The Stacks project

36.38 K-groups

A tiny bit about $K_0$ of various categories associated to schemes. Previous material can be found in Algebra, Section 10.55, Homology, Section 12.11, Derived Categories, Section 13.28, and More on Algebra, Lemma 15.119.2.

Analogous to Algebra, Section 10.55 we will define two $K$-groups $K'_0(X)$ and $K_0(X)$ for any Noetherian scheme $X$. The first will use coherent $\mathcal{O}_ X$-modules and the second will use finite locally free $\mathcal{O}_ X$-modules.

Lemma 36.38.1. Let $X$ be a Noetherian scheme. Then

\[ K_0(\textit{Coh}(\mathcal{O}_ X)) = K_0(D^ b(\textit{Coh}(\mathcal{O}_ X)) = K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X)) \]

Proof. The first equality is Derived Categories, Lemma 13.28.2. We have $K_0(\textit{Coh}(\mathcal{O}_ X)) = K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X))$ by Derived Categories, Lemma 13.28.5. This proves the lemma. (We can also use that $D^ b(\textit{Coh}(\mathcal{O}_ X)) = D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Proposition 36.11.2 to see the second equality.) $\square$

Here is the definition.

Definition 36.38.2. Let $X$ be a scheme.

  1. We denote $K_0(X)$ the Grothendieck group of $X$. It is the zeroth K-group of the strictly full, saturated, triangulated subcategory $D_{perf}(\mathcal{O}_ X)$ of $D(\mathcal{O}_ X)$ consisting of perfect objects. In a formula

    \[ K_0(X) = K_0(D_{perf}(\mathcal{O}_ X)) \]
  2. If $X$ is locally Noetherian, then we denote $K'_0(X)$ the Grothendieck group of coherent sheaves on $X$. It is the is the zeroth $K$-group of the abelian category of coherent $\mathcal{O}_ X$-modules. In a formula

    \[ K'_0(X) = K_0(\textit{Coh}(\mathcal{O}_ X)) \]

We will show that our definition of $K_0(X)$ agrees with the often used definition in terms of finite locally free modules if $X$ has the resolution property (for example if $X$ has an ample invertible module). See Lemma 36.38.5.

Lemma 36.38.3. Let $X = \mathop{\mathrm{Spec}}(R)$ be an affine scheme. Then $K_0(X) = K_0(R)$ and if $R$ is Noetherian then $K'_0(X) = K'_0(R)$.

Proof. Recall that $K'_0(R)$ and $K_0(R)$ have been defined in Algebra, Section 10.55.

By More on Algebra, Lemma 15.119.2 we have $K_0(R) = K_0(D_{perf}(R))$. By Lemmas 36.10.7 and 36.3.5 we have $D_{perf}(R) = D_{perf}(\mathcal{O}_ X)$. This proves the equality $K_0(R) = K_0(X)$.

The equality $K'_0(R) = K'_0(X)$ holds because $\textit{Coh}(\mathcal{O}_ X)$ is equivalent to the category of finite $R$-modules by Cohomology of Schemes, Lemma 30.9.1. Moreover it is clear that $K'_0(R)$ is the zeroth K-group of the category of finite $R$-modules from the definitions. $\square$

Let $X$ be a Noetherian scheme. Then both $K'_0(X)$ and $K_0(X)$ are defined. In this case there is a canonical map

\[ K_0(X) = K_0(D_{perf}(\mathcal{O}_ X)) \longrightarrow K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X)) = K'_0(X) \]

Namely, perfect complexes are in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ (by Lemma 36.10.3), the inclusion functor $D_{perf}(\mathcal{O}_ X) \to D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ induces a map on zeroth $K$-groups (Derived Categories, Lemma 13.28.3), and we have the equality on the right by Lemma 36.38.1.

Lemma 36.38.4. Let $X$ be a Noetherian regular scheme. Then the map $K_0(X) \to K'_0(X)$ is an isomorphism.

