# Orbit and stabilizer of the representation of multiplication decomposition of $U(3)$

We know that in SU(2), we have the multiplication of 2-dimensional representation (rep) decomposed as $$2 \times 2= 1+3 \tag{a}$$ where 1 is the singlet of SU(2). And 3 is the adjoint of SU(2) and vecto rep of SO(3). We can ask what is the orbit and stabilizer of each element.

1. For the 3 in the above eq. (a), we should have the base space $S^2$ as the orbit and the fiber $S^1$ as the stabilizer, with the following relations: $$S^1 \hookrightarrow S^3 \to S^2$$ $$\text{stabilizer}\hookrightarrow \text{total space} \to \text{orbit}$$ Naively, I write $$U(1) \hookrightarrow SU(2) \to SO(3),$$ as the SO(3) is the orbit that each object in $3$ can move around in the SO(3) space, while the stabilizer (a certain action of U(1)) makes the object invariant. The more proper way to write $SU(2)/U(1)=\mathbf{CP}^1$ as complex protective space. However, if we consider the total space as U(2), then the relations become: $$U(1) \times \mathbb{Z}_2 \hookrightarrow U(2) \to \frac{U(2)}{U(1) \times \mathbb{Z}_2},$$

2. For the 1 in the above eq. (a), which is a trivial representation of SU(2), thus we have the object invariant under the full SU(2), thus we have, $$SU(2) \hookrightarrow SU(2) \to pt,$$ the orbit is a single point. If we consider the full U(2) as the total space that can act on the SU(2) fundamentals, we have $$SU(2) \hookrightarrow U(2) \to U(1)/\mathbb{Z}_2,$$

What are the orbits and stabilizers of the right hand side objects in the multiplication of 3-dimensional representation of SU(3): $$3 \times 3= \bar{3}+6 \tag{b}$$ $$3 \times \bar{3}= 1+8 \tag{c}$$

question: What are the orbits and stabilizers of $\bar{3}$, $6$ and $1$, $8$ in the above decompositions, if we view the total space as SU(3) or U(3)?

Namely, what is $$\text{stabilizer}\hookrightarrow SU(3) \to \text{orbit},$$ $$\text{stabilizer}\hookrightarrow U(3) \to \text{orbit},$$ for each of $\bar{3}$, $6$ and $1$, $8$ in the above decomposition?

1. $$3 \times 3= \bar{3}+6 \tag{b}$$
(1). Given the $$\bar{3}$$ as the pair of U(3) or SU(3) fundamentals, we have the leftover invariant subgroup as $$SU(2) \times U(1)$$. For example, we can choose the $$\bar{3}$$ as the first 2-component out of 3-component of fundamentals, in order to form an anti-symmetric pair. In this case, we can write $$SU(2) \times U(1) =SU(2)_{1,2} \times U(1)_{3}$$ which indicates that the $$SU(2)$$ is rotating in the first 2-component subspace, and the $$U(1)$$ is acting on the 3rd component. We use a similar notations below. We then have $$SU(2)_{1,2} \hookrightarrow SU(3) \to \mathbf{CP}^2 \times (S^1)_{1,2,3},$$ $$SU(2)_{1,2} \times U(1)_{3} \hookrightarrow U(3) \to (\mathbf{CP}^2 \rtimes S^1) .$$ The notation $$\mathbf{CP}^2 \rtimes S^1$$ is meant to say that the $$\mathbf{CP}^2$$ has a nontrivial fibration by the $$S^1$$.
(2). Given the $$6$$ as the pair of U(3) or SU(3) fundamentals, we have the left over invariant subgroup as $$U(1) \times U(1)$$. For example, we can choose the $$6$$ as the first 2-component out of 3-component of fundamentals, in order to form an anti-symmetric pair. In this case, we can write $$U(1) \times U(1) =U(1)_{1,2} \times U(1)_{3}$$ which indicates that the $$U(1)_{1,2}$$ is rotating in the first 2-component subspace, and the $$U(1)_3$$ is acting on the 3rd component. We then have $$U(1)_{1,2} \hookrightarrow SU(3) \to (\mathbf{CP}^2 \rtimes \mathbf{CP}^1) \times (S^1)_{1,2,3},$$ $$U(1)_{1,2} \times U(1)_{3} \hookrightarrow U(3) \to ((\mathbf{CP}^2 \rtimes \mathbf{CP}^1)\rtimes S^1) .$$
2. $$3 \times \bar{3}= 1+8 \tag{c}$$ (1) Given the $$1$$ as the pair of U(3) or SU(3) fundamental and anti-fundamental, we have $$SU(3) \hookrightarrow SU(3) \to 1,$$ $$SU(3) \times \mathbb{Z}_2 \hookrightarrow U(3) \to S^1/\mathbb{Z}_6.$$
(2) Given the $$8$$ as the pair of U(3) or SU(3) fundamental and anti-fundamental, we have $$U(1)_{1,2} \hookrightarrow SU(3) \to (\mathbf{CP}^2 \rtimes \mathbf{CP}^1) \times (S^1)_{1,2,3},$$ $$U(1)_{1,2} \times U(1)_{3} \hookrightarrow U(3) \to ((\mathbf{CP}^2 \rtimes \mathbf{CP}^1)\rtimes S^1).$$