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Math 215A Homework 5 Enya Hsiao Problem 3 (Finite group actions on manifolds) Let M be a manifold with properly discontinuous action of a finite group G. Show that the quotient M/G is again a manifold. To find a non-abelian example, consider the quaternions H which is a skew-field, with R-basis {1, i, j, k} such that i2 = j 2 = k 2 = i · j · k = −1. Then the group Q8 := {±1, ±i, ±j, ±k} of order 8 acts on S 3 ⊆ H by left multiplication. Show that this action is properly discontinuous and compute π1 (S 3 /Q8 ). Sketch solutions. Recall from lecture that for any topological space X with strongly discontinuous action of any group G, the map p : X → X/G is open and a covering map. Furthermore, if X is simply connected, π1 (X/G) ∼ = G. (a) To show that M/G is again a manifold, we check that it is second countable, locally Euclidean and Hausdorff . Second countability: This property follows from the fact that the quotient map π : M → M/G is continuous and open. The countable base is thus mapped from M to M/G. Locally Euclidean: Given x ∈ M/G, let V be an evenly covered neighborhood and let p ∈ M be a point in the fibers, π(p) = x. Since M is locally Euclidean, there is a neighborhood U of p homeomorphic e = U ∩ Vp is homeomorphic to Rn by φ : U → Rn . Restricting U to the slice Vp ⊆ π −1 (V ) comtaining p, U e ) = π(U ) ∩ V is a neighborhood of x to its image. Composing homeomorphisms, we conclude that π(U homeomorphic to Rn . The homeomorphism is given by φ ◦ π −1 . Hausdorff : Let x, y be distinct points in M/G. Since G is finite, the fibers π −1 (x), π −1 (y) are finite in M and can be separated by disjoint neighborhoods Uα , Vα respectively, where α = |G|. Now define U, V ∈ M/G by: \ \ U := π(Uα ), V := π(Vα ). α α To begin, x ∈ π(Uα ) and y ∈ π(Vα ) for all α. Further, U and V are open since they are under finite intersections, so they are indeed neighborhoods of x and y. Now note that π −1 (U ) and π −1 (V ) are both G-invariant, i.e. for any u ∈ π −1 (U ), gu ∈ π −1 (U ) for all g ∈ G. In other words, they are a union of orbits, so in particular since they are disjoint in M , they must contain disjoint orbits. Hence under π, U ∩ V = ∅. . (b) For this problem, let us consider a more general proposition. Proposition. Let G be a group of homeomorphisms of X. The action of G on X is said to be fixed-point free(or simply, free) if no element of G other than the identity e has a fixed point. If M is Hausdorff and G is a finite group whose action is fixed-point free, then the action of G is properly discontinuous. Proof. Since G is finite and its action is fixed-point free, the orbit of any x ∈ M is a set of finite points in M . By premise M is Hausdorff, so these points can be separated by disjoint neighborhoods Vα , α = |G|. T ` Let V := α π(Vα ) ∈ M/G. Then π −1 (V ) = Veα is a disjoint union of sets such that no two elements are in the same orbit. In particular, if x ∈ Vβ , gVβ ∩ Vβ = ∅ for any g ∈ G, g 6= e. Hence, G acts properly discontinuously on M . . Using the proposition above, it now suffices to check that Q8 acts freely on S 3 . But H acting on R4 an associative skew-field is free, and so is its restriction to Q8 . It is worthy to check that Q8 preserves length, i.e. it indeed acts on S 3 . Since −1 is length preserving, for the Quaternion relations: i2 = j 2 = k 2 = −1 to hold, it must be that i, j, k is also length preserving. Finally, since S 3 is simply connected, by the remarks above π1 (S 3 /Q8 ) ∼ = Q8 . 1