SU(2) in nLab
Proof
Consider the canonical linear basis
(1)4≃⟨{q 0,q 1,q 2,q 3}⟩ ℝ \mathbf{4} \;\simeq\; \big\langle \{q_0, q_1, q_2, q_3\}\big\rangle_{\mathbb{R}}
q 0,q 1,q 2,q 3∈ℍ q_0, q_1, q_2, q_3 \;\in\; \mathbb{H}
q 0=1, q iq i=−1=−q 0fori∈{1,2,3}, q σ(1)q σ(3)=sgn(σ)q σ(3),forσ∈Sym(3) \begin{aligned} & q_0 = 1\,, \\ & q_i q_i = -1 = - q_0 \;\; \text{for} i \in \{1,2,3\} \,, \\ & q_{\sigma(1)} q_{\sigma(3)} = \mathrm{sgn}(\sigma) q_{\sigma(3)} \,, \text{for} \sigma \in \mathrm{Sym}(3) \end{aligned}
Sp(1)≃{q∈ℍ|qq¯=1} \mathrm{Sp}(1) \;\simeq\; \{ q \in \mathbb{H} \;\vert\; q \bar q = 1 \}
is by left quaternion multiplication.
Consider then the following linear basis of 4∧4\mathbf{4}\wedge \mathbf{4}:
4∧4≃⟨{a 1 +:=q 0∧q 1+q 2∧q 3, a 2 +:=q 0∧q 2+q 3∧q 1, a 3 +:=q 0∧q 3+q 1∧q 2,a 1 −:=q 0∧q 1−q 2∧q 3, a 2 −:=q 0∧q 2−q 3∧q 1, a 3 −:=q 0∧q 3−q 1∧q 2}⟩ \mathbf{4}\wedge \mathbf{4} \;\simeq\; \left\langle \left\{ \array{ a_1^+ := q_0 \wedge q_1 + q_2 \wedge q_3, \\ a_2^+ := q_0 \wedge q_2 + q_3 \wedge q_1, \\ a_3^+ := q_0 \wedge q_3 + q_1 \wedge q_2, } \array{ a_1^- := q_0 \wedge q_1 - q_2 \wedge q_3, \\ a_2^- := q_0 \wedge q_2 - q_3 \wedge q_1, \\ a_3^- := q_0 \wedge q_3 - q_1 \wedge q_2 } \right\} \right\rangle
with induced 𝔰𝔭(1)\mathfrak{sp}(1) Lie algebra action given by
q i⋅(q j∧q k)=(q iq j)∧q k+q j∧(q iq k). q_i \cdot ( q_j \wedge q_k ) \;=\; (q_i q_j) \wedge q_k + q_j \wedge (q_i q_k) \,.
From this we find
q 1⋅a 1 ± =q 1⋅(q 0∧q 1±q 2∧q 3) =(q 1∧q 1⏟=0+q 0∧(−q 0)⏟=0)±(q 3∧q 3⏟=0+q 2∧(−q 2)⏟=0) =0 \begin{aligned} q_1 \cdot a_1^{\pm} & = \; q_1 \cdot \big( q_0 \wedge q_1 \pm q_2 \wedge q_3 \big) \\ & =\; \big( \underset{ = 0 }{ \underbrace{ q_1 \wedge q_1 } } + \underset{ = 0 }{ \underbrace{ q_0 \wedge (-q_0) } } \big) \pm \big( \underset{ = 0 }{ \underbrace{ q_3 \wedge q_3 } } + \underset{ = 0 }{ \underbrace{ q_2 \wedge (- q_2) } } \big) \\ & =\; 0 \end{aligned}
and
q 1⋅a 2 ± =q 1⋅(q 0∧q 2±q 3∧q 1) =(q 1∧q 2⏟=q 1∧q 2+q 0∧q 3⏟=q 0∧q 3)±((−q 2)∧q 1⏟=q 1∧q 2+q 3∧(−q 0)⏟=q 0∧q 3) ={2a 3 + 0 \begin{aligned} q_1 \cdot a_2^{\pm} & =\; q_1 \cdot \big( q_0 \wedge q_2 \pm q_3 \wedge q_1 \big) \\ & = \; \big( \underset{ = q_1 \wedge q_2 }{ \underbrace{ q_1 \wedge q_2 } } + \underset{ = q_0 \wedge q_3 }{ \underbrace{ q_0 \wedge q_3 } } \big) \pm \big( \underset{ = q_1 \wedge q_2 }{ \underbrace{ (- q_2) \wedge q_1 } } + \underset{ = q_0 \wedge q_3 }{ \underbrace{ q_3 \wedge (- q_0) } } \big) \\ & =\; \left\{ \begin{array}{l} 2 a_3^+ \\ 0 \end{array} \right. \end{aligned}
and
q 1⋅a 3 ± =q 1⋅(q 0∧q 3±q 1∧q 2) =(q 1∧q 3⏟=−q 3∧q 1+q 0∧(−q 2)⏟=−q 0∧q 2)±((−q 0)∧q 2⏟=−q 0∧q 2+q 1∧q 3⏟=−q 3∧q 1) ={−2a 2 + 0. \begin{aligned} q_1 \cdot a_3^{\pm} & =\; q_1 \cdot \big( q_0 \wedge q_3 \pm q_1 \wedge q_2 \big) \\ & = \; \big( \underset{ = - q_3 \wedge q_1 }{ \underbrace{ q_1 \wedge q_3 } } + \underset{ = - q_0 \wedge q_2 }{ \underbrace{ q_0 \wedge (-q_2) } } \big) \pm \big( \underset{ = - q_0 \wedge q_2 }{ \underbrace{ (- q_0) \wedge q_2 } } + \underset{ = - q_3 \wedge q_1 }{ \underbrace{ q_1 \wedge q_3 } } \big) \\ & =\; \left\{ \begin{array}{l} -2 a_2^+ \\ 0 \end{array} \right. \end{aligned} \,.