Proof. Follows immediately from Lemma 36.11.8 and our construction of the map $K_0(X) \to K'_0(X)$ above. $\square$

Let $X$ be a scheme. Let us denote $\textit{Vect}(X)$ the category of finite locally free $\mathcal{O}_ X$-modules. Although $\textit{Vect}(X)$ isn't an abelian category in general, it is clear what a short exact sequence of $\textit{Vect}(X)$ is. Denote $K_0(\textit{Vect}(X))$ the unique abelian group with the following properties1:

  1. For every finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ there is given an element $[\mathcal{E}]$ in $K_0(\textit{Vect}(X))$,

  2. for every short exact sequence $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ of finite locally free $\mathcal{O}_ X$-modules we have the relation $[\mathcal{E}] = [\mathcal{E}'] + [\mathcal{E}'']$ in $K_0(\textit{Vect}(X))$,

  3. the group $K_0(\textit{Vect}(X))$ is generated by the elements $[\mathcal{E}]$, and

  4. all relations in $K_0(\textit{Vect}(X))$ among the generators $[\mathcal{E}]$ are $\mathbf{Z}$-linear combinations of the relations coming from exact sequences as above.

We omit the detailed construction of $K_0(\textit{Vect}(X))$. There is a natural map

\[ K_0(\textit{Vect}(X)) \longrightarrow K_0(X) \]

Namely, given a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ let us denote $\mathcal{E}[0]$ the perfect complex on $X$ which has $\mathcal{E}$ sitting in degree $0$ and zero in other degrees. Given a short exact sequence $0 \to \mathcal{E} \to \mathcal{E}' \to \mathcal{E}'' \to 0$ of finite locally free $\mathcal{O}_ X$-modules we obtain a distinguished triangle $\mathcal{E}[0] \to \mathcal{E}'[0] \to \mathcal{E}''[0] \to \mathcal{E}[1]$, see Derived Categories, Section 13.12. This shows that we obtain a map $K_0(\textit{Vect}(X)) \to K_0(D_{perf}(\mathcal{O}_ X)) = K_0(X)$ by sending $[\mathcal{E}]$ to $[\mathcal{E}[0]]$ with apologies for the horrendous notation.

Lemma 36.38.5. Let $X$ be a quasi-compact and quasi-separated scheme with the resolution property. Then the map $K_0(\textit{Vect}(X)) \to K_0(X)$ is an isomorphism.

Proof. This lemma will follow in a straightforward manner from Lemmas 36.37.2, 36.37.3, and 36.37.4 whose results we will use without further mention. Let us construct an inverse map

\[ c : K_0(X) = K_0(D_{perf}(\mathcal{O}_ X)) \longrightarrow K_0(\textit{Vect}(X)) \]

Namely, any object of $D_{perf}(\mathcal{O}_ X)$ can be represented by a bounded complex $\mathcal{E}^\bullet $ of finite locally free $\mathcal{O}_ X$-modules. Then we set

\[ c([\mathcal{E}^\bullet ]) = \sum (-1)^ i[\mathcal{E}^ i] \]

Of course we have to show that this is well defined. For the moment we view $c$ as a map defined on bounded complexes of finite locally free $\mathcal{O}_ X$-modules.

Suppose that $\mathcal{E}^\bullet \to \mathcal{F}^\bullet $ is a surjective map of bounded complexes of finite locally free $\mathcal{O}_ X$-modules. Let $\mathcal{K}^\bullet $ be the kernel. Then we obtain short exact sequences of $\mathcal{O}_ X$-modules

\[ 0 \to \mathcal{K}^ n \to \mathcal{E}^ n \to \mathcal{F}^ n \to 0 \]

which are locally split because $\mathcal{F}^ n$ is finite locally free. Hence $\mathcal{K}^\bullet $ is also a bounded complex of finite locally free $\mathcal{O}_ X$-modules and we have $c(\mathcal{E}^\bullet ) = c(\mathcal{K}^\bullet ) + c(\mathcal{F}^\bullet )$ in $K_0(\textit{Vect}(X))$.

Suppose given a bounded complex $\mathcal{E}^\bullet $ of finite locally free $\mathcal{O}_ X$-modules which is acyclic. Say $\mathcal{E}^ n = 0$ for $n \not\in [a, b]$. Then we can break $\mathcal{E}^\bullet $ into short exact sequences

\[ \begin{matrix} 0 \to \mathcal{E}^ a \to \mathcal{E}^{a + 1} \to \mathcal{F}^{a + 1} \to 0, \\ 0 \to \mathcal{F}^{a + 1} \to \mathcal{E}^{a + 2} \to \mathcal{F}^{a + 3} \to 0, \\ \ldots \\ 0 \to \mathcal{F}^{b - 3} \to \mathcal{E}^{b - 2} \to \mathcal{F}^{b - 2} \to 0, \\ 0 \to \mathcal{F}^{b - 2} \to \mathcal{E}^{b - 1} \to \mathcal{E}^ b \to 0 \end{matrix} \]

