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).
Let be the binary tetrahedral group. This group appears as the double cover of the group of rotations of the tetrahedron (under ), as a group of units in an appropriate -form of the quaternion group, or as .
Consider a local field of residue characteristic . Now consider the Galois group of a finite Galois extension. It has a large pro- part, together with two cyclic parts corresponding to the tamely ramified part and the unramified part. This structure alone shows that cannot be such a group unless .
Alternatively, every 2-dimensional irreducible representation of the Galois group of a local field of odd residue characteristic is induced, and has no index two subgroups, so again cannot occur as a Galois group when is odd.
So what about when is even? By considering the fixed field of a Sylow-2-subgroup, we see that if appears as a Galois group, then it is the Galois closure of a degree 8 extension.
Now the degree 8 extensions of 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 is not a Galois group over , but is a Galois group over the unramified quadratic extension of .
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).
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 , and for A, B and C respectively. Let G be . It is easy then to check that the midpoint D of AB is 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
To prove this, note that in the case of concurrence
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 is a map of the form where is an invertible matrix and 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.
There are about six of them at the domestic terminal, not under cover, on the ground floor, near the rental cars and the lift to the overpass to the terminals. Here is a map with their location, together with the cycle routes in and out.
As you can see from the map, openstreetmap also knows where these bike racks are.
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 by 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
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
hat tip: this question
In the appendix here, we make the claim that the semigroup
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
generates a free semigroup.
Now, consider the action of our semigroup on the interval by fractional linear transformations:
In particular, note that
Now suppose that
Apply both sides to some . The LHS lies in and the RHS lies in . Therefore and the rest is an easy induction.