How can one maximize the following function:


$ f(V) = || V \otimes V - U\_1 \otimes U\_2 ||$


where $U\_1, U\_2 \in SU(n)$ are given and we seek to maximize over $V \in SU(n)$.


Both the maximum value of $f$ and the $V$ which maximizes it are interesting.



Maximize inner product of a tensor of unitary matrices
$\newcommand{\diag}{\operatorname{diag}}%
$The maximum distance achieved is 2, except possibly when $n=2$. By left multiplying by the unitary $U\_1^{-1} \otimes U\_1^{-1}$, we are reduced to $U\_1 = I$ and $U\_2 = U$, a unitary. 


Rewrite $V\otimes V - I \otimes U = (V \otimes VU^{-1} - I)(I \otimes U)$. Since the latter factor is unitary, it suffices to find $V$ such that $-1$ is an eigenvalue of $V \otimes VU^{-1}$ [this will show that the norm is at least 2, but of course it cannot exceed 2 in any case], that is, there is an eigenvalue $a$ of $V$ and an eigenvalue $b$ of $VU^{-1}$ such that $ab = -1$.


Let $v$ be an eigenvector for $U$, with corresponding eigenvalue $\lambda$ (obviously of modulus one), and let $w$ be orthogonal to $v$. By Gram-Schmidt, we can find $V$ in SU($n$) such that $Vv = \lambda v$ and $Vw = -w$ (here is where we use $n \geq 3$; for $n = 2$, all we can get is a unitary, not necessarily of determinant one). Then $v$ is an eigenvector for $VU^{-1}$, with eigenvalue $1$, and $-1$ is an eigenvalue of $V$, so $V \otimes VU^{-1}$ has $-1$ as an eigenvalue, and we are done.


For the case that $n=2$, I imagine there is a brute force argument; we can diagonalize $U$, and try to find the corresponding $V$ so that the set of products of the eigenvalues includes $-1$. If the result fails, it shouldn't be too difficult to optimize.


*Edit*: Actually, the $n=2$ case is easier than I thought. Reduce to $U = \diag(a,1/a)$, find $d$ in the unit circle such that $d^2 = -a$, and set $V = \diag (d, 1/d)$. Then the eigenvalues of $VU^{-1}$ are $d/a$ and $a/d$, so among the products are $d^2/a = -1$. 


So in all cases, we can find optimal $V$ that commutes with $U = U\_1^{-1}U\_2$.


