The canonical rational form helps us to parametrize the conjugacy classes in $GL(n)$ over any commutative field. 



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> How can we parametriize the conjugacy classes in $SL\_n(k)$, where $k$ is an arbitrary locally compact field or a global field?
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Canonical rational form for $SL(n)$
The results in Section 3 of my [joint paper](http://arxiv.org/abs/1001.0811) with John Britnell are relevant. They give a complete parametrization if $k$ is a finite field. 


Let $g \in GL\_n(k)$ be an element with rational canonical form labelled by $f\_1^{\lambda\_1} \ldots f\_r^{\lambda\_r}$ where the $f\_i \in k[x]$ are irreducible polynomials, and the $\lambda\_i$ are partitions. We define the *part-size invariant* of $g$ to be the highest common factor of all the parts of the partitions $\lambda\_1, \ldots, \lambda\_r$. 


Suppose that $g \in SL\_n(k)$ has part-size invariant $m$. Then by Proposition 3.10 in the paper (which is stated for finite fields, but holds more generally) the determinant of any matrix in $\mathrm{Cent}\_{GL\_n(k)}(g)$ is an $m$th power. If the subgroup of $k^\times$ generated by $m$th powers has index $t$ in $k^\times$ then $g^{GL\_n(k)}$ is the union of at least $t$ conjugacy classes of $SL\_n(k)$.


If $k = \mathbb{F}\_q$ then $t = \mathrm{hcf}(m,q-1)$ and, by Corollary 3.2 in the paper, $g^{GL\_n(\mathbb{F}\_q)}$ splits into exactly $t$ conjugacy classes of $SL\_n(\mathbb{F}\_q)$. If the determinant of $x \in GL\_n(\mathbb{F}\_q)$ is a generator of $\mathbb{F}\_q^\times$, then the conjugates of $g$ by $1,x,\ldots,x^{t-1}$ are a set of representatives for the conjugacy classes of $SL\_n(\mathbb{F}\_q)$ contained in $g^{GL\_n(q)}$. It follows that the conjugacy classes of $SL\_n(\mathbb{F}\_q)$ are parametrized by pairs 


$$(f\_1^{\lambda\_1} \ldots f\_r^{\lambda\_r}, i)$$


where $f\_1^{\lambda\_1} \ldots f\_r^{\lambda\_r}$ is the label of a rational canonical form of a conjugacy class of $GL\_n(\mathbb{F}\_q)$ contained in $SL\_n(\mathbb{F}\_q)$ and $0 \le i < \mathrm{hcf}(m,q-1)$ where $m$ is the corresponding part-size invariant.


In general I think this question quickly leads to difficult arithmetic problems. To take one very special case, if $\gamma \in \mathbb{Q}$ and


$$g = \left( \begin{matrix} 0 & -1 \newline 1 & \gamma \end{matrix} \right) \in SL\_2(\mathbb{Q}),$$ 


then there is a matrix commuting with $g$ of determinant $d$ if and only if $d = a^2 + ab\gamma + b^2$ for some $a,b \in \mathbb{Q}$. Hence $g^{GL\_n(\mathbb{Q})}$ splits into $s$ classes of $SL\_n(\mathbb{Q})$ where $s$ is the index in $\mathbb{Q}^\times$ of the subgroup of $\mathbb{Q}^\times$ consisting of all elements of the form $a^2 + ab\gamma + b^2$.


