# Lie Algebra of Automorphism Group of $\mathbb{P}_k^1$

Let $$X$$ be a scheme over an algebraically closed field $$k$$ and let $$\operatorname{Aut}(X)$$ denote the functor sending a $$k$$-scheme $$T$$ to the group $$\operatorname{Aut}_T(X \times_k T)$$ of automorphisms of $$X \times_k T$$ over $$T$$.

My goal is to have a better grasp of the equality $$\operatorname{Lie}(\operatorname{Aut}(X))= H^0(X, \mathcal{T} X)$$.
Therefore I am trying to work through the example where $$X = \mathbb{P}_k^1$$ so that $$\operatorname{Aut}(X)= PGL(2,k)$$.

The global sections of $$\mathcal{T} X$$ are are of the form $$a_0 \partial_z + a_1 z \partial_z+ a_2 z^2 \partial_z$$ where $$z=v/u$$ is a choice of homogeneous coordinates on $$X$$. On the other hand, I know every $$\phi \in \operatorname{Aut}_k(X)$$ is given by the following map of $$k$$-algebras. $$z \mapsto \frac{az+b}{cz + d}.$$

What is the identification between the global sections of $$\mathcal{T} X$$ and the $$k$$-algebra maps $$z \mapsto \frac{az+b}{cz + d}$$?

Solving for the integral curve I end up with the equation $$z'(t) = a_0 + a_1 z(t) + a_2z^2(t)$$. If $$a_0=0$$, this would be a Bernoulli differential equation and I can solve it to find $$z(t)= \frac{a_1 z_0 e^{a_1 t}}{a_1 -a_2 z_0 e^{a_1 t}}$$. I think that this corresponds to the $$k$$-algebra map $$z \mapsto (a_1 a^*) z /(a_1 - a_2 a^* z )$$ where $$a^* \in k^*$$. This is close but not exactly right.

However, the affine subset itself has automorphisms given by $$z \mapsto \alpha + \beta z$$. If I compose these maps with the maps $$z \mapsto (a_1 a^*) z /(a_1 - a_2 a^* z )$$ I get from integrating the tangent space I do get the Mobius tranformation. Is this the correct approach?

Elements of $$\operatorname{Lie}(\operatorname{Aut}(X))$$ are not $$k$$-algebra maps, but rather maps over the ring $$k[\epsilon]/(\epsilon^2)$$ that reduce to the identity $$k$$-algebra map under $$\epsilon\mapsto 0$$. For $$X=\mathbb{P}^1$$, such maps can be identified with elements of the kernel of $$PGL_2(k[\epsilon]/(\epsilon^2))\to PGL_2(k)$$, or in other words maps $$z\mapsto \frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)},$$ with $$a$$, $$b$$, $$c$$, $$d\in k$$. We can compute $$\frac{(1+a\epsilon)z+b\epsilon}{c\epsilon z + (1+d\epsilon)}=\big((1+a\epsilon)z+b\epsilon\big)\big(1-d\epsilon-c\epsilon z\big)=z+\epsilon\big( b+(a-d)z-cz^2 \big),$$ and the vector field corresponding to the map above is $$b\partial_z+(a-d)z\partial_z-cz^2\partial_z$$.
• Thank you for the answer. How do I now use this to write down the automorphisms of $X$? It seems like we set $t=\epsilon$ and then evaluate at $t=1$, i.e something akin to the exponential map. – user7090 May 29 '19 at 4:54
• In general, I'm not sure you can get an automorphism of $X$ from a global vector field. Setting $\epsilon=1$ won't work because we need $\epsilon^2=0$. Exponentiation is a transcendental operation, and might not make sense algebraically (e.g. the exponential of $z\partial_z$ wants to be multiplication by $e$, but why should $e$ be an element of $k$?). – Julian Rosen May 29 '19 at 15:07
• Thanks again. Do you know why the equality $\operatorname{Lie}(G)=H^0(X, \mathcal{T}X)$ doesn't hold when $X= \mathbb{A}_k^1$ ? Here, global sections are of the form $f(z) \partial_z$ which is an infinite dimensional $k$ vector space. However, $G= \operatorname{Aut}(X)$ is only two-dimensional, I think isomorphic to $k^* \times k$. – user7090 Jun 1 '19 at 22:19
• The map $\mathbb{G}_m\times\mathbb{G}_a\to \operatorname{Aut}(\mathbb{A}^1)$ is a monomorphism which is an isomorphism on $k$ points, but this map is not an isomorphism because, for example, $x\mapsto x+\epsilon x^2$ is an automorphism of $\mathbb{A}^1\times\operatorname{Spec}\, k[\epsilon]/\epsilon^2$ that does not come from a $k[\epsilon]/\epsilon^2$ point of $\mathbb{G}_m\times\mathbb{G}_a$. – Julian Rosen Jun 2 '19 at 15:05
• Now $\operatorname{Lie}(\operatorname{PGL}(2,k))= sl_{2,k}$ is isomorphic to the group of automorphisms of the trivial deformation $\mathcal{X}$ of $\mathbb{P}_k^1$ over $k[\epsilon]$. Locally automorphism of $\mathcal{X}$ are given by $z \mapsto z + \epsilon(a_0 + a_1z + a_2 z^2)$. Unfortunately, there is no obvious global description of these automorphism. However, knowing that $\operatorname{Aut}(\mathcal{X}) \cong sl_{2,k}$, is there a way to determine $\operatorname{Aut}(\mathcal{X})$ from the given local description? – user7090 Jun 8 '19 at 23:07