Can the the monodromy matrix of the action of a pseudo-Anosov homeomorphism of a surface on it's homology be of the form of a reducible or block matrix? Any such reference can be helpful. Thanks.
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$\begingroup$ "some space" is a bit unprecise... $\endgroup$– YCorJun 8, 2017 at 10:43
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$\begingroup$ Sorry. I meant "can it be possible"? Apologies again. $\endgroup$– hakunamatataJun 8, 2017 at 18:28
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$\begingroup$ I was asking about the meaning of space: vector space, topological space, etc, especially because "surface" appears in the title and not in the question. $\endgroup$– YCorJun 8, 2017 at 21:18
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$\begingroup$ Yes, I meant surfaces. Sorry for the lame writing. $\endgroup$– hakunamatataJun 9, 2017 at 6:53
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$\begingroup$ The question needs rewriting to make much sense. Let's assume the previous suggested correction: "surface" instead of "space". A surface does not have a "pseudo-Anosov monodromy matrix", but perhaps a homeomorphism of a surface does; should your question be "can the monodromy matrix of a pseudo-Anosov homemorphism of some surface be..."? If so, then what do you mean by "the monodromy matrix of a homeomorphism of a surface"? Do you mean the monodromy matrix of the action of that homeomorphism on the homology of the surface (as assumed in the answer of @IgorRivin)? $\endgroup$– Lee MosherJul 8, 2017 at 13:51
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Every symplectic matrix arises as the monodromy of a pseudo-Anosov homeomorphism. The canonical reference is Farb-Margalit:
Farb, Benson; Margalit, Dan, A primer on mapping class groups, Princeton Mathematical Series. Princeton, NJ: Princeton University Press (ISBN 978-0-691-14794-9/hbk; 978-1-400-83904-9/ebook). xiv, 492 p. (2011). ZBL1245.57002.
Where you can find the facts that there exists pseudo-Anosovs in the Torelli group (trivial monodromy), and the map from the mapping class group to the Symplectic group is surjective. Taking some element with your favorite matrix, and multiplying it by a high power of a Torelli pseudo-anosov will give you what you want.
answered Jun 8, 2017 at 13:17
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$\begingroup$ Thanks. I had read it previously, but was not sure overall. It is indeed a great book. Thanks again. $\endgroup$ Jun 8, 2017 at 18:29