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Sam Nead
Sam Nead
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Edited to make it right:

The “capping off” map indeed is a fibre bundle map. However, the fibre is not the configuration space you suggest. For, consider the case of $n = 1$. Here the configuration space would be a copy of the surface, and the larger Teichmuller space would have nontrivial fundamental group! This is obviously wrong.

When $n = 1$, what is actually going on is that the fibre over $X$ is a copy of the hyperbolic plane. This plane is the space of points of $X$, equipped with a choice of homotopy class of path back to a (fixed earlier) base point. Thus the fibre over $X$ is the universal cover of $X$.

Edited to make it right:

The “capping off” map indeed is a fibre bundle map. However, the fibre is not the configuration space you suggest. For, consider the case of $n = 1$. Here the configuration space would be a copy of the surface, and the larger Teichmuller space would have nontrivial fundamental group! This is obviously wrong.

When $n = 1$, what is actually going on is that the fibre is a copy of the hyperbolic plane.

Edited to make it right:

The “capping off” map indeed is a fibre bundle map. However, the fibre is not the configuration space you suggest. For, consider the case of $n = 1$. Here the configuration space would be a copy of the surface, and the larger Teichmuller space would have nontrivial fundamental group! This is obviously wrong.

When $n = 1$, what is actually going on is that the fibre over $X$ is a copy of the hyperbolic plane. This plane is the space of points of $X$, equipped with a choice of homotopy class of path back to a (fixed earlier) base point. Thus the fibre over $X$ is the universal cover of $X$.

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Sam Nead
Sam Nead
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Edited to make it right:

The “capping off” map $\Phi$indeed is a homeomorphismfibre bundle map. It doesHowever, the fibre is not have further propertiesthe configuration space you suggest. For, becauseconsider the case of $n = 1$. Here the configuration space is justwould be a manifold - it does notcopy of the surface, and the larger Teichmuller space would have nontrivial fundamental group! This is obviously wrong.

When $n = 1$, what is actually going on is that the fibre is a preferred holomorphic structurecopy of the hyperbolic plane.

The map $\Phi$ is a homeomorphism. It does not have further properties, because the configuration space is just a manifold - it does not have a preferred holomorphic structure.

Edited to make it right:

The “capping off” map indeed is a fibre bundle map. However, the fibre is not the configuration space you suggest. For, consider the case of $n = 1$. Here the configuration space would be a copy of the surface, and the larger Teichmuller space would have nontrivial fundamental group! This is obviously wrong.

When $n = 1$, what is actually going on is that the fibre is a copy of the hyperbolic plane.

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Sam Nead
Sam Nead
  • 28.1k
  • 5
  • 72
  • 131

The map $\Phi$ is a homeomorphism. It does not have further properties, because the configuration space is just a manifold - it does not have a preferred holomorphic structure.