Category Archives: maths

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).

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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.

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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.


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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).

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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.

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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.

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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.

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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

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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.

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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.

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