10
$\begingroup$

Let $M$ be a closed topological manifold, and let $\operatorname{MCG}(M):=\operatorname{Homeo}(M)/\operatorname{Homeo}_0(M)$ denote the topological mapping class group of $M$ ($\operatorname{Homeo}_0(M)$ denotes the identity component of $\operatorname{Homeo}(M)$). More generally, we could let $M$ be compact with boundary and consider homeomorphisms fixing the boundary.

What known methods/invariants can be used to certify that a given element $\varphi\in\operatorname{MCG}(M)$ has infinite order?

Of course, $\varphi$ might have infinite order for homotopical reasons, but I am primarily interested in finer invariants. So, for the sake of this question, let's assume $\varphi$ is homotopic to the identity as maps $M\to M$. Of course, if $M$ has boundary, then this means homotopic through maps fixing the boundary.

For instance, the kernels in this answer are not detected by their action on homotopy, however unfortunately they are purely torsion (though I would still like to know how they are detected).

The focus of this question is on the high-dimensional case, though an answer in any dimension $\geq 4$ would be interesting.

$\endgroup$
  • $\begingroup$ If $M$ has boundary, then a homeomorphism will induce a mapping class on the boundary, and that can lead to interesting obstructions. (Consider a three-dimensional handlebody and a Dehn twist about a disk). If $M$ is non-compact, then a homeomorphism will induce a permutation of the ends of $M$. (Consider a thickened tree.) So - do you have side conditions on $M$ to eliminate these kinds of "dimension-reducing" techniques? $\endgroup$ – Sam Nead Apr 11 '16 at 2:45
  • $\begingroup$ @Sam Nead: The easiest way to state the question so its answer is interesting is just to restrict to closed manifolds (edited accordingly). $\endgroup$ – John Pardon Apr 11 '16 at 3:19
  • 1
    $\begingroup$ If $G(M)$ is the monoid of homotopy self-equivalences of $M$, and $H(M)=Homeo(M)$, then your question is about the kernel of $\pi_1 BH(M)\to\pi_1 G(M)$ which by the homotopy exact sequence of the fibration comes from $\pi_1 G(M)/H(M)$. As in Hatcher's survey in the linked answer one understands $G(M)/H(M)$ is two stages via the intermediate space $\tilde{H}(M)$ of blocked homeomorphisms. Rationally one can estimate $G(M)/\tilde H(M)$ by the surgery exact sequence. The other stage $\tilde H(M)/H(M)$ is sort of controlled by the concordance space of $M$ via Hatcher's spectral sequence. $\endgroup$ – Igor Belegradek Apr 11 '16 at 3:54
  • 1
    $\begingroup$ (cont.) A slower explanation of the above (for the analogous case of diffeomorphisms) can be found in section 8 of arxiv.org/abs/1501.03475. $\endgroup$ – Igor Belegradek Apr 11 '16 at 3:56
  • 1
    $\begingroup$ typo in my comment: $\pi_1 G(M)$ should have been $\pi_1 BG(M)$. $\endgroup$ – Igor Belegradek Apr 11 '16 at 4:08
4
$\begingroup$

Let me suppose that $M$ is a closed manifold of dimension $d$, and let $\varphi : M \to M$ be a diffeomorphism / homeomorphism which is homotopic to the identity, and choose such a homotopy $h_t$. The mapping torus $X_\varphi$ of $\varphi$ is a smooth / topological manifold fibering over $S^1$, and our choice of homotopy $h_t$ yields a preferred homotopy equivalence $$ H : X_\varphi \longrightarrow S^1 \times M.$$

For each cohomology class $v \in H^{d-4i+1}(M;\mathbb{Q})$ we may form $$\int_{X_\varphi} p_i(TX_\varphi - H^*TM) \smile H^*(1 \otimes v) \in \mathbb{Q}$$ where $p_i$ is the (rational) Pontrjagin class. This defines a linear functional $H^{d-4i+1}(M;\mathbb{Q}) \to \mathbb{Q}$ which is Poincare dual to a class $\xi_i \in H^{4i-1}(M;\mathbb{Q})$.

Morally this class measures the following: on the one hand $\varphi$, by virtue of being a homeomorphism, preserves the rational Pontrjagin class $p_i(M)$, so preserves a given cocycle representative $c$ up to a "preferred" coboundary. On the other hand, the homotopy $h_t$ gives another coboundary for $c-\varphi^*c$; the difference between these cobounding cochains is thus a cycle, and it represents the cohomology class $\xi_i$.

In principle $\xi_i$ is an invariant of the pair $(\varphi, h_t)$. However, a different choice of homotopy $h_t'$ yields a homotopy equivalence $H'$ which differs from $H$ by postcomposition by a homotopy automorphism of $S^1 \times M$ which i) is over $S^1$ and, ii) fixes a fibre $M \times \{*\}$. This means that for each $v$ we have $$(H')^*(1 \otimes v) = H^*(1 \otimes v) + H^*(u \otimes \bar{v})$$ for some $\bar{v}$, where $u \in H^1(S^1;\mathbb{Q})$ is the canonical class. But $$\int_{X_\varphi} p_i(TX_\varphi - H^*TM) \smile H^*(u \otimes \bar{v})$$ can be written as $$\int_{S^1 \times M} (H^{-1})^*(p_i(TX_\varphi-H^*TM)) \smile (u \otimes \bar{v}) = \int_M p_i(TM-TM) \cup \bar{v}=0$$ so in fact $\xi_i$ only depends on $\varphi$.

As $X_{\varphi \circ \psi}$ is cobordant to $X_{\varphi} \sqcup X_{\psi}$, one sees that $\xi_i(\varphi \circ \psi) = \xi_i(\varphi) + \xi_i(\psi)$. In particular if $\xi_i(\varphi) \neq 0$ for some $i$ then $\varphi$ has infinite order.

There is a theorem due to Sullivan that a simply-connected high-dimensional manifold is determined "up to finite ambiguity" by its homotopy type and rational Pontrjagin classes. The analogous statement for automorphisms of simply-connected (or perhaps 2-connected?) manifolds says that elements in $$\mathrm{Ker}(\pi_0(Top(M)) \to \pi_0(G(M)))$$ (or the smooth analogue) are determined up to finite ambiguity by their associated $\xi_i$'s. This can (surely?) be proved using the surgery exact sequence.

$\endgroup$
  • $\begingroup$ It seems your argument is similar to pp.322-323 of Sullivan's "Infinitesimal computations in topology" where he discusses the kernel of the map that sends isotopy classes of diffeomorphisms to their homotopy classes. The link is archive.numdam.org/article/PMIHES_1977__47__269_0.pdf. $\endgroup$ – Igor Belegradek Apr 12 '16 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.