Projective Special Linear Group is an Linear Algebraic Group

Question

So I was wondering why the group $PSL(2,K)$ is a linear algebraic group, in the case that the characteristic of $K$ is not equal to $2$.

Actually there is a description of $PSL(2,K)$, namely:

$PSL(2,K) = SL(2,L)/C,$

whereas $C$ is the center of $SL(2,K)$, namely the subrgoup $x \cdot I$, with $I$ the identity and $x$ any solution of $x^2 = 1$.

We can embedd $SL(2,K)$ in the affine space $\mathbb{A}^4$ where the coordinate ring is just $K[x_1,x_2,x_3,x_4]/(x_1x_3 - x_2x_4 - 1)$. We can simply extend the action of $C$ on the affine space $\mathbb{A}^4$. But I dont know houw to embedd $\mathbb{A}^4/C$ in some bigger affine space, s. th. we get a description of $PSL(2,K)$ i.e. s. th. $PSL(2,K)$ is itself an linear algebraic group.

I would appreciate some hints.

Answer 1

The general result is that the quotient of a linear algebraic group by a closed normal subgroup is a linear algebraic group. Surprisingly this doesn't appear to be in Humphries, but it's definitely in Springer. The proof doesn't work by quotienting all of the ambient affine space, though; IIRC the idea of all these results is to let a group act on a ring of functions, extract a finite-dimensional faithful subrepresentation, and use that to embed the group as a subgroup of some $GL_n$.

Answer 2

I haven't checked everything, but I think the following works.

Let $\{x_1,x_2\}$ be a basis for the space that $SL_2(K)$ acts on naturally. That is $$ x_1=\binom10,\qquad x_2=\binom01, $$ and $$ A=\left(\begin{array}{rr}a&b\\c&d\end{array}\right) $$ acts as $A.x_1=ax_1+cx_2$, $A.x_2=bx_1+dx_2$.

The group $SL_2(K)$ then also acts on the 3-dimensional space of homogeneous quadratic polynomials, spanned by $\{x_1^2,x_1x_2,x_2^2\}$. The action goes by the above linear substitutions, so for example $A.x_1^2=(ax_1+cx_2)^2$ et cetera. This gives a homomorphism from $f:SL_2(K)\to GL_3(K)$ - a 3-dimensional representation of $SL_2(K)$ if you wish.

The point is that the matrix $-I_2$ is in the kernel of $f$. Because the center is the only non-trivial normal subgroup of $SL_2$ we have, in fact, $\ker f=\langle -I_2\rangle$.

So the remaining task is to show that $\operatorname{Im} f\cong PSL_2(K)$ is an algebraic subgroup of $GL_3(K)$.

Answer 3

Despite being 6 years too late to answer the question, I really enjoyed going through the details of Jyrki's suggestion, and I would like to share my results here.

Let $\mathbb{k}$ be any base field of characteristic $\neq 2$ (actually, any base ring containing $1/2$ would do). Jyrki's group action leads to a homomorphism $\varphi \colon \mathrm{SL}_2 \to \mathrm{SL}_3$ $$ \left( \begin{array}{cc} a & b\\ c & d \end{array} \right) \mapsto \left( \begin{array}{cc} a^2 & ab & b^2\\ 2ac & ad+bc & 2bd\\ c^2 & cd & d^2\end{array} \right) $$ of algebraic groups over $\mathbb{k}$. The kernel of $\varphi$ is canonically isomorphic to $\mu_2$ (the algebraic group given by the equation $x^2=1$). My aim is to find the image of $\varphi$ as an algebraic group in the naive sense, i.e. the smallest functor of subgroups of $\mathrm{SL}_3$ given by polynomial equations containing the set theoretic image of $\varphi$.

