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Hello everyone,

I have been trying (without success) to determine the following. Let $P$ denote the space of monic polynomials of degree $n$ with complex coefficients, which have distinct roots. It is known that the fundamental group of this space is the braid group $B_n$. Let $X \rightarrow P$ denote the universal cover. Then $B_n\subset Aut (X)$ where $Aut (X)$ is the group of holomorphic automorphisms of $X$. For $n\geq 4$, is it true that $Aut (X)=B_n$? I do not see any other automorphisms (but I am a novice in the area).

edited Feb 6 at 12:03

quid
13.3k●33●54

asked Feb 4 at 15:08

Aakumadula
2,885●1●6●14

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The space $X$ is very close to the Teichmuller space $T=T_{0,n+1}$, but they are a bit different: You can still act on $P$ (and, hence, on $X$) by complex affine transformations. Once you mod out $P$ (and $X$) by such action, you get the space $T$. It is Royden's theorem that for $n\ge 4$ the full group of biholomorphic automorphisms of $T_{0,n+1}$ is the mapping class group of the corresponding $n+1$-punctured sphere. The same holds for other finite type Riemann surfaces with few low-complexity exceptions (exclude torus, once punctured torus and 4 times punctured sphere). – Misha Feb 4 at 18:17

Thanks, Misha. This really answers my question. I forgot about the action of affine transformations. – Aakumadula Feb 5 at 0:30

Aakumadula · Answer 1 · 2013-02-05 01:07:15Z UTC

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Misha has answered my question completely.

answered Feb 5 at 1:07

Aakumadula
2,885●1●6●14

holomorphic automorphisms of universal cover of configuration spaces

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