MATH 0052 GROUPS AND GEOMETRY FINAL ONLINE 2020
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MATH 0052 GROUPS AND GEOMETRY FINAL ONLINE
Instructions:
(a) Attempt all questions.
(b) Work neatly. Points may be deducted for messy work. It is strongly recommended that you type your solutions (LATEXor Word).
(c) Your answers must be carefully but succinctly explained, using complete sentences. A correct but unjustified answer may attract only half-marks. An incorrect and unjustified answer will probably be worth zero.
(d) When explaining your answers you may make reference to the results in the printed notes (but you must be precise, providing the theorem or page number).
Question 1. The diagram shows a cube with each face bisected by a line parallel to two of its sides:
|
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Use the orbit-stabiliser theorem to calculate the order of the symmetry group of this figure (reflections allowed).
Question 6. Consider the action of the group PSL(2, C) on the Riemann sphere S2 by M¨obius transformations. Suppose that C c S2 is a circle and that g e PSL(2, C) maps C to itself, g(C) = C . Show that the trace of g must be real or imaginary.
Question 7.
(a) Let R1 and R2 be reflections in R| in orthogonal mirrors Π 1 and Π2 , which you may take to contain the origin. Show that R1R2 = R2R1 . Explain why this is a half-turn about the axis Π 1 ∩ Π2 .
(b) In R3, describe, as precisely as you can, the composite of two half- turns about distinct axes, distinguishing carefully the different pos- sible cases. [If the two axes do not meet and are not parallel, you may find it helpful to make use of the unique line which meets both axes at 90↓ .]
In answering this question, bear in mind that the composite of two half- turns must be an isometry. If, for example, this isometry comes out to be a rotation, when you ‘describe’ it you will need to give its axis and angle. Similar information needs to be given for other types of isometries.
One approach is to write the half-turns as products of reflections in ap- propriately chosen pairs of orthgonal planes—but you may use any method.
Answer 1. Applying orbit-stabliser to a face, the size of the orbit is the number of faces, 6, and the stabiliser is of order 4 (identity, two reflections, half-turn). The order of the symmetry group is therefore 24.
[Straightforward unseen application of the orbit-stabiliser theorem.]
Answer 2. The symmetry group is the dihedral group of order 14, D14 , generated by a reflection and a rotation through 2π/7.
The problem is equivalent to counting the number N of G-orbits on X , where G = D14 and X is the set of 2-colourings of P . We may use the
formula
N = | L |X扌| (1)
· There are 27 colourings in total. These are fixed by the identity element.
· For any of the six non-trivial rotations through angle an integer multiple of 2π/7, the only fixed colourings are the ones which colour all edges the same colour. This is essentially because 7 is prime. Thus there is a contribution of 6 x 2 to the sum from these group elements.
· The mirrors of the reflections go through a vertex and the midpoint of the opposite edge. For such a reflection, the number of fixed colorings is 23 x 2 (there are six edges which are exchanged in pairs by the reflection, and for each pair, the two edges must be coloured the same. The last factor of 2 is because the edge which is bisected by the mirror line can be coloured arbitarily). There are 7 such reflections, so these group elements contribute 7 x 16 to the sum.
Hence
N = ╱ 27 + 12 + 7 x 16、= (64 + 6 + 56) = 18.
Thus there are 18 inequivalent two-colourings of the edges of our heptagon.
Answer 3. (a) Suppose that our glide reflection is given by x -1 Rx+b,
where
R = ┐ , b = ┌ ┐0(b1) .
Note this can also be written R(x + b), since Rb = b. Now we write the translation as a product of two reflections R1 , R2, in planes parallel to the y-axis, separated by a distance b1 /2.
(b) This is false. Any rotation of the form
-!
!│
does not have an eigenvalue 1 if the rotation matrices are not the identity.
Answer 4.
