# The Lie algebra of $PGL(2)$ and its correspondence to automorphisms of the projective line.

Let $$X$$ denote $$\mathbb{P}^1$$ and let $$\textbf{Aut}(X)$$ denote the functor sending an affine scheme $$A$$ over $$\mathbb{C}$$ to the group $$Aut_A(X \times A)$$ of automorphisms of $$X \times A$$ over $$A$$.

We know that $$\textbf{Aut}(X)$$ is representable by the Lie group $$PGL(2)$$.

Let us now consider $$\textbf{Aut}(X)(D)$$, where $$D=\mathbb{C}[\epsilon]/(\epsilon)^2$$. We have that $$\textbf{Aut}(X)(D)= Aut_D(X \times D)=PGL(2)(D)$$.

On the other hand, we know that the set of automorphisms of $$Aut_D(X \times D)$$ which restrict to the identity morphism on the fiber are given by $$Lie(Aut(X))=Lie(PGL(2))$$. These automorphisms locally look like $$z \mapsto z + \epsilon(a_0 + a_1 z + a_2 z^2)$$. (Somehow this corresponds to $$sl_{2}$$ but I don't know how exactly.)

But, these automorphism should also correspond to elements in $$PGL(2)(D)$$? Which is matrix with entries from $$D$$. I don't see how this is possible given the local form I have written down.

How do automorphisms of the type $$z \mapsto z + \epsilon(a_0 + a_1 z + a_2 z^2)$$ corresponds to $$PGL(2)(D)$$?

From the identification, $$\frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}=\big((1+a\epsilon)z+b\epsilon\big)\big(1-d\epsilon-c\epsilon z\big)=z+\epsilon\big( b+(a-d)z-cz^2 \big),$$ we have that, $$z\mapsto \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)},$$
which corresponds to, $$u \mapsto (1 + a \epsilon)v + b \epsilon u$$ $$v \mapsto (1 + d \epsilon) v + d \epsilon u$$
which is clearly corresponds to an element in $$PGL(2)(D)$$.