# Does a deck transformation have a homotopy that lifts to it?

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?

Edit:

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:

$$$$G: I^2 \rightarrow X \\ G(s,t) = F(f(s),t)$$$$

$$F$$ 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 $$\pi_1(X)$$.

• 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 – Max Feb 12 at 10:23
• 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. – WallE Feb 12 at 11:06
• 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 – Max Feb 12 at 11:24
• 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. – WallE Feb 12 at 15:29
• 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 – 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 $$id$$.
• 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) – Max Feb 10 at 21:44
• $-id$ is the only non-trivial deck transformation. – Paul Frost Feb 10 at 23:53