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

Feedback on the 2017 exam:

You can see the exam here

Question 1

(a) was done well by all, though a few omitted to give the generators of the symmetry groups, thereby losing a mark.

(b) This was mostly done well. A few just gave one or two transformations, whereas the question asked to give the full symmetry group (eg, you could have specified the generators, as many did). The second graph was more problematic; some missed that there was a glide-reflection (reversing the sign of the function) in the plane which preserves the graph. [If it doesn't preserve the graph, it's not a symmetry!]

Question 2

(a) (i) Almost everyone got the definition of orbit and stabilizer correct and most correctly proved that stabilizers are subgroups (a few forgot to show they were non-empty, but weren't penalized).

(ii) Some did this well, but a few tried to show \(G_x=gG_yg^{-1}\) without saying what \(g\) was, so couldn't get very far. If they had used \(k\) instead of \(g\) it would (might?) have worked. Some wrote what I can only call rubbish: eg, using expressions like \(g\cdot x\cdot h\), but \(x\cdot h\) is not defined, only \(h\cdot x\). And even worse was expressions like \(gx^{-1}\), but \(x\) is in \(X\) (not in \(G\)) so has no inverse (eg, what is the inverse of an edge of a cube???)

(b) This had many parts:

b(i) A few wrote that the orbit \(D_6\cdot S_1 = \{A,B,C,D,E,F\}=V\), but this is wrong. Each \(g\cdot S_1\) is a set with 2 elements, such as \(\{A,B\}\) or \(\{E,F\}\) etc., and in fact, \[D_6\cdot S_1 = \{\{A,B\},\{B,C\},\{C,D\},\{D,E\},\{E,F\},\{F,A\}\}.\] Everyone (almost?) remembered the orbit-stabilizer theorem (and the theorem worked for the wrong orbit above, which was lucky). The stabilizer of \(S_1\) is \(\{e,r_{\pi/2}\}\)

b(ii) A popular element was \(\{A,D\}\) which has stabilizer of order 4 (not 3 as several wrote!). But other diagonals were also ok.

b(iii) Some showed \(\sigma:\mathcal{P}(V)_2 \to\mathcal{P}(V)_4\) is a bijection, which is good, but to show it is an isomorphism of group actions, you need to involve the group actions - namely, you need to show that it's equivariant. A few did that well. There were even a few who couldn't write down \(\sigma(S)\), I don't know why. It's just the complement of the set, so for \(S_1=\{B,C\}\), we get \(\sigma(S_1)=\{A,D,E,F\}\).

b(iv) Many gave up on this, although it is rather similar to one question in the coursework. A few did it very well.

Question 3

On the whole this was done well.

(a) This was done well by most. A few (very few) left it blank.

(b) Several parts:

b(i) This was done well.

b(ii) Many did this well, showing both inclusions (something pointed out on last year's feedback). A few only showed one inclusion, and a few thought they had shown both, but in fact showed the same inclusion twice by slightly different arguments. NB. When the question says "prove carefully", it is not sufficient to give a short answer without any details. For example, a few said that for their choice \(\mathbf{a}\) was the shortest vector in the lattice and \(\mathbf{b}\) the next shortest, and therefore they generated the lattice. This might be true, but we didn't prove it in the course, and in "showing carefully" you would need to prove this as well. (Note: we used the shortest and second-shortest vectors to classify the lattices into their 5 types, not to prove what the generators were.)

b(iii) Most did this well, although quite a few couldn't remember the definition of the point group of a lattice.

b(iv) The transformation \((r_0\mid\mathbf{v})\in\mathcal{W}\) has for example \(\mathbf{v}=(1,1)^T\) (others are possible too). This gives a glide-reflection (question says, "describe the transformation"), and if you didn't point that out, you would have lost a mark.

Question 4

(a) A large majority did this ok. A few tried (unwisely!) to solve the ODE \(\dot x = x\cos(x)\), but the question didn't ask that (...read the question!!!)

(b) Several parts to this:

  • some lost 2 marks by not giving the axial subgroups (presumably they forgot the definition), but they were still able to find the equilibria just by solving the equations.
  • Well done to those who used the fact that to check invariance, you only need check it for the generators, which are (for example) \(r_0\) and \(R_{\pi/2}\}\) (though one could also use \(r_0\) and \(r_{\pi/4}\)).
  • Some made the A-level mistake of saying \(2x-2xy^2=0\) implies \(y=\pm1\), thereby losing the solution \(x=0\).
  • Many did well finding the equilibria, taking advantage of the fact that some of the 4 axial subgroups are conjugate so reducing the number of calculations required. (There were 5 equilibria in all.)
  • the ODE at the end of this part had initial condition \((1,0)\) which lies on a line of reflection (Fix(\(r_0\))), so the solution has \(y=0\), and the equation becomes simply \(\dot x = 2x\), which is easy to solve.
  • Finally, one candidate complained that "axial subgroups were introduced in Section 4.4, which we were told on several occasions was not examinable". This would be valid, except for the fact that they were re-introduced (defined again) in Sec 5.2 which was examinable. The same student also complained that there were no examples in lectures like the ODE (see above), but again he/she was mistaken.

(c) I was pleased most remembered the definition of the symmetry group of a periodic orbit correctly (a few floundered and wrote rubbish, but I expect they knew that and were just hoping to gain a mark somewhere - they didn't!).

Most correctly wrote , eg, for \(\gamma(t)\), that \[\textstyle\gamma(t+\frac14T)=R_{\pi/2}\,\gamma(t).\] This was meant to be a challenging question, and more explicit information in terms of \(x\) and \(y\), would have earned more marks - namely that \[\textstyle x(t+\frac14T)=-y(t)\quad\text{and}\quad y(t+\frac14T)=x(t).\] For \(\widetilde{D_2}\) that would be \[\textstyle x(t+\frac12T)=x(t)\quad\text{and}\quad y(t+\frac12T)=-y(t).\qquad(*)\] Doing this would have made the last part clearer: I think only 2 candidates correctly identified (with explanation) the solution as having \(\widetilde{D_2}\) symmetry, with the dashed curve being \(x(t)\) (it has period \(T/2\) (see \((*)\)) and the solid curve being \(y(t)\).