Lifting theorem for n operators

Question

I am aware of the following statement of the lifting theorem. For $i\in \{1,2\}$ let $B_i$ be a contraction on a Hilbert space $H_i$ and let $A_i$, acting on the Hilbert space $K_i$, be the minimal unitary dilation of $B_i$. Let $P_i$ be the orthogonal projection of $K_i$ onto $H_i$. Then an operator $X$ from $H_1$ to $H_2$ satisfies $B_2 X=X B_1$ if and only if there exists an operator $Y$ from $K_1$ to $K_2$ such that

$$A_2 Y=Y A_1, \quad \|X\|=\|Y\|, \quad P_2 Y P_1 = X P_1 \ .$$

I am looking for a similar theorem but for $i \in \{1,2,\dots n\}$. So now $X$ satisfies $n$ operator equations and I want a lift $Y$ of $X$ which will further satisfy $n$ equations. Any reference or suggestion is most welcome for the above question.

Answer 1

This will not happen in general:

Consider the example: $$C=\left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right], D =\left[\begin{matrix}0 & 1/\sqrt2 \\ 0 & 0\end{matrix}\right], \ and \ X = \left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right] $$ Then $CX = XC$ and $DX=XD$.

Now the minimal unitary dilation of $C$ is the bilateral shift $$C' = \left[\begin{matrix}\ddots &\ddots \\ & 0 & 1 \\ & & 0& 1 \\ &&0 & 0 & 1 \\ &&&&0 & \ddots \\ &&&&&\ddots \end{matrix}\right]$$ and the minimal unitary dilation of $D$ is $$D' = \left[\begin{matrix}\ddots &\ddots&1 \\ & 0 & 0 & 1/\sqrt2 & -1/\sqrt2 \\ & & 0& 1/\sqrt2 & 1/\sqrt2\\ &&0 & 0 & 0 & 1 \\ &&&&0&\ddots&\ddots \\ \end{matrix}\right]$$

Now suppose $X'$ is a dilation of $X$ such that $\|X'\| = \|X\|$ and $$ C'X' = X'C' \ and \ D'X' = X'D' $$ The only operators that commute with $C'$, the bilateral shift, are Toeplitz operators (constant (super/sub) diagonals) hence this forces $X'$ to be $C'$ since we cannot add any other super or sub diagonal nonzero entries without increasing the norm.

However, $C'D' \neq D'C'$ which is a contradiction.