Since everything here is invariant under cyclic permutation of the three non-zero indices it follows generally that
(12q i)⋅a j +=∑kϵ ijka k +,(12q i)⋅a j −=0for alli,j∈{1,2,3} (\tfrac{1}{2} q_i) \cdot a_j^+ \;=\; \underset{k}{\sum} \epsilon_{i j k} a_k^+ \,, \;\; (\tfrac{1}{2} q_i) \cdot a_j^- \;=\; 0 \;\; \text{for all} i,j \in \{1,2,3\}
But this means that
⟨{a 1 +,a 2 +,a 3 +}⟩≃3,aa⟨{a i −}⟩≃1∈RO(Sp(1)). \big\langle \{a^+_1, a^+_2, a^+_3\} \big\rangle \;\simeq\; \mathbf{3} \,, \phantom{aa} \big\langle \{a^-_i\} \big\rangle \;\simeq\; \mathbf{1} \;\;\;\in \mathrm{RO}(\mathrm{Sp}(1)) \,.
Proof
Consider the following linear basis
∧ 34≃ ℝ⟨{b 0:=+q 1∧q 2∧q 3 b 1:=−q 0∧q 2∧q 3 b 2:=+q 0∧q 1∧q 3 b 3:=−q 0∧q 1∧q 2}⟩. \wedge^3 \mathbf{4} \;\simeq_{\mathbb{R}}\; \left\langle \left\{ \begin{array}{l} b_0 := + q_1 \wedge q_2 \wedge q_3 \\ b_1 := - q_0 \wedge q_2 \wedge q_3 \\ b_2 := + q_0 \wedge q_1 \wedge q_3 \\ b_3 := - q_0 \wedge q_1 \wedge q_2 \end{array} \right\} \right\rangle \,.
We claim that in terms of these basis elements and those of (1) the isomorphism is given by b 0↦q 0b_0 \mapsto q_0, b i↦q ib_i \mapsto q_i. This follows by direct inspection. For instance for the Lie algebra action of q 1q_1 we find:
q 1⋅b 0 =q 1⋅(q 1∧q 2∧q 3) =(−q 0)∧q 2∧q 3⏟=b 1+q 1∧q 3∧q 3⏟=0+q 1∧q 2∧(−q 2)⏟=0 =b 1 \begin{aligned} q_1 \cdot b_0 & \; = \; q_1 \cdot ( q_1 \wedge q_2 \wedge q_3 ) \\ & \; = \; \underset{ = b_1 }{ \underbrace{ (- q_0) \wedge q_2 \wedge q_3 } } \;+\; \underset{ = 0 }{ \underbrace{ q_1 \wedge q_3 \wedge q_3 } } \;+\; \underset{ = 0 }{ \underbrace{ q_1 \wedge q_2 \wedge (- q_2) } } \\ & \;=\; b_1 \end{aligned}
q 1⋅b 1 =q 1⋅(−q 0∧q 2∧q 3) =−q 1∧q 2∧q 3⏟b 0−q 0∧q 3∧q 3⏟=0−q 0∧q 2∧(−q 2)⏟=0 =−b 0 \begin{aligned} q_1 \cdot b_1 & \;=\; q_1 \cdot ( - q_0 \wedge q_2 \wedge q_3 ) \\ & \;=\; - \underset{ b_0 }{ \underbrace{ q_1 \wedge q_2 \wedge q_3 } } - \underset{ = 0 }{ \underbrace{ q_0 \wedge q_3 \wedge q_3 } } - \underset{ = 0 }{ \underbrace{ q_0 \wedge q_2 \wedge (-q_2) } } \\ & \;=\; - b_0 \end{aligned}
q 1⋅b 2 =q 1⋅(q 0∧q 1∧q 3) =q 1∧q 1∧(−q 2)⏟=0+q 0∧(−q 0)∧q 3⏟=0+q 0∧q 1∧(−q 2)⏟=b 3 =b 3 \begin{aligned} q_1 \cdot b_2 & \;=\; q_1 \cdot (q_0 \wedge q_1 \wedge q_3) \\ & \;=\; \underset{ = 0 }{ \underbrace{ q_1 \wedge q_1 \wedge (- q_2) } } + \underset{ = 0 }{ \underbrace{ q_0 \wedge (- q_0) \wedge q_3 } } + \underset{ = b_3 }{ \underbrace{ q_0 \wedge q_1 \wedge (- q_2) } } \\ & \;=\; b_3 \end{aligned}
q 1⋅b 3 =q 1⋅(−q 0∧q 1∧q 2) =−q 1∧q 1∧q 2⏟=0−q 0∧(−q 0)∧q 2⏟=0−q 0∧q 1∧q 3⏟=b 2 =−b 2 \begin{aligned} q_1 \cdot b_3 & \;=\; q_1 \cdot (- q_0 \wedge q_1 \wedge q_2) \\ & \;=\; - \underset{ = 0 }{ \underbrace{ q_1 \wedge q_1 \wedge q_2 } } - \underset{ = 0 }{ \underbrace{ q_0 \wedge (- q_0) \wedge q_2 } } - \underset{ = b_2 }{ \underbrace{ q_0 \wedge q_1 \wedge q_3 } } \\ & \;=\; - b_2 \end{aligned}