# How manifold-like is Aut(C^n) in the holomorphic category?

This question is similar to, but not the same as this one. Take the space of automorphisms of $\mathbb{C}^n$ in the holomorphic category, with the compact-open topology. For $n=1$ this is just $\mathbb{C} \times \mathbb{C}^*$, but for larger $n$ is very complicated. It clearly contains $GL(n,\mathbb{C})$, and translations.

To give a taste of how big $Aut(\mathbb{C}^n)$ is, there is a theorem that given any two countable dense subsets $X,Y \subset \mathbb{C}^n$, $n >1$, there is a volume preserving automorphism taking $X$ to $Y$. But I have no idea about what this space is like. Is it some sort of infinite dimensional manifold? Analytic space? Does it contain an infinite dimensional Lie group as a (closed) subgroup?

• I don't know what counts as an infinite-dimensional Lie group, but you can make lots of closed subgroups, for example the stabilizer of a point or a line or a function $\mathbb C^n\to \mathbb C$. – Tom Goodwillie Dec 1 '10 at 3:27
• Say a Frechet Lie group. A Banach Lie group might be too much to ask. – David Roberts Dec 1 '10 at 4:23
• In the algebraic case it's horrifying: not representable for any $n > 1$. Indeed, suppose rep'td by some $G$. Then Grothendieck's functorial criterion for being locally of finite presentation forces $G$ to be loc. of finite type, then smooth (by Cartier), so $G^0$ is a finite type connected smooth group variety. This forces the points of $G^0$ to act by automorphisms represented by polynomials of bounded degree, which is absurd for $n > 1$ since for two distinct variables $(x,y)$ can use $(x,y) \mapsto (x + t y^n, y)$ to get a connected family of auts joining id to arb. big degree. – BCnrd Dec 1 '10 at 5:06
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