How to describe $G/U$?

Question

Let $G=SL_2(\mathbb{C})$ and let $U = \{\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C}\}$. We have an action of $U$ on $G$ by right multiplication. By definition, $G/U$ is the set of all $U$-orbits under this action. How to describe $G/U$ as a set?

Take $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$. The orbit of $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$ is the set $$ \{ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C} \}. $$

But I don't know how to classify orbits in $G/U$. Thank you very much.

Edit: $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$ and $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right) \in G$ are in the same orbit. Indeed, Suppose that $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)$ = $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right)$. Then $x = (b'-b)/a$. We also have $cx+d=(cb'+ad-bc)/a=(cb'+1)/a$. Since $ad'-b'c=1$, $(cb'+1)/a=d'$. Therefore $x = (b'-b)/a$ is a solution of $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)$ = $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right)$. Hence $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$ and $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right) \in G$ are in the same orbit.

I think that $G/U = \{ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right): a, c \in \mathbb{C}, b,d \text{ are fixed }\}$. Is this correct? Thank you very much.

Answer 1

For any $a \ne 0,$ and $c \in \mathbb{C},$ the orbit of $\begin{pmatrix} a & 0 \\ c & a^{-1} \end{pmatrix}$ under $U$ gives you $$\Big\{\begin{pmatrix} a & ax \\ c & cx+a^{-1} \end{pmatrix}: \; x \in \mathbb{C}\Big\},$$ so these are all distinct orbits ($a$ and $c$ are uniquely determined by the left column) and run through all matrices in $SL_2(\mathbb{C})$ with nonzero upper-left element.

For any $c \ne 0,$ the orbit of $\begin{pmatrix} 0 & -c^{-1} \\ c & 0 \end{pmatrix}$ is $$\Big\{\begin{pmatrix} 0 & -c^{-1} \\ c & cx \end{pmatrix}: \; x \in \mathbb{C} \Big\},$$ so these are distinct orbits and run through all matrices in $SL_2(\mathbb{C})$ with left column $\begin{pmatrix} 0 \\ c \end{pmatrix}.$

In other words, the orbit is determined uniquely by the left column of the matrix

Answer 2

Consider the tautological action of $G:= SL_2(\mathbb C)$ on $\mathbb C^2$. It is transitive on the points of $\mathbb C^2 \setminus \{0\}$, and the stabilizer of the vector $(1,0)$ is precisely $U$.

So the quotient $G/U$ is naturally identified with $\mathbb C^2\setminus \{0\}$. As user165670 notes in their answer, the map is given explicitly by sending a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ to the vector $(a,c)$ (since this is the image of $(1,0)$ under the action of this matrix).