Why is the dimension of $SL(2,\mathbb{H})$ equal to $15$?
Let me ask a very basic question which is inspired by reading M. Atiyah's "Geometry and physics of knots". Could you explain me (or give a reference to) the definition of the special linear group $SL(2,\mathbb{H})$?
What I don't understand is how to compute the determinant of a matrix with noncommutative entries. Further, even if there was some way to define it, the naive counting gives $12$ for the dimension of $SL(2,\mathbb{H})$, whereas it is clear from the context that it should be equal to $15$.
To solve the contradiction I thought that, probably, $SL(2,\mathbb{H})$ is not defined as a matrix group with determinant equal to $1$, but rather as a group of fractional linear transformations of $\mathbb{HP}^1$. Could someone please confirm if this is correct?
Thank you in advance.
Lie algebra $sl_2(\mathbb H)$ is easier to define: it's the Lie algebra generated by traceless matrices. In non-commutative case the space of traceless matrices is not closed under Lie bracket — that's why dim>12. In fact, $sl_2(\mathbb H)$ is the Lie algebra of matrices with pure imaginary trace.
$SL(2,\mathbb R)\cong Spin(2,1)$, $SL(2,\mathbb C)\cong Spin(3,1)$. Now $Spin(5,1)$ is connected and 1-connected, has Lie algebra $sl_2(\mathbb H)$ (and acts by conformal transformations on $\mathbb HP^1$, btw) — so $SL(2,\mathbb H)\cong Spin(5,1)$ (and is 15-dimensional). (All this is explained somewhere in Baez's Octonions, I believe.)
It seems (after googling) that $SL_2(\mathbb H)$ can also be defined as the commutator subgroup of $GL_2(\mathbb H)$ and this is exactly the subgroup of matrices with Dieudonné determinant equal to 1.