Does the Grothendieck group detect the Picard group? Let $C$ be a curve (=smooth projective curve) of genus $g$ over an algebraic closed field $\mathbb{k}$.
It is well known that the Grothendieck group $K_0(\operatorname{coh} C)$ of the category of coherent sheaves on $C$ is the following (see, for example, Exercise 6.11 of Hartshorne):
$$
K_0(\operatorname{coh} C)\simeq \operatorname{Pic} C \times \mathbb{Z} 
\simeq \operatorname{Pic}_0 C \times \mathbb{Z}^2,
$$
where $\operatorname{Pic} C$ is the Picard group of $C$ and
$\operatorname{Pic}_0 C$ is the kernel of the degree map $\mathrm{deg} : \operatorname{Pic} C\to \mathbb{Z}$.
In the case $\mathbb{k} = \mathbb{C}$, we also know that
$$
\operatorname{Pic}_0 C \simeq \mathbb{C}^g/\Lambda,
$$
where $\Lambda$ is a $\mathbb{Z}$-submodule in $\mathbb{C}^g$ of rank $2g$.
My Questions:

*

*For two curves $C_1$ and $C_2$, if $K_0(\operatorname{coh} C_1) \simeq K_0(\operatorname{coh} C_2)$ as abstract groups,
then do we conclude that $\operatorname{Pic}_0 C_1 \simeq \operatorname{Pic}_0 C_2$
(or $\operatorname{Pic} C_1 \simeq \operatorname{Pic} C_2$)
as abstract groups ?
Here abstract groups means groups without additional structure, such as structure of topology, variety and so on.

*Does an isomorphism $\operatorname{Pic}_0 C_1 \simeq \operatorname{Pic}_0 C_2$ of abstract groups contain some geometric information about $C_1$ and $C_2$?

 A: I don't know about 2., but the answer to 1. is yes.
More generally, suppose $(S^1)^g\times\mathbb Z^k \cong (S^1)^h \times \mathbb Z^m$ as abstract groups, then $g=h, k=m$ (in fact, you only need $g=h$ for question 1., because the abstract isomorphism type of $\mathbb C^g/\Lambda$ only depends on $g$, and because you already know $k=m=2$)
(Note that this gives part of the answer for 2., namely that $C_1$ and $C_2$ have the same genus)
Let's prove this statement.
First, if there is an isomorphism between the two groups, then their torsion parts are isomorphic. Their torsion parts are $(\mathbb{Q/Z})^g$ and $(\mathbb{Q/Z})^h$ respectively. But you can recover $g$ from $(\mathbb{Q/Z})^g$ in the following way : take the $2$-torsion (not $2$-power torsion, $2$-torsion) part as an $\mathbb F_2$-vector space, this has dimension $g$. (I chose $2$ but could have chosen any $p$, in fact for any integer $n\geq 2$, the $n$-torsion part is a free $\mathbb Z/n$-module of rank $g$, and these rings have the invariant basis property).
Therefore $g=h$ - this is all that's relevant to your question, but let me still prove $k=m$.
Note that to get $k=m$, we cannot rationalize and take dimensions because $(S^1)^g\otimes \mathbb Q$ is an uncounatbly dimensional rational vector space.
We now observe that we can mod out the divisible part - I don't know the standard name, but let me define it : $\mathrm{Div}(A) = \{a\in A\mid \forall n \neq 0, \exists b \in A, nb = a\}$. This is clearly a subgroup of $A$, and $\mathrm{Div}(A\times B)\cong \mathrm{Div}(A)\times \mathrm{Div}(B)$. In particular $\mathrm{Div}((S^1)^g\times \mathbb Z^k) = (S^1)^g\times \{0\}$, and so modding out $A/\mathrm{Div}(A)$ gives, in our case, an isomorphism $\mathbb Z^k\cong \mathbb Z^m$, and therefore $k=m$.
