It is a classical result of Quinn that for a simply-connected closed $4$-manifold $X$ the isometries of its intersection form are in one-to-one correspondence with $\pi_0 \text{Homeo}(X)$. (Isotopy of 4-manifolds, 1986)

Let $X$ have some simple fundamental group, say $\mathbb{Z}_2$, and let $h\colon X \to X$ be a diffeomorphism which acts trivially on $H_2(X;\mathbb{Z})$.

Is $h$ isotopic to the identity? (through $\text{Homeo}(X)$)

Edit: let $\tilde{X}$ denote the universal cover of $X$, and let $s \colon \tilde{X} \to \tilde{X}$ be the covering involution so that $\tilde{X}/s \cong X$. One can take a lift $\tilde{h} \colon \tilde{X} \to \tilde{X}$ of $h$ (there are two, but take any of them.) Another reasonable assumption on $h$ is that we want $\tilde{h}$ to be isotopic to either $\text{id}$ or to $s$.

Edit: The second question has been removed.

  • $\begingroup$ I suspect that at you need to look also at the induced action of $h$ on $H_2(X;\Bbb Z_-)$, where this denotes twisted coefficients with fiber $\Bbb Z$ and nonzero $\pi_1$ action. (Whether this action being trivial is enough for (2) I don't know off the top of my head.) $\endgroup$ – Mike Miller Jan 15 at 12:49
  • $\begingroup$ I don't quite understand what you say, yet I’m going to rephrase it. One can take the universal cover $\tilde{X} \to X$ and consider a lift $\tilde{h} \colon \tilde{X} \to \tilde{X}$. If $h$ is isotopic to $\text{id}$ then at least $\tilde{h}$ is isotopic to either $\text{id}$ or to the covering involution. What you suggest is to check this property for $\tilde{h}$? $\endgroup$ – Enumerator Jan 15 at 13:01
  • $\begingroup$ That is not precisely what I said, homology with local coefficients is not the same as homology of a covering space. However, suppose $h: X \to X$ is a diffeomorphism which induces an isomorphism on integer homology. Then any two of the following three conditions implies the third. (1) $h$ induces an isomorphism on $\Bbb Z/2$ homology. (2) The cover $\tilde h$ induces an isomorphism on the integer homology of $\tilde X$. (3) $h$ induces an isomorphism on the local coefficient homology $H_*(X;\Bbb Z_-)$. If $X$ is non-orientable then (3) I believe is automatic by Poincare duality and UCT. $\endgroup$ – Mike Miller Jan 15 at 17:46

Let $X = (S^2\times S^2)/\mathbb{Z}_2$ where the $\mathbb{Z}_2$ action is generated by $(x, y) \mapsto (-x, -y)$. Note that $H_2(X; \mathbb{Z}) \cong \mathbb{Z}_2$, so every diffeomorphism acts trivially.

Consider the diffeomorphism $f : S^2\times S^2 \to S^2\times S^2$ given by $(x, y) \mapsto (x, -y)$. This descends to a diffeomorphism $g : (S^2\times S^2)/\mathbb{Z}_2 \to (S^2\times S^2)/\mathbb{Z}_2$. Letting $\pi : S^2\times S^2 \to (S^2\times S^2)/\mathbb{Z}_2$ be the universal covering map, we have a commutative diagram

$$\require{AMScd} \begin{CD} S^2\times S^2 @>{f}>> S^2\times S^2\\ @V{\pi}VV @VV{\pi}V \\ (S^2\times S^2)/\mathbb{Z}_2 @>{g}>> (S^2\times S^2)/\mathbb{Z}_2 \end{CD}$$

Taking $\pi_2$ of this diagram, we get a commutative diagram of abelian groups

$$\require{AMScd} \begin{CD} \mathbb{Z}\oplus\mathbb{Z} @>{f_*}>> \mathbb{Z}\oplus\mathbb{Z}\\ @V{\pi_*}VV @VV{\pi_*}V \\ \mathbb{Z}\oplus\mathbb{Z} @>{g_*}>> \mathbb{Z}\oplus\mathbb{Z} \end{CD}$$

Note that $\pi_* = \operatorname{id}$ as $\pi$ is a covering map, but $f_*$ is given by $(a, b) \mapsto (a, -b)$. By commutativity, the same is true of $g_*$. In particular, $g_* \neq \operatorname{id}$ and therefore $g$ is not homotopic to the identity map.

Alternatively, note that $(S^2\times S^2)/\mathbb{Z}$ is orientable. As $\pi\circ f = g\circ\pi$, the maps $f$ and $g$ have the same degree, and it is easy to see that $f$ has degree $-1$. Again we see that $g$ is not homotopic to the identity map.

Geometrically, $X = \operatorname{Gr}(2, 4)$, the Grassmannian of unoriented two-planes in $\mathbb{R}^4$, and $S^2\times S^2 = \operatorname{Gr}^+(2, 4)$ the corresponding oriented Grassmannian. The diffeomorphisms $f$, $g$ are the maps given by $P \mapsto P^{\perp}$.

  • $\begingroup$ Thank you very much for your answer! You posted your counter-example while I was editing my question (the very same minute!). In the edited version, I want my homeomorphism to have a homotopically trivial lift. $\endgroup$ – Enumerator Jan 15 at 13:42
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    $\begingroup$ @Enumerator: You might not get much attention given that you accepted my answer. I think it would be better to either (1) unaccept my answer, or (2) revert your question back to the original form and then ask a new question in the form it is now, linking to this one. $\endgroup$ – Michael Albanese Jan 15 at 22:30

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