# Dimension of the full automorphism

Let $\mathbb P_1$ be the one dimensional complex projective space. What is the connected component of the full automorphism of $\mathbb C^*\times \mathbb P_1$. Is it a complex Lie group? I mean is it finite dimensional?

We know that ${\rm Aut}(\mathbb C)^\ast$ is ${\mathbb C}^\ast$, and $Aut^\circ(\mathbb P_1)$ is $PSL(2,\mathbb C)$.

## 2 Answers

It is not a finite dimensional Lie group. For example, all of the maps $$\Phi\bigl(z,[a,b]\bigr) = \bigl(z, [a+p(z)b,\ b]\bigr),$$ where $p:\mathbb{C}^\ast\to \mathbb{C}$ is holomorphic, belong to this group. More generally, the automorphism group is essentially the semi-direct product of $\mathbb{C}^\ast$ with the group of holomorphic mappings $p:\mathbb{C}^\ast\to PSL(2,\mathbb{C})$.

To see this, note that the foliation of $\mathbb{C}^\ast\times \mathbb{P}^1$ by the fibers $\{z\}\times \mathbb{P}^1$ for $z\in\mathbb{C}^\ast$ must be preserved by any automorphism, since these are the only compact complex curves that represent a generator of the second homology group (with the orientation given by the complex structure). Thus, every automorphism $\Phi$ has a corresponding automorphism $\phi:\mathbb{C}^\ast\to\mathbb{C}^\ast$ that intertwines $\Phi$ and $\phi$ with the projection $\mathbb{C}^\ast\times\mathbb{P}^1\to\mathbb{C}^\ast$ onto the first factor. Of course $\phi$ is represented by scalar multiplication by a nonzero complex number $\lambda$, and the map $\Phi\mapsto\lambda$ is a homomorphism from this automorphism group to $\mathbb{C}^\ast$. The kernel of this homomorphism is clearly the group of holomorphic mappings $p:\mathbb{C}^\ast\to PSL(2,\mathbb{C})$, since $PSL(2,\mathbb{C})$ is the full group of automorphisms of $\mathbb{P}^1$.

An historical note: Of course, Lie (and Cartan) would certainly have regarded this automorphism group as a Lie group (albeit not one of finite dimension), since it can be defined as the set of solutions of a system of differential equations. By contrast, this is not so for the group $\text{Aut}(\mathbb{C}^2)$, which cannot be defined as the set of solutions of a system of differential equations.

I have troublesome understanding Robert's answer, but since I can not give comments, I write my comments here. If the maps $\Phi : (z,[a,b])\rightarrow (z,[a+p(z),b])$ were an isomorphism, then the maps $[a,b]\mapsto [a+p(z),b]$ must be an isomorphism of $\mathbb{P}^1$. But I think the latter can not be a map from $\mathbb{P}^1$ to itself, except when $p(z)$ is the zero function. For example, the map is not homogeneous in a and b.

• Oh, you are right. That's a typo that I will fix. It should be $[a,b]\mapsto [a + p(z)b,b]$. – Robert Bryant Oct 16 '12 at 17:27
• Why can't you give comments? – Robert Bryant Oct 16 '12 at 17:28
• That requires sufficient reputation. But answering doesn't. – Allen Knutson Oct 16 '12 at 17:42
• @Allen: Ah! I didn't know that. I really must read the MO directions some time; they probably explain why we have this rule, which, apparently, doesn't allow questions from beginners about someone's answer, except as another answer (which seems weird). – Robert Bryant Oct 16 '12 at 20:37