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.


  • $\begingroup$ You might want to define the Lefschetz number... $\endgroup$ – Igor Rivin Aug 11 '13 at 23:16
  • $\begingroup$ @IgorRivin The Lefschetz number is quite standard, it is defined to be the alternating sum over dimension of the trace of the induced map on homology. See mathworld.wolfram.com/LefschetzNumber.html for instance. $\endgroup$ – Vidit Nanda Aug 12 '13 at 3:04
  • $\begingroup$ @IgorRivin added the definition for the case being considered. $\endgroup$ – user38496 Aug 12 '13 at 3:45
  • $\begingroup$ For any $(g,n)$, there is a universal lower bound on $|\lambda|>1$ - does that imply what you want? $\endgroup$ – Ian Agol Aug 12 '13 at 4:52
  • $\begingroup$ @IanAgol This is my intuition but I want to show that the integer $m$ depends only on (g,n) and so it seems to me a universal gap between the first and second eigenvalues modulus is necessary. $\endgroup$ – user38496 Aug 12 '13 at 5:17

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 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.

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. 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, $k\leq n(n+1)$.

  • $\begingroup$ Why are there only finitely many monic integer polynomials of degree $n$ in $C_2$? $\endgroup$ – user38496 Sep 3 '13 at 23:48
  • $\begingroup$ @user38496: The elementary symmetric functions of $\lambda_1,\ldots,\lambda_n$ are rational functions of the power symmetric values, and give the coefficients of the monic polynomial with those roots. So given a compact set of power-symmetric functions, there are finitely many associated polynomials. $\endgroup$ – Ian Agol Sep 3 '13 at 23:56

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