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We know that there is an isomorphism $\mathsf{SU}(2) \otimes \mathsf{SU}(2) \to \mathsf{SO}(4)$ given explicitly by $M \mapsto Q^\dagger M Q$ where $$ Q=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 0 & 0 & i \\ 0 & i & 1 & 0 \\ 0 & i & -1 & 0 \\ 1 & 0 & 0 & -i \end{bmatrix}. $$ Note that $Q \in \mathsf{U}(4)$.

Is there an isomorphism $\mathsf{SU}(2) \otimes \mathsf{SU}(2) \otimes \mathsf{SU}(2) \to G$, where $G$ is a subgroup of $\mathsf{SO}(8)$? If so, what is $G$? Also, is there a unitary $Q \in \mathsf{U}(8)$ such that the isomorphism is given by conjugation $M \mapsto Q^\dagger M Q$? If so, what is $Q$ explicitly as a matrix?

Eric Kubischta
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  • This tensor product notation is not standard in mathematics so you should define it. There is no such thing as a tensor product of Lie groups but I believe your notation means the following: if $G$ is a group of $n \times n$ matrices and $H$ is a group of $m \times m$ matrices then I guess $G \otimes H$ denotes the group of $nm \times nm$ matrices consisting of the tensor (Kronecker) products of matrices in $G$ and matrices in $H$. Is that correct? – Qiaochu Yuan Jan 25 '23 at 00:05
  • @QiaochuYuan Yes, sorry about that. I guess I am thinking about these as matrix groups. I am basically following the notation in this question: https://math.stackexchange.com/questions/4397313/so-4-mathbbr-and-su-2-otimes-su-2-subgroups-of-su-4 – Eric Kubischta Jan 25 '23 at 02:18
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    Perhaps one approach to this problem is to classify all the $ \mathfrak{su}_2 \oplus \mathfrak{su}_2 \oplus \mathfrak{su}_2 $ subalgebras of $ \mathfrak{so}_8 $ up to conjugacy. Then determine all the corresponding 9 dimensional connected subgroups of $ SO_8 $ and moreover find the size of the center of all these 9d groups (computer algebra. The group $ SU_2 \otimes SU_2 \otimes SU_2 $ that you are interested in has center of size $ 2 $ so if all the 9d groups you find have center of a different size then we can be sure $ SO_8 $ has no subgroup conjugate to $ SU_2 \otimes SU_2 \otimes SU_2 $. – Ian Gershon Teixeira Jan 27 '23 at 02:18

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