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

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – user7090 May 29 '19 at 4:54
  • 1
    $\begingroup$ 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 $k^* \times k$. $\endgroup$ – user7090 Jun 1 '19 at 22:19
  • 1
    $\begingroup$ 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? $\endgroup$ – user7090 Jun 8 '19 at 23:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.