Arguing by descending induction we see that $\mathcal{F}^{b - 2}, \ldots , \mathcal{F}^{a + 1}$ are finite locally free $\mathcal{O}_ X$-modules, and

\[ c(\mathcal{E}^\bullet ) = \sum (-1)[\mathcal{E}^ n] = \sum (-1)^ n([\mathcal{F}^{n - 1}] + [\mathcal{F}^ n]) = 0 \]

Thus our construction gives zero on acyclic complexes.

It follows from the results of the preceding two paragraphs that $c$ is well defined. Namely, suppose the bounded complexes $\mathcal{E}^\bullet $ and $\mathcal{F}^\bullet $ of finite locally free $\mathcal{O}_ X$-modules represent the same object of $D(\mathcal{O}_ X)$. Then we can find quasi-isomorphisms $a : \mathcal{G}^\bullet \to \mathcal{E}^\bullet $ and $b : \mathcal{G}^\bullet \to \mathcal{F}^\bullet $ with $\mathcal{G}^\bullet $ bounded complex of finite locally free $\mathcal{O}_ X$-modules. We obtain a short exact sequence of complexes

\[ 0 \to \mathcal{E}^\bullet \to C(a)^\bullet \to \mathcal{G}^\bullet [1] \to 0 \]

see Derived Categories, Definition 13.9.1. Since $a$ is a quasi-isomorphism, the cone $C(a)^\bullet $ is acyclic (this follows for example from the discussion in Derived Categories, Section 13.12). Hence

\[ 0 = c(C(f)^\bullet ) = c(\mathcal{E}^\bullet ) + c(\mathcal{G}^\bullet [1]) = c(\mathcal{E}^\bullet ) - c(\mathcal{G}^\bullet ) \]

as desired. The same argument using $b$ shows that $0 = c(\mathcal{F}^\bullet ) - c(\mathcal{G}^\bullet )$. Hence we find that $c(\mathcal{E}^\bullet ) = c(\mathcal{F}^\bullet )$ and $c$ is well defined.

A similar argument using the cone on a map $\mathcal{E}^\bullet \to \mathcal{F}^\bullet $ of bounded complexes of finite locally free $\mathcal{O}_ X$-modules shows that $c(Y) = c(X) + c(Z)$ if $X \to Y \to Z$ is a distinguished triangle in $D_{perf}(\mathcal{O}_ X)$. Details omitted. Thus we get the desired homomorphism of abelian groups $c : K_0(X) \to K_0(\textit{Vect}(X))$.

It is clear that the composition $K_0(\textit{Vect}(X)) \to K_0(X) \to K_0(\textit{Vect}(X))$ is the identity. On the other hand, let $\mathcal{E}^\bullet $ be a bounded complex of finite locally free $\mathcal{O}_ X$-modules. Then the the existence of the distinguished triangles of “stupid truncations” (see Homology, Section 12.15)

\[ \sigma _{\geq n}\mathcal{E}^\bullet \to \sigma _{\geq n - 1}\mathcal{E}^\bullet \to \mathcal{E}^{n - 1}[-n + 1] \to (\sigma _{\geq n}\mathcal{E}^\bullet )[1] \]

and induction show that

\[ [\mathcal{E}^\bullet ] = \sum (-1)^ i[\mathcal{E}^ i[0]] \]

in $K_0(X) = K_0(D_{perf}(\mathcal{O}_ X))$ with apologies for the notation. Hence the map $K_0(\textit{Vect}(X)) \to K_0(D_{perf}(\mathcal{O}_ X)) = K_0(X)$ is surjective which finishes the proof. $\square$

Remark 36.38.6. Let $X$ be a scheme. The K-group $K_0(X)$ is canonically a commutative ring. Namely, using the derived tensor product

\[ \otimes = \otimes ^\mathbf {L}_{\mathcal{O}_ X} : D_{perf}(\mathcal{O}_ X) \times D_{perf}(\mathcal{O}_ X) \longrightarrow D_{perf}(\mathcal{O}_ X) \]

and Derived Categories, Lemma 13.28.6 we obtain a bilinear multiplication. Since $K \otimes L \cong L \otimes K$ we see that this product is commutative. Since $(K \otimes L) \otimes M = K \otimes (L \otimes M)$ we see that this product is associative. Finally, the unit of $K_0(X)$ is the element $1 = [\mathcal{O}_ X]$.

