Let L be a line bundle on a smooth affine variety X (say, over complex numbers). Is it true that L always admits a FLAT algebraic connection?
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$\begingroup$ Is this motivated the Atiyah class, which lives in coherent cohomology and thus vanishes for affines? The vanishing of this class implies the existence an algebraic connection. Moreover, one can compute the Chern classes in Dolbeault cohomology from the Atiyah class. If the variety is complete, so that Dolbeault cohomology is pretty close to de Rham cohomology, does the vanishing of the Atiyah class imply the vanishing of the rational Chern classes? or is something lost in the extension data? $\endgroup$– Ben WielandCommented Apr 13, 2010 at 22:52
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1$\begingroup$ Yes, a line bundle on a compact Kähler variety has a connection iff it has a flat connection. This follows from the Hodge decomposition of de Rham cohomology, the obstruction for finding a flat connection lies in $H^1(X,\Omega^{\ge1})$ and the obstruction for finding a connection is its image in $H^1(X,\Omega^1)$. By the Hodge decomposition (and the fact that the obstruction is of type $(1,1)$), if the latter vanishes so does the former. $\endgroup$– Torsten EkedahlCommented Apr 14, 2010 at 5:31
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2 Answers
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No, any line bundle with a flat connection has a trivial rational Chern class. Now, take any smooth connected projective variety $X$ for which the Chern classes of line bundles form a group of rank $r$ larger than $1$. Removing an irreducible ample divisor $D$ from $X$ gives a smooth affine variety for which the Chern classes form a group of rank $r-1$. A specific example is $\mathbb P^1\times\mathbb P^1$ but there are lots of others of any dimension $>1$.
answered Apr 13, 2010 at 21:39
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$\begingroup$ Dear Torsten Ekedahl, could you please explain why the Chern classes group of the affine variety $X \setminus D$ is of rank $r-1$, because I don't really get it... $\endgroup$– HenriCommented May 8, 2011 at 21:33
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1$\begingroup$ This is a standard argument (found in Hartshorne for instance): There is a section of $\mathcal O_X(D)$ which vanishes exactly at $D$ and hence gives a trivialisation of it in the complement. This (and its tensor powers) is the only line bundles killed and every line bundle on the complement extends to $X$. $\endgroup$ Commented May 9, 2011 at 4:17
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No, there is no reason for an an $O$-module, even locally free rank 1, to be a $D$-module.
answered Apr 13, 2010 at 21:27
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1$\begingroup$ Um, care to provide a counterexample? My intuition is the same as yours, but I couldn't come up with a counter-example on an affine variety (of course, there are lots on quasi-projective ones). $\endgroup$– Ben Webster ♦Commented Apr 13, 2010 at 21:34
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$\begingroup$ Why should I? It is obviously wrong and the guy is trying to prove something using it and he really should not:-)) And, anyway, Torsten has done it already and got my point for it... $\endgroup$ Commented Apr 13, 2010 at 21:44
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9$\begingroup$ For the edification of humanity? I mean, what's the point of writing the answer otherwise? $\endgroup$– Ben Webster ♦Commented Apr 13, 2010 at 22:08