it suffices to prove that the set

\left\{ \begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\mid 1\leq a\leq A\right \}

generates a free semigroup.

Now, consider the action of our semigroup on the interval (1,\infty) by fractional linear transformations:

\begin{pmatrix}a & b \\ c & d \end{pmatrix}\cdot x=\frac{ax+b}{cx+d}

In particular, note that

\begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\cdot x=a+\frac{1}{x}.

Now suppose that

\begin{pmatrix}y_1 & 1 \\ 1 & 0 \end{pmatrix}\cdots\begin{pmatrix}y_m & 1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix}z_1 & 1 \\ 1 & 0 \end{pmatrix}\cdots\begin{pmatrix}z_n & 1 \\ 1 & 0 \end{pmatrix}

Apply both sides to some x\in(1,\infty). The LHS lies in (y_1,y_1+1) and the RHS lies in (z_1,z_1+1). Therefore y_1=z_1 and the rest is an easy induction.

Some easy singular Schubert varieties

I write w for both an element of the Weyl group and the corresponding point in G/B. Let X_w be a Schubert variety. Then T_eX_w\subset T_e(G/B) is an inclusion of \mathfrak{b} representations. The latter is generated by the -\theta-weight space, where \theta is the highest root.

Suppose w\geq s_\theta, where s_\theta is the reflection corresponding to \theta. Then the \mathbb{P}^1 connecting e and s_\theta lies in X_w (think of the corresponding SL2), so the -\theta-weight space lies in T_eX_w. Since this generates T_e(G/B), we have T_eX_w=T_e(G/B).

Therefore if w_0\neq w\geq s_\theta, then X_w is singular. This includes some of the first examples of singular Schubert varieties, for example the B2 singular Schubert variety and one of the A3 ones.