Is $ SU_2 \otimes SU_2 $ conjugate to $ SO_4(\mathbb{R}) $ in $ SU_4 $?

Question

Let $ A,B $ be matrix groups (with entries in the same field). Then the tensor/Kronecker product $ A \otimes B $ is a matrix group and $$ \pi: A \times B \to A \otimes B $$ is a group homomorphism. Taking $ A=B=SU_2 $ we have a map $ \pi: SU_2 \times SU_2 \to SU_4 $ given by $$ (A,B) \mapsto A \otimes B $$ The only nontrivial element of the kernel is $ (-1,-1) $. So the image $ SU_2 \otimes SU_2 $ of $ \pi $ is a subgroup of $ SU_4 $ isomorphic to $$ SU_2 \times SU_2/ (-1,-1) \cong SO_4(\mathbb{R}) $$

Is $ SU_2 \otimes SU_2 $ conjugate to $ SO_4(\mathbb{R}) $ in $ SU_4 $? If so what is a matrix conjugating one to the other? (this part was unanswered for a while but is now answered in the update)

Since matrices in $ SU_2 $ have real trace then all matrices in $ SU_2 \otimes SU_2 $ have real trace. So it is at least plausible that $ SU_2 \otimes SU_2 $ and $ SO_4(\mathbb{R}) $ are conjugate.

Also what is the normalizer in $ SU_4 $ of $ SU_2 \otimes SU_2 $ / the normalizer in $ SU_4 $ of $ SO_4(\mathbb{R}) $? Certainly $ iI $ normalizes $ SO_4(\mathbb{R}) $ and $$ T= \zeta_8 \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$ normalizes $ SU_2 \otimes SU_2 $.

EDIT: Just to reiterate what Jason DeVito both $ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ are normalized by $$ T= \zeta_8 \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} $$ where $$ \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0& 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}=SWAP \in O_4(\mathbb{R}) $$ is the $ SWAP $ operator and $ \zeta_8 $ just normalizes the determinant. Both $ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ contain $ -I $ and both do not contain $ iI $. Since \begin{align*} (\zeta_8 SWAP)^2&=iI\\ (\zeta_8 SWAP)^4&=-I \end{align*} we can combine this with the results from https://arxiv.org/pdf/math/0605784.pdf to conclude that both $ SO_4(\mathbb{R}) $ and $ SU_2 \otimes SU_2 $ have cyclic $ 4 $ component group generated by $ \zeta_8 SWAP $.

Answer 1

Yes, $SU(2)\otimes SU(2)$ is conjugate to the usual $SO(4)$ in $SU(4)$.

There's probably a direct way to see it, but here's some theory. Suppose $G = G_1\times G_2$ is a product of compact Lie groups. Then an irreducible representation (irrep) of $G$ is a tensor product of irreps of $G_1$ and $G_2$. As $G$ is compact, we may assume it acts isometrically, so a rep of $G$ is simply a homomorphism $G\rightarrow U(n)$ for some $n$. If $G$ has no codimension one normal subgroups and it is connected, the $G\rightarrow U(n)$ factors through $SU(n)$: the composition $G\rightarrow U(n)\xrightarrow{\det} S^1$ has connected image, so is either trivial or surjective. If surjetive, the kernel is a codim 1 normal subgroup.

Now, let's specialize to the case $G = SU(2)\times SU(2)$. We are looking for $4$-dim reps of $G$. Such a rep is a sum of irreps. The irreps of $SU(2)$ are well known: there is a a unique irrep of each dimension $n\geq 1$. I'll use the notation $(n\times m)$ to refer to the tensor product of the $n$-dim irrep of $SU(2)$ with the $m$-dim irrep of $SU(2)$.

Then the possible $4$-dim reps of $SU(2)\times SU(2)$ are (up to permuting the factors):

1. $4\cdot (1\otimes 1)$
2. $2\cdot (1\otimes 1) + (2\otimes 1)$
3. $(1\otimes 1) + (3\otimes 1)$
4. $2\cdot(2\otimes 1)$
5. $(2\otimes 1) + (1\otimes 2)$
6. $(4\otimes 1)$
7. $(2\otimes 2)$.

Now, the reps given by $SU(2)\otimes SU(2)$ and $SO(4)$, when thought of as reps of $SU(2)\times SU(2)$, both have kernel precisely $\pm(1,1)$. In the above list, apart from entries $5$ and $7$, the kernel contains an entire factor of $SU(2)$. Entry $5$ is injective.

