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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?

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  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 $$ U(1)_{1,2} \hookrightarrow SU(3) \to (\mathbf{CP}^2 \rtimes \mathbf{CP}^1) \times (S^1)_{1,2,3}, $$ $$ U(1)_{1,2} \times U(1)_{3} \hookrightarrow U(3) \to ((\mathbf{CP}^2 \rtimes \mathbf{CP}^1)\rtimes S^1). $$

p.s. The above is the tentative answer (so far), if I understand your question correctly.

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