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 $U$.
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1$\begingroup$ I don't understand your notion of 'closeness' since the function you give doesn't really measure a distance in any sense. You seem to be claiming that one can define a notion of distance on $\mathrm{SU}(m)$ using the function $d(g,h) = |\mathrm{Tr}(gh^{-1})|=|\mathrm{Tr}(gh^\dagger)|$. But, for example, if $\omega^m=1$, then $d(I,\omega\,I) = m = d(I,I)$, even though $\omega\,I$ is not close to $I$ when $\omega\not=1$. Wouldn't you rather use something like $d(g,h) = \|g-h\|$ instead? $\endgroup$– Robert BryantCommented Feb 18, 2017 at 11:27
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1$\begingroup$ A further caution is that 'the closest point' may well not be unique. Generally, if $G$ is a (connected) Lie group with closed subgroup $K$ and $d:G\times G\to[0,\infty)$ is a biïnvariant distance function, then there is a function $f:G\to[0,\infty)$ such that $f(g)$ is the infimum of the numbers $d(g,k)$ for $k\in K$. This function $f$ will satisfy $f(kg)=f(gk)=f(g)$ for all $k\in K$ so it descends to the space $K\backslash G/K$ (not usually a manifold). This function may be easier to compute than finding a closest point, which is probably best done by either Newton's method or gradient flow. $\endgroup$– Robert BryantCommented Feb 19, 2017 at 13:48
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1$\begingroup$ When $n=2$, $G=\mathrm{SU}(4)=\mathrm{Spin}(6)$ and $K=\rho_-(\mathrm{SU}(2))\rho_+(\mathrm{SU}(2))$, both of which contain the element $-I_4$, so the problem descends to $G=\mathrm{SO}(6)$ and $K=\mathrm{SO}(3){\times}\mathrm{SO}(3)$. In this case, $G/K$ is a symmetric space, so the quotient $K\backslash G/K$ is easy to understand; it's a ($3$-dimensional) tetrahedron and the descended function $f$ is probably relatively easy to compute. When $f$ (or, rather $f^2$, which, with your distance function, will be smooth) is small, Newton's method should work quite well to find the closest point. $\endgroup$– Robert BryantCommented Feb 19, 2017 at 14:03
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$\begingroup$ I've found that I can min $| \log(V_1 \otimes V_2) - \log(U) |$, I wondered if the argmin of this new function would exponentiate to the min of the original distance on the group? The min value is actually not important, only the minimizing $V_1 \otimes V_2$ is. $\endgroup$– BenjaminCommented Feb 19, 2017 at 15:51
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$\begingroup$ I mean exponentiate to the argmin of the original function of course. $\endgroup$– BenjaminCommented Feb 19, 2017 at 16:16
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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).
answered Feb 21, 2017 at 15:02
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$\begingroup$ I see. Very clarifying thanks. Does this relationship between $SU(2) \otimes SU(2)$, $SO(4)$ and $SU(4)$ have an analogue for higher $n$? $\endgroup$– BenjaminCommented Feb 21, 2017 at 16:59
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$\begingroup$ @Benjamin: In higher dimensions, it's more complicated. The factorization isn't as simple as (1) in the discussion of $n=2$, although there is, in principle, such a factorization, and the story is roughly similar. The case $n=2$ is especially nice because $\mathrm{SU}(4)/\mathrm{SO}(4)$ is a symmetric space. Unfortunately, that property fails for higher $n$. $\endgroup$ Commented Feb 21, 2017 at 17:43
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$\begingroup$ When you say there is a factorization in principle, what do you mean? A factorization into two orthogonal matrices and a diagonal one? $\endgroup$– BenjaminCommented Feb 22, 2017 at 1:31
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$\begingroup$ @Benjamin: Not in general. When $H\subset G$ is a closed subgroup of the compact, connected Lie group $G$, and $\frak{g} = \frak{h}\oplus\frak{m}$ is an $H$-invariant orthogonal direct sum decomposition of $\frak{g}$, there will be a subspace $\frak{d}\subset \frak{m}$ of minimal dimension that meets each $\mathrm{Ad}(H)$-orbit in $\frak{m}$. Then every $g\in G$ can be factored in the form $g = h_1e^\delta h_2$ with $h_i\in H$ and $\delta\in \frak{d}$. Unfortunately, this factorization doesn't have very good properties unless $(G,H)$ is a symmetric pair. $\endgroup$ Commented Feb 25, 2017 at 13:24