2
$\begingroup$

Let $X$ be a Riemannian surface. Suppose $f:X\to X$ is a Teichmüller map with respect to a quadratic differential $q$ on $X$. This means that, if $q=dz^2$ in local coordinates in a neighborhood of nonzero point of $q$, then $f=Kx+\frac1Ky$ for $z=x+iy$ and for a constant $K$ which does not depend on local coordinates.

My question is: Why is $f$ not a pseudo-Anosov map for horizontal and vertical foliations of $q$ (defined as $q(v)>0$ and $q(v)<0$)? It seems that $f$ multiplies ther transversal meashuers by $\frac1K$ and $K$. Maybe whe should suppose $k\ne1$, are there any other reasons for a Teichmüller map not to be a pseudo-Anosov?

The question is motivated by comparing the Teichmüller existence theorem and the Nielsen-Thurston classification. On the one hand, every map $X\to X$ is homotopic to a Teichmüller map. On the over hand, periodic and reducible maps shouldn't be pseudo-Anosov?

Improve this question
$\endgroup$
8
  • $\begingroup$ I think you may be confused. Usually, a Teichmuller map is between distinct (marked) Riemann surfaces, and not from a Riemann surface to itself. Perhaps you have two different markings on the given surface X? $\endgroup$
    – Sam Nead
    Commented Nov 9, 2020 at 14:02
  • $\begingroup$ @SamNead i think we need to assume the markings for $X$ coincide, and $f$ to be the map not homotopic to identity $\endgroup$
    – Andrey Ryabichev
    Commented Nov 9, 2020 at 15:04
  • 2
    $\begingroup$ Even if the map is from $X$ to itself, to be pA, it needs to preserve the vertical/horizontal foliations on $X$, while Teichmuller map only preserves them after you choose two sets of conformal coordinates on $X$ (regarded as both domain and the range of the map), effectively replacing $f$ with compositions $\phi \circ f \circ \psi$. This pre/post composition messes up VH foliations completely. $\endgroup$
    – Moishe Kohan
    Commented Nov 9, 2020 at 17:03
  • $\begingroup$ @MoisheKohan in other words, $f$ may have initial differential $q_1$ and terminal differential $q_2$, which may not coincide, even up to isotopy (do i understand your argument correctly?) $\endgroup$
    – Andrey Ryabichev
    Commented Nov 10, 2020 at 13:19
  • $\begingroup$ @AndreyRyabichev - That does not work. The Teichmuller map is required to be homotopic to the "change of marking" map. For an overview, see Section 11.1.3 of "A primer on mapping class groups". $\endgroup$
    – Sam Nead
    Commented Nov 10, 2020 at 14:15

3 Answers 3

Reset to default
1
$\begingroup$

You have misunderstood the definition of Teichmuller space. You might want to look at "A primer on mapping class groups" by Farb and Margalit (in particular Chapters 10 through 14).

Improve this answer
$\endgroup$
0
$\begingroup$

The map $f$ is a Teichmuller map with respect to a pair of quadratic differentials $q_1,q_1$ on $X$ --- initial differential and terminal differential. More precisely, $f$ maps the horisontal/vertical foliation of $q_1$ to the horisontal/vertical foliation of $q_2$.

To verify that $f$ is pseudo-anosov we need to constuct one pair of transversal measured foliations which is mapped by $f$ to itself. But $q_1$ and $q_2$ may not coincide up to isotopy, so their pairs of horisontal and vertical foliations will be different.

Improve this answer
$\endgroup$
0
$\begingroup$

If the initial and terminal quadratic differentials of a Teichmuller map coincide, then it is a pseudo-anosov map.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .