Consider $SL\_2$ embedded into $SL\_3$ as upper left block matrices. The quotient $SL\_3/SL\_2$ is an affine variety, as is any quotient of reductive groups. How does one describe $SL\_3/SL\_2$? What are the equations for it in some affine space?


(One can also pose the same question more generally for $SL\_{n}/SL\_{n - 1}$.)



What are the equations for $SL_3/SL_2$?
In general, $SL\_n/SL\_{n - 1}$ is isomorphic to an affine subvariety $X$ of $\mathbb{A}^{2n}$ with coordinate equations given by $\sum\_i x\_i y\_i = 1$.


The map $SL\_n/SL\_{n-1} \rightarrow X$ is given by: the coordinates $y\_i$ are given by the "last" vector (i.e. the last column), which is unaffected by $SL\_{n-1}$, and the coordinates $x\_i$ are given by the $(n-1)$-minors which ignore the last column (and the $i$th row), with some sign switching. Note that both of these are invariant, and so this is a well-defined map. The equation then comes from the equation for determinants from expansion by minors.


This map can also be reversed: given coordinates $x\_i, y\_i$ such that $\sum\_i x\_i y\_i = 1$, we can find a matrix $A \in SL\_n$ such that the last column of $A$ is $y\_i$, and such that the $(n-1)$-minors ignoring the last column are $x\_i$. Clearly, we can "fill in" the last column as-is, so we can focus on the minors, using the matrix $A\_{n-1}$ with $n - 1$ columns. At least one of the $x\_i$ must be nonzero. Fill in the rows of $A\_{n-1}$ other than $i$ with a diagonal matrix with first entry equal to $x\_i$, and the rest equal to $1$. We then only need to fill the $i$th row. But clearly the equations for each minor tell us one coordinate in that row - try it for yourself to see why. That "fills in" the coordinates, and each of the minors works, so we now have such a matrix $A$.


I claim that this map is well-defined - that it gives us the same class of $SL\_n/SL\_{n-1}$ no matter which nonzero $x\_i$ we choose. Assume we have some $B$ with the same $(n-1)$-minors ignoring the last column (and the same last column) as the $A$ constructed above. Then its $i$th $(n-1)$-submatrix must have determinant $x\_i$, which is nonzero, and so must be invertible. Therefore, there is a unique element $M \in SL\_{n-1}$ such that $B\_{n-1}M = A\_{n-1}$ for all rows but the $i$th row. But by checking the other minors, we can then see that $B\_{n-1}M = A\_{n-1}$ - again, try and see for yourself. Therefore, this map is well-defined. We therefore have that if $M'$ is the extension of $M$ by adding a single $1$-block to $M$, then $BM' = A$.




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Incorporating an idea from YCor's comment:


In general, a homogenous space $G/H$ is isomorphic to the $G$-orbit of a point with stabilizer $H$. As such, we only need to find an affine space $X$ with a $SL\_n$-action and a point $x \in X$ such that $SL\_{n-1}$ is the stabilizer of $x$. We can consider $SL\_{n-1}$ as the intersection of two subgroups of $SL\_n$: the subgroup that leaves the last column fixed (which consists of all matrices where the last column is $\begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix}$), and the subgroup that leaves the $(n-1)$-minors fixed (which consists of all matrices where the last row is $\begin{pmatrix} 0 & 0 & \dots & 1\end{pmatrix}$). Let $V$ be the natural representation of $SL\_n$, and $V^\*$ be its dual. The first of these subgroups can be considered the stabilizer of the vector $\vec{v} := \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} \in V$, while the second of the subgroups can be considered the stabilizer of the vector $\vec{v}^\* := \begin{pmatrix} 0 & 0 & \dots & 1\end{pmatrix} \in V^\*$. As such, $SL\_{n-1}$ is the intersection of two stabilizers, and therefore is the stabilizer of $(\vec{v}, \vec{v^\*}) \in V \oplus V^\*$. Correspondingly, its orbit is isomorphic to $SL\_n/SL\_{n-1}$. 