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As a possible candidate for the image of $\varphi$, consider the functor $G \colon \mathbb{k}\text{-}\mathrm{Alg} \to \mathrm{Set}$ sending any $\mathbb{k}$-algebra $A$ to the subset of ($3\times3$)-matrices $(x_{ij}) \in A^{3 \times 3}$ satisfying the following equations.

$$ \begin{align} x_{11} x_{13} &= x_{12}^2 \tag{1}\label{eq:1}\\ x_{31} x_{33} &= x_{32}^2 \tag{2}\label{eq:2}\\ x_{11} x_{33} - 2x_{12}x_{32} + x_{13}x_{31} &= 1 \tag{3}\label{eq:3}\\ 2(x_{32} x_{11} - x_{31}x_{12}) &= x_{21} \tag{4}\label{eq:4}\\ 2(x_{33} x_{12} - x_{32}x_{13}) &= x_{23} \tag{5}\label{eq:5}\\ 2(x_{32} x_{12} - x_{31}x_{13}) &= x_{22}-1 \tag{6}\label{eq:6}\\ 2(x_{33} x_{11} - x_{32}x_{12}) &= x_{22}+1 \tag{7}\label{eq:7}\\ \end{align} $$

It is easy to check that the matrices produced by $\varphi$ all satisfy \eqref{eq:1} - \eqref{eq:7}, so $\varphi$ gives rise to a natural transformation $\varphi' \colon \mathrm{SL}_2 \to G$ of functors. The hard part is to show that these equations imply all polynomial equalities satisfied by those matrices. For example, it is easy to see that all matrices produced by $\varphi$ have determinant one, but how does this follow from \eqref{eq:1} - \eqref{eq:7}? It is also unclear at the moment whether $G$ is a group functor, i.e. whether matrices satisfying \eqref{eq:1} - \eqref{eq:7} are closed under multiplication and inversion.

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All those questions can be answered at once by the following "surjectivity" condition.

Claim: For any $\mathbb{k}$-algebra $A$ and for any matrix $X \in G(A)$ there is an algebra extension $B/A$ such that $X$ (regarded as an element of $G(B)$) can be pulled back to $\mathrm{SL}_2(B)$.

Proof: Let $A$ be any $\mathbb{k}$-algebra, and let $X = (x_{ij}) \in G(A)$ be some element. We construct the extension algebra $B$ in two steps: First, we define an $A$-module M, then we pick two elements $a,b \in \mathrm{End}_A(M)$ and define $B$ as the $A$-sub-abgebra generated by $a$ and $b$. Finally, we pick certain elements $c,d \in B$ such that $ad-bc=1$ and $\varphi \left( \begin{array}{cc} a&b\\c&d \end{array}\right) = X.$

We define $M$ by 3 generators and 2 relations: $$ M = \left\langle \hat{1}, \hat{a}, \hat{b} \mid x_{12}\hat{a}=x_{11}\hat{b}, x_{13}\hat{a}=x_{12}\hat{b} \right\rangle $$ The endomorphisms $a,b \in \mathrm{End}_A(M)$ are defined by their actions on the generators: $$ a \colon \quad \hat{1} \mapsto \hat{a}, \quad \hat{a} \mapsto x_{11}\hat{1}, \quad \hat{b} \mapsto x_{12} \hat{1} $$ $$ b \colon \quad \hat{1} \mapsto \hat{b}, \quad \hat{a} \mapsto x_{12}\hat{1}, \quad \hat{b} \mapsto x_{13} \hat{1} $$ Checking $a$ and $b$ are well-defined is immediately done by looking at equation $\eqref{eq:1}$. It is easy to check (again, by looking at the generators of $M$) that $$ a^2 = x_{11}, \quad b^2 = x_{13}, \quad ab = ba = x_{12} \tag{*} \label{eq:b_rels} $$ holds.

