What are the equations for $SL_3/SL_2$?

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

• 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. – LSpice Dec 28 '18 at 2:20
• @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. – user44191 Dec 28 '18 at 2:37
• @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.) – LSpice Dec 28 '18 at 4:13

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

• 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. – YCor Dec 28 '18 at 8:08