What is the relationship between the Lie algebras $su(4)$, $so(4)$ and $su(2)\oplus su(2)$ (if any)? I have read that $so(4)=su(2)\oplus su(2)$ but what is their relationship to $su(4)$?
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1$\mathfrak{su}(4)\cong \mathfrak{so}(6)$ has higher dimension, so is different from the others. See here. – Dietrich Burde May 14 '20 at 16:50
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Yes, $\mathfrak{su}(2)\bigoplus\mathfrak{su}(2)\simeq\mathfrak{so}(4,\Bbb R)$. This comes from the fact that, if you see $SU(2)$ as the group of quaternions with norm $1$ and if $\Bbb H$ is the space of quaternions, then the image of the map$$\begin{array}{ccc}SU(2)\times SU(2)&\longrightarrow&\operatorname{Aut}(\Bbb H)\\(q,r)&\mapsto&\left(\begin{array}{ccc}\Bbb H&\longrightarrow&\Bbb H\\h&\mapsto&qhr^{-1}\end{array}\right)\end{array}$$is isomorphic to $SO(4,\Bbb R)$ (it is the set of the norm-preserving automorphims of $\Bbb H$), and its kernel is discrete (it is equal to $\{\pm(e,e)\}$).
And $SO(4,\Bbb R)$ is a strict subgroup of $SU(4)$.
José Carlos Santos
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It also comes from the fact that Lie algebras of compact semisimple groups are classified by their root system, and the root system $A_1 \times A_1$ (of $\mathfrak{su}_2 \oplus \mathfrak{su}_2$) is $D_2$ (of $\mathfrak{so}_4$). Whereas $\mathfrak{su_4}$ has root system $A_3$, which is $D_3$, which naturally belongs to $\mathfrak{so}_6$, which explains the correspondence given in Dietrich Burde's comment's link. – Torsten Schoeneberg May 14 '20 at 17:04
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@TheQuantumMan Other than the fact that $\mathfrak{su}(2)$ is isomorphic to a subalgebra of $\mathfrak{su}(4)$, I am not aware of any. – José Carlos Santos May 14 '20 at 17:12