Actual Classification re Nielsen-Thurston Theorem (how to)?

Question

according to Nielsen -Thurston Classification:

http://en.wikipedia.org/wiki/Nielsen%E2%80%93Thurston_classification

If $S$ is compact and orientable surface, then any homeomorphism is isotopic to (at least one) of three classes of maps:

i ) Periodic

ii) Reducible

iii) Pseudo Anosov.

But the method does not seem to be constructive, i.e., how do we tell, given a

specific homeomorphism $ h: S \rightarrow S$ , whether $S$ belongs to one or more

of the classes? Do we actually test whether $h$ itself is periodic, i.e., $h^n=Id$ , for some integer $n$ , or whether $h$ itself fixes non-essential curves, etc.? If not, how do we otherwise determine this fact? Are these properties "robust" under isotopy, i.e., if $h$ satisfies $h^n=id$ , is $h$ necessarily periodic, and if g fixes a set of inessential curves, is g then necessarily reducible?

Thanks.

I imagine if we have a nice expression for $MCG(S)$ , e.g., like in the case of the 2-tours $T^2:= S^1 \times S^1 $, maybe we can use some linear algebraic techniques, but how do we classify homeomorphisms otherwise? Or maybe there are canonical choices for representatives of each $h$ in $MCG(S)$ ?

Answer 1

There is an algorithm by Bestiva and Handel to determine the Nielsen-Thurston representative from other descriptions. See these two papers.

The robust descriptions which are preserved by isotopy are: (1) periodic means $h^n \simeq id$ for some $n$ (isotopic, not equal), and (2) reducible means (not periodic and) $h$ fixes the homotopy class of some essential closed curve, and then (3) pseudo-Anosov is everything else. This last is not particularly illuminating, and one way of viewing the content of Thurston's theorem is that all of these have a nice representative that has a particular geometric structure. The papers I linked above show how to take a description of a mapping class and determine the nice representative (whether it be the exactly periodic map with $h^n = id$, or a map exactly fixing a curve (as a set), or the nice pseudo-Anosov representative).

On $T^2$, the classification is simple. The (orientation-preserving) mapping class group of the torus is isomorphic to $SL_2(\mathbb{Z})$, and given $A \in SL_2(\mathbb{Z})$, the trace of $A$ determines whether it is periodic ($|\operatorname{tr}(A)| < 2$, i.e. complex eigenvalues; also $\pm I$ would fit here), reducible (not $\pm I$ and $|\operatorname{tr}(A)| = 2$), or Anosov ($|\operatorname{tr}(A)| > 2$, i.e. two distinct real eigenvalues).

Answer 2

As pointed out in the answer of @aes, your guesses at robusteness properties are too strong to be true. However, there are weaker robustness properties that are quite useful.

Here is a weaker robustness property which is satisfied by a pseudo-Anosov mapping class.

It is not true, of course, that if $g$ is a pseudo-Anosov homeomorphism and $g$ is isotopic to $h$ then $h$ is a pseudo-Anosov homeomorphism.

However, what is true is that if $g$ and $h$ are both pseudo-Anosov homeomorphisms of $S$, and if $g$ and $h$ are isotopic to each other, then they are isotopic through pseudo-Anosov homeomorphisms. In other words, there is an isotopy from $g$ to $h$ such that at all moments of the isotopy the homeomorphism is pseudo-Anosov. In fact it is even stronger than that: there exists an isotopy $K_t(x)$, $(x,t) \in S \times [0,1]$, such that $K_0$ is the identity map and such that $h(x) = K_1^{-1} \circ g \circ K_1(x)$. The isotopy through pseudo-Anosov homeomorphisms is then given by $$(x,t) \mapsto K_t^{-1} \circ g \circ K_t(x) $$ This is proved in the book on the Orsay seminar of Thurston's work, which is an excellent source to address many foundational issues related to your overall question.