Let $X$ be a space, $x \in X$, and $(\wt{X}, p)$ a covering space. We determine the group of automorphisms of each of the following particular covering spaces.

(a) Consider $X = S^1$, $\wt{X} = S^1$, and $p : z \mapsto z^n$.

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Note $\pi_1(X) \cong \mathbb{Z}$. We have that $p_* \pi(\wt{X})$ is isomorphic to $n\mathbb{Z}$. Since $\mathbb{Z}$ is abelian, the normalizer of this subgroup is the whole group $\mathbb{Z}$. So we have:

$$A(\wt{X}) = \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$$

Each element $k \in \mathbb{Z}_n$ corresponds to the homeomorphism of the circle given by a rotation of $2\pi k /n$. Denote it $h_k$ -- then we have $h_k(z) = z \zeta_n^k$ where $\zeta_n$ is a primitive $n$'th root of unity. Note it is indeed a covering space automorphism because:

$$(p \circ h_k)(z) = (z\zeta_n^k)^n = z^n \zeta_n^{kn} = z^n = p(z)$$

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(a.2) Consider $X = S^1$, $\wt{X} = \mathbb{R}$ and $p: x \mapsto \exp(2\pi i x)$.

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We discuss this here before doing part (b). It's relatively easy to show (and shown in Massey) that $A(\wt{X}) \cong \mathbb{Z}$, and in fact $A(\wt{X}) = \setbuilder{t_n : x \mapsto x+2\pi n}{n \in \mathbb{Z}}$, the group of integer translations of the line.

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(b) Consider $X = T^2 = S^1 \times S^1$.

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Since this is just a product, each possible covering space corresponds to a product of covering spaces of $S^1$. 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.

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(c.1) Consider $X$ as the figure 8 curve, and $\wt{X} = \setbuilder{(x,y) \in \mathbb{R}^2}{x \in \mathbb{Z} \tor y \in \mathbb{Z}}$ (the union of all the vertical and horizontal integer lines). The map $p$ is the map that wraps each horizontal line around the right circle and each vertical line around the left circle ($X$ and $p$ are specifically described in $\mathbb{R}^2$ in Massey).

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We have $\pi_1(X) = \la a, b \ra$. To determine $\pi_1(\wt{X})$, consider the following:

$$\wt{X}$$ has the same homotopy type as $$\mathbb{R}^2 - \mathbb{Z}^2$$, 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 $$\wt{X}$$. In fact, call it $$\la \gamma_{m,n} \ra$$, where $$\gamma_{m,n}$$ is the path that starts at $$(0,0)$$, goes $$n$$ units vertically, $$m$$ units horizontally, traverses the square cell with lower left corner $$(m,n)$$, and then retraces its steps back to $$(0,0)$$. Let $$K$$ be the commutator subgroup of $$\pi_1(X)$$. We claim that $$\im(p_*) = K$$. Consider what $$p_*$$ does to a generator of $$\pi_1(\wt{X})$$:$$

p*(\gamma{m,n}) = a^n b^m (ab a^{-1} b^{-1}) b^{-m} a^{-n}

$$$$p_*$$ sends $$\gamma_{m,n}$$ to the conjugate of a commutator $$aba^{-1}b^{-1} \in K$$. Since the commutator subgroup is normal, its stable under conjugation, so $$p_*(\gamma_{m,n}) \in K$$. Since this happens for all the generators, we conclude that $$\im(p_*) \subseteq K$$. Next, let $$x = (a^{n_1} b^{m_1} a^{-n_1} b^{-m_1}) \dots (a^{n_r} b^{m_r} a^{-n_r} b^{-m_r}) \in K$$ be an arbitrary element in the commutator subgroup. Then, let $$\alpha_{m,n}$$ be the path that starts at the origin, travels $$n$$ units up, $$m$$ units right, $$n$$ units down, and $$m$$ units left to trace out a rectangle -- note this returns to the basepoint so it is indeed a loop. Then we have $$p_*(\alpha_{n_1,m_1} \dots \alpha_{n_r,m_r}) = x$$

Since $x$ was arbitrary, we have $K \subseteq \im(p_*)$ and thus $\im(p_*) = K$.