Let $B$ be the (commutative!) $A$-sub-algebra of $\mathrm{End}_A(M)$ generated by $a$ and $b$. We define the elements $c, d \in B$ by $$ c = x_{32}a - x_{31}b, \quad d = x_{33}a - x_{32}b. $$

Now $c^2 = x_{31}$ follows by $\eqref{eq:2}$ and $\eqref{eq:3}$, $ac = x_{21}/2$ follows by $\eqref{eq:4}$, and $bc = (x_{22}-1)/2$ follows by $\eqref{eq:6}$.

Likewise, $d^2 = x_{33}$ follows by $\eqref{eq:2}$ and $\eqref{eq:3}$, $ad = (x_{22}+1)/2$ follows by $\eqref{eq:7}$, and $bd = x_{23}/2$ follows by $\eqref{eq:5}$.

Finally, $cd = x_{32}$ follows by $\eqref{eq:2}$ and $\eqref{eq:3}$, and $ad+bc = x_{22}$ as well as $ad-bc=1$ are immediate.

In summary, we have found a matrix $\left( \begin{array}{cc} a&b\\c&d \end{array}\right) \in \mathrm{SL}_2(B)$ sent to $X \in G(B)$ by $\varphi'$. $\square$

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Let's see how this property answeres the previous questions. To begin with, let's see why all matrices in $G$ have determinant one.

Let $A$ be any $\mathbb{k}$-algebra, and let $X \in G(A)$. By "surjectivity" we can extend $A$ to an algebra $B$ such that $X$ regarded as an element of $G(B)$ can be written as $X = \varphi'(X')$ for some matrix $X' \in \mathrm{SL}_2(B)$. Since we already know that all matrices produced by $\varphi'$ have determinant one, the same must be true for $X$ (even as an element of $G(A)$).

Exactly the same argument shows that all polynomial equations satisfied by the set theoretic image of $\varphi$ are also satisfied by the elements of $G$. In this way, $G$ is "as small as possible".

Analogously, we can show that $G$ is an algebraic group. Let $A$ be any $\mathbb{k}$-algebra, and let $X, Y \in G(A)$ arbitrary matrices. Of course the product matrix $X \cdot Y$ has coefficients in $A$, so in order to prove $X \cdot Y \in G(A)$, we only have to show that this product satisfies the equations $\eqref{eq:1} - \eqref{eq:7}$. This is easily done by "surjectivity": We may extend $A$ (twice if necessary) to an algebra $B$ and pick matrices $X', Y' \in \mathrm{SL}_2(B)$ such that $\varphi'(X') = X$ and $\varphi(Y') = Y$. Therefore, $XY = \varphi'(X'Y')$ satisfies all defining equations for $G$, whence $XY \in G(A)$. In the same way we conclude $X^{-1} \in G(A)$.

In summary, we see that $\varphi$ factors into a composition $$ \mathrm{SL}_2 \to G \to \mathrm{SL}_3, $$ and $G$ is the smallest closed subgroup of $\mathrm{SL}_3$ with that property.

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Using the basics of the theory of algebraic quotient groups, one can easily show that $G$ (together with $\varphi')$ is the quotient of $\mu_2 \to \mathrm{SL}_2$. However, our construction shows that it would be a bad idea to call this quotient $\mathrm{PSL}_2$, since $\varphi'$ is not surjective for all fields $\mathbb{K}$. More concretely, $G(\mathbb{K})$ in general contains strictly more elements than $\mathrm{PSL}_2(\mathbb{K})$. For example, the matrix $$ \left(\begin{array}{cc} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1 \end{array}\right) \in G(\mathbb{Q}) $$ is not in the image of $\varphi' \colon \mathrm{SL}_2(\mathbb{Q}) \to G(\mathbb{Q})$, and therefore does not correspond to an element of $\mathrm{PSL}_2(\mathbb{Q})$. (However, it corresponds to an element of $\mathrm{PSL}_2(\mathbb{Q(i)})$, as it can be written by $ \varphi' \left(\begin{array}{cc} i & 0\\ 0&-i \end{array}\right)$). Surprisingly, it turns out that $G$ is isomorphic to $\mathrm{PGL}_2$!