I have a closed connected manifold $X$, consider the universal cover $p: \tilde X \rightarrow X$. If I recall correctly, any homotopy $F: X \times I \rightarrow X$ with $F(\cdot, 0) = id_X$ lifts to a homotopy $\tilde F : \tilde X \times I \rightarrow \tilde X$ (again with $\tilde F(\cdot,0) = id_{\tilde X}$).

Given any deck transformation $\psi: \tilde X \rightarrow \tilde X$ (i.e. $p \circ \psi = p$), is there a homotopy $F: X \times I \rightarrow X$ with $F(\cdot,0) = id_X$ that lifts to $\tilde F$ with $\tilde F(\cdot, 1) = \psi$?

My motivating example is $X = \mathbb T^2 = \mathbb R^2 / \mathbb Z^2$, where this is true: $\tilde X = \mathbb R^2$, and the group of deck transformations is $\mathbb Z^2 \simeq \pi_1 (X)$. Any deck transformation $\tilde X \rightarrow \tilde X : x \mapsto x + a$ (for $a \in \mathbb Z^2$) can be obtained by the homotopy $F_a: X \times I \rightarrow X : (x,t) \mapsto x + ta$ which lifts to $\tilde F_a : \tilde X \times I \rightarrow \tilde X : x \mapsto x + ta$. Can this be done for any deck transformation of higher-genus surfaces? Orientable manifolds? Arbitrary manifolds? Is there a "nice" sufficient condition for a manifold that makes it have this property?


Recall that deck transformations of $p: \tilde X \rightarrow X$ can be identified with $\pi_1(X)$; denote the isomorphism by $\alpha: Deck(p) \stackrel \sim \rightarrow \pi_1(X)$. I show the following claim:

If $\psi \in Deck(p)$ such that there is a homotopy $F: X \times I \rightarrow X$ with $F(\cdot,0) = id_X$ that lifts to $\tilde F$ with $\tilde F(\cdot,1) = \psi$, then $\alpha (\psi)$ is in the center of $\pi_1(X)$. Specifically, if we want this property to hold for all deck transformations, $\pi_1(X)$ is abelian.

Proof: denote the basepoint of $X$ by $x_0$. Let $g: I \rightarrow X$ be $g(t) = F(x_0,t)$, a representative of the class $\alpha(\psi) \in \pi_1(X,x_0)$, with $g(0)=g(1)=x_0$, and let $f: I \rightarrow X$ be a representative of any class $a \in \pi_1(X,x_0)$. Consider the following:

\begin{equation} G: I^2 \rightarrow X \\ G(s,t) = F(f(s),t) \end{equation}

$F$ lifts to $\tilde F$ which has $\tilde F(\cdot,1) = \psi \in Deck(p)$, so $F(\cdot, 1) = id_X$, and $F(\cdot, 0) = id_X$. Thus, $G(\cdot, 0) = G(\cdot, 1) = f$. $G(0, \cdot) = G(1, \cdot) = g$, since $f(0)=f(1)=x_0$, and $F(x_0, \cdot) = g$.

Going around the square Im(G), we see that $a \alpha(\psi) a^{-1} \alpha(\psi)^{-1} = 1$ in $\pi_1(X)$. This was for arbitrary $a \in \pi_1(X)$, therefore $\alpha(\psi)$ is in the center of $\pi_1(X)$.

  • $\begingroup$ The thing that makes it work for $\mathbb{T}^2$ is that it's a $K(G,1)$, i.e. that its universal cover is contractible; not that it's orientable I think. The problem is that I don't know many noncontractible simply connected manifolds that aren't spheres (spheres of even dimension will yield nonorientable quotients, and sphere of odd dimension have only finite subgroups of $SO(n+1)$ that act on them, and that's path connected anyway). I would say that if you find an example of such an orientable manifold, some counterexamples should come too $\endgroup$ – Max Feb 12 at 10:23
  • $\begingroup$ I've been thinking about this and while I haven't reached a conclusion, my intuition currently tells me that "most" manifolds shouldn't have this property. I think that manifolds have this property ("for every deck transformation...") iff they're orientable and their fundamental group is abelian. If that's correct, then any surface of genus >1 doesn't have this property, but has a contractible universal cover. $\endgroup$ – WallE Feb 12 at 11:06
  • $\begingroup$ If the universal cover is contractible, then any two maps $\tilde{X}\to \tilde{X}$ are homotopic, so any deck transformation is homotopic to the identity; there can't be counterexamples with contractible universal cover. Moreover, if $G$ is a nonabelian subgroup of $SO(n+1)$ that acts freely on $S^n$, $n\geq 3$ odd, then $\pi_1(S^n/G) = G$ is nonabelian, but any deck transformation is in $SO(n+1)$ so homotopic to the identity. So the abelianness of the fundamental group doesn't come into play. As I said, I don't know yet whether orientability plays a role $\endgroup$ – Max Feb 12 at 11:24
  • $\begingroup$ But a homotopy between two maps $\tilde X \rightarrow \tilde X$ doesn't necessarily project to a homotopy of the maps on $X$. I'll edit the question to include what I (now) know about conditions of the manifold for this property. $\endgroup$ – WallE Feb 12 at 15:29
  • $\begingroup$ Oh right, it's not $\tilde{X}\times I \to X$, my bad ! Then it should be easier, I'll think about it when I have the time $\endgroup$ – Max Feb 12 at 15:58

The answer is no, with counterexamples being given by the projective planes : $S^n\to \mathbb{R}P^n$ is the universal cover of $\mathbb{R}P^n$ for $n\geq 2$, and $-id : S^n\to S^n$ is the only nontrivial deck transformation of this cover.

However, when $n$ is even (e.g. $n=2$), this has degree $-1$ and is thus not homotopic to $id$.

  • $\begingroup$ Good point, do you have a counterexample for an orientable manifold? (Edited question to include this) $\endgroup$ – WallE Feb 10 at 20:51
  • $\begingroup$ I'm not sure yet, but I think for orientable ones your property will work. What makes me think that is that if $\omega$ is a volume form below, then $p^*\omega$ will be a volume form above, that is invariant under deck transformation; in particular if $\psi$ is a deck transformation, $\mathrm{d}\psi$ will have positive "determinant", so it'll look like the identity. I don't know yet if this can lead to something though (I'm not fluent in differential geometry/topology) $\endgroup$ – Max Feb 10 at 21:44
  • $\begingroup$ $-id$ is the only non-trivial deck transformation. $\endgroup$ – Paul Frost Feb 10 at 23:53
  • $\begingroup$ @Paulfrost : yes of course, I added that, thanks for pointing it out ! $\endgroup$ – Max Feb 11 at 7:51

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