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Oct 23, 2011 at 21:48 vote accept Veen
Oct 16, 2011 at 13:58 vote accept Veen
Oct 23, 2011 at 21:43
Oct 6, 2011 at 12:56 answer added Donu Arapura timeline score: 2
Oct 5, 2011 at 20:27 comment added Veen @Donu: thanks for your comment I don't exactly understand what you mean with $d+ \alpha$ as you need a connection on $E$. Would be glad if you could explain that to me.
Oct 5, 2011 at 15:12 comment added Damian Rössler would like to point out that the Hodge to de Rham spectral sequence gives an algebraic surjective map $H^1_{\rm dR}(X)\to H^1(X,{\cal O}_X)$. The inclusion $H^1(X,{\cal O}_X)\to H^1_{\rm dR}(X)$ you mention is given by the Hodge splitting, which is not algebraic. So the description of the mysterious connection must be analytic (as hinted at by Donu Arapura). On the other hand, the natural inclusion $H^0(X,\Omega_X)\to H^1_{\rm dR}(X)$ (also given by Hodge to de Rham) has a natural algebraic description in terms of connections.
Oct 5, 2011 at 14:55 comment added Donu Arapura I've got to run, but wouldn't it just be $d+\alpha$, where $\alpha$ is the harmonic $(0,1)$-form representing the class of $H^1(X,\mathcal{O}_X)$?
Oct 5, 2011 at 14:39 history asked Veen CC BY-SA 3.0