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.



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> What known methods/invariants can be used to certify that a given element $\varphi\in\operatorname{MCG}(M)$ has infinite order?
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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](https://mathoverflow.net/q/35909) 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.



Elements of infinite order in the topological mapping class group
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.


