Let $T^2$ be a 2torus and $f:T^2\rightarrow T^2$ a smooth map. Let $f_*:\pi_1(T^2)\rightarrow\pi_1(T^2)$ be the induced map on the fundamental group $\pi_1$. If $f_*$ has no eigenvalue greater than 1, then all balls of $T^2$ have subexponential growth by $ f $. Why?
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closed as offtopic by Chris Gerig, Stefan Kohl, Anthony Quas, Igor Belegradek, Fernando Muro Oct 10 '14 at 22:02
 This question does not appear to be about research level mathematics within the scope defined in the help center.

1$\begingroup$ What makes you think this is true? Also what do you mean by a ball having subexponential growth? $\endgroup$ – Anthony Quas Oct 10 '14 at 20:23

$\begingroup$ This statement is in a paper. Let $B\subset T^2$ be a ball, $B$ have subexponential growth by $f$ If $Vol(f^n(B))<\lambda^nVol(B)$, $\forall\lambda>1$. $\endgroup$ – Jose Santana Oct 10 '14 at 21:03

1$\begingroup$ But the measure is finite. $\endgroup$ – Anthony Quas Oct 10 '14 at 21:19

$\begingroup$ "This statement is in a paper." Cite? $\endgroup$ – Gerry Myerson Oct 10 '14 at 22:34

$\begingroup$ This statement is in the paper [M.Brin, D. Burago, S Ivanov. ON PARTIALLY HYPERBOLIC DIFFEOMORPHISSMS OF 3MANIFOLDS WITH COMMUTATIVE FUNDAMENTAL GROUP.] in the proposition 2.1 $\endgroup$ – Jose Santana Oct 13 '14 at 19:51
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You should check out the paper of R. Bowen: ENTROPY FOR GROUP ENDOMORPHISMS AND HOMOGENEOUS SPACES (Transactions of the AMS, 1971)