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This is may be obvious, but I am not comfortable with ind-schemes.

I have an ind-scheme $X$ over $\mathbb{C}$. Every point has a neighborhood which can be written as an ascending union of regular varieties, which is about as smooth as an ind-scheme can be.

I have an unipotent ind-group $U$. More precisely, $U$ is a group object in the category of ind-schemes, and $U$ has a descending filtration all of whose quotients are $\mathbb{G}_a$'s. The group $U$ acts freely on $X$.

There is a quasi-projective variety $Y$, and a map $f: X \to Y$, which is a principal $U$-bundle. (Meaning that there is a cover $Y = \bigcup V_i$ and $f^{-1}(V_i)$ is isomorphic to $U \times V_i$.)

Is $X(\mathbb{C}) \to Y(\mathbb{C})$ a weak homotopy equivalence, using the analytic topology on both sides?

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My recollection is that when you turn these into analytic spaces you get something which is locally contractible topologically. In this case what you are describing is a principal bundle for locally contractible spaces in which the fiber is contractible. If the base is paracompact then this will indeed be a weak equivalence, in fact the space corresponding to X will be a topological product space $X = U \times Y$.

This follows because you can build a global section (trivialization). How do you do this? You start with you local trivializations, and you choose a partition of unity subordinate to this cover. You also choose a contraction of U. You can then patch these together to obtain a global section. The exact method and formula is explained, for example, in the appendix of this paper. (This is probably not the only/first/best reference).

Segal, G. Cohomology of topological groups. 1970 Symposia Mathematica, Vol. IV (INDAM, Rome, 1968/69) pp. 377--387 Academic Press, London

So the real question is whether your space Y is paracompact. I'm pretty sure that your conditions (that Y is quasi-projective) ensure that this is the case.

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  • $\begingroup$ btw, you can relax the local contractibility assumption as well, but it gets technical as to what conditions you need. I don't remember off the top of my head and don't know a reference. $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 26, 2010 at 14:53
  • $\begingroup$ Thanks! The following question may be addressed in your reference, but I'll ask here in case you have a quick answer: how does one make a partition of unity argument when dealing with a noncommutative group? If U were finite dimensional, I could use the partition of unity argument for G_a and then the long exact sequence of cohomology for noncommutative sheaves, but when U is infinite dimensional the induction doesn't end. $\endgroup$
    – David E Speyer
    Commented Feb 26, 2010 at 14:55
  • $\begingroup$ Well first of all, you only use a partition of unity on Y, not on U or X. U can be very very big and not necessarily paracompact. You just need Y to be reasonable. The way this works is roughly that you can choose some ordering of the cover (huge!) then you can form a formal expression for the global section (also huge!). This uses the partition of unity, the local sections, and the contraction of U. This huge expression, which is basically an infinite product over all the covers, makes sense because (by paracompactness) the cover is locally finite (I guess I forgot to mention this). $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 26, 2010 at 17:56
  • $\begingroup$ (cont.) In particular no induction is necessary. The expression works for non-commutative groups, but in that case depends on the choice of ordering. Also one more thing. In the reference I cite, Segal only states this in the case that Y is also a group. However if you look at the proof you see that it applies exactly to the case you were considering. $\endgroup$
    – Chris Schommer-Pries
    Commented Feb 26, 2010 at 18:29

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