So, $SU(2)\otimes SU(2)$ and $SO(4)$ are isomorphic representations. By definition, then, there is a matrix $A\in Gl(4,\mathbb{C})$ where $A$ conjugates $SU(2)\otimes SU(2)$ to $SO(4)$. But can we take $A$ to be in $SU(4)$?

Yes, and here's how. First, we know that for any $k\in SO(4)$, that $$(A^{-1}ka)(\overline{A^{-1}kA})^t = I$$ since $ASO(4)A^{-1}= SU(2)\otimes SU(2) \subseteq SU(4)$. This equation can be rearranged to read $ k A\overline{A}^t \overline{k}^t = A\overline{A}^t$, which means that the $SO(4)$ action preserves the Hermiatian forms represented by $I$ and by $A\overline{A}^t$. But the $SO(4)$ rep is irreducible, so from this MSE question, we must have $A\overline{A}^t = \lambda I$ for some non-zero $\lambda \in \mathbb{C}$. As the diagonal entries of $A\overline{A}^t$ are of the form $\lambda =\sum_j |a_{ij}|^2\geq 0$, we conclude $\lambda$ must be a positive number. Then the matrix $B:=\frac{1}{\sqrt{\lambda}}A$ is in $U(4)$ and conjugates $SU(2)\otimes SU(2)$ to $SO(4)$. Finally, to find an example in $SU(4)$, use $C:= \overline{\mu} B$ where $\mu$ is any $4$th root of $\det(B)$.

I'm not sure what the normalizer is. I'll keep working on it and update this answer if I figure it out.

Answer 2

The $\mathbb{C}$-vector space isomorphism $\mathbb{C}^2\otimes\mathbb{C}^2\cong\mathbb{C}(2)$ (the latter being the space of $2\times2$ complex matrices) given by $u\otimes v\mapsto uv^T$ allows us to transport the $\mathrm{SU}(2)\otimes\mathrm{SU}(2)$ action over to $\mathbb{C}(2)$, with the action $(A\otimes B)X:=AXB^T$ respecting the Frobenius norm $\|X\|_F^2:=\mathrm{tr}(X^\dagger X)$.

We know $\mathbb{H}\cong\mathbb{C}^2$ as $\mathbb{C}$-vector spaces, and thus there is a norm-preserving algebra homomorphism $\mathbb{H}\hookrightarrow\mathbb{C}(2)$; the image is an isomorphic copy of $\mathbb{H}$ which I'll call $\overline{\mathbb{H}}$, and the image of $S^3$ is none other than $\mathrm{SU}(2)$. Thus, $\overline{\mathbb{H}}$ is a subrep of $\mathbb{C}(2)$, and moreover we can check $\mathbb{C}(2)\cong\overline{\mathbb{H}}\oplus i\overline{\mathbb{H}}$ is true as $\mathrm{SU}(2)\otimes\mathrm{SU}(2)$-reps. This allows us to identify $\mathrm{SU}(2)\otimes\mathrm{SU}(2)$ with the matrices arising from $\mathrm{SO}(\overline{\mathbb{H}})\cong\mathrm{SO}(4)$ within $\mathrm{SU}(4)$. (All isometries of $\mathbb{H}$ arise from bimultiplications by versors, so the same must be true of $\overline{\mathbb{H}}$ and $\mathrm{SU}(2)\otimes\mathrm{SU}(2)$.)

This is conjugate to the "obvious" copy $\mathrm{SO}(\mathbb{R}(2))$ of $\mathrm{SO}(4)$ in $\mathrm{SU}(4)$ via any $\mathbb{R}$-linear isometry $\overline{\mathbb{H}}\leftrightarrow\mathbb{R}(2)$ extended $\mathbb{C}$-linearly (since $\mathbb{C}(2)=\overline{\mathbb{H}}\oplus i\overline{\mathbb{H}}=\mathbb{R}(2)\oplus i\mathbb{R}(2)$ as $\mathbb{R}$-vector spaces) to an element of $\mathrm{U}(4)$ which in turn can be rescaled to be an element of $\mathrm{SU}(4)$ if need be.

