SOME MORE COVERING SPACE EXAMPLES Example 1: Coverings of the M¨obiusstrip. This is a formal version of the construction I tried to give in class. Let X be the M¨obiusstrip, X = ([0; 1] × [0; 1])=∼ with (0; y) ∼ (1; 1 − y). Then the quotient space Y = ([0; n] × [0; 1])=∼ with (0; y) ∼ (n; 1 − y) for n odd and (0; y) ∼ (n; y) for n even is an n-sheeted covering space of X. The covering map p is given by p([(k + x; y)]) = [(x; y)] for k even, x; y 2 [0; 1], and p([(k + x; y)]) = [(x; 1 − y)] for k odd, x; y 2 [0; 1]. I will leave it up to you to convince yourselves that this is the correct answer|never mind what was said in class. Just to confirm: The corresponding model of the n-sheeted connected covering of X = S1 really is Y = [0; n]=∼ with 0 ∼ n, and p([k + x]) = x for k 2 f0; :::; n − 1g and x 2 [0; 1]. Example 2: Action on fibers. Let p : Y ! X be a covering map such that Y is path-connected. −1 Fix any point x 2 X and define Fx = p (x). Then every automorphism Φ of the covering satisfies Φ(Fx) = Fx, so Aut(Y : X) acts on Fx. Now Fx is in bijection with Fx¯ for every other pointx ¯ 2 X by HW 6, and I want to argue that the Aut(Y : X)-actions on Fx and Fx¯ are \the same". Connect x andx ¯ by a path f. Now for any y 2 Fx lety ¯ 2 Fx¯ be the endpoint of the unique lift of f with initial point y. Then setting Mf (y) =y ¯ defines a bijection Mf : Fx ! Fx¯, and I claim that ~ ~ Mf ◦ ΦjFx = ΦjFx¯ ◦ Mf . But this is just the statement that if f is a lift of f then so is Φ ◦ f. The precise choice of f wasn't important here. This is remarkable because even if x =x ¯ then Mf need not be the identity (this phenomenon is called \monodromy") unless π1(X) = f1g. Example 3: Lifting of maps. Here is a difficult problem: Find the sheet number of the covering 1 1 1 1 a b c d p : S × S ! S × S given by p(z; w) = (z w ; z w ), where a; b; c; d 2 Z and ad − bc 6= 0. Thinking about this should be quite instructive. One useful idea is to lift p to the universal cover 2 2 of the torus, i.e. to findp ~ : R ! R withp ~(0) = 0 such that the following diagram commutes: 2 2 R p~ - R puniv puniv ? ? T 2 p - T 2 The lifting criterion from lectures applied to p ◦ puniv tells us thatp ~ exists, but you probably want to computep ~ explicitly here. Also it may help to look at the easier example p : S1 ! S1, p(z) = za, where we know in advance that the sheet number is jaj (provided that a 6= 0). Example 4: Spin. You may find this interesting. I assume here that you are somewhat comfortable with the quaternions H = fq = a + bi + cj + dk : a; b; c; d 2 Rg. (i) Let G ⊂ H denote the subset of quaternions of length one, which is obviously homeomorphic to S3. Then G is closed under quaternion multiplication and inversion. −1 (ii) G acts on the 3-dimensional subspace Im H ⊂ H of imaginary quaternions via (g; q) 7! g qg. −1 The matrix Ag representing the linear map q 7! g qg with respect to fi; j; kg lies in SO(3). (iii) The group homomorphism G ! SO(3), g 7! Ag, is a 2-sheeted covering. In particular, SO(3) 3 is homeomorphic to RP . (This is not so easy. If A 2 SO(3) represents rotation through an oriented angle of θ around the oriented axis spanned by a unit vector v 2 S2 [using the left-hand rule], then θ θ A = Ag, where g = cos( 2 ) + sin( 2 )v with v = (v1; v2; v3) = v1i + v2j + v3k 2 Im H.) (iv) The group G is isomorphic to SU(2) as a topological group. An explicit isomorphism is given α −β¯ by sending ( β α¯ ) 2 SU(2) to Re(α) + Im(α)i + Re(β)j − Im(β)k 2 G..