5

Let $X$ be a smooth algebraic surface over $\mathbb{C}$ i.e a smooth and connected scheme of finite type over $\operatorname{spec}(\mathbb{C})$ of dimension $=2$. I denote with $X^{[n]}$ the Hilbert scheme of $0$-dimensional subscheme of length $=n$. This is known to be a smooth scheme of dim$=2n$ which moreover admits the Hilbert-Chow morphism: $$\chi:X^{[n]} \to X^{(n)} $$ where $X^{(n)}=X \times \cdots X/S_n$ is the symmetric product.

Among the properties of $\chi$ it is frequently stated that this morphism restricts to a fiber bundle (in the analytic topology) when considering $$\chi^{-1}(X) \rightarrow X $$ where $X \hookrightarrow X^{(n)} $ via the diagonal embedding. If I got it correctly the idea should be that every fibre $\chi^{-1}(x) \in \chi^{-1}(X)$ is isomorphic to the punctual Hilbert scheme and one would use this to obtain the property.

I was not able however to actually describe explicit trivialization maps neither to get which open subsets to use in the trivializing cover nor why one should use the analytic and not the Zariski topology. Moreover, is there an analogous statement over other algebraically closed field where the world analytic is substituted by etale?

KReiser
  • 74,746
Tommaso Scognamiglio
  • 1,468
  • Try this definition Hilbert_scheme. Specially the section universality. This deals with algebraic cycles chow spaces and hilbert chow morphisms. This give some (diagonal+embedding)[https://mathoverflow.net/search?q=diagonal+embedding]. Trivialization – Steffen Jaeschke Nov 15 '20 at 08:20
  • zariski open sets are dense in analytic topology and etale algebraically closed field – Steffen Jaeschke Nov 15 '20 at 08:25
  • I'm sorry I wasn't able to get how to locally trivialize the map after looking at those answers. Is there an intuitive way to understand the local triviality? – Tommaso Scognamiglio Nov 16 '20 at 13:06
  • Not sure what the previous commenter was doing; link spamming is not good commenting and suspiciously their previous posts are all on calculus questions and similar level. – Tabes Bridges Nov 18 '20 at 02:27
  • 1
    But here's a non-spam link that might be of interest: https://indigo.uic.edu/articles/thesis/Effective_Divisors_on_Kontsevich_Moduli_Spaces/10783163/1 In my thesis I (rather trivially) trivialized the Kontsevich space (which is isotrivial over projective space) in order to compute the Picard group of the fibers of the evaluation map. If I understand your question correctly (in which case all fibers are isomorphic, the space of length $n$ structures on a smooth point in a surface) then the same idea should work; just use the standard cover of $\mathbb P^r$ for some projective embedding of $X$. – Tabes Bridges Nov 18 '20 at 02:31
  • It seems actually to work to me: the only point which is non clear is that seems to be Zariksi open sets and not just in the analytic topology. This is actually stronger, but it just seems strange as all the references strongly emphasize the analytic part (without proving it ahaha). – Tommaso Scognamiglio Nov 18 '20 at 14:59
  • Perhaps the following standard observation is useful --- any smooth scheme is etale-locally isomorphic to the affine space (to see this just embed a Zariski neighborhood of a point into an affine space, then a general projection to an affine space of the same dimension is regular at the point, hence is etale on a smaller neighborhood). Consequently, the morphism $\chi$ is etale-locally the same as the same morphism for $\mathbb{A}^2$, and the latter is a fiber bundle, e.g., because of the transitive group action. – Sasha Nov 19 '20 at 08:18
  • Have a look at (Trivialization)[https://mathworld.wolfram.com/Trivialization.html] and in (Vector_bundle)[https://en.wikipedia.org/wiki/Vector_bundle] – Steffen Jaeschke Nov 20 '20 at 07:47
  • @Sasha I think more or less I understood the reduction to the affine plane case, but I do not manage actually to end up the proof. Also, this actually seems similar to what the other user suggested which seems still not using the etale/analytic topology but the Zariski one. What am I missing? – Tommaso Scognamiglio Nov 20 '20 at 09:24
  • @TommasoScognamiglio: I am not sure I understand what the other user has suggested. I also don't know why you can't end up the proof. So, how can I tell what are you missing? – Sasha Nov 20 '20 at 09:38
  • Sorry I was being super imprecise. Actually I think I've worked out the affine plane case ahahahah. Thank you in any case :) – Tommaso Scognamiglio Nov 20 '20 at 09:41

0 Answers0