Closest point in $SU(n) \otimes SU(n)$ to $SU(n^2)$ What is the closest $V_1 \otimes V_2 \in SU(n)\otimes SU(n)$ in the  squared trace inner product to a given $U \in SU(n^2)$? I.e. minimize over $V_1, V_2$:
$\min_{V_1, V_2} | V_1 \otimes V_2 - U|$ in terms of a given $U$.
 A: Maybe an example will clarify things a bit:  If you think of $\mathrm{SU}(2)$ as the group of complex $2$-by-$2$ matrices of the form
$$
q = \begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix}
$$
such that $a\bar a + b \bar b = 1$, and you think of $\mathbb{C}^2\otimes\mathbb{C}^2=\mathbb{C}^4$ as the space of $2$-by-$2$ complex matrices, then the representation of $\mathrm{SU}(2)\times \mathrm{SU}(2)$ into $\mathrm{SU}(4)$ can be thought of as the action
$$
(q_1,q_2)\cdot m = q_1\,m\,q_2^\dagger = q_1\,m\,{q_2}^{-1}.
$$
This action preserves the $4$-dimensional real subspace $\mathbb{H}\subset \mathbb{C}^4$ consisting of matrices of the form
$$
p = \begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix},
$$
and, in fact, as is well-known, the above action of $\mathrm{SU}(2)\times \mathrm{SU}(2)$ on $\mathbb{H}$ is identical with the action of $\mathrm{SO}(4)$ acting on $\mathrm{H}=\mathbb{R}^4$.  
Thus, in the case $n=2$ of the OP's question, the subgroup being denoted by $\mathrm{SU}(2)\otimes\mathrm{SU}(2)\subset\mathrm{SU}(4)$ is just $\mathrm{SO}(4)\subset\mathrm{SU}(4)$.  The problem then is how to find 'the' (or rather, 'a') closest point in $\mathrm{SO}(4)$ to a given element of $\mathrm{SU}(4)$.
Now, as is known, any element $g\in\mathrm{SU}(4)$ can be factored as
$$
g = h_1\,\mathrm{e}^{i\delta}\,h_2\tag 1
$$
with $h_1, h_2\in \mathrm{SO}(4)$ and $\delta$ a real diagonal matrix with trace zero.  If $h_\delta\in\mathrm{SO}(4)$ is a closest element of $\mathrm{SO}(4)$ to $\mathrm{e}^{i\delta}\in\mathrm{SU}(4)$ (i.e., $|\mathrm{e}^{i\delta}-h_\delta|\le |\mathrm{e}^{i\delta}-h|$ for all $h\in\mathrm{SO}(4)$), then $h_1\,h_\delta\,h_2\in\mathrm{SO}(4)$ will be a closest point in $\mathrm{SO}(4)$ to $g = h_1\,\mathrm{e}^{i\delta}\,h_2$.
Unfortunately, $h_\delta$ cannot be chosen to be continuous with respect to $\delta$.  For example, if $\delta = \mathrm{diag}(t,-t,0,0)$ then, for $|t|<\pi/2$, one can show that $h_\delta = I_4$ is the closest point in $\mathrm{SO}(4)$ to $\mathrm{e}^{i\delta}$.  When $|t|=\pi/2$, there is a whole circle of points in $\mathrm{SO}(4)$ that are at minimum distance from $\mathrm{e}^{i\delta}$.  When $\pi/2<|t|\le \pi$, though, the closest point to $\mathrm{e}^{i\delta}$ in $\mathrm{SO}(4)$ is $h_\delta= \mathrm{diag}(-1,-1,1,1)$.
Meanwhile, for all $\delta$ sufficiently small (in the sense that $\mathrm{tr}(\delta^2)$ is sufficiently small), one has $h_\delta = I_4$ is the unique closest element in $\mathrm{SO}(4)$ to $\mathrm{e}^{i\delta}$, so, in that case, the mapping 
$$
g = h_1\,\mathrm{e}^{i\delta}\,h_2 \mapsto h_1h_2 = h(g)
$$
gives the (unique) closest point in $\mathrm{SO}(4)$ to $g$.  This takes care of an open set in $\mathrm{SU}(4)$ for which your problem has a stable solution, provided you know how to perform the factorization (1).
