One more question about mapping quaternionic matrices into real matrices

Question

Real matrices that lie in the image of the inclusion homomorphism $\rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R)$ are called complex linear real matrices. It is easy to see that a real matrix is complex linear if and only if it commutes with $I = \rho_n(iI)$.

In analogy to this I am now studying the quaternionic inclusion $M_n(\mathbb H) \to M_{4n}(\mathbb R)$ using the inclusion $\psi_n: M_n(\mathbb H) \to M_{2n}(\mathbb C)$. If $i,j,k$ denote the unit quaternions then I want to find matrices $I$ and $J$ such that a real matrix is quaternionic linear if and only if it commutes with $I$ and $J$.

In $1$ dimension I tried $I=\rho_{2n} \circ \psi_n (iI)$ and $J=\rho_{2n} \circ \psi_n (jI)$ but the problem then is that $I^2 \neq -1$ and $J^2 \neq -1$.

Why does $I=\rho_{2n} \circ \psi_n (iI), J=\rho_{2n} \circ \psi_n (jI)$ not work in the quaternionic case? Is there an insightful geometric (or other) explanation? For the inclusion of complex matrices into real matrices setting $J=\rho_{2n} (iI)$ worked.

Edit For a definition of $\rho_n$:

define $\rho_n : M^n(\mathbb C) \to M^{2n}(\mathbb R)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -b_{ij} & a_{ij} \end{array}$ if $A_{ij}=(a_{ij} + i b_{ij})$

and $\color{blue}{\psi_n}: M^n(\mathbb H) \to M^{2n}(\mathbb C)$ as $A_{ij}\mapsto \begin{array}{cc} a_{ij} & b_{ij} \\ -\overline{b_{ij}} & \overline{a_{ij}} \end{array}$ if $A_{ij}=(a_{ij} + b_{ij}j)$

Edit 2 (in response to the anwer)

Let $$I = \rho(\color{blue}{\psi(i)})= \rho\left ( \begin{array}{cc} \color{blue}{i} & \color{blue}{0} \\ \color{blue}{0} & \color{blue}{-i} \end{array}\right)$$ Then $$ I = \left ( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array} \right )$$

so that

$$ I^2 = -1$$

Edit 3

After the discussion with Incnis Mrsi I calculated $J^2=I^2 =-1$ for $J=\rho_{2}(\psi_1(j))$ and $I=\rho_{2}(\psi_1(i))$. I am confused that this seems to work. The reason why I asked this quetion is the following passage in Tapp's matrix groups for undergraduates:

In particular,

why is $I\neq\rho_{2n}(\psi_n(i))$ and $J\neq\rho_{2n}(\psi_n(j))$ for $n>1$? (for $n=1$, apparently it works as I just verified).

Answer 1

First of all, if your representation of quaternionic matrices gives $I^2≠−1$, it indicates you made a mistake and need to check which 4 × 4 real (or 2 × 2 complex) blocks do you use. You can consult this Wikipedia discussion for working examples (although there are many possible choices).

Second, your analogy between ℂ and ℍ is flawed because of poor algebraic design. Complex numbers form a field, that requires all element to commute, specifically $$∀z∈ℂ: i\,z = z\,i\,,$$ that implies said commutation requirement for matrix representations.

The algebra of quaternions isn’t a (true) field, it is only a skew field (a division algebra). $i\,z = z\,i$ isn’t an identity anymore, it’s an equation of ℂ (a two-dimensional real subalgebra of ℍ). Consequently, real (or complex) representation of quaternions, or quaternionic matrices, will not generally commute with any of $I, J, K$. You have to use other kind of matrix equation, such as ones using complex conjugate transpose (look again at the en.WP link for some insights).

Answer 2

Since I've not the book cited in OP and I'm not sure to well understand the notation that was used, I begin by fixing notation. We have two homomorphisms: $$ \rho: \mathbb{C} \rightarrow M_2(\mathbb{R}) \qquad \rho(z)=\rho(a+ib)= \left[ \begin{array}{ccccc} a&b \\ -b &a \end{array} \right] $$ and: $$ \psi: \mathbb{H} \rightarrow M_2(\mathbb{C}) \qquad \psi(x)=\psi(x_0+\mathbf{i}x_1+ \mathbf{j}x_2+\mathbf{k}x_3)= \left[ \begin{array}{ccccc} x_0+ix_1&x_2 +ix_3 \\ -x_2+ix_3 &x_0-ix_1 \end{array} \right] = X_{(2)} $$ that are homomorphisms between rings, so that: $$ \rho(wz)=\rho(w)\rho(z) \qquad \psi(xy)=\psi(x)\psi(y) $$ (Note that in second case the order is important because $\mathbb{H}$ is a noncommutative ring.)

We can genaralize $\rho$ as: $$ \rho_2: M_2(\mathbb{C}) \rightarrow M_4(\mathbb{R}) $$ $$ \rho_2(A) = \rho_2 \left( \left[ \begin{array}{ccccc} z_{1,1}&z_{1,2} \\ z_{2,1} &z_{2,2} \end{array} \right] \right) = \left[ \begin{array}{ccccc} \rho(z_{1,1})&\rho(z_{1,2}) \\ \rho( z_{2,1}) &\rho(z_{2,2}) \end{array} \right] $$ and more general: $$ \rho_n: M_n(\mathbb{C}) \rightarrow M_{2n}(\mathbb{R}) $$ $$ \rho_n(A) = \rho_n \left( \left[ \begin{array}{ccccc} z_{1,1}& \cdots &z_{1,n} \\ \cdots \\ z_{n,1} & \cdots &z_{n,n} \end{array} \right] \right) =\left[ \begin{array}{ccccc} \rho(z_{1,1})&\cdots &\rho(z_{1,n}) \\ \cdots \\ \rho( z_{n,1})&\cdots &\rho(z_{2,n}) \end{array} \right] $$ and analogously we can do for $\psi$ to define $\psi_n$:

Now, working in n=1 for simplicity, we have the homomorphisms: $$ \rho_2 \circ \psi : \mathbb{H} \rightarrow M_4(\mathbb{R}) $$ so defined: $$ \begin{split} \rho_2 \circ \psi(x)=\rho_2(\psi(x))=\rho_2(X_{(2)}) &= \\ = \left[ \begin{array}{ccccc} \rho(x_0+ix_1)&\rho(x_2 +ix_3) \\ \rho( -x_2+ix_3) &\rho(x_0-ix_1) \end{array} \right] &=\\ = \left[ \begin{array}{ccccc} x_0&x_1&x_2 &x_3 \\ -x_1&x_0&-x_3 &x_2 \\ -x_2&x_3&x_0 &-x_1 \\ -x_3&-x_2&x_1 &x_0 \end{array} \right] = X_{(4)} \end{split} $$ note that the rows of this matrix are the ordered quadruples of the components of the quaternionic vector $$ \left[ \begin{array}{ccccc} 1x \\ \mathbf{i}x\\ \mathbf{j}x \\ \mathbf{k}x \end{array} \right] $$ so we adopt the useful notation: $$ \rho_2 \circ \psi(x)= \left[ \left[ \begin{array}{ccccc} 1x \\ \mathbf{i}x\\ \mathbf{j}x \\ \mathbf{k}x \end{array} \right] \right] $$ now it's easy to see that $$ \rho_2 \circ \psi(\mathbf{i})=I_{(4)}= \left[ \left[ \begin{array}{ccccc} \mathbf{i} \\ -1\\ -\mathbf{k} \\ \mathbf{j} \end{array} \right] \right] = \left[ \begin{array}{ccccc} 0&1&0 &0 \\ -1&0&0 &0 \\ 0&0&0 &-1 \\ 0&0&1 &0 \end{array} \right] $$ as in OP and, in the same way, we can find $J_{(4)}=\rho_2 \circ \psi(\mathbf{j})$ and $K_{(4)}=\rho_2 \circ \psi(\mathbf{k})$.

And, more important, we easily see that $$ \rho_2 \circ \psi(\mathbf{i}x)= \left[ \left[ \begin{array}{ccccc} 1\mathbf{i}x \\ \mathbf{i}\mathbf{i}x\\ \mathbf{j}\mathbf{i}x \\ \mathbf{k}\mathbf{i}x \end{array} \right] \right]= \left[ \left[ \begin{array}{ccccc} \mathbf{i}x \\ -x\\ -\mathbf{k}x \\ \mathbf{j}x \end{array} \right] \right] $$ is different from $$ \mathbf{i} \left(\rho_2 \circ \psi(x)\right)= \mathbf{i} \left[ \left[ \begin{array}{ccccc} 1x \\ \mathbf{i}x\\ \mathbf{j}x \\ \mathbf{k}x \end{array} \right] \right]= \left[ \left[ \begin{array}{ccccc} \mathbf{i}x \\ -x\\ \mathbf{k}x \\ -\mathbf{j}x \end{array} \right] \right] $$ but we have: $$ \rho_2 \circ \psi(\mathbf{i}x)=\rho_2\left[\psi(\mathbf{i})\psi(x)\right]=\rho_2\left(\psi(\mathbf{i})\right)\rho_2\left(\psi(x)\right) $$ so that the diagram in OP commutes if we take, on the right side, $I_{(4)}=\rho_2\left(\psi(\mathbf{i})\right)$. I think that, with a bit of work, we can extend this reasoning to the case $n>1$.

I hope this can be an answer to the question. I'm not sure in my understanding of the page of the book, but I suspect that there is a mistake in the matrices named $\mathcal{I}_4$ and $\mathcal{J}_4$.