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 $SL_{n}/SL_{n - 1}$.)

  • $\begingroup$ I think that it can just be cut out of $\mathbb A^6$ by thinking of it as $\begin{pmatrix} a & 0 & c \\ 0 & 1 & f \\ g & h & i \end{pmatrix}$, with $a$ constrained so that the determinant is 1. $\endgroup$
    – LSpice
    Dec 28, 2018 at 2:20
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    $\begingroup$ @LSpice I don't think it can; consider the matrix $\begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$. Additionally, even on that set, the "constraint" isn't exactly "on $a$"; if the bottom-right 2 by 2 has determinant $0$, then $a$ doesn't affect the determinant. $\endgroup$
    – user44191
    Dec 28, 2018 at 2:37
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    $\begingroup$ @user44191, thanks; I knew that it was a good decision to leave it as a comment rather than an answer. (Maybe a better decision still would have been to wait to comment until after some computation.) $\endgroup$
    – LSpice
    Dec 28, 2018 at 4:13

1 Answer 1


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 $BM' = A$.

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 $SL_n/SL_{n-1}$.

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    $\begingroup$ In a more group-theoretic point of view: $SL(V)$ acts linearly on $V\oplus V^*$. If $\dim(V)\ge 3$, this action has exactly 5 orbits: $\{(0,0)\}$, $(V-\{0\})\oplus\{0\}$, $\{0\}\oplus (V^*-\{0\})$, the set of pairs $(v,\ell)$ such that $v\neq 0$, $\ell(v)=0$, and finally the the set of pairs $(v,\ell)$ such that $\ell(v)\neq 0$. For this latter orbit, the point stabilizer of $(v,\ell)$ preserves the decomposition $Kv\oplus \mathrm{Ker}(\ell)$, fixes $v$ and acts on the hyperplane as an element of determinant 1, so the orbit of $(v,\ell)$ is the required homogeneous space. $\endgroup$
    – YCor
    Dec 28, 2018 at 8:08

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