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!



Iterated Lefschetz numbers
Your question may be recast as: 


Is there an $N(n)$ such that for any integral matrix $A\in SL(n,\mathbb{Z})$, there is $k\leq N(n)$ with $tr(A^k)>2 $ (with a few small exceptional cases)?
(this is not quite equivalent, since the image of $Mod(S\_{g,n})$ is not
all of $SL(H\_1(S\_{g,n}))$, but it certainly implies what you want)


Let the singular values of $A$ be denoted $\lambda\_1,\ldots, \lambda\_n$, so that $det(xI-A)=\prod\_{i=1}^n (x-\lambda\_i)$. Then $tr(A^k)=\sum\_{i=1}^n \lambda\_i^k =p\_k(\lambda\_1,\ldots,\lambda\_n)=p\_k$, where $p\_k$ is the [power symmetric polynomial](http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial) of degree $k$ in $n$ variables. Now, we observe that $p\_k$ is a polynomial in $p\_1,\ldots, p\_n$ for $k>n$. I found [another mathoverflow question which computed the first few of these](https://mathoverflow.net/a/37235/1345). 


Consider $n=2$ (note, $SL(2)=Sp(2)$), and suppose $|p\_1(\lambda\_1,\lambda\_2)|=|\lambda\_1+\lambda\_2|=|\lambda\_1+\lambda\_1^{-1}| \leq 2$. It turns out then that $|tr(A^k)|\leq 2$ for all $k$, and $A$ is either parabolic or elliptic (and similarly if $|p\_2|\leq 2$). So assume $p\_1,p\_2 \leq -3$. Then $p\_3=-p\_1^3 +\frac32 p\_1p\_2 \geq \frac{81}{2}$. Thus, $N(2)$ exists (where one excludes $|tr(A)|\leq 2$.


Now, consider $n=3$. One has the relation $p\_4=\frac16 p\_1^4−p\_1^2p\_2+\frac12 p\_2^2+\frac43 p\_1p\_3$. Suppose that $p\_1\leq -R$, for some $R>>0$, and $p\_2,p\_3\leq 2$. 
Then one has $p\_4 = \frac16 p\_1^4-p\_1^2p\_2 +\frac12 p\_2^2+\frac43 p\_1p\_3 \geq \frac16 R^4 -2R^2 -\frac43 2R$. Then choose $R$ large enough that $\frac16 R^4 -2R^2 -\frac83 R >2 $, then one has $p\_4 >2$ if $p\_1\leq -R$. Now,
suppose $p\_2\leq -R$ or $p\_3 \leq -R$. Then we similarly conclude that $p\_8 >2$
or $p\_{12} >2$. Otherwise, $ -R\leq p\_i\leq 2$ for $1\leq i\leq 3$, and there are only finitely many characteristic polynomials $det(xI-A)$ of such matrices, since the elementary symmetric polynomials are determined by the power sum symmetric polynomials. Thus, there is a uniform $N(3)$ such that for some $k\leq N(3)$, $tr(A^k)>2$. 


Similarly, for $n=4$, $p\_5=−\frac{1}{24} p\_1^5+\frac{5}{12} p\_1^3p\_2−\frac58 p\_1p\_2^2−\frac56 p\_1^2p\_3+\frac56 p\_2p\_3+\frac54 p\_1p\_4$. The leading order
term here is $-\frac{1}{24} p\_1^5$, so if $p\_1\leq -R$ for $R>>0$, and $p\_2,p\_3,p\_4\leq 2$,
one concludes that $p\_5 \geq \frac{1}{24} R^5 -\frac{5}{6} R^3 +\frac58 R p\_2^2 -\frac56 R^2 p\_3 +\frac56 p\_2 p\_3 - \frac52 R$. If $p\_3 < 0, 0<p\_2<2$, then the term $\frac56 p\_2p\_3$ will be dominated by the term $-\frac56 R^2 p\_3$ for $R>>0$. If $p\_2 <0, 0<p\_3<2$, then $\frac56 p\_2p\_3$ will be dominated by $\frac58 R p\_2^2$, and otherwise $\frac56 p\_2 p\_3 \geq 0$. So we conclude that for $R>>0$ and $p\_1\leq -R$, $p\_5>2$. Similar to before, we may replace $p\_1$ with $p\_2, p\_3$, or $p\_4$ if any of these is $\leq -R$, to conclude that $p\_{5i}>2$ for $i=2,3$ or $4$, and otherwise there will be only finitely many possibilities.
So $N(4)$ exists. 


**Addendum:** I now have a proof for general $n$. 


There's a similar pattern to the general formula for $p\_{n+1}$ in terms of $p\_1,\ldots,p\_n$, which is given in a comment by Darij Grinberg to [this question](https://mathoverflow.net/q/36771/1345). He gives the formula:
$$ p\_{n+1}= (n+1)\sum\_{i=2}^{n+1} \frac{(-1)^i}{i!} \sum\_{j\_1,\ldots,j\_i\geq 1; j\_1+\cdots+j\_i=n+1} \frac{1}{j\_1 \cdots j\_i} p\_{j\_1}\cdots p\_{j\_i}= 0.$$
This gives the same leading order asymptotics $p\_{n+1}=(-1)^np\_1^{n+1} + O(p\_1^n)$. This sum has the property that if $p\_1,\ldots,p\_n <0$,
then all the terms are positive, since each term is a product of the
form $(-1)^i p\_{j\_1}\cdots p\_{j\_i}$. 


**Theorem**
Consider monic degree $n$ polynomials $q(x)\in \mathbb{R}[x]$, $q(x)=\prod\_{i=1}^n (x-\lambda\_i)$. There is a compact subset $C\_\delta$ of the space of monic degree $n$ polynomials so that if $q(x)\notin C\_\delta$, then $p\_k(\lambda\_1,\ldots,\lambda\_n) > \delta$ for some $k \leq n(n+1)$. 


**Proof** Assume $p\_1,\ldots,p\_n \leq \delta$. We'll assume $p\_1 \leq -R$,
for some $R>>0$, to be determined. If $p\_2,\ldots,p\_n \leq 0$, then 
$p\_{n+1} \geq \frac{1}{n!} R^n > \delta$ for $R>>0$. So assume some
subset $0< p\_{l\_1},\ldots ,p\_{l\_m} \leq \delta$. Consider all the terms
of the expression of $p\_{n+1}$ in which an odd number of $p\_{l\_1},\ldots, p\_{l\_m}$ appear (with multiplicity), then these terms will be $\leq 0$, and
the rest of the terms will be positive. However, these terms will be dominated
by a term replacing each $p\_{l\_i}$ with $p\_1^{l\_i}$. This correspondence
is not 1-1, and the coefficients may differ, but by making $R>>0$ large
enough, we may guarantee that all the negative terms will be bounded
above by corresponding large positive terms. Thus, in this case we may
show that $p\_{n+1}>\delta$ for $R>>0$.


Now, suppose $p\_k \leq -R$ for some $1\leq k\leq n$. Then by the same
reasoning, $p\_{k(n+1)} >\delta$, essentially replacing $A$ with $A^k$. 


So our compact set $C\_\delta$ consists of $-R\leq p\_i \leq \delta, 1\leq i\leq n$. 


Now, consider monic degree $n$ polynomials with integer coefficients. 
There are only finitely many in $C\_2$, so for all these, there will be 
some uniform power $N(n)$ so that $p\_k >2$ for some $k\leq N(n)$. 
In fact, we see that for all but finitely many polynomials, $k\leq n(n+1)$. 


