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I want to show that if $X$ is a smooth complex projective variety, then analytification induces an equivalence of categories between algebraic integrable connections on $X$ and analytic integrable connections on $X^\text{an}$. It does not follow immediately from GAGA that every analytic connection $(M, \nabla)$ comes from an algebraic connection, since $\nabla: M \to M \otimes \Omega^1_{X^\text{an}}$ is not $\mathcal{O}_{X^\text{an}}$-linear. A fix mentioned in this question is to view an integrable connection on $X^\text{an}$ as an $\mathcal{O}$-linear splitting of the Atiyah sequence

$$0 \mapsto \operatorname{End}(M) \to \operatorname{At}(M) \to \mathcal{T}_{X^\text{an}} \to 0$$

and apply GAGA.

I can't figure out what this exact sequence should be, and I couldn't find any references. What is $\operatorname{At}(\bullet)$, and what precisely is this exact sequence?

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    $\begingroup$ Why don't you look at Atiyah's original paper (Complex analytic connections in fibre bundles, Trans. Amer. Math. Soc. 85 (1957), 181-207)? It is very clearly written. $\endgroup$
    – abx
    Commented Apr 14 at 4:12
  • $\begingroup$ for a modern treatment you could try [[0807.4374] Hermitian vector bundles and extension groups on arithmetic schemes. II. The arithmetic Atiyah extension](arxiv.org/abs/0807.4374) $\endgroup$
    – Niels
    Commented Apr 14 at 14:22
  • $\begingroup$ @abx Thank you for this! It seems that 0.4 is the dual of what I wrote here. What I don't get is the following: in the context of my question, I think $\operatorname{At}(M)$ should be the sheaf of differential operators on $M$ of degree $\leq 1$. But then what is the map to the tangent sheaf $T_{X^\text{an}}$ be? $\endgroup$ Commented Apr 14 at 16:25
  • $\begingroup$ Nevermind, I think I got it. It should be $D \mapsto ( f \mapsto \operatorname{tr}([D,f]))$, right? $\endgroup$ Commented Apr 14 at 16:48
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    $\begingroup$ The point is that $\operatorname{At}(M) $ is the sheaf of differential operators on $M$ of degree $\leq 1$ with scalar symbol. $\endgroup$
    – abx
    Commented Apr 15 at 5:44

1 Answer 1

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Here's how the argument goes, thanks to @abx. $\operatorname{At}(M)$ is given by $\mathbb{C}$-linear endomorphisms $D: M \to M$ for which the commutator $[D,f]$ given by $[D,f](m) := D(fm) - fD(m)$ acts via left multiplication by some element $h_{D,f} \in \mathcal{O}$. The map to the tangent sheaf $T$ is given by $$D \mapsto (f \mapsto h_{D,f})$$ and it's straightforward to check that this is a derivation. It's clear that the kernel is $\operatorname{End}_{\mathcal{O}}(M)$, and surjectivity follows by defining $$D_P\left( \sum_{i=1}^n f_i m_i\right) := \sum_{i=1}^n P(f_i) m_i$$ in some local basis $m_i$ of $M$, where $P \in T_X$. It is now immediate that $\mathcal{O}$-linear right splittings correspond to integrable connections on $M$.

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