Let $X$ be a riemannian surface. Suppose $f:X\to X$ is a Teihmuller 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 $f$ is 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 Teichmuller map not to be a pseudo-anosov?


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



Why a Teichmuller map is not a pseudo-anosov?
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.


