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.



Lifting theorem for n operators
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.


