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I am referring to page 300 of Okonek et al. book "Vector Bundles on Complex Projective Spaces" (1988.)

Let $G=GL_n(\mathbb{C})$ act holomorphically and freely on a complex manifold $X$. Define the mapping

$\gamma:X\times G\longrightarrow X\times X$

by sending $(x,g)$ to $(x,gx)$. Given that $\gamma$ is an isomorphism onto a closed analytic subspace of $X\times X$,

Why is $X/G$ also a complex manifold?

Why is $X\rightarrow X/G$ a $G$-principal bundle (in the complex analytic category)?

Apparently, this result can be found in the two papers:

  • Holmann, H.: Komplexe Raume mit komplexen Transformationsgruppen. Math Ann. 150, 1963 (p.359),

  • Holmann, H.: Quotienten komplexer Raume. Math Ann 142, 1961 (p.433.)

However, Germans is so hard for me and I would like to understand at least the rough idea. My feeling is that this should be something standard nowadays?

I know the condition on the image of $\gamma$ is basically to ensure separatedness/Hausdorffness of $X/G$ so maybe the first question is not hard but for the second question I really have no idea.

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    $\begingroup$ If you prefer French (resp. English) see Prop. 10, section 1.5, Chapter III of Bourbaki's Groupes et Alg`ebres de Lie (resp. Lie groups and Lie algebras) for such an action by any real or complex Lie group; also see Prop. 13 there. Also see section 12 in Chapter III of Part II of Serre's book "Lie groups and Lie algebras" (Springer LNM 1500) for a wider context (also treated elsewhere in Bourbaki). $\endgroup$
    – nfdc23
    Commented Mar 26, 2017 at 2:39
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    $\begingroup$ Since the $G$-action on $X$ is free, the quotient $X/G$ is a complex manifold [see Complex Tori and Abelian Varieties, by Olivier Debarre, page 75]. Since the $G$-action preserves the fibers of $X/G$, $X\rightarrow X/G$ is a principal $G$-bundle. $\endgroup$
    – Mee Seong Im
    Commented Mar 26, 2017 at 3:56

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