Let $\mathcal{P}_3$ denote the space of homogeneous cubics in three variables, and let $\mathcal{H}_3$ denote the (seven-dimensional) subspace of harmonic cubics, an irreducible $\mathrm{SO}(3)$-module. Let $H : \mathcal{P}_3 \to \mathcal{H}_3$ denote the orthogonal projection map. Then it turns out that any unit norm $p \in \mathcal{H}_3$ can be factorized as $p = H( \ell_1 \ell_2 \ell_3),$ where the $\ell_i$ are linear functions on $\mathbb{R}^3$ of unit norm. The linear factors $\ell_i$ are unique up to permutation and replacing two of them at a time by their negatives. Denote this equivalence relation by $\sim.$
The resulting map $(S^2 \times S^2 \times S^2) / {\sim} \to S^6$ is $\mathrm{SO}(3)$ equivariant, where $\mathrm{SO}(3)$ acts diagonally on $(S^2 \times S^2 \times S^2) / {\sim}.$ Therefore, to find an element $v \in \mathcal{H}_3$ with trivial $\mathrm{SO}(3)$-stabiliser, it suffices to find a collection of three points on $S^2$ for which there are no nontrivial $\mathrm{SO}(3)$ elements sending these elements to another member of their equivalence class under $\sim.$ The generic triple of points on $S^2$ should fit this requirement.
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):
- $4\cdot (1\otimes 1)$
- $2\cdot (1\otimes 1) + (2\otimes 1)$
- $(1\otimes 1) + (3\otimes 1)$
- $2\cdot(2\otimes 1)$
- $(2\otimes 1) + (1\otimes 2)$
- $(4\otimes 1)$
- $(2\otimes 2)$.
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.
Best Answer
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.