Tensor identities for the real division algebras

Define $\varphi :\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{C}\to \mathbb{C}\oplus \mathbb{C}$ by $\varphi (z\otimes w)=(zw,z\bar{w})$. The formula is real-bilinear in $(z,w)$, so [F2] extends it linearly. Since $\mathbb{C}$ is commutative, componentwise multiplication gives $\varphi ((z\otimes w)(z^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }}))=\varphi (z\otimes w)\varphi (z^{\mathrm{\prime }}\otimes w^{\mathrm{\prime }})$. The images of $1\otimes 1$, $i\otimes 1$, $1\otimes i$, and $i\otimes i$ are $(1,1)$, $(i,i)$, $(i,-i)$, and $(-1,1)$. If a real linear combination with coefficients $a,b,c,d$ vanishes, its real coordinates give $a-d=a+d=0$ and its imaginary coordinates give $b+c=b-c=0$, so all coefficients vanish. They form a basis of the four-dimensional target by [F3], hence $\varphi$ is an isomorphism.

Regard $\mathbb{H}$ as a two-dimensional left $\mathbb{C}$-vector space with basis $1,j$, so $k=ij$ and $jz=\bar{z}j$ for $z\in \mathbb{C}$, using the real basis of $\mathbb{H}$ from [F1] and the field $\mathbb{C}$ from [F3]. Define $\psi :\mathbb{C}{\otimes }_{\mathbb{R}}\mathbb{H}\to {\mathrm{End}}_{\mathbb{C}}(\mathbb{H})$ by $\psi (y\otimes z)(x)=yx\bar{z}$. The formula is real-bilinear and therefore defines a linear map by [F2]. It is multiplicative because $\psi (y\otimes z)\psi (y^{\mathrm{\prime }}\otimes z^{\mathrm{\prime }})(x)=y{y}^{\mathrm{\prime }}x\overline{z{z}^{\mathrm{\prime }}}$. In the basis $1,j$, the maps $\psi (1\otimes z)$ for $z=1,i,j,k$ have matrices $I$, $\left(\begin{array}{cc}-i& 0\\ 0& i\end{array}\right)$, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, and $\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)$. Their diagonal entries force the coefficients of the first two matrices in a linear relation to vanish, and their upper-right and lower-left entries force the other two coefficients to vanish; hence they are linearly independent over $\mathbb{C}$. They form a basis of the four-dimensional complex vector space ${\mathrm{End}}_{\mathbb{C}}(\mathbb{H})$, so $\psi$ is an algebra isomorphism with $M_2(\mathbb{C})$.

Define $\theta :\mathbb{H}{\otimes }_{\mathbb{R}}\mathbb{H}\to {\mathrm{End}}_{\mathbb{R}}(\mathbb{H})$ by $\theta (a\otimes b)(x)=ax\bar{b}$. The real-bilinearity gives a well-defined linear map by [F2]. It is multiplicative since $\theta (a\otimes b)\theta (c\otimes d)(x)=acx\overline{bd}$. Set $T_{u,v}(x)=ux\bar{v}$ for basis units $u,v\in \{1,i,j,k\}$. Direct multiplication using [F1] gives the trace matrix $(\mathrm{Tr}{T}_{u,v}{)}_{u,v=1,i,j,k}=\left(\begin{array}{cccc}4& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$, with rows and columns ordered $1,i,j,k$. Equip $\mathbb{H}$ with the real inner product making $1,i,j,k$ orthonormal. Each $T_{a,b}$ is a signed permutation of this basis, hence orthogonal with $T_{a,b}^T=T_{\bar{a},\bar{b}}$. The composition formula and trace matrix give $⟨T_{a,b},T_{c,d}{⟩}_{\mathrm{H}\mathrm{S}}=\mathrm{Tr}(T_{a,b}^T{T}_{c,d})=4{\delta }_{a,c}{\delta }_{b,d}$. The sixteen nonzero images are linearly independent; since ${\mathrm{End}}_{\mathbb{R}}(\mathbb{H})$ has dimension $16$, $\theta$ is an isomorphism with $M_4(\mathbb{R})$.

Choose real bases of $D$ and $E$; these exist for $\mathbb{R},\mathbb{C},\mathbb{H}$ by their displayed coordinate descriptions. The corresponding matrix units with basis coefficients form bases of $M_m(D)$ and $M_n(E)$, and [F4] makes their elementary tensors a basis of the tensor product. They multiply by the two Kronecker-delta rules in their matrix indices and by $(d\otimes e)(d^{\mathrm{\prime }}\otimes e^{\mathrm{\prime }})=d{d}^{\mathrm{\prime }}\otimes e{e}^{\mathrm{\prime }}$ in their coefficients. Sending such a basis tensor $E_{ij}(d)\otimes E_{kl}(e)$ to the matrix unit indexed by row $(i,k)$ and column $(j,l)$ with coefficient $d\otimes e$ sends this basis bijectively to the corresponding basis of $M_{mn}(D{\otimes }_{\mathbb{R}}E)$ and is multiplicative. This proves the stated matrix identity.