# Topological mapping class groups of 4-manifolds

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.

• 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.) – Mike Miller Jan 15 at 12:49
• 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 $\tilde{h}$? – Enumerator Jan 15 at 13:01
• 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. – 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 $$P \mapsto P^{\perp}$$.

• 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. – Enumerator Jan 15 at 13:42
• @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. – Michael Albanese Jan 15 at 22:30