Geometric interpretation of $H^1_{Zar}(X,\mathcal O_X)$

Question

Let $X/k$ be a smooth projective geometrically integral variety, perhaps over $k$ algebraically closed. What is the geometric interpretation of $H^1_{Zar}(X,\mathcal{O}_X)$? Does it have something to do with the tangent space? Does $H^1_{Zar}(X,\mathcal{O}_X) = 0$ imply $Pic^0(X) = 0$?

Edit: Another interpretation: $H^1_{Zar}(X,\mathcal{O}_X) = Ext^1(\mathcal{O}_X,\mathcal{O}_X)$.

Answer 1

These notes by Brian Conrad show (assuming $X$ has a $k$-rational point) that $H^1(O_X)$ can be identified with the tangent space to $Pic$ at the identity.

So $H^1$ "has something to do with the tangent space", if we're talking about the tangent space to the Picard scheme. Moreover, if $H^1=0$, then $Pic$ must be reduced and zero-dimensional, hence $Pic^0=0$.