Coverings and Their Symmetries
A covering space can have symmetries that move points in a fiber while leaving every point of the base fixed. These symmetries are constrained: they must fit together continuously across the covering space, not merely permute one fiber. The key tool is uniqueness of lifts. We first prove the version needed here, with continuity stated explicitly, and then use it to understand what a deck transformation can do.
Let \(E\) and \(B\) be topological spaces, and let \(p:E\to B\) be a covering map. This means that \(p\) is a continuous surjection and every \(b\in B\) has an open neighborhood \(U\) such that \(p^{-1}(U)\) is a union of pairwise disjoint open sets, each mapped homeomorphically onto \(U\) by \(p\). Such a neighborhood is evenly covered. A lift of a continuous map \(F:X\to B\) is, by definition, a continuous map \(f:X\to E\) such that \(p\circ f=F\).
The equation \(p\circ h=p\) says that \(h\) moves points only within their fibers. It does not say that every permutation of a fiber extends to a deck transformation; continuity and the covering structure impose additional restrictions.
Uniqueness of Lifts on a Connected Domain
Proof. Define \(A=\{x\in X:f(x)=g(x)\}\). We show that both \(A\) and \(X\setminus A\) are open. Fix \(x\in X\). Choose an evenly covered open neighborhood \(U\) of \(F(x)\). Since \(F\) is continuous, \(F^{-1}(U)\) is an open neighborhood of \(x\). Let \(V_f,V_g\) be the sheets over \(U\) containing \(f(x),g(x)\), respectively. By continuity of \(f\) and \(g\), there is an open neighborhood \(W\) of \(x\), contained in \(F^{-1}(U)\), for which \(f(W)\subseteq V_f\) and \(g(W)\subseteq V_g\). If \(f(x)=g(x)\), then \(V_f=V_g\), because the sheets are disjoint and both contain that point. On \(W\), both maps are the inverse of the same homeomorphism \(p|_{V_f}:V_f\to U\), composed with \(F|_W\). Thus \(f=g\) throughout \(W\), so \(A\) is open near each of its points. If \(f(x)\ne g(x)\), then \(V_f\ne V_g\), so these sheets are disjoint; consequently \(f(y)\ne g(y)\) for every \(y\in W\). Thus \(X\setminus A\) is open as well. The set \(A\) is nonempty by the assumed agreement at \(x_0\), and it is both open and closed. Connectedness of \(X\) gives \(A=X\), proving \(f=g\). \(\square\)
Continuity is essential in this argument: it is what provides a neighborhood \(W\) on which each lift stays in its chosen sheet. A map satisfying only the equation \(p\circ f=F\) need not be a lift in the definition above, and the theorem does not apply to such discontinuous maps.
Deck Transformations Act on Fibers
For \(b\in B\), the fiber over \(b\) is \(p^{-1}(\{b\})=\{e\in E:p(e)=b\}\). A deck transformation sends every point of this fiber to another point of the same fiber. In fact, it permutes the fiber.
Proof. If \(e\in p^{-1}(\{b\})\), then \(p(h(e))=p(e)=b\), so \(h(e)\) is in the same fiber. To check that the inverse also preserves fibers, write \(y=h(e)\). From \(p(h(e))=p(e)\) we get \(p(y)=p(h^{-1}(y))\), hence \(p\circ h^{-1}=p\). Thus \(h^{-1}\) also maps each fiber into itself, and the restriction of \(h\) to each fiber is bijective. For the last claim, view two deck transformations \(h_1,h_2:E\to E\) as lifts of \(p:E\to B\) through the covering map \(p\). If they agree at one point and \(E\) is connected, the uniqueness theorem gives \(h_1=h_2\). \(\square\)
Composition and inverses preserve the equation \(p\circ h=p\), so the deck transformations do form a group: the identity is a deck transformation; if \(h,g\) are deck transformations, then \(p\circ(h\circ g)=(p\circ h)\circ g=p\circ g=p\); and the inverse case was verified in the proof above. Restriction to a fiber therefore gives a group action by permutations. When \(E\) is connected, the action on any one nonempty fiber is faithful: if two transformations have the same restriction there, they agree at a point and hence are equal. Faithful does not mean that every permutation of the fiber occurs.
Worked Example: A Three-Point Fiber over One Point
Let \(E=\{a_1,a_2,a_3\}\) have the discrete topology, let \(B=\{b\}\) have the discrete topology, and define \(p(a_i)=b\) for \(i=1,2,3\). This is a covering map: the only neighborhood \(\{b\}\) of \(b\) has preimage \(E\), the disjoint union of the open singletons \(\{a_1\},\{a_2\},\{a_3\}\), each mapped homeomorphically to \(\{b\}\).
Every permutation \(h\) of \(E\) is a homeomorphism, since \(E\) is discrete, and satisfies \(p(h(a_i))=b=p(a_i)\). Thus every permutation is a deck transformation. There are \(3\cdot2\cdot1=6\) such permutations, so \(\operatorname{Deck}(p)\) is the symmetric group on three points. Its action on the sole fiber \(p^{-1}(\{b\})=E\) consists of all six permutations.
Check: for the transposition \(h(a_1)=a_2,\ h(a_2)=a_1,\ h(a_3)=a_3\), each input \(a_i\) and its image both map to \(b\), so \(p\circ h=p\). This example also shows why connectedness matters: \(E\) is disconnected, and the nonidentity transposition fixes \(a_3\).
