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Let $\mathbb{H}$ be the algebra of quaternions. It can be proven that each derivation $D:\mathbb{H}\to \mathbb{H}$ is inner that is of the form $\mathrm{ad}x$ for some $x\in \mathbb{H}$. I am to prove that the Lie algebra of derivations of $\mathbb{H}$ is a Lie algebra $\mathbb{R}^3$ of usual "geometrical" vectors with the usual vector product.

Okay, we have a Lie-algebra homomorphism $f:L\mathbb{H}\to \mathrm{Der}(\mathbb{H})$ where $L\mathbb{H}$ is $\mathbb{H}$ endowed with the bracket $[x,y]=xy-yx$. And we know that $f$ is surjective (since each derivation is inner). Also the kernel of $f$ is the center of $L\mathbb{H}$ that is $\mathbb{R}$. Okay, $L\mathbb{H}/\mathbb{R}$ is a Lie algebra, generated by $i,j,k$ such that $[i,j]=2k, [j,k]=2i, [k,i]=2j$. But this is not what we need. How to fix it?

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If we denote $Q\in\mathbb H$ by $q_0+q$, where $q_0$ is the scalar part and $q$ is the vector part, then: $$ \begin{split} [P,Q]&=PQ-QP\\ &=\underbrace{p_0q_0-p\cdot q}_{\text{scalar}}+\underbrace{p_0q+q_0p+p\times q}_{\text{vector}}-(\underbrace{q_0p_0-q\cdot p}_{\text{scalar}}+\underbrace{q_0p+p_0q+q\times p}_{\text{vector}})\\ &=2p\times q \end{split} $$ This indeed turns the pure quaternions $L\mathbb H/\mathbb R$ into a Lie algebra with vector product (since only the vector part is left after the quotient): $$ [p,q]=2p\times q $$

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  • $\begingroup$ In the notation of quaternion algebras, $q_0$ would typically denote the pure quaternion part (equivalently non-scalar part, equivalently "vector part") of $q$. For instance $\mathbb{H}_0:=\{xi+yj+zij\mid x,y,z\in\mathbb{R}\}\subset\mathbb{H}$. In general we can write $q=\frac{\mathrm{tr}(q)}{2}+q_0$. $\endgroup$ – j0equ1nn Mar 1 '17 at 0:35

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