Massey Exercise 5.7.2

Let be a space, , and a covering space. We determine the group of automorphisms of each of the following particular covering spaces.

(a) Consider , , and .

Note . We have that is isomorphic to . Since is abelian, the normalizer of this subgroup is the whole group . So we have:

Each element corresponds to the homeomorphism of the circle given by a rotation of . Denote it – then we have where is a primitive ‘th root of unity. Note it is indeed a covering space automorphism because:

(a.2) Consider , and .

We discuss this here before doing part (b). It’s relatively easy to show (and shown in Massey) that , and in fact , the group of integer translations of the line.

(b) Consider .

Since this is just a product, each possible covering space corresponds to a product of covering spaces of . Furthermore, then the automorphism group for each is the product of the automorphism groups of each factor. Since we’ve determined the possible covering spaces of the circle and their automorphism groups in the last two sections, the problem is also completely solved for the torus.

(c.1) Consider as the figure 8 curve, and (the union of all the vertical and horizontal integer lines). The map is the map that wraps each horizontal line around the right circle and each vertical line around the left circle ( and are specifically described in in Massey).

We have . To determine , consider the following: has the same homotopy type as , the plane with integer points removed (the latter deformation retracts to the former). Since this group is the plane with countably many points removed, the fundamental group is the free group with countably many generators, each corresponding to the traversal of one square cell in . In fact, call it , where is the path that starts at , goes units vertically, units horizontally, traverses the square cell with lower left corner , and then retraces its steps back to .

Let be the commutator subgroup of . We claim that . Consider what does to a generator of :

sends to the conjugate of a commutator . Since the commutator subgroup is normal, its stable under conjugation, so . Since this happens for all the generators, we conclude that .

Next, let be an arbitrary element in the commutator subgroup. Then, let be the path that starts at the origin, travels units up, units right, units down, and units left to trace out a rectangle – note this returns to the basepoint so it is indeed a loop. Then we have Since was arbitrary, we have and thus .

Since is certainly normal, its normalizer is the whole group and we can finally conclude:

In fact, each corresponds to the integer translation of the grid . Using the explicit description of the map in Massey its easy to show that (the trig functions are periodic with period 1), so these homeomorphisms are actually covering space automorphisms.

(c.2) Take to be the figure eight curve, and to be the union of a vertical line and a countable collection of disjoint circles tangent to the line. The map is the map that maps each circle homeomorphically onto the right circle, and wraps the line periodically around the left circle (again, maps and spaces are explicitly described in Massey).

Again, . This time, is homotopy equivalent to the wedge product of countably many circles, so is the free group on countably many generators. Let it be where is the path that travels up the line, traverses the ‘th circle, and returns.

Let be the normal closure of in . We claim that . We have , so clearly .

Next, let where , so is an arbitrary element of . We can construct a path that will map to this element in the naive way. Start a the basepoint, and simply read off the word . When encountering , traverse the line units. When encountering , loop the circle in the current location times. For each , every power of in will be eventually undone by the opposite power of in , and since the only other element in each is a pure power of , the sum of powers of in all of is zero – this means that will eventually end on its basepoint, guaranteeing that is actually a loop. So is indeed an element in that maps to . Then, and .

Note that is a normal subgroup of . Let be the same arbitrary element as above:

So, its normalizer is the whole group and we have . Since modding out by , the normal closure of is the same as adding as a relation, we get:

In fact, each corresponds to the homeomorphism shifting the space vertically units. Indeed, it can be checked that each of these homeomorphisms is a covering space automorphism.

(d): Consider ( without the origin) and . The covering map is .

is contractible, so this is a universal cover and , which is isomorphic to (deformation retract onto the circle). Each corresponds to the homeomorphism that shifts along the imaginary axis by units.

(e): Consider , , and .

This is entirely analogous to part (a). We have and for each , the homeomorphism is the rotation of by .