Worked Example: The Identity Covering
Give \(E=B=[0,1]\) the standard subspace topology from \(\mathbb{R}\), and let \(p=\operatorname{id}_{[0,1]}\). This is a covering map: for every open neighborhood \(U\) in \([0,1]\), its preimage is \(U\) itself, which is one sheet mapped homeomorphically onto \(U\).
If \(h\) is a deck transformation, then \(\operatorname{id}_{[0,1]}\circ h=\operatorname{id}_{[0,1]}\). Evaluating at any \(x\in[0,1]\) gives \(h(x)=x\). Therefore the only deck transformation is the identity. Its restriction to every fiber is the identity permutation; each fiber has exactly one point because \(p^{-1}(\{b\})=\{b\}\).
Check: the identity map is continuous, bijective, and its own continuous inverse, and it satisfies \(p\circ\operatorname{id}_{[0,1]}=p\). Thus it is indeed a deck transformation, and the pointwise equation rules out any others.
A Connected Covering: The Threefold Circle Cover
Worked Example: Deck Transformations of a Threefold Circle Cover
Give \(S^1=\{z\in\mathbb{C}:|z|=1\}\) its subspace topology from \(\mathbb{C}\), and define \(p:S^1\to S^1\) by \(p(z)=z^3\). This map is a covering. To check this, write \(q(t)=e^{2\pi i t}\), choose \(b=q(t)\), and take the open arc \(U=q((t-1/12,t+1/12))\). Its three sheets are \(q(((t+k)/3-1/36,(t+k)/3+1/36))\), for \(k=0,1,2\). They are disjoint arcs: their centers differ by \(1/3\) around the circle and each has coordinate radius \(1/36\). On each arc, cubing maps homeomorphically onto \(U\), as its coordinate expression is \(x\mapsto 3x\) modulo integers. These three arcs exhaust the preimage, since \(q(3x)\in U\) exactly when \(x\) lies in one of them modulo integers.
Let \(\omega=e^{2\pi i/3}=-1/2+i\sqrt{3}/2\). The fiber over \(1\) is \(\{1,\omega,\omega^2\}\): the equation \(z^3=1\) has precisely these three roots. For \(j=0,1,2\), define \(r_j(z)=\omega^jz\). Each is a homeomorphism, with inverse \(r_{3-j}\) when \(j=1,2\), and inverse \(r_0\) when \(j=0\). Also, \(p(r_j(z))=(\omega^jz)^3=\omega^{3j}z^3=z^3=p(z)\), since \(\omega^3=1\). So all three maps are deck transformations.
There are no others. If \(h\) is any deck transformation, then \(p(h(1))=p(1)=1\), so \(h(1)\) is one of \(1,\omega,\omega^2\). Choose \(j\in\{0,1,2\}\) such that \(h(1)=\omega^j=r_j(1)\). Both \(h\) and \(r_j\) are continuous lifts of \(p:S^1\to S^1\), and they agree at \(1\). The circle is connected, so uniqueness of lifts gives \(h=r_j\). Hence the deck group consists exactly of \(r_0,r_1,r_2\).
On the fiber over \(1\), the action is \(r_j(\omega^k)=\omega^{j+k}\), with exponents reduced modulo \(3\). For example, \(r_2(\omega)=\omega^3=1\), and \(r_1(\omega^2)=\omega^3=1\). The identity \(r_0\) fixes all three points, while \(r_1\) and \(r_2\) cyclically permute them.
Check: each listed map preserves the projection by the displayed cube calculation, and the uniqueness argument proves that any deck transformation must be on the list. The three resulting permutations are distinct because they send \(1\) to three different points.
Common Pitfall: A Fiber Permutation Need Not Extend
A common error is to assume that a deck transformation can be specified by choosing any permutation of a single fiber. The threefold circle cover provides a concrete correction. The transposition that swaps \(1\) and \(\omega\) while fixing \(\omega^2\) is a permutation of the fiber over \(1\), but it is not the restriction of any deck transformation. The only possible restrictions are the three cyclic permutations \(r_0,r_1,r_2\), as proved above. In contrast, the discrete three-point cover has all six permutations as deck transformations. Thus the fiber alone does not determine which permutations are allowed; the topology of the whole covering space matters.
Connectedness gives a precise determination principle, not a general existence principle. For a connected covering, specifying the image of one point determines at most one deck transformation, because any two transformations with that value agree at that point. It does not follow that every point in the same fiber can be the image of that point under a deck transformation. The existence question requires additional hypotheses and is not part of the uniqueness argument.
Check Your Understanding
Use the definitions and uniqueness result to justify each answer.
- Why must a deck transformation map each fiber into itself, and why is its restriction to that fiber bijective?
- In the threefold circle cover, compute \(r_2(\omega^2)\) and verify that its image is still in the fiber over \(1\).
- Which hypothesis in the uniqueness theorem rules out the nonidentity transformation fixing \(a_3\) in the three-point discrete example?
- For a connected covering, explain why two deck transformations agreeing on one point must agree on every fiber.
- Does faithfulness of the action on one fiber imply that every permutation of that fiber is induced by a deck transformation? Give the counterexample from this tutorial.