By definition, for $X$ a compact Kähler manifold, $$ Pic^0(X)=H^1(X,\mathcal O_X)/H^1(X,\mathbb Z) $$ and $H^1(X,\mathbb Z)$ is a lattice of maximal rank. Thus the vanishing of $Pic^0(X)$ is equivalent to the vanishing of $H^1(X,\mathcal O_X)$, which is, by Hodge theory, purely topological: $H^1(X,\mathcal O_X)\simeq H^{0,1}(X;\mathbb C)$ and $h^{1,0}(X)=h^{0,1}(X)=b_1(X)/2$.

Hence you are asking for Kähler manifolds whose first Betti number vanishes.

Now, $H^1(X,\mathbb Z)$ is the abelianization of $\pi_1(X)$. So a compact Kähler manifold has zero Jacobian variety if and only if its fundamental group is totally non abelian.

A first rich class of examples is given, as Jim said, by the Lefschetz hyperplane theorem. Take an $n$-dimensional smooth projective variety $X$ with zero $H^1(X,\mathcal O_X)$ (for instance a simply connected smooth projective manifold, e.g. $\mathbb P^n$), and $D\subset X$ any smooth ample divisor. Then one has isomorphisms $$ H^k(X,\mathbb Z)\to H^k(D,\mathbb Z),\quad k<n-1 $$ and it is injective when $k=n-1$. In particular $H^1(D,\mathbb Z)=0$ and all such hypersurfaces will have zero Jacobian variety, provided $n>2$. You can figure out a similar construction in the smooth complete intersection case.

Fano varieties are known to be rationally connected and hence simply connected, therefore provide other examples (to see it more elementary if $-K_X$ is ample then $H^q(X,O_X)=0$ if $q>0$ by Kodaira's vanishing. Thus $H^1(X,\mathcal O_X)=0$).

By definition, for $X$ a compact Kähler manifold, $$ Pic^0(X)=H^1(X,\mathcal O_X)/H^1(X,\mathbb Z) $$ and $H^1(X,\mathbb Z)$ is a lattice of maximal rank. Thus the vanishing of $Pic^0(X)$ is equivalent to the vanishing of $H^1(X,\mathcal O_X)$, which is, by Hodge theory, purely topological: $H^1(X,\mathcal O_X)\simeq H^{0,1}(X;\mathbb C)$ and $h^{1,0}(X)=h^{0,1}(X)=b_1(X)/2$.

Hence you are asking for Kähler manifolds whose first Betti number vanishes.

Now, $H^1(X,\mathbb C)$ is the complexification of the abelianization of $\pi_1(X)$. So a compact Kähler manifold has zero Jacobian variety if and only if the abelianization of its fundamental group vanishes after complexification, i.e. it is completely torsion.

A first rich class of examples is given, as Jim said, by the Lefschetz hyperplane theorem. Take an $n$-dimensional smooth projective variety $X$ with zero $H^1(X,\mathcal O_X)$ (for instance a simply connected smooth projective manifold, e.g. $\mathbb P^n$), and $D\subset X$ any smooth ample divisor. Then one has isomorphisms $$ H^k(X,\mathbb Z)\to H^k(D,\mathbb Z),\quad k<n-1 $$ and it is injective when $k=n-1$. In particular $H^1(D,\mathbb Z)=0$ and all such hypersurfaces will have zero Jacobian variety, provided $n>2$. You can figure out a similar construction in the smooth complete intersection case.

Fano varieties are known to be rationally connected and hence simply connected, therefore provide other examples (to see it more elementary if $-K_X$ is ample then $H^q(X,O_X)=0$ if $q>0$ by Kodaira's vanishing. Thus $H^1(X,\mathcal O_X)=0$).

By definition, for $X$ a compact Kähler manifold, $$ Pic^0(X)=H^1(X,\mathcal O_X)/H^1(X,\mathbb Z) $$ and $H^1(X,\mathbb Z)$ is a lattice of maximal rank. Thus the vanishing of $Pic^0(X)$ is equivalent to the vanishing of $H^1(X,\mathcal O_X)$, which is, by Hodge theory, purely topological: $H^1(X,\mathcal O_X)\simeq H^{0,1}(X;\mathbb C)$ and $h^{1,0}(X)=h^{0,1}(X)=b_1(X)/2$.

Hence you are asking for Kähler manifolds whose first Betti number vanishes.

Now, $H^1(X,\mathbb Z)$ is the abelianization of $\pi_1(X)$. So a compact Kähler manifold has zero Jacobian variety if and only if its fundamental group is totally non abelian.

By definition, for $X$ a compact Kähler manifold, $$ Pic^0(X)=H^1(X,\mathcal O_X)/H^1(X,\mathbb Z) $$ and $H^1(X,\mathbb Z)$ is a lattice of maximal rank. Thus the vanishing of $Pic^0(X)$ is equivalent to the vanishing of $H^1(X,\mathcal O_X)$, which is, by Hodge theory, purely topological: $H^1(X,\mathcal O_X)\simeq H^{0,1}(X;\mathbb C)$ and $h^{1,0}(X)=h^{0,1}(X)=b_1(X)/2$.

Hence you are asking for Kähler manifolds whose first Betti number vanishes.

Now, $H^1(X,\mathbb Z)$ is the abelianization of $\pi_1(X)$. So a compact Kähler manifold has zero Jacobian variety if and only if its fundamental group is totally non abelian.

A first rich class of examples is given, as Jim said, by the Lefschetz hyperplane theorem. Take an $n$-dimensional smooth projective variety $X$ with zero $H^1(X,\mathcal O_X)$ (for instance a simply connected smooth projective manifold, e.g. $\mathbb P^n$), and $D\subset X$ any smooth ample divisor. Then one has isomorphisms $$ H^k(X,\mathbb Z)\to H^k(D,\mathbb Z),\quad k<n-1 $$ and it is injective when $k=n-1$. In particular $H^1(D,\mathbb Z)=0$ and all such hypersurfaces will have zero Jacobian variety, provided $n>2$. You can figure out a similar construction in the smooth complete intersection case.

Fano varieties are known to be rationally connected and hence simply connected, therefore provide other examples (to see it more elementary if $-K_X$ is ample then $H^q(X,O_X)=0$ if $q>0$ by Kodaira's vanishing. Thus $H^1(X,\mathcal O_X)=0$).