(a) If T is spherical, then we must have
+ + _ π > 0
+ > . (2)
Wlog p > q > 2, so 1/p < 1/q < 1/2. There is no solution if q > 4, so q = 2 or 3. If q = 2 we can have any value of p > 2. If q = 3, the only possibilities are p = 3, 4, 5.
It T is euclidean, then in place of (2) we have
1 + 1 = 1
Keeping p > q, we must have q > 3 and the only option with q = 3 is (p, q) = (6, 3). If q = 4, then (p, q) = (4, 4) solves, and if q > 5, 1/p + 1/q < 1/2, so these are the only two euclidean triangles. All other pairs (p, q) appear as hyperbolic triangles.
(b) For such a tesselation by equilateral spherical triangles, suppose that r > 2 come together at each vertex. It must be the same number because the triangles are congruent. Thus we have α = 2π/r . By the angle excess formula it follows that
6π 4π 6 4 4r
The RHS is > 1 so 2 < r < 5. Each of these values gives an integer value for N ,
r = 2 ÷ N = 2; r = 3 ÷ N = 4; r = 4 ÷ N = 8; r = 5 ÷ N = 20.
The case (r, N) = (2, 2) should now be excluded, because it was specified that α < π .
If we choose axes so that n1 = (1, 0, 0, . . . , 0) and n2 = (0, 1, 0, . . . , 0), then from the above formula
R1R2(x) = (_x1 , _x2 , x3 , . . . , x|)
which is a half-turn in the (x1 , x2 )-plane, i.e. with axis {x3 = . . . = x| = 0}, the intersection of the two mirrors.
(b) There are three cases: the axes α 1 and α2 are parallel, meet in a point or are skew (not parallel, don’t meet).
In each case, we shall express the half-turn about αj as a product of reflections in a cunningly chosen pair of orthogonal planes which contain αj . Other methods are possible.
Case 1, parallel: We have a vector v orthogonal to both lines and connecting a point on α 1 to a point on α2 . Then the composite is translation through 2v . To see this, choose Π to contain both the αj , and Πj to be orthogonal to Π, containing αj . Let the corresponding reflections be R and Rj . Then the product of half-turns can be written (RR2)(RR1) = (R2R)(RR1) = R2R1 . This is the composite of reflections in parallel mirrors, which gives translation through twice the orthogonal vector connecting them, as claimed.
Case 2, intersecting: Let Π be the plane which contains α 1 and α2 , and let Πj be the plane through αj orthogonal to Π . Denoting the corresponding reflections R, R1 and R2, our half-turns are RR1 and R2R. The composite (doing the rotation about α 1 first) is then R2RRR1 = R2R1 . Being a product of two reflections, this is a rotation with axis Π 1 ∩ Π2 through twice the angle between α 1 and α2 . [In fact the argument for this case is the same as for the previous one in terms of choice of the mirrors, just the composite R2R1 is a translation if Π 1 and Π2 are parallel and a rotation otherwise.] Case 3, skew lines: Let l be the line which cuts α 1 and α2 at right-angles. This line is unique. Let v be the vector P1P2, where Pj = l ∩ αj . Construct planes as follows: Π 1 containing α 1 and l, Π2 containing α2 and l, Π orthogonal to Πj and containing αj . Let Rj and R be the corresponding reflections. The half-turn about αj is then RjR and so the composite is
R2RR1R = R2(RR)R1
Then Π and Π are parallel (they are both perpendicular to l) and so the composite in the middle is translation through 2v, where v is the vector connecting α 1 to α2 along l. Now we can further rear- range the product as (R2R1)(RR) because a reflection commutes with translation in a direction parallel to the mirror. R2R1 is ro- tation through twice the angle between α 1 and α2 , with axis l. So we have an isometry sometimes called a ‘screw’, or perhaps ‘rotary translation’, the composite of a rotation with a translation in the direction of the axis of rotation. We have already given the vector, axis and angle of rotation.
2022-08-12