Let's say I have a unitary matrix $U$, how do I know if $U$ is the resultant of a tensor product of two other unitary matrices $U_1$ and $U_2$ (dim$\geq$2), such that $U_1\otimes U_2=U$?
More specifically I am concerned with the almost trivial case of where the dimensions of $U_1$ and $U_2$ are 2×2 ($U$ is 4×4).
Do I necessarily need to find the basis where $U$ is block diagonal? Or do I need to fully diagonalize $U$ and find common factors? Are there any other strategies to verify without doing these?
I tried to find some direction, but I do not know what is the right term here, is it reducibility? separability?
Best Answer
In the smallest interesting dimension you can use the exceptional morphism $\mathrm{SL}_2(\mathbb{C}) \times \mathrm{SL}_2(\mathbb{C}) \to \mathrm{SO}_4(\mathbb{C})$ obtained by viewing $\mathrm{SL}_2$ as a symplectic group, and using the fact that the tensor product of two vector spaces endowed with alternate forms is naturally endowed with a symmetric form. This morphism is onto and has kernel $\{(1,1),(-1,-1)\}$. I believe that the image of $\mathrm{SU}(2) \times \mathrm{SU}(2)$ is isomorphic to $\mathrm{SO}(4)$ (the compact Lie group). This may be deduced from $\mathrm{SU}(2) = \mathrm{SL}_2(\mathbb{C})^\sigma$ where $\sigma(g) = A \overline{g} A^{-1}$ and $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. Working out a basis of $\mathbb{C}^2 \otimes \mathbb{C}^2$ in which the image is the usual $\mathrm{SO}(4)$ is an exercise in Galois cohomology (for $\mathbb{C}/\mathbb{R}$). Choosing as basis of $\mathbb{C}^2 \otimes \mathbb{C}^2$ the family $(e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2)$ where $(e_1, e_2)$ is the standard basis of $\mathbb{C}^2$, I find that the image of the tensor product map $\mathrm{SU}(2) \times \mathrm{SU}(2) \to \mathrm{GL}_4(\mathbb{C})$ is $h \mathrm{SO}(4) h^{-1}$ where $$ \mathrm{SO}(4) = \{ g \in \mathrm{SL}_4(\mathbb{R}) \,|\, g^T = g^{-1} \}$$ and $$h = \begin{pmatrix} 1 & 0 & 0 & -i \\ 0 & i & 1 & 0 \\ 0 & i & -1 & 0 \\ 1 & 0 & 0 & i \end{pmatrix}.$$
So the image of $\mathrm{U}(2) \times \mathrm{U}(2)$ in $\mathrm{U}(4)$ is $\mathrm{U}(1) \cdot h \mathrm{SO}(4) h^{-1}$. To check if $g \in \mathrm{U}(4)$ is in the image, first find out if there exists $z \in \mathrm{U}(1)$ such that $z^{-1} h^{-1} g h$ has real entries (since $g \neq 0$ there is at most one such $z$ up to multiplication by $-1$), and if so check if $z^{-1} h^{-1} g h$ belongs to $\mathrm{SO}(4)$.
Edit: a few indications on how $h$ is found: $A \in \mathrm{SL}_2(\mathbb{C})$ corresponds to a $1$-cocycle $c: \mathrm{Gal}(\mathbb{C}/\mathbb{R}) \to \mathrm{PGL}_2(\mathbb{C})$ defined by $c(1) = 1$ and $c(\overline{\cdot}) = A$ (well, its image in $\mathrm{PGL}_2(\mathbb{C})$) because we have $A \overline{A} = A^2 = -1 \in \ker(\mathrm{SL}_2(\mathbb{C}) \to \mathrm{PGL}_2(\mathbb{C}))$. Similarly $A \otimes A \in \mathrm{SO}_4(\mathbb{C})$ (with the choice of basis above, this $\mathrm{SO}_4$ is the one for the symmetric matrix $S = \begin{pmatrix} & & & 1 \\ & & -1 & \\ & -1 & & \\ 1 & & & \end{pmatrix}$, which also happens to be $A \otimes A$) defines a $1$-cocycle $c': \mathrm{Gal}(\mathbb{C}/\mathbb{R}) \to \mathrm{SO}_4(\mathbb{C})$. Now by Hilbert 90 we know that any $1$-cocycle $\mathrm{Gal}(\mathbb{C}/\mathbb{R}) \to \mathrm{GL}_4(\mathbb{C})$ is a coboundary, in other words $A \otimes A$ may be written as $h \overline{h}^{-1}$ for some $h \in \mathrm{GL}_4(\mathbb{C})$. For the explicit computation we can use the standard proof of Hilbert 90: $x$ may be chosen of the form $\sum_{\tau \in \mathrm{Gal}(\mathbb{C}/\mathbb{R})} c'(\tau) \tau(a)$ for "generic" $a \in \mathrm{GL}_4(\mathbb{C})$. In fact we can make the computation slightly simpler by replacing $A \otimes A$ by $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ because $A$ is made up of this $2 \times 2$ matrix and its opposite. Finally, translating the above (split) $\mathrm{SO}_4$ for $S$ via the change of basis corresponding to $h$, we find the "usual" special orthogonal group for the symmetric matrix $I_4$.