If $\textit{Vect}(X)$ and $K_0(\textit{Vect}(X))$ are as above, then it is clearly the case that $K_0(\textit{Vect}(X))$ also has a ring structure: if $\mathcal{E}$ and $\mathcal{F}$ are finite locally free $\mathcal{O}_ X$-modules, then we set

\[ [\mathcal{E}] \cdot [\mathcal{F}] = [\mathcal{E} \otimes _{\mathcal{O}_ X} \mathcal{F}] \]

The reader easily verifies that this indeed defines a bilinear commutative, associative product. Details omitted. The map

\[ K_0(\textit{Vect}(X)) \longrightarrow K_0(X) \]

constructed above is a ring map with these definitions.

Now assume $X$ is Noetherian. The derived tensor product also produces a map

\[ \otimes = \otimes ^\mathbf {L}_{\mathcal{O}_ X} : D_{perf}(\mathcal{O}_ X) \times D^ b_{\textit{Coh}}(\mathcal{O}_ X) \longrightarrow D^ b_{\textit{Coh}}(\mathcal{O}_ X) \]

Again using Derived Categories, Lemma 13.28.6 we obtain a bilinear multiplication $K_0(X) \times K'_0(X) \to K'_0(X)$ since $K'_0(X) = K_0(D^ b_{\textit{Coh}}(\mathcal{O}_ X))$ by Lemma 36.38.1. The reader easily shows that this gives $K'_0(X)$ the structure of a module over the ring $K_0(X)$.

Remark 36.38.7. Let $f : X \to Y$ be a proper morphism of locally Noetherian schemes. There is a map

\[ f_* : K'_0(X) \longrightarrow K'_0(Y) \]

which sends $[\mathcal{F}]$ to

\[ [\bigoplus \nolimits _{i \geq 0} R^{2i}f_*\mathcal{F}] - [\bigoplus \nolimits _{i \geq 0} R^{2i + 1}f_*\mathcal{F}] \]

This is well defined because the sheaves $R^ if_*\mathcal{F}$ are coherent (Cohomology of Schemes, Proposition 30.19.1), because locally only a finite number are nonzero, and because a short exact sequence of coherent sheaves on $X$ produces a long exact sequence of $R^ if_*$ on $Y$. If $Y$ is quasi-compact (the only case most often used in practice), then we can rewrite the above as

\[ f_*[\mathcal{F}] = \sum (-1)^ i[R^ if_*\mathcal{F}] = [Rf_*\mathcal{F}] \]

where we have used the equality $K'_0(Y) = K_0(D^ b_{\textit{Coh}}(Y))$ from Lemma 36.38.1.

Lemma 36.38.8. Let $f : X \to Y$ be a proper morphism of locally Noetherian schemes. Then we have $f_*(\alpha \cdot f^*\beta ) = f_*\alpha \cdot \beta $ for $\alpha \in K'_0(X)$ and $\beta \in K_0(Y)$.

Proof. Follows from Lemma 36.22.1, the discussion in Remark 36.38.7, and the definition of the product $K'_0(X) \times K_0(X) \to K'_0(X)$ in Remark 36.38.6. $\square$

Remark 36.38.9. Let $X$ be a scheme. Let $Z \subset X$ be a closed subscheme. Consider the strictly full, saturated, triangulated subcategory

\[ D_{Z, perf}(\mathcal{O}_ X) \subset D(\mathcal{O}_ X) \]

consisting of perfect complexes of $\mathcal{O}_ X$-modules whose cohomology sheaves are settheoretically supported on $Z$. The zeroth $K$-group $K_0(D_{Z, perf}(\mathcal{O}_ X))$ of this triangulated category is sometimes denoted $K_ Z(X)$ or $K_{0, Z}(X)$. Using derived tensor product exactly as in Remark 36.38.6 we see that $K_0(D_{Z, perf}(\mathcal{O}_ X))$ has a multiplication which is associative and commutative, but in general $K_0(D_{Z, perf}(\mathcal{O}_ X))$ doesn't have a unit.

[1] The correct generality here would be to define $K_0$ for any exact category, see Injectives, Remark 19.9.6.

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