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Feb 24 at 20:21 comment added Sam Nead @Dinisaur - In dimension three the argument is extremely similar (including the "shapes" of the matrices).
Feb 24 at 18:10 comment added Dinisaur Is it possible to give a similarly elementary argument to prove the same statement in dim>2?
Aug 9, 2022 at 20:10 comment added Sam Nead Yes, some pA maps arise in this way. In particular, if you have a flat surface that decomposes nicely (in two ways) as a union of annuli with rationally related moduli, then the right twist (aka shear) on one, followed by the left twist on the other, will be pA. The Thurston--Veech construction is usually phrased as a special case of this, where the flat surface is tiled by squares.
Aug 9, 2022 at 19:59 comment added John Rached I see. But you can produce pA diffeomorphisms from a product of two transverse parabolic elements, using the Thurston-Veech construction, correct ?
Aug 9, 2022 at 18:34 comment added Sam Nead It is not true that all pA maps arise from the Thurston-Veech construction. See arxiv.org/abs/1410.6974
Aug 9, 2022 at 13:23 comment added John Rached Great, I was wanting something like this, as pseudo-Anosov diffeomorphisms have derivatives with trace larger than 2, and you can produce any one of these (Thurston) from a product of parabolic elements
Aug 9, 2022 at 13:14 vote accept John Rached
Aug 9, 2022 at 8:29 history edited Sam Nead CC BY-SA 4.0
third proof
Aug 9, 2022 at 8:23 history answered Sam Nead CC BY-SA 4.0