2
$\begingroup$

I believe the following statement is true:

Given a complex analytic map $f:\Delta\to V/G$, where $\Delta$ is a disc in $\mathbb{C}$, $V$ a finite dimensional complex vector space and $G$ a finite subgroup of $GL(V)$, then $f$ admits an analytic lift $\tilde f:\Delta'\to V$ up to a ramified cover. More precisely, there exists a ramified cover $r:\Delta'\to\Delta$ such that $f\circ r = \pi\circ \tilde f$ where $\pi:V\to V/G$ is the canonical projection.

I think I have a relatively elementary proof. However, this statement sounds very much like a "classical" result, but I have been unable to find a reference. Does anyone knows to whom I should attribute this result ?

Thanks a lot

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Salut Yann!

You can refer to the general notion of fiber product, for example. This is a construction that works in the complete generality. It might happen that in the situation you consider the fiber product will not be irreducible, (this happen when the preimage of $f(\Delta)$ in $V$ is not irreducible), then you just take one of the irreducible components of this fiber product.

Here is a reference for fiber products : http://planetmath.org/fibreproduct .
So according to the notations (I believe they are standard), this fiber product should be denoted as $\Delta\times _{V/G} V$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hi Dima, can you be somewhat more specific about which fiber product you are looking at ? I guess you are talking about $\Delta\times_f V \to \Delta$ ? $\endgroup$
    – Yann
    Commented Jun 5, 2011 at 11:52
  • $\begingroup$ Thanks Dima, this leads to a nice proof. Slicker, although equivalent to the one I had at hand. $\endgroup$
    – Yann
    Commented Jun 8, 2011 at 9:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .