7
$\begingroup$

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 $U$.

$\endgroup$
7
  • 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? $\endgroup$ Feb 18 '17 at 11:27
  • 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. $\endgroup$ Feb 19 '17 at 13:48
  • 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. $\endgroup$ Feb 19 '17 at 14:03
  • $\begingroup$ 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. $\endgroup$
    – Benjamin
    Feb 19 '17 at 15:51
  • $\begingroup$ I mean exponentiate to the argmin of the original function of course. $\endgroup$
    – Benjamin
    Feb 19 '17 at 16:16
4
$\begingroup$

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).

$\endgroup$
4
  • $\begingroup$ I see. Very clarifying thanks. Does this relationship between $SU(2) \otimes SU(2)$, $SO(4)$ and $SU(4)$ have an analogue for higher $n$? $\endgroup$
    – Benjamin
    Feb 21 '17 at 16:59
  • $\begingroup$ @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 $n$. $\endgroup$ Feb 21 '17 at 17:43
  • $\begingroup$ When you say there is a factorization in principle, what do you mean? A factorization into two orthogonal matrices and a diagonal one? $\endgroup$
    – Benjamin
    Feb 22 '17 at 1:31
  • $\begingroup$ @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. $\endgroup$ Feb 25 '17 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.