So on a curve, $\text{Pic}^0(X)$ is just the Jacobian variety, and just correspond to degree $0$ divisors. One way to extend the notion of divisors corresponding to a vector bundle is taking the first Chern class; the image of a line bundle under $c_1$ is an element of $A^1(X)$, the Chow group in codimension $1$, so we have a well-defined notion of degree, the kernel of the degree map gives a way to construct the Jacobian. Does this extend in any reasonable way to more general situations? Perhaps as the kernel of the self-intersection number of the first Chern class? The notation $\text{Pic}^0$ is quite suggestive, but I don't see it.
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2$\begingroup$ Not sure how helpful this is (I'm not an algebraic geometer), but I thought I would mention it. In Huybrechts' Complex Geometry: An Introduction, the Jacobian of a complex manifold $X$, denoted $\operatorname{Pic}^0(X)$, is defined as the kernel of the map $c_1 : \operatorname{Pic}(X) \to H^2(X, \mathbb{Z})$. Using the exponential sequence, one can show that $\operatorname{Pic}^0(X) \cong H^1(X, \mathcal{O})/H^1(X, \mathbb{Z})$. $\endgroup$– Michael AlbaneseCommented Jul 22, 2015 at 3:46
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$\begingroup$ This is explained on Wikipedia: en.wikipedia.org/wiki/Néron–Severi_group $\endgroup$– Ben Webster ♦Commented Jul 22, 2015 at 6:42
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Even if we might not be able to speak of degree without reference to a given ample divisor, we can still speak of codimension-1 cycles being algebraically or rationally equivalent to 0 on a smooth projective variety $X$ of any positive dimension. In this context, ${\rm Pic}^{0}(X)$ is the group of codimension-1 cycles algebraically equivalent to 0 modulo the subgroup of those rationally equivalent to 0.
EDIT: As suggested by the above, ${\rm Pic}^{0}(X)$ is the kernel of the map from divisors mod rational equivalence to divisors mod algebraic equivalence. When $X$ is a smooth projective curve, divisors mod algebraic equivalence is just $\mathbb{Z}.$
