SOME MORE 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 ∈ [0, 1], and p([(k + x, y)]) = [(x, 1 − y)] for k odd, x, y ∈ [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 ∈ {0, ..., n − 1} and x ∈ [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 ∈ 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 with Fx¯ for every other pointx ¯ ∈ 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 ∈ Fx lety ¯ ∈ Fx¯ be the endpoint of the unique of f with initial point y. Then setting Mf (y) =y ¯ defines a bijection Mf : Fx → Fx¯, and I claim that ˜ ˜ Mf ◦ Φ|Fx = Φ|Fx¯ ◦ 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 “”) unless π1(X) = {1}.

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 ∈ Z and ad − bc 6= 0. Thinking about this should be quite instructive. One useful idea is to lift p to the universal 2 2 of the , 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 |a| (provided that a 6= 0).

Example 4: Spin. You may find this interesting. I assume here that you are somewhat comfortable with the H = {q = a + bi + cj + dk : a, b, c, d ∈ R}. (i) Let G ⊂ H denote the subset of quaternions of length one, which is obviously homeomorphic to S3. Then G is closed under 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 {i, j, k} 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 ∈ SO(3) represents through an oriented angle of θ around the oriented axis spanned by a unit vector v ∈ 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 ∈ Im H.) (iv) The group G is isomorphic to SU(2) as a . An explicit isomorphism is given α −β¯ by sending ( β α¯ ) ∈ SU(2) to Re(α) + Im(α)i + Re(β)j − Im(β)k ∈ G.