Let $X$ be a scheme and it's Picard scheme $\underline{Pic}(X)$ exist. Denote by $\underline{Pic}^0(X)$ the connected component of the origin of the Picard scheme. My question is what the geometric intuition behind the fact the the tangent space of $\underline{Pic}^0(X)$ at the origin is canonically identified to $H^1(X,O_X)$?

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    $\begingroup$ Think about what the exponential map does in differential geometry and then think about what the exponential map coming from the long exact sequence in cohomology of the exponential sequence does. $\endgroup$ – Samir Canning Jun 30 '19 at 6:06
  • $\begingroup$ @SamirCanning: yes, you refer to the theory of Lie groups,right? There you can endow the tangent space at neutral element $e$ with Lie algebra structure and the exp-map provides a map from the tangent space at $e$ (= induced Lie algebra) to the group back. On the other hand the exponential sequence induces a map $H^1(X, O_X) \to H^1(O, O_X^*)$ where the right object is the Picard group. So intuitively $H^1(X, O_X)$ "sits" on the place of tangent space. Is this the picture you refer? $\endgroup$ – KarlPeter Jul 1 '19 at 16:00
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    $\begingroup$ Yes. It may be helpful for you to work out the example of a complex torus/abelian variety $\endgroup$ – Samir Canning Jul 1 '19 at 16:40
  • $\begingroup$ The first two pages of math.stanford.edu/~conrad/248BPage/handouts/pic.pdf may be helpful. $\endgroup$ – PrimeRibeyeDeal Dec 2 '19 at 17:28

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