$5$ dimensional space over $\mathbb{R}$

Question

When coming up with a double cover of $SO(5)$, I used conjugation by matrices of the form $$\begin{pmatrix} r & q\\ \overline{q} & r \end{pmatrix}$$ where $r\in\mathbb{R}$ and $q$ is a quaternion. These matrices are clearly $5$ dimensional over $\mathbb{R}$, but I'm wondering if someone can identify this space by name so that I can find more information.

Edit: I should have also added that for any matrix $A$ of this form, $A=A^*$ where $*$ denotes conjugate transpose. However, this requirement follows from how $A$ was defined, so mentioning it again doesn't add information, but rather is a (potentially) useful observation.

Answer 1

There are a few other interpretations, but I believe you are describing a system very similar to the root system $D_5$.

From the Wikipedia definition of $SO_5(\mathbb{R})$, we have that :

SO(5) is a simple Lie group of dimension 10

which takes us to the Poincaré Group.

The page on Simple Lie Groups leads to Dynkin Diagrams and the next chapter defines several relations between Dynkin diagrams and various groups.

In the list we have $B_2$ corresponds to $SO(5)$, and we can find a description of $B_2$ at the link to Root Systems.

Neither of these seem to answer your question.

Further down the list, however, we also find that $D_5$ corresponds to $SO(10)$, which seems to be consistent with your development of a double covering of $SO(5)$.

I have found two references which explore the $D_5$ root system.

• Conformal Invariance and String Theory, page 428 onwards.
• A FAMILY OF MARKED CUBIC SURFACES AND THE ROOT SYSTEM D5, by Elisabetta Colombo and Bert Van Geemen

Answer 2

I feel compelled to write this answer, based on the fact that the answer which was accepted makes no sense.

What the OP mentions seems to be an embedding of $\mathbb{R}^5$ inside the space $M$ of hermitian $2\times 2$ matrices over the quaternions, which is $Sp(2)$-equivariant: where $Sp(2)$ acts on $\mathbb{R}^5$ via the canonical double cover $Sp(2) \to SO(5)$ and acts on $M$ by the conjugation.

This is analogous to the following well-known construction.

You can exhibit the double cover $SL(2,\mathbb{C}) \to SO(3,1)_0$ as follows. You identify $\mathbb{R}^4$ with the real vector space $H$ of hermitian $2\times 2$ matrices. (Minus) the determinant of a hermitian $2 \times 2$ matrix defines a quadratic form on $H$ coming from an inner product of signature $(3,1)$. Define an action of $SL(2,\mathbb{C})$ on $H$ by $h \mapsto s h s^\dagger$, for $h \in H$ and $s \in SL(2,\mathbb{C})$. This is a linear transformation of $H$ which preserves the determinant (since $s$ has unit determinant) and hence defines an element of $O(3,1)$. Since $SL(2,\mathbb{C})$ is connected, it actually defines a surjective map $SL(2,\mathbb{C}) \to SO(3,1)_0$ to the identity component. The kernel of this map is the group of order $2$ generated by $-I$.

The analogy breaks down in that there is no quaternionic determinant in general, but perhaps for matrices in $M$ it can be defined. I have not checked.

There are similar double covers which can be explicitly described in this way: e.g., $SL(2,\mathbb{R}) \to SO(2,1)_0$ and $SL(2,\mathbb{H}) \to SO(1,5)_0$.