## 13.28 K-groups

A tiny bit about $K_0$ of a triangulated category.

Definition 13.28.1. Let $\mathcal{D}$ be a triangulated category. We denote $K_0(\mathcal{D})$ the zeroth $K$-group of $\mathcal{D}$. It is the abelian group constructed as follows. Take the free abelian group on the objects on $\mathcal{D}$ and for every distinguished triangle $X \to Y \to Z$ impose the relation $[Y] - [X] - [Z] = 0$.

Observe that this implies that $[X[n]] = (-1)^ n[X]$ because we have the distinguished triangle $(X, 0, X[1], 0, 0, -\text{id}[1])$.

Lemma 13.28.2. Let $\mathcal{A}$ be an abelian category. Then there is a canonical identification $K_0(D^ b(\mathcal{A})) = K_0(\mathcal{A})$ of zeroth $K$-groups.

Proof. Given an object $A$ of $\mathcal{A}$ denote $A[0]$ the object $A$ viewed as a complex sitting in degree $0$. If $0 \to A \to A' \to A'' \to 0$ is a short exact sequence, then we get a distinguished triangle $A[0] \to A'[0] \to A''[0] \to A[1]$, see Section 13.12. This shows that we obtain a map $K_0(\mathcal{A}) \to K_0(D^ b(\mathcal{A}))$ by sending $[A]$ to $[A[0]]$ with apologies for the horrendous notation.

On the other hand, given an object $X$ of $D^ b(\mathcal{A})$ we can consider the element

$c(X) = \sum (-1)^ i[H^ i(X)] \in K_0(\mathcal{A})$

Given a distinguished triangle $X \to Y \to Z$ the long exact sequence of cohomology (13.11.1.1) and the relations in $K_0(\mathcal{A})$ show that $c(Y) = c(X) + c(Z)$. Thus $c$ factors through a map $c : K_0(D^ b(\mathcal{A})) \to K_0(\mathcal{A})$.

We want to show that the two maps above are mutually inverse. It is clear that the composition $K_0(\mathcal{A}) \to K_0(D^ b(\mathcal{A})) \to K_0(\mathcal{A})$ is the identity. Suppose that $X^\bullet$ is a bounded complex of $\mathcal{A}$. The existence of the distinguished triangles of “stupid truncations” (see Homology, Section 12.15)

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

and induction show that

$[X^\bullet ] = \sum (-1)^ i[X^ i[0]]$

in $K_0(D^ b(\mathcal{A}))$ (with again apologies for the notation). It follows that the composition $K_0(\mathcal{A}) \to K_0(D^ b(\mathcal{A}))$ is surjective which finishes the proof. $\square$

Lemma 13.28.3. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of triangulated categories. Then $F$ induces a group homomorphism $K_0(\mathcal{D}) \to K_0(\mathcal{D}')$.

Proof. Omitted. $\square$

Lemma 13.28.4. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor from a triangulated category to an abelian category. Assume that for any $X$ in $\mathcal{D}$ only a finite number of the objects $H(X[i])$ are nonzero in $\mathcal{A}$. Then $H$ induces a group homomorphism $K_0(\mathcal{D}) \to K_0(\mathcal{A})$ sending $[X]$ to $\sum (-1)^ i[H(X[i])]$.

Proof. Omitted. $\square$

Lemma 13.28.5. Let $\mathcal{B}$ be a weak Serre subcategory of the abelian category $\mathcal{A}$. Then there are canonical maps

$K_0(\mathcal{B}) \longrightarrow K_0(D^ b_\mathcal {B}(\mathcal{A})) \longrightarrow K_0(\mathcal{B})$

whose composition is zero. The second arrow sends the class $[X]$ of the object $X$ to the element $\sum (-1)^ i[H^ i(X)]$ of $K_0(\mathcal{B})$.

Proof. Omitted. $\square$

Lemma 13.28.6. Let $\mathcal{D}$, $\mathcal{D}'$, $\mathcal{D}''$ be triangulated categories. Let

$\otimes : \mathcal{D} \times \mathcal{D}' \longrightarrow \mathcal{D}''$

be a functor such that for fixed $X$ in $\mathcal{D}$ the functor $X \otimes - : \mathcal{D}' \to \mathcal{D}''$ is an exact functor and for fixed $X'$ in $\mathcal{D}'$ the functor $- \otimes X' : \mathcal{D} \to \mathcal{D}''$ is an exact functor. Then $\otimes$ induces a bilinear map $K_0(\mathcal{D}) \times K_0(\mathcal{D}') \to K_0(\mathcal{D}'')$ which sends $([X], [X'])$ to $[X \otimes X']$.

Proof. Omitted. $\square$

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