Stabilizers and Subgroups from Covering Spaces

The Fiber Action and Its Stabilizers

A loop in the base of a covering space can be lifted from any point above its basepoint. The endpoint records how that loop moves the chosen fiber point. A central question is which loop classes leave the point fixed. The answer is expressed entirely in terms of loops upstairs: a loop class fixes the point exactly when it comes from the fundamental group of the covering space based there.

Let $p:E\to B$ be a covering map, and fix $b_0\in B$. A covering map means that each point of $B$ has an open neighborhood $U$ for which $p^{-1}(U)$ is a union of disjoint open sets, each mapped homeomorphically onto $U$ by $p$. Give $E$ and $B$ their stated topologies, and let $F=p^{-1}(\{b_0\})$ be the fiber. For any $y\in F$, the groups $\pi_1(E,y)$ and $\pi_1(B,b_0)$ consist of based-loop homotopy classes, with multiplication given by traversing the first loop and then the second.

Definition: For $g=[\gamma]\in\pi_1(B,b_0)$ and $y\in F$, define $y\cdot g$ to be the endpoint of the lift of $\gamma$ that starts at $y$. The stabilizer of $y$ is $\operatorname{Stab}(y)=\{g\in\pi_1(B,b_0):y\cdot g=y\}$. The induced homomorphism is $p_*:\pi_1(E,y)\to\pi_1(B,b_0)$, given by $p_*([\widetilde{\gamma}])=[p\circ\widetilde{\gamma}]$.

Here we use a right action. Path lifting gives a unique lift of each path with a specified starting point. Homotopy lifting implies that homotopic based loops have lifts from $y$ with the same endpoint, so the endpoint depends only on $[\gamma]$. If $\alpha$ and $\beta$ are based loops, lifting their concatenation first lifts $\alpha$, then lifts $\beta$ from the endpoint of that first lift. Consequently $y\cdot([\alpha][\beta])=(y\cdot[\alpha])\cdot[\beta]$. These are the path-lifting, uniqueness, and endpoint-invariance results for covering maps.

The Stabilizer Theorem

Theorem: For every covering map $p:E\to B$, every $b_0\in B$, and every $y\in p^{-1}(\{b_0\})$, $$ \operatorname{Stab}(y)=\operatorname{im}\bigl(p_*:\pi_1(E,y)\to\pi_1(B,b_0)\bigr). $$ In particular, the stabilizer is a subgroup of $\pi_1(B,b_0)$.

Proof. First suppose $g=[\gamma]\in\operatorname{Stab}(y)$. Let $\widetilde{\gamma}$ be the lift of $\gamma$ starting at $y$. The stabilizer condition says that its endpoint is $y$, so $\widetilde{\gamma}$ is a loop in $E$ based at $y$. Since $p\circ\widetilde{\gamma}=\gamma$, we have $g=p_*([\widetilde{\gamma}])$. This proves that the stabilizer is contained in the image.

Conversely, suppose $g$ is in the image. Then there is a loop $\widetilde{\gamma}$ in $E$ based at $y$ such that $g=[p\circ\widetilde{\gamma}]$. The path $\widetilde{\gamma}$ is the lift of $p\circ\widetilde{\gamma}$ starting at $y$, by uniqueness of path lifts. It ends at $y$, so $g\in\operatorname{Stab}(y)$. Thus the two sets are equal.

Finally, $p_*$ is a homomorphism: composing a concatenation of loops with $p$ gives the concatenation of their projected loops, and composing a reversed loop gives the reversed projected loop. The image of a homomorphism is a subgroup: it contains the identity, and if $p_*(a)$ and $p_*(b)$ are in the image, their product is $p_*(ab)$, while the inverse of $p_*(a)$ is $p_*(a^{-1})$. Hence the stabilizer is a subgroup. $\square$

The basepoint in the theorem matters. The map $p_*$ used for $y$ has domain $\pi_1(E,y)$, not $\pi_1(E,e_0)$ for some other point $e_0$. A loop upstairs must begin and end at the particular fiber point whose stabilizer is being considered.

Worked Example: A Two-Sheet Product Cover

Let $B$ be any topological space, choose $b_0\in B$, and give $\{0,1\}$ the discrete topology. Give $E=B\times\{0,1\}$ the product topology and define $p(b,i)=b$. This is a covering map: for any open neighborhood $U$ in $B$, its preimage is the disjoint union $U\times\{0\}$ and $U\times\{1\}$, and each sheet maps homeomorphically onto $U$.

Choose $y=(b_0,0)$. A loop $\gamma$ at $b_0$ lifts from $y$ as $\widetilde{\gamma}(t)=(\gamma(t),0)$, so it ends at $y$. Thus every element of $\pi_1(B,b_0)$ stabilizes $y$. The map $p_*$ is also onto: every based loop $\gamma$ in $B$ is the projection of the loop $t\mapsto(\gamma(t),0)$ based at $y$. Therefore both sides of the theorem are all of $\pi_1(B,b_0)$.

Check: the lift remains in the sheet with second coordinate $0$ throughout, and its endpoint is $(\gamma(1),0)=(b_0,0)=y$, as required.

Worked Example: The Real-Line Cover of the Circle

Give $\mathbb{R}$ its standard topology and $S^1=\{z\in\mathbb{C}:|z|=1\}$ the subspace topology from $\mathbb{C}$. Consider $p(t)=e^{2\pi i t}$. This is a covering map: arcs of the form $\{e^{2\pi i s}:a<s<b\}$, with $b-a<1$, have preimages that are disjoint intervals $(a+k,b+k)$, $k\in\mathbb{Z}$, each mapped homeomorphically onto the arc. Take $b_0=1$ and $y=0$, so $p(y)=1$.

Every loop $\widetilde{\gamma}$ in $\mathbb{R}$ based at $0$ is homotopic to the constant loop through based loops: set $H(s,u)=(1-u)\widetilde{\gamma}(s)$, for $s,u\in[0,1]$. For every $u$, $H(0,u)=H(1,u)=0$, and $H$ continuously deforms $\widetilde{\gamma}$ to the constant loop. Hence $\pi_1(\mathbb{R},0)$ is trivial and $p_*(\pi_1(\mathbb{R},0))$ consists only of the identity class. By the theorem, the stabilizer of $0$ is trivial: only the identity loop class fixes $0$.

Check: the loop in $S^1$ given by $\gamma(s)=e^{2\pi i s}$ lifts from $0$ as $\widetilde{\gamma}(s)=s$, since $p(s)=e^{2\pi i s}=\gamma(s)$. Its endpoint is $1$, not $0$, so it does not fix $0$, in agreement with the theorem.

Changing the Basepoint Within the Fiber

Proposition: Suppose $y,z\in p^{-1}(\{b_0\})$ can be joined by a path $\delta$ in $E$ from $y$ to $z$. Set $c=[p\circ\delta]\in\pi_1(B,b_0)$. Then $$ p_*\bigl(\pi_1(E,z)\bigr)=c^{-1}\,p_*\bigl(\pi_1(E,y)\bigr)\,c. $$

Proof. Let $H_y=p_*(\pi_1(E,y))$ and $H_z=p_*(\pi_1(E,z))$. If $\eta$ is a loop in $E$ based at $z$, then $\delta*\eta*\delta^{-1}$ is a loop based at $y$. Its projection has class $c\, [p\circ\eta]\,c^{-1}$. Thus $cH_zc^{-1}\subseteq H_y$. For the reverse inclusion, take any loop $\xi$ based at $y$. The loop $\delta^{-1}*\xi*\delta$ is based at $z$, and its projected class is $c^{-1}[p\circ\xi]c$. Therefore $c^{-1}H_yc\subseteq H_z$. These inclusions give the stated equality. $\square$

This formula specifies how the subgroup depends on the chosen point of the fiber. It applies when a path in $E$ joins the two points; it does not assert that every pair of points in the fiber can be joined. The conjugating element is determined by the projected path, so changing that path can change the displayed conjugate.

A Starting-Point Pitfall

Worked Example: A Loop Lifted from the Wrong Fiber Point

Use $S^1$ with its subspace topology and the covering map $p:S^1\to S^1$, $p(z)=z^2$. To check that it is a covering, take a sufficiently short open arc $U$ in the target. Its preimage consists of two disjoint open arcs, each mapped homeomorphically onto $U$. Let $b_0=1$, and consider the loop $\gamma(s)=e^{2\pi i s}$, based at $1$. The fiber over $1$ contains both $1$ and $-1$.

Starting at $-1$, a lift is $\widetilde{\gamma}_{-1}(s)=e^{\pi i(1+s)}$. Indeed, its initial value is $e^{\pi i}=-1$, and $p(\widetilde{\gamma}_{-1}(s))=e^{2\pi i(1+s)}=e^{2\pi i s}=\gamma(s)$. Its endpoint is $e^{2\pi i}=1$, so this loop class does not stabilize $-1$.

Starting instead at $1$, the lift is $\widetilde{\gamma}_{1}(s)=e^{\pi i s}$, which projects to $\gamma$ and ends at $e^{\pi i}=-1$. This lift also does not return to its starting point, but it is the lift for the other fiber point. Common mistake: using the lift from $1$ to decide whether the loop fixes $-1$. The correction is to start every lift at the specific point whose stabilizer is being tested; uniqueness then makes the endpoint test valid for that point.

Check: at $s=0$, the two lifts start at $-1$ and $1$, respectively; at $s=1$, they end at $1$ and $-1$, respectively. The endpoints follow directly from substituting $s=0$ and $s=1$ into the displayed formulas.

The stabilizer theorem is useful because it turns an action question into a based-loop question upstairs. To determine which base loops fix a fiber point, identify the loops that lift to closed loops at that point. Their projected homotopy classes form exactly the image subgroup, with no need to assume that the covering space is connected or that the action on the fiber is transitive.

Check Your Understanding

Use the right-action convention and keep track of the specified basepoints.

Write the induced homomorphism whose image is the stabilizer of $z\in p^{-1}(\{b_0\})$.
Explain why a loop class in the image of $p_*:\pi_1(E,y)\to\pi_1(B,b_0)$ fixes $y$.
In the two-sheet product cover, describe the lift of a based loop starting at $(b_0,1)$, and determine its endpoint.
In the real-line cover of $S^1$, why does the loop $s\mapsto e^{2\pi i s}$ fail to fix $0$?
If a path $\delta$ joins $y$ to $z$ in the fiber, what element conjugates the image subgroup at $z$ to the image subgroup at $y$?

Stabilizers and Subgroups from Covering Spaces

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