*algebra embedding from complex matrices into real matrices

Question

I want algebra embeddings from $\text{Mat}_n(\mathbb{C})$ into $\text{Mat}_{2n}(\mathbb{R})$ which preserve self-adjoitness and positive definiteness. So, this embedding should be involutive. Two natural embeddings are $A \mapsto \begin{bmatrix} A_R & A_I \\ -A_I& A_R\end{bmatrix}$ and $A \mapsto \begin{bmatrix} A_R & -A_I \\ A_I& A_R\end{bmatrix}$, where $A=A_R+iA_I$. So, all the *-embedding that I know has the form $A \mapsto O^T \begin{bmatrix} A_R & A_I \\ -A_I& A_R\end{bmatrix}O$ where $O$ is an orthogonal matrix. Is there any other *-embedding? If not how can we show that?

Answer 1

This answer is incomplete; I'm just posting it to ensure it stays saved on the site and in case it is useful in its incomplete form

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We aim to show that for any convolutive algebraic embedding $\Phi:\mathcal M_n(\Bbb C) \to \mathcal M_{2n}(\Bbb R)$, there is an associated orthogonal matrix $O$.

To that end, begin by considering the element $J = \Phi(i\,\text{id}_{\Bbb C^n})$. Because $\Phi$ is adjoint-preserving, the fact that $[i\,\text{id}_{\Bbb C^n}]^* = -i\,\text{id}_{\Bbb C^n}$ implies that $$ J^* = \Phi([i\,\text{id}_{\Bbb C^n}]^*) = \Phi(-[i\,\text{id}_{\Bbb C^n}]) = - \Phi(i\,\text{id}_{\Bbb C^n}) = -J. $$ Because $J$ is skew self-adjoint with $J^2 = -\text{id}_{\Bbb R^{2n}}$, we can conclude that there exists an orthogonal matrix $O_1$ such that $$ J = O_1\pmatrix{0 & -I\\ I & 0}O_1^T. $$ Indeed, the spectral theorem ensures that the matrices $J$ and $\left[ \begin{smallmatrix}0&-I\\I&0\end{smallmatrix}\right]$ are unitarily similar (since they are unitarily diagonalizable with the same eigenvalues), and it generally holds that two unitarily similar real matrices are necessarily orthogonally similar.

Conclude from the fact that $\Phi(A)$ commutes with $J$ (for all $A \in \mathcal M_n(\Bbb C)$) that we must have the form $$ \Phi(A) = O_1\pmatrix{f(A) & -g(A)\\ g(A) & f(A)}O_1^T $$ for some maps $f,g:\mathcal M_n(\Bbb C) \to \mathcal M_n(\Bbb R)$.

Now, note that for any $A \in \mathcal M_n(\Bbb R)$, $$ O_1\pmatrix{f(iA) & -g(iA)\\ g(iA) & f(iA)}O_1^T = \\ \Phi(iA) = J\Phi(A) = \\ O_1\pmatrix{-g(A) & -f(A)\\ f(A) & -g(A)}O_1^T. $$ So, we have $$ f(iA) = - g(A)\\ g(iA) = f(A). $$

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A relevant note:

Your "natural" algebra embeddings arise from an embedding of the underlying spaces. In particular, consider the map $\phi_0:\Bbb C^n \to \Bbb R^{2n}$ given by $\phi_0(x + iy) = \left[ \begin{smallmatrix} x\\y \end{smallmatrix}\right]$. To compute the adjoint of $\phi_0$ relative to the usual inner product on $\Bbb R^{2n}$ and the real inner product $\langle w,z\rangle = \operatorname{Re}[w^*z]$ over $\Bbb C^n$, note that if we define $\psi:\left[ \begin{smallmatrix} x\\y \end{smallmatrix}\right] \mapsto x + iy$, then $$ \begin{align*} \left\langle \phi_0(x + iy), \begin{bmatrix}a\\b\end{bmatrix}\right\rangle &= \left\langle \begin{bmatrix}x\\y\end{bmatrix}, \begin{bmatrix}a\\b\end{bmatrix}\right\rangle = x^Ta + y^Tb\\ \left\langle x + iy, \psi\left(\begin{bmatrix}a\\b\end{bmatrix}\right)\right\rangle &= \langle x + iy, a + ib \rangle = x^Ta + y^Tb, \end{align*} $$ So, that $\psi = \phi_0^*$. With that, we have $$ \pmatrix{A_R & -A_I\\A_I & A_R} \pmatrix{x\\y} = (\phi_0 \circ A \circ \phi_0^*) (x + iy). $$ Note that for any orthogonal matrix $O$, $\phi = O^T \circ \phi_0$ induces the embedding $$ O^T\pmatrix{A_R & -A_I\\A_I & A_R}O $$ in the same way.