Since $K$ is certainly normal, its normalizer is the whole group $\pi_1(X)$ and we can finally conclude:

$$A(\wt{X}) \cong \la a, b \ra / K \cong \mathbb{Z}^2$$

In fact, each $(j, k) \in \mathbb{Z}^2$ corresponds to the integer translation of the grid $t_{j,k} : (x,y) \mapsto (x+j, y+k)$. Using the explicit description of the map in Massey its easy to show that $(p \circ t_{j,k}) = p$ (the trig functions are periodic with period 1), so these homeomorphisms are actually covering space automorphisms.

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(c.2) Take $X$ to be the figure eight curve, and $\wt{X}$ to be the union of a vertical line and a countable collection of disjoint circles tangent to the line. The map $p$ 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).

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Again, $\pi_1(X) = \la a, b \ra$. This time, $\wt{X}$ is homotopy equivalent to the wedge product of countably many circles, so $\pi_1(\wt{X})$ is the free group on countably many generators. Let it be $\la \gamma_n \ra$ where $\gamma_n$ is the path that travels up the line, traverses the $n$'th circle, and returns.

Let $K$ be the normal closure of $\la b \ra$ in $\la a, b \ra$. We claim that $\im(p_*) = K$. We have $p_*(\gamma_n) = a^n b a^{-n}$, so clearly $\im(p_*) \subseteq K$.

Next, let $x = x_1 \dots x_r$ where $x_i = (g_i b^{s_i} g_i^{-1})$, so $x$ is an arbitrary element of $K$. 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 $x$. When encountering $a^n$, traverse the line $n$ units. When encountering $b^m$, loop the circle in the current location $m$ times. For each $x_i$, every power of $a$ in $g_i$ will be eventually undone by the opposite power of $a$ in $g_i^{-1}$ , and since the only other element in each $x_i$ is a pure power of $b$, the sum of powers of $a$ in all of $x$ is zero -- this means that $\alpha$ will eventually end on its basepoint, guaranteeing that $\alpha$ is actually a loop. So $\alpha$ is indeed an element in $\pi_1(\wt{X})$ that maps to $x$. Then, $K \subseteq \im(p_*)$ and $\im(p_*) = K$.

Note that $K$ is a normal subgroup of $\pi_1(X)$. Let $x$ be the same arbitrary element as above:

$$\begin{align*} gxg^{-1} &= g(g_1 b^{s_1} g_1^{-1})(g_2 b^{s_2} g_2^{-1}) \dots (g_r b^{s_r} g_r^{-1}) g^{-1} \\ &= (g g_1 b^{s_1} g_1^{-1} g^{-1}) (g g_2 b^{s_2} g_2^{-1} g^{-1}) \dots (g g_r b^{s_r} g_r^{-1} g^{-1}) \in K \end{align*}$$

So, its normalizer is the whole group and we have $A(\wt{X}) \cong \la a, b \ra / K$. Since modding out by $K$, the normal closure of $\la b \ra$ is the same as adding $b$ as a relation, we get:

$$A(\wt{X}) \cong \la a, b \ra / K = \la a, b \mid b \ra = \la a \ra \cong \mathbb{Z}$$

In fact, each $k \in \mathbb{Z}$ corresponds to the homeomorphism shifting the space vertically $k$ units. Indeed, it can be checked that each of these homeomorphisms is a covering space automorphism.

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(d): Consider $X = \mathbb{C}^{\times}$ ($\mathbb{C}$ without the origin) and $\wt{X} = \mathbb{C}$. The covering map is $p : z \mapsto \exp(z)$.

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$\wt{X}$ is contractible, so this is a universal cover and $A(\wt{X}) \cong \pi_1(X)$, which is isomorphic to $\mathbb{Z}$ (deformation retract onto the circle). Each $k \in \mathbb{Z}$ corresponds to the homeomorphism that shifts $\mathbb{C}$ along the imaginary axis by $2\pi k$ units.

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(e): Consider $X = \mathbb{C}^{\times}$, $\wt{X} = \mathbb{C}^{\times}$, and $p : z \mapsto z^n$.

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This is entirely analogous to part (a). We have $A(\wt{X}) \cong \mathbb{Z}_n$ and for each $k \in \mathbb{Z}_n$, the homeomorphism is the rotation of $\mathbb{C}^{\times}$ by $2 \pi k / n$.