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How is $PGL_2(\Bbb R)$ a scheme? Here is my thought process

  1. $GL_2(\Bbb R)=Spec(\Bbb{R}[w,x,y,z,q]/((wz-xy)q-1))$
  2. We want $PGL_2(\Bbb R)=GL_2(\Bbb R)/\Bbb{G}_m(\Bbb R)$ somehow.
  3. We can find $PGL_2(\Bbb R)$ as the open subset of $\Bbb RP^3$ $$\{[w:x:y:z]\in\Bbb RP^3\mid wz-xy\ne 0\}$$
  4. We can find $PGL_2(\Bbb R)$ as the closed subset of $\Bbb RP^5$ given by $$\{[w:x:y:z:q:a]\in\Bbb RP^5\mid (wz-xy)q-a^3=0\}$$
  5. Perhaps then we can conclude that $PGL_2(\Bbb R)$ is the scheme: $$Proj(\Bbb{R}[w,x,y,z,q,a]/((wz-xy)q-a^3)))$$

Is this correct? Or do I need to make sense of $PGL_2(\Bbb R)$ as a categorical quotient or something else?

user469689
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    $PGL_2(\Bbb R)$ is not a scheme but only the set (or group) of the $\Bbb R$-points of an affine group scheme denoted $PGL_2$. – Watson Oct 24 '18 at 06:48
  • Related: https://math.stackexchange.com/questions/1376597, https://math.stackexchange.com/questions/2312208, https://math.stackexchange.com/questions/979937, https://math.stackexchange.com/questions/1811373 – Watson Oct 24 '18 at 07:04

1 Answers1

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$PGL_n$ is given as a group-scheme by $\operatorname{Spec} \Bbb Z[x_{ij},\frac{1}{\det}]^{\Bbb G_m}$ with $1\leq i,j\leq n$, where the superscript denotes taking invariants. It is easy to check that the invariants are exactly the homogeneous degree-zero elements of the above ring.

This is equivalent to your statement (3), but your statements (4) and (5) aren't quite right. For example, $[0:0:0:0:1:0]$ is an element of (4) but obviously not an element in $PGL_2$.

KReiser
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