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This is an interesting question from a class I'm taking that I'm not really sure how to approach.

It is known that we can define many embeddings of the quaternion algebra $\mathbb{H}$ in the real matrix algebra $M(4, \mathbb{R})$. The standard such embedding is defined by the map

$$ a + bi + cj + zk \mapsto \begin{bmatrix} a & -b & -c & -d \\ b & a & -d & c \\ c & d & a & -b \\ d & -c & b & a \\ \end{bmatrix} $$

Clearly, such matrices are invertible (such that their inverse is the image of the multiplicative inverse of the quaternion), satisfy $A^{T} = -A$, among other useful properties. If we define $\phi : \mathbb{H} \mapsto M(4, \mathbb{R})$ to be the above embedding, the problem is to describe the subspace $S \in M(4, \mathbb{R})$ of elements commuting with the image $\phi(\mathbb{H})$. That is,

$$S = \{A \in M(4, \mathbb{R}) : HAH^{-1} = A, \forall H \in \phi(\mathbb{H})\}$$

I can provide a few properties of such a subspace, but I'm not entirely sure how to classify it nicely. What can be said of such a subspace? Is there a way to view this subspace as a set of rotations or some other sort of geometric object?

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With some work, one can write this subspace as $$\left\{\begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix}\colon a,b,c,d\in\mathbb R\right\}.$$ (The way I did this is to note that a matrix $M$ commutes with $\phi(\mathbb H)$ if and only if it commutes with $\phi(i)$ and $\phi(j)$, since such a matrix then commutes with $\phi(k)=\phi(ij)=\phi(i)\phi(j)$ and every linear combination thereof. I imagine there are plenty of other ways to do the algebra.) This looks an awful lot like the definition of $\phi$ -- a few negative signs are in different places, but it's pretty similar. Can you show that this is another embedding of $\mathbb H$ into $M(4,\mathbb R)$?

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  • $\begingroup$ Oh wow this is very interesting... how can one derive such an explicit form for the subspace? Certainly you could do it by hand, but I assume there's some sort of trick you use here. I've looked at these matrices for a while, and I don't seem to see it... I certainly agree such matrices are in fact another embedding of H in M(4, R), though. $\endgroup$
    – beeselmane
    May 13, 2022 at 7:18
  • $\begingroup$ @beeselmane This was pretty much by hand (I cheated a little bit, using WolframAlpha). I'll add a bit about how I found the subspace in the answer. $\endgroup$ May 13, 2022 at 7:47

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