# Grothendieck Group of a Smooth Projective Complex Curve

Our goal for this blog post is to describe, for a smooth projective complex curve $C$, the group $K_0(C)$. In this regard, it is a natural continuation of the last post I wrote which gives calculations of the Grothendieck group of $\mathbb{A}^n, \mathbb{P}^n$ for all $n$. This blog mainly consists of two big proofs (which probably should be skipped unless you’re looking for the proofs specifically). After these two proofs, we conclude with some explicit examples describable by (arbitrarily chosen) polynomials.

1. Theorem: Let $C$ be a smooth, projective curve. Then $K_0(C)\cong \mathbb{Z}\oplus \text{Pic}(C)$.

Proof. Our computation will rely on the Brown-Quillen spectral sequence of K-theory: for a regular scheme of finite type over a field, there is a convergent spectral sequence

$E_1^{p,q}=\coprod\limits_{x\in X^{(p)}} K_{-p-q}(k(X))\implies K'_{-p-q}(X)$

which satisfies $E^{p,-p}_2(X)\cong \text{CH}^p(X)$ (here the notation $X^{(p)}$ means the set of points in codimension $p$; the filtration defined on the right hand side is the coniveau filtration, where the $F^i$th step of the filtration is generated by sheaves with support in codimension $\geq i$).

For a curve there are no codimension 2 points thus $E_\infty^{1,-1}\cong E_2^{1,-1}$ and $E_\infty^{0,0}\cong E_2^{0,0}$. This gives isomorphisms $\text{CH}^1(X)\rightarrow F^1K'_0(X)/F^2K'_0(X)$, $\text{CH}^0\rightarrow K'_0(X)/F^1K'_0(X)$ where, for a curve we have $F^2K'_0(X)=0$ and for a smooth scheme we have $K_0(X)=K'_0(X)$. Using that $\text{Pic}(X)=\text{CH}^1(X)$ (which is essentially definition) and $\mathbb{Z}=\text{CH}^0(X)$ (which is because our curve has only one irreducible component) we need only show there is an isomorphism $K_0'(X)\cong K_0'(X)/F^1(K_0'(X))\oplus F^1(K_0'(X))$.

This is manageable: we have a short exact sequence of abelian groups

$0\rightarrow F^1(K'_0(X))\rightarrow K_0'(X)\rightarrow K_0'(X)/F^1(K'_0(X))\rightarrow 0$

with $K_0'(X)/F^1(K'_0(X))\cong \mathbb{Z}$. Since this map surjects onto $\mathbb{Z}$, we need only choose a preimage of $1,-1$ and it follows this sequence splits by sending a generator of $\mathbb{Z}$ to its preimage. This can also be summarized: $\text{Ext}^1(\mathbb{Z},G)=0$ for every group $G$.$\square$

We now specialize to the case where $C$ is defined over $\mathbb{C}$, which allows us to give an explicit description of $\text{Pic}(C)$. We use analytic methods, which are described succinctly in [Algebraic Geometry – Hartshorne, Appendix B].

2. Proposition: For a smooth, projective, complex curve $C$, $\text{Pic}(C)\cong (\mathbb{C}^g/\Lambda)\oplus \mathbb{Z}$ where $g$ is the genus of $C$, and $\Lambda$ is a lattice in $\mathbb{C}^g$.

Proof. We will use the complex manifold $C(\mathbb{C})$ and obtain, using an application of Serre’s GAGA, results for the algebraic variety $C$. Let $\mathbb{Z}_{C(\mathbb{C})}$ be the constant sheaf $U\mapsto \mathbb{Z}$, $\mathcal{O}_{C(\mathbb{C})}$ the sheaf of holomorphic functions $U\mapsto \mathcal{O}_{C(\mathbb{C})}(U)$, and $\mathcal{O}_{C(\mathbb{C})}^\times$ the sheaf of invertible regular functions $U\mapsto \mathcal{O}_{C(\mathbb{C})}(U)^\times$. There is then a short exact sequence of sheaves

$0\rightarrow \mathbb{Z}_{C(\mathbb{C})}\rightarrow \mathcal{O}_{C(\mathbb{C})}\rightarrow \mathcal{O}_{C(\mathbb{C})}^\times\rightarrow 0$

with arrows $\mathbb{Z}_{C(\mathbb{C})}\rightarrow \mathcal{O}_{C(\mathbb{C})}$  and $\mathcal{O}_{C(\mathbb{C})}\rightarrow \mathcal{O}_{C(\mathbb{C})}^\times$ defined by $1\mapsto 1$ and $f\mapsto e^{2\pi i f}$ respectively. This induces a long exact Čech cohomology sequence

$0\rightarrow H^0(C(\mathbb{C}),\mathbb{Z}_{C(\mathbb{C})})\rightarrow H^0(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})})\rightarrow H^0(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})}^\times)\rightarrow H^1(C(\mathbb{C}),\mathbb{Z}_{C(\mathbb{C})})\rightarrow H^1(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})})\rightarrow H^1(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})}^\times)\rightarrow H^2(C(\mathbb{C}),\mathbb{Z}_{C(\mathbb{C})})\rightarrow 0$

Above we used the isomorphism $H^2(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})})\cong H^2(C,\mathcal{O}_C)$, an application of GAGA, along with the equivalence of Čech cohomology and the derived functor cohomology for coherent sheaves, and Grothendieck vanishing to conclude these groups are 0.

