The canonical rational form helps us to parametrize the conjugacy classes in $GL(n)$ over any commutative field.
How can we parametriize the conjugacy classes in $SL_n(k)$, where $k$ is an arbitrary locally compact field or a global field?
The canonical rational form helps us to parametrize the conjugacy classes in $GL(n)$ over any commutative field.



The results in Section 3 of my joint paper 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 partsize 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 partsize 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 $t = \mathrm{hcf}(m,q1)$ 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 $1,x,\ldots,x^{t1}$ 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 $0 \le i < \mathrm{hcf}(m,q1)$ where $m$ is the corresponding partsize 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 $a^2 + ab\gamma + b^2$. 

