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 and, moreover, this doesn't include the case when $a_0 \neq 0$.

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?

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 $a^* \in k^*$ and $z \neq 0$. This is close but not exactly right and, moreover, this doesn't include the case when $a_0 \neq 0$.

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 and, moreover, this doesn't include the case when $a_0 \neq 0$.