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Ian Agol
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Given a pseudo Anosov mapping class $f:S_{g,n}\rightarrow S_{g,n}$ is the Lefschetz number for $f^m$ negative for some $m$ depending only on $(g,n)$?

The Lefschetz number of a mapping class $f$ can be defined as $2-Tr(f^{\star})$ where $f^{\star}$ is the induced map on $H_1(S_{g,n},\mathbb{Z})$.

If the leading eigenvalue for the action on homology is 1 then we are done. If the leading eigenvalue $\lambda\neq1$ then the sequence of Lefschetz numbers is dominated $\lambda^m$ and so the Lefschetz number will eventually be negative.

Thanks!

Given a pseudo Anosov mapping class $f:S_{g,n}\rightarrow S_{g,n}$ is the Lefschetz number for $f^m$ negative for some $m$ depending only on $(g,n)$?

If the leading eigenvalue for the action on homology is 1 then we are done. If the leading eigenvalue $\lambda\neq1$ then the sequence of Lefschetz numbers is dominated $\lambda^m$ and so the Lefschetz number will eventually be negative.

Thanks!

Given a pseudo Anosov mapping class $f:S_{g,n}\rightarrow S_{g,n}$ is the Lefschetz number for $f^m$ negative for some $m$ depending only on $(g,n)$?

The Lefschetz number of a mapping class $f$ can be defined as $2-Tr(f^{\star})$ where $f^{\star}$ is the induced map on $H_1(S_{g,n},\mathbb{Z})$.

If the leading eigenvalue for the action on homology is 1 then we are done. If the leading eigenvalue $\lambda\neq1$ then the sequence of Lefschetz numbers is dominated $\lambda^m$ and so the Lefschetz number will eventually be negative.

Thanks!

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Iterated Lefschetz numbers

Given a pseudo Anosov mapping class $f:S_{g,n}\rightarrow S_{g,n}$ is the Lefschetz number for $f^m$ negative for some $m$ depending only on $(g,n)$?

If the leading eigenvalue for the action on homology is 1 then we are done. If the leading eigenvalue $\lambda\neq1$ then the sequence of Lefschetz numbers is dominated $\lambda^m$ and so the Lefschetz number will eventually be negative.

Thanks!