# 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.

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There is a notion of a determinant for matrices with coefficients in a noncommutative ring due to Dieudonné. Also the quaternions can be seen as a subring of $M_{2\times 2}(\Bbb C)$, so that $M_{2\times 2}(\Bbb H)\subset M_{4\times 4}(\Bbb C)$ as a subring. That gives you at least one possible definition of a $\Bbb C$-valued determinant. You could search "quaternionic determinant" on google, and "Dieudonné determinant". As for dimension, I would expect that the above determinant is a submersion onto $\Bbb C^*$ so that $SL(2,\Bbb H)$ should have complex dimension 15. –  Olivier Bégassat May 23 '13 at 17:38
The dimesion statement in my comment is false. –  Olivier Bégassat May 23 '13 at 17:45

1. 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.
2. $SL(2,\mathbb R)\cong Spin(2,1)$, $SL(2,\mathbb C)\cong Spin(3,1)$. 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.)
3. 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.