This argument also shows how $\mathrm{SL}_2\mathbb{R}\otimes\mathrm{SL}_2\mathbb{R}$, $\,\mathrm{SU}(1,1)\otimes\mathrm{SU}(1,1)$, $\,\mathrm{SL}_2\mathbb{R}\otimes\mathrm{SU}(1,1)$ are all conjugate to the identity component $\mathrm{SO}(2,2)_0$ respectively within $\mathrm{SL}_4\mathbb{R}$, $\,\mathrm{SU}(2,2)$, $\,\mathrm{SL}_4\mathbb{C}$ respectively. Note $\mathrm{SL}_4\mathbb{R}$ and $\mathrm{SU}(2,2)$ are $\mathrm{Spin}(3,3)$ and $\mathrm{Spin}(4,2)$ respectively, and $\mathrm{SL}_2\mathbb{R}\sim\mathrm{SU}(1,1)$ are conjugate within $\mathrm{SL}_2\mathbb{C}$ via the Cayley transform (notice how we use them as Moebius transformations for the corresponding models of the hyperbolic plane).

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Now for the normalizer. If $g\in N_{\mathrm{U}(4)}(\mathrm{SO}(4))$ then $g\mathbb{R}^4$ must be an $\mathrm{SO}(4)$-subrep of $\mathbb{C}^4$, but all such subreps are of the form $\lambda\mathbb{R}^4$, so $g\mathbb{R}^4=\lambda\mathbb{R}^4$ for $\lambda\in S^1/S^0$, from which we get $\overline{\lambda}g\in\mathrm{O}(4)$ and therefore $N_{\mathrm{U}(4)}(\mathrm{SO}(4))=\mathrm{U}(1)\mathrm{O}(4)$. When we intersect with $\mathrm{SU}(4)$ and mod out by $\mathrm{SO}(4)$ we indeed get a $\mathbb{Z}_4$ generated cyclically by any coset represented by $\zeta_8R$ with $R\in\mathrm{O}(4)\setminus\mathrm{SO}(4)$.

(The fact an $\mathrm{SO}(4)$-subrep of $\mathbb{C}^4$ is of the form $\lambda\mathbb{R}^4$ is a special case of a more general fact: if $U\cong V^{\oplus m}$ is an isotypic rep of $G$ over a field $\mathbb{F}$, then $U\cong V\otimes\mathbb{F}^m$ with $\mathbb{F}^m$ trivial, and every $G$-subrep of $V\otimes\mathbb{F}^m$ is expressible as $V\otimes W$ for some $\mathbb{F}$-subspace $W\le\mathbb{F}^m$.)

Answer 3

Theorem 1 of Optimal Quantum Circuits for General Two-Qubit Gates states that the unitary change of basis matrix $$ \mathcal{M}= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & i & 0 & 0 \\ 0 & 0 & i & 1 \\ 0 & 0 & i & -1 \\ 1 & -i & 0 & 0 \end{bmatrix} $$ conjugates $ SO_4(\mathbb{R}) $ to $ SU_2 \otimes SU_2 $. Explicitly, for any $ U \in SO_4(\mathbb{R}) $ then $ \mathcal{M}U \mathcal{M}^{-1} \in SU_2 \otimes SU_2 $.

The matrix $ \mathcal{M} $ is the change of basis from standard basis to a Bell basis. This theorem is an interesting and short read for anyone interested in quantum information or just Lie groups. The argument proceeds by concocting a definite form that is preserved by local unitaries $ SU_2 \otimes SU_2 $, instantiating the claims in the answers below that such a form must exist.

Essentially the same claim is made in Theorem 1 of Nonlocal properties of two-qubit gates and mixed states and optimization of quantum computations about the matrix $$ Q=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 0 & 0 & i \\ 0 & i & 1 & 0 \\ 0 & i & -1 & 0 \\ 1 & 0 & 0 & -i \end{bmatrix} $$ which is the same change of basis matrix as $ \mathcal{M} $ except with the 4th 2nd and 3rd column vectors cyclicly permuted.

And even earlier than both of those references, the fact that changing between the standard basis and a particular Bell basis (in some early papers this particular Bell basis was called the magic basis, though now that would probably conflict with modern terminology about the totally unrelated concept of "magic states") conjugates $ SO(4) $ to $ SU(2) \otimes SU(2) $ is mentioned in Entanglement of a Pair of Quantum Bits , c.f. "The set of unitary transformations that are real when expressed in the magic basis (or real except for an overall phase factor) is the same as the set of transformations that act independently on the two qubits."