Canonical rational form for $SL(n)$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:50:11Z http://mathoverflow.net/feeds/question/91651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91651/canonical-rational-form-for-sln Canonical rational form for $SL(n)$ Marc Palm 2012-03-19T18:46:16Z 2012-03-21T15:17:11Z <p>The canonical rational form helps us to parametrize the conjugacy classes in $GL(n)$ over any commutative field. </p> <blockquote> <p>How can we parametriize the conjugacy classes in $SL_n(k)$, where $k$ is an arbitrary locally compact field or a global field?</p> </blockquote> http://mathoverflow.net/questions/91651/canonical-rational-form-for-sln/91752#91752 Answer by Mark Wildon for Canonical rational form for $SL(n)$ Mark Wildon 2012-03-20T21:12:52Z 2012-03-21T13:14:27Z <p>The results in Section 3 of my <a href="http://arxiv.org/abs/1001.0811" rel="nofollow">joint paper</a> with John Britnell are relevant. They give a complete parametrization if $k$ is a finite field. </p> <p>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 <i>part-size invariant</i> of $g$ to be the highest common factor of all the parts of the partitions $\lambda_1, \ldots, \lambda_r$. </p> <p>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 $SL_n(k)$.</p> <p>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 </p> <p>$$(f_1^{\lambda_1} \ldots f_r^{\lambda_r}, i)$$</p> <p>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 $0 \le i &lt; \mathrm{hcf}(m,q-1)$ where $m$ is the corresponding part-size invariant.</p> <p>In general I think this question quickly leads to difficult arithmetic problems. To take one very special case, if $\gamma \in \mathbb{Q}$ and</p> <p>$$g = \left( \begin{matrix} 0 &amp; -1 \newline 1 &amp; \gamma \end{matrix} \right) \in SL_2(\mathbb{Q}),$$ </p> <p>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 $a^2 + ab\gamma + b^2$.</p>