The data in $H^1(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})}^\times)$ exactly classifies isomorphism classes of line bundles over $C(\mathbb{C})$, giving an isomorphism with $\text{Pic}(C(\mathbb{C}))$. The equivalence of categories of line bundles given by GAGA gives an isomorphism $\text{Pic}(C(\mathbb{C}))\cong\text{Pic}(C)$.

Since any complex curve is locally contractible in the complex topology, we have isomorphisms $H^i_{sing}(C(\mathbb{C}),\mathbb{Z})\cong H^i(C(\mathbb{C}),\mathcal{O}_{C(\mathbb{C})})$. If $g=\dim H^1(C,\mathcal{O}_C)$ is the genus of $C$, we can rewrite the above long exact sequence as

$0\rightarrow \mathbb{Z}\rightarrow \mathbb{C}\rightarrow \mathbb{C}^\times \rightarrow \mathbb{Z}^{2g}\rightarrow \mathbb{C}^g \rightarrow \text{Pic}(C)\rightarrow \mathbb{Z}\rightarrow 0$

from which we also observe $\mathbb{C}\rightarrow \mathbb{C}^\times$ is a surjection so that $\mathbb{Z}^{2g}\rightarrow \mathbb{C}^g$ is an injection (hence a lattice). This implies there is a short exact sequence $0\rightarrow \mathbb{C}^g/\mathbb{Z}^{2g}\rightarrow Pic(C)\rightarrow \mathbb{Z}\rightarrow 0$. Finally, since $\text{Ext}^1(\mathbb{Z},\mathbb{C}^g/\mathbb{Z}^{2g})=0$ and the correspondence with elements of $\text{Ext}^1$ and extensions, we find $Pic(C)\cong (\mathbb{C}^g/\mathbb{Z}^{2g})\oplus \mathbb{Z}$ as claimed. Of course, it is also possible here, as it was before, to define an explicit splitting of this sequence.$\square$

We’ll conclude by giving three examples of smooth projective curves with varying genus.

3. Example: Let $C=\text{Proj}(\mathbb{C}[x,y,z]/(y^2z=x^3-xz^2))$, which is the curve $V(y^2z=x^3-xz^2)\subset \mathbb{P}^2$. It is smooth since it has nowhere vanishing Jacobian matrix, and it has genus $1=1/2(3-1)(3-2)$ by the degree-genus formula. Then for some nonzero $\tau\in \mathbb{C}$ we have $\text{Pic}(C)\cong (\mathbb{C}/\mathbb{Z}+\tau\mathbb{Z})\oplus \mathbb{Z}$ and $K_0(C)\cong \mathbb{Z}\oplus(\mathbb{C}/\mathbb{Z}+\tau\mathbb{Z})\oplus\mathbb{Z}$.

Remark: I’m sure with more work one can find an explicit $\tau$ for this isomorphism. Partly because, the curve provided is a complex elliptic curve – these curves are in bijection with quotients of the complex line by lattices $\mathbb{C}/\mathbb{Z}+\tau\mathbb{Z}$ – and the map described above embedding the torus $T$ into the Picard group is actually a well known map which, I think, identifies $T$ as the Jacobian variety of the elliptic curve. If you want to look up more, some keywords are: Jacobian variety, Albanese variety, Abel-Jacobi theorem.

Remark 2: Using resultants, one can show a polynomial of the form $y^2=x^3+ax+b$ is smooth if and only if $4a^3+27b^2\neq 0$. One can then check, as I just did, the Jacobian matrix for the polynomial $y^2z=x^3+axz^2+bz^3$ doesn’t vanish whenever it is defined over a field of characteristic not $2,3$.

4. Example: Let $C=\text{Proj}(\mathbb{C}[x,y,z]/(x^5+y^5-z^5)$. The curve is irreducible, which can be seen here using the Eisenstein criterion (we make one adjustment: if $f(x,y,z)g(x,y,z)=x^5+y^5-z^5$ then at $z=1$ we find one of $f,g$ to be a unit, hence one of $f,g$ is the monomial $z^\alpha$ which is absurd). Using the Jacobian criterion, it follows this is a smooth curve. The degree-genus formula shows it is a genus 6 curve. Our principal result then says $K_0(C)\cong \mathbb{Z}\oplus (\mathbb{C}^6/\mathbb{Z}^{12})\oplus \mathbb{Z}$.

5. Example: Let $C=\text{Proj}(\mathbb{C}[x,y,z]/(yz+x^2-2xz+z^2))$. The degree-genus formula tells us this is a degree 0 curve. It is irreducible by the criterion here (using $y=y, f=x-1, u=1, v=2$ and then arguing by homogeneity as in example 4). Smootheness is checked by the nonvanishing of the Jacobian matrix $(2x-2z,z,y-2x+2z)$. Theorem 1 implies $K_0(C)=\mathbb{Z}\oplus \mathbb{Z}$.