\( \def\Fix{\mathrm{Fix}} \def\D{\mathsf{D}} \def\C{\mathsf{C}} \def\D{\mathsf{D}} \def\GL{\mathsf{GL}} \def\OO{\mathsf{O}} \def\SO{\mathsf{SO}} \def\RR{\mathbb{R}} \def\ZZ{\mathbb{Z}} \def\Aut{\mathrm{Aut}} \def\xx{\mathbf{x}} \)
jam - wiki

Feedback on the 2016 exam:

You can see the exam here

Question 1

(a) This was straightforward, and almost everyone did this correctly.

(b) This was more challenging, The first graph was fairly clear: the symmetry group is generated by a reflection in the y-axis, and a translation by 2 units. The second was less clear: in addition to the translation by 2 units there is a glide reflection \((x,y)\mapsto(x+1,-y)\) (this was the symmetry allowing reversing of the sign of the function referred to in the question: \(g(x+1)=-g(x)\)). I was pleased to see how many did spot the glide-reflection, although just writing 'there is a glide-reflection' without making it explicit was not enough to get the marks. [Each of the graphs was marked out of 4 points.] Note that saying reflection in the \(x\)-axis reverses the sign of the function is correct (for any function!) but irrelevant - it's not a symmetry of the function. A transformation that reverses the sign would be a transformation \(T(x)\) for which \(g(T(x))=-g(x)\), and here that is \(T(x)=x+1\), as written above.

Question 2

(a) Almost all got (i) including the proof. For (ii),there was some confusion between the given element \(h\) and some arbitrary element \(g\) which you introduced. It was much easier if you realized that the coset involved was \(hG_x\). (iii) This question said 'Deduce ...', so those who only used the orbit-stabilizer theorem to show there was a bijection (because the sets had the same cardinality) did not get many marks (1 out of a possible 3). The correct way to do this was to write down a map between the sets and show it's a bijection; quite a good number of you succeeded in that.

(b) Some did this directly in x, y coordinates, some used complex numbers. Both approaches are valid of course. However, a few tried to show \(f\) was invariant rather than equivariant, which of course they couldn't do (because it's not!).

Question 3

(a) Most did this well. A very small number of students didn't write the action, and then the rest was unclear (or incoherent) - and they got 0, 1 or 2 out of 4 depending on how close their argument came to the correct one.

(b) (i) Many of you did this well. However there were some surprising lapses: for example not being able to show that \(\mathbf{u}_1\) and \(\mathbf{u}_2\) belong to \(L\): it was just a matter of showing they satisfied the definition of \(L\) (ie, \(\mathbf{u}_1=(2,0)\) has \(x=2,y=0\) so \(x=2\in\mathbb{Z}\) and \(y=0\in2\mathbb{Z}\) and importantly \(x+\frac12 y=2+0=2\in2\mathbb{Z}\) - it was that simple!). Several of you didn't get the diagram right, missing some of the points out and showing a rectangular lattice (it should have been a centred-rectangular lattice). That makes it impossible to do part (ii) correctly - though partial marks were given for a good attempt at a proof.

(ii) Full marks were obtained if you showed both \(\ZZ\{\mathbf{a},\,\mathbf{b}\}\subset L\) (the easy part) and \(L\subset \ZZ\{\mathbf{a},\,\mathbf{b}\}\). Those who used an argument based on the shortest element only got partial marks; the shortest elements were useful for distinguishing between the types of lattice, not for showing whether the lattice in question was generated by those elements. And finally, a few suggested \(L = \ZZ\{\mathbf{a},\,\mathbf{b}\}\) with \(\mathbf{a}=(1,0)\) and \(\mathbf{b}=(0,2)\), but since neither of these vectors belong to the lattice that was no good.

(iii) was straightforward and worked out even with the wrong lattice.

Question 4

Very few did this question at all well. Many just got 3 or 4 marks out of 30! [See note on scaling below].

(a) To make this equivariant, you need to have \(\mathbb{Z}_3\) acting by cyclic permutations (that is, the group generated by the permutation \((1\;2\;3)\)). Several had the group acting by rotations, generated by \(R_{2\pi/3}\) - while not exactly right I gave the benefit of the doubt. Most got the diagram right, although quite a few had the arrows in the wrong direction and lost a point for that.

(b) this part was about taking a solution \(\gamma\) with \(\gamma(0)\) on the diagonal, then deducing that \(\gamma(t)\) also lies on the diagonal (= invariant under the evolution of the system). This was because the diagonal is the fixed point set of the \(\mathbb{Z}_3\) action.

(c) This should have been the easiest 4 points on the exam: just put all the right-hand-sides equal to 0, and \(x=y=z\). There are two points \((0,0,0)\) and \((1,1,1)\).

(d) This requires solving an ODE! Namely \(\dot x = x-x^2\), which you can solve by separation of variables. Only a few got that right. Some did something strange: substituting the initial values before solving the ODE, so getting for example \(\dot x = x-y^2=\frac14\), and then solving to get \(x=\frac12+\frac14t\). Wrong!! (and zero marks)

(e) Quite a few could define the symmetry group of a periodic orbit correctly - but quite a few couldn't! (For example, writing a definition that didn't depend on the periodic orbit.) Some realized correctly they needed to classify homomorphisms from \(\mathbb{Z}_3\) to \(S^1\), so well done if you did that (some got partial marks for that if they realized that was what was needed, even though their explanation of the classification wasn't right).

Scaling

Since there were quite a lot of poor marks (mostly due to the challenges of Question 4, and this being the first year the course was run) the marks were scaled up significantly. This excuse won't hold in the future.