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The 2d irrep of $ SU_2 $ is just the defining irrep $$ SU_2:=\left\{ \begin{bmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha} \end{bmatrix} : |\alpha|^2+|\beta|^2=1 \right\} $$ The 3d irrep of $ SU_2 $ is the subgroup $$ SO_3(\mathbb{R}) $$ of $ U_3 $. This is the adjoint representation of $ SU_2 $.

I want to better understand the $ 4d $ irrep of $ SU_2 $. Is there some well known subgroup of $ U_4 $ that is the image of a $ 4d $ irrep of $ SU_2 $? For example can we think of the $ 4d $ irrep of $ SU_2 $ as a subgroup of $ SU_2 \otimes SU_2 $? Or perhaps as a subgroup of the $ 10 $ dimensional symplectic subgroup $ Sp_2 $ of $ U_4 $?

Another User
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Ian Gershon Teixeira
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1 Answers1

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In general, the $(n+1)$-dimensional irrep of $SU(2)$ is the $n^{th}$ symmetric power $S^n(\mathbb{C}^2)$, which can be described explicitly as the space of homogeneous polynomials of degree $n$ in two variables. So the $4$-dimensional irrep is $S^3(\mathbb{C}^2)$, with basis $\{ x^3, x^2 y, xy^2, y^3 \}$, where $\left[ \begin{array}{cc} \alpha & \beta \\ - \overline{\beta} & \overline{\alpha} \end{array} \right]$ acts via extending the action $x \mapsto \alpha x - \overline{\beta} y$ and $y \mapsto \beta x + \overline{\alpha} y$ by multiplication. (Here I'm using the convention that $x$ and $y$ correspond to the standard basis of $\mathbb{C}^2$. There's another convention where we consider polynomial functions on $\mathbb{C}^2$ rather than the symmetric power, which is the dual of this representation, but all these reps are self-dual anyway.)

This representation is quaternionic; to see this we use the fact that the defining representation $\mathbb{C}^2$ is quaternionic, so admits an antilinear map $J : \mathbb{C}^2 \to \mathbb{C}^2$ satisfying $J^2 = -1$. Then $J^{\otimes 3} : (\mathbb{C}^2)^{\otimes 3} \to (\mathbb{C}^2)^{\otimes 3}$ is again an antilinear map squaring to $-1$ and it induces a map with that same property on $S^3(\mathbb{C}^2)$. So the corresponding image of $SU(2)$ in $SU(4)$ is contained in a conjugate of $Sp(2)$ but not of $SO(4)$. Presumably the corresponding map $Sp(1) \to Sp(2)$ has some nice quaternionic description but I don't know it.

Qiaochu Yuan
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    Dmytro Yeroshkin wrote code for computing some of these representations in coordinates, just to see them. The map $Sp(1)\rightarrow Sp(2)$ sends an element $a+bj$ (with $a,b\in\mathbb{C})$ to $\begin{bmatrix} a^3 & b^3 & \sqrt{3}ab^2 & \sqrt{3}a^2b\ -\overline{b}^3 & \overline{a}^3 & -\sqrt{3}\overline{a}^2\overline{b} & \sqrt{3}\overline{a}\overline{b}^2\ \sqrt{3}a\overline{b}^2 & \sqrt{3}b\overline{a}^2 & |a|^2 \overline{a} -2|b|^2\overline{a} & |b|^2 \overline{b} - 2|a|^2\overline{b}\ -\sqrt{3}a^2\overline{b} & \sqrt{3}b^2\overline{a} & 2|a|^2b - |b|^2b & |a|^2a-2a|b|^2\end{bmatrix}.$ – Jason DeVito - on hiatus Sep 22 '22 at 03:09
  • (I think I typed the matrix correctly, but could have made a mistake with signs/complex conjugates. Any mistakes are mine, not Dmytro's. Also, I don't recall if one uses elements of the form $a+jb$ or $a+bj$....) – Jason DeVito - on hiatus Sep 22 '22 at 03:11
  • @JasonDeVito is this code publicly available? – Ian Gershon Teixeira Sep 22 '22 at 11:02
  • @IanGershonTeixeira: He didn't make it available to me, but I also didn't ask. Using Qiaochu's description, it shouldn't be too hard to implement in maple... – Jason DeVito - on hiatus Sep 22 '22 at 14:58
  • Do you have any idea what the normalizer of this $ SU_2 $ subgroup of $ SU_4 $ is? Is it just $ SU_2 \times SU_2 $? – Ian Gershon Teixeira Feb 26 '23 at 13:47
  • Another interesting perspective is that $ SU(2){irr} \subset Sp(2) \subset SU(4) $ is the lift of $ SO(3){irr} \subset SO(5) \subset SO(6) $ through the double cover $ SU(4) \to SO(6) $. Here $ SU(2){irr} $ is the image of the 4d irrep of $ SU(2) $ and $ SO(3){irr} $ is the image of the 5d irrep of $ SO(3) $ – Ian Gershon Teixeira Feb 26 '23 at 16:16
  • Does this imply in general that the even dimensional irreps of $ SU(2) $ are always symplectic? – Ian Gershon Teixeira Dec 14 '23 at 23:23