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Let me recall the definition which seems the most standard of Fricke polynomials. Let $G$ be the free group with two generators $u,v$. It is not very hard to prove that there exists a unique application $t : G \rightarrow \mathbb{Z}[X,Y,Z]$ such that

(1) $t(u)=X,t(v)=Y,t(uv)=Z$,

(2) $t$ is the character (i.e. the trace) of a representation $r : G \rightarrow \rm{SL}_2(K)$ where $K$ is some field containing $\mathbb{Z}[X,Y,Z]$.

A Fricke polynomial is just an element of $\mathbb{Z}[X,Y,Z]$ which is in the mage of $t$.

What is the image of $t$ ?


Edit: As requested, I give here the proof of the statement about $t$. I believe that this statement is due in some form to Fricke in the nineteenth century:

If $t:G \rightarrow K$ is the trace of a representation $r: G \rightarrow {\rm{SL}}_2(K)$, then one has $t(gh)+t(gh^{-1}) = t(g)t(h)$ for all $g,h$ in $G$ (since this formula holds when $t$ is the trace, and $g,h$ are matrices in ${\rm SL}_2(K)$, as easily checked). Now this formula, and an easy induction on the length of a word $g$ in $G$ shows that $t(g)$ can be expressed as a polynomial (with integral coefficients) in $t(u)$, $t(v)$ and $t(uv)$. Hence a $t$ satisfying (1) and (2) is unique if it exists.

For the existence, let $U$ and $V$ be two matrices in ${\rm SL}_2(\mathbb{C})$ such that tr$U$, tr$V$, and tr$UV$ are algebraically independent (this is really easy). Then see $\mathbb{Z}[X,Y,Z]$ as a subring of $\mathbb{C}$ by sending $X$ on tr$U$, $Y$ on tr$V$, $Z$ on tr$UV$. Let $r$ be the representation $G \rightarrow \rm{SL}_2(\mathbb{C})$ sending $u$ on $U$ and $v$ on $V$, and $t=$tr $r$. Then by definition $t$ satisfies (2), with $K=\mathbb{C}$, one has $t(u)=X$, $t(v)=Y$, $t(uv)=Z$ by construction and $t$ takes value in the subring $\mathbb{Z}[X,Y,Z]$ of $\mathbb{C}$ by what we have said earlier. Hence the existence of $t$.

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In this comment I sketch the proof of the existence and uniqueness of $t$. If $t: G \rightarrow K$ satisfies (2), that $t(gh)+t(gh^{-1})=t(g)t(h)$ for all $g,h$ in $G$, since this relation is satisfied for the trace of matrices in $\rm{SL}_2(K)$. With the "initial conditions" given in (1), one checks easily that $t$ is unique, and take values in $\mathbb{Z}[X,Y,Z]$. –  Joël Jul 24 '12 at 17:55
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For the existence, ind two matrices $a,b$ in $SL_2(\mathbb{C})$ such tr$(a)$,tr$(b)$ and tr$(ab)$ are algebraically independent (easy). Embed $\mathbb{Z}[X,Y,Z]$ into $\mathbb{C}$ by sending $X,Y,Z$ to these traces. Define $r: \rightarrow mathbb{Sl}_2(\mathbb{C})$ by sending $u$ on $a$ and $v$ and $b$. Then $t = tr \, r$ satisfies (1) and (2). –  Joël Jul 24 '12 at 17:57
    
Instead of adding comments, could you please edit the question? –  Bruce Westbury Jul 24 '12 at 18:13
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Done. Do you agree that $t$ is well-defined now ? –  Joël Jul 24 '12 at 19:12
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This question has been considered by Troels Jorgensen: dx.doi.org/10.2307/2043555 –  Ian Agol Jul 24 '12 at 21:42
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2 Answers 2

up vote 2 down vote accepted

I do not think that there is a complete answer to this question. However, one can give some necessary conditions (which show that any answer must be complicated).

One can show that a triple $(x,y,z) \in \mathbb C^3$ comes from traces of matrices in $SU(2)$ (i.e. there exist $u,v \in SU(2)$ such that $(x,y,z)=(t(u),t(v),t(uv))$) if and only if $(x,y,z) \in \mathbb [-2,2]^3$ (this is obviously necessary) and $$x^2 + y^2 + z^2 - xyz \leq 4.$$

Hence, a necessary condition on a polynomial $p \in \mathbb Z[X,Y,Z]$ to be a Fricke polynomial is that for all $(x,y,z) \in [-2,2]^3$ we have the implication $$x^2 + y^2 + z^2 - xyz \leq 4 \quad \Rightarrow \quad p(x,y,z) \in [-2,2].$$

Indeed, the word evaluated on $u$ and $v$ will give again a matrix in $SU(2)$ and hence its trace is in the interval $[-2,2]$.

The semi-algebraic set $$S:=[-2,2]^3 \cap \lbrace (x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 - xyz \leq 4 \rbrace$$ is a spectrahedron (that means it can be defined by a linear matrix inequality) and is called the Elliptope $E_3$.

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Thanks. I didn't see this analytic condition coming ! Because of it, it seems indeed very hard to characterize the image. Here I must confess that the initial problem I met in my research was different: a version mod $p$ (and actually for $p=2$) of the same question. That is, what polynomial in $\mathbb{F}_p[x,y,z]$ is the reduction of a Fricke polynomial. And here, computations seems to show that the image is pretty big. –  Joël Jul 25 '12 at 0:10
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I think your new (or initial) question is really interesting. Why don't you ask this as a separate question? –  Andreas Thom Jul 25 '12 at 8:22
    
Thanks. I will in a few days, when I have thought at the best way to formulate it. –  Joël Jul 25 '12 at 12:26
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You have not shown that your map $t$ is well-defined. I suspect you are defining the quotient of $SL_2\times SL_2$ by simultaneous conjugation. This was studied by Procesi.

Edit: A quick search found the paper http://www.mathematics.jhu.edu/brown/Documents/FrickeCharAutos.pdf which confirms Andreas comment. It also gives the action of $Out(F_2)$ in Example 6.1. This raises the question: does $Out(F_2)$ act transitively on this set of polynomials?

In particular they give the reference:

`Fricke, R., and Klein, F., Vorlesungen uber die Theorie der automorphem Functionen, Vol. 1, pp. 365-370. Leipzig: B.G. Teubner 1897. Reprint: New York Juhnson Reprint Corporation (Academic Press) 1965

This can be found at http://archive.org/details/vorlesungenber01fricuoft

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Fricke showed (I think) that every polynomial function on $SL_2 \times SL_2$ which is conjugation invariant, is a polynomial in $X,Y$ and $Z$ as above. This applies to traces of words. The question which such polynomials are coming from words in the free group is more complicated. I do not think that this subset is structured in any obvious way. –  Andreas Thom Jul 24 '12 at 18:58
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