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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 .
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