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 $b\partial_z+(a-d)z\partial_z-cz^2\partial_z$.

  • $\begingroup$ 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. $\endgroup$ – Nadia May 29 '19 at 4:54
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    $\begingroup$ 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 $k$?). $\endgroup$ – Julian Rosen May 29 '19 at 15:07
  • $\begingroup$ 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 $k^* \times k$. $\endgroup$ – Nadia Jun 1 '19 at 22:19
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    $\begingroup$ 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 $\mathbb{G}_m\times\mathbb{G}_a$. $\endgroup$ – Julian Rosen Jun 2 '19 at 15:05
  • $\begingroup$ 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? $\endgroup$ – Nadia Jun 8 '19 at 23:07

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