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 $[Y] - [X] - [Z] = 0$.

## 13.28 K-groups

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

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

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)

and induction show that

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 $\mathcal{A}$. There is a canonical isomorphism

The inverse sends the class $[X]$ of $X$ to the element $\sum (-1)^ i[H^ i(X)]$.

**Proof.**
We omit the verification that the rule for the inverse gives a well defined map $K_0(D^ b_\mathcal {B}(\mathcal{A})) \to K_0(\mathcal{B})$. It is immediate that the composition $K_0(\mathcal{B}) \to K_0(D^ b_\mathcal {B}(\mathcal{A})) \to K_0(\mathcal{B})$ is the identity. On the other hand, using the distinguished triangles of Remark 13.12.4 and an induction argument the reader may show that the displayed arrow in the statement of the lemma is surjective (details omitted). The lemma follows.
$\square$

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

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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)