The Grothendieck group of the derived category

R. Virk

Department of Mathematics
University of Wisconsin
Madison, WI 53706

virk@math.wisc.edu

Let $\mathcal{C}$ be an abelian category, let $\mathcal{D}^b(\mathcal{C})$ be the associated bounded derived category. Denote by $K_0(\mathcal{C})$ the Grothendieck group of $\mathcal{C}$, this is the free abelian group on isomorphism classes $[M]$ of objects in $\mathcal{C}$ modulo the relations $[M]=[N]+[L]$ for every exact sequence $0 \to N \to M \to L \to 0$. Similarly, $K_0(\mathcal{D}^b(\mathcal{C}))$ is defined as the free abelian group on isomorphism classes $[X]$ of complexes in $\mathcal{D}^b(\mathcal{C})$ modulo the relations $[X]=[Y]+[Z]$ for every distinguished triangle $Y\to X \to Z \leadsto Y[1]$. We are using the convention that for a complex $X=(X^i, d_X^i)$, the complex $X[n]$ is given by $(X[n])^i=X^{n+i}$ and $d^i_{X[n]}=(-1)^nd_X^{i+n}$.

Lemma 0.1   With notation as above, let $X$ be a complex in $\mathcal{D}^b(\mathcal{C})$ with cohomology concentrated in exactly one degree, then in $K_0(\mathcal{D}^b(\mathcal{C}))$ we have that

$$[X]+[X[-1]] =0.$$

Proof. We may assume that $X$ has exactly one non-zero component, say $X^i$. Let $Y$ be the complex

$$\cdots \to 0 \to 0 \to X^i {\smash{\mathop {\to}\limits^{\mathrm{id}}}} X^i \to 0 \to 0 \to \cdots$$

where the first $X^i$ is in degree $i$. Then $Y \simeq 0$ in $\mathcal{D}^b(\mathcal{C})$. Furthermore, we have a commutative diagram

$$\xymatrix{ \cdots \ar[r]& 0 \ar[r]& X^i \ar[d]_{\mathrm{id}}\ar[r... ...ots\\ \cdots \ar[r]&0\ar[r]&X^i\ar[r]^{\mathrm{id}}&X^i\ar[r]&0\ar[r]&\cdots }$$

The mapping cone of the resulting map is isomorphic to $X[-1]$. $\qedsymbol$

Remark 0.2   The argument in the latter part of the proof of the next proposition shows that the above lemma is in fact true for an abitrary complex $X$ in $\mathcal{D}^b(\mathcal{C})$.

Proposition 0.3   With notation as before

$$K_0(\mathcal{C}) \simeq K_0(\mathcal{D}^b(\mathcal{C})).$$

Proof. Let $X$ be a complex in $\mathcal{D}^b(\mathcal{C})$. We define an element in $K_0(\mathcal{C})$ associated to $X$, the Euler characteristic of $X$ as

$$\chi(X)=\sum_{i\in \mathbb{Z}}(-1)^i [H^i(X)].$$

If $X\sim X'$ in $D^b(\mathcal{C})$, then $H^i(X)\sim H^i(X')$, hence $\chi(X)=\chi(X')$. Moreover, if $Y\to X \to Z \leadsto Y[1]$ is a distinguished triangle in $\mathcal{D}^b(\mathcal{C})$, then in $\mathcal{C}$ we have the long exact sequence of cohomology

$$\cdots \to H^i(Y^i) \to H^i(X^i) \to H^i(Z^i) \to H^{i+1}(Y^{i+1}) \to \cdots$$

which gives that $\chi(X)=\chi(Y)+\chi(Z)$. Thus, we have a well defined group homomorphism

$$\alpha: K_0(\mathcal{D}^b(\mathcal{C})) \to K_0(\mathcal{C}),$$

$$[X] \mapsto \chi(X).$$

Using the canonical embedding, $\iota: \mathcal{C}\to \mathcal{D}^b(\mathcal{C})$ given by viewing an object of $\mathcal{C}$ as a complex with cohomology concentrated in degree 0, we get a group homomorphism

$$\beta: K_0(\mathcal{C}) \to K_0(\mathcal{D}^b(\mathcal{C})),$$

$$[X] \mapsto [\iota(X)].$$

It is clear that $\alpha \circ \beta =\mathrm{id}_{K_0(\mathcal{C})}$. On the other hand, for any complex $X\in \mathcal{D}^b(X)$, we claim that

$$[X]=\sum_{i\in\mathbb{Z}}(-1)^i[\iota(H^i(X))].$$

This is seen as follows: let $X^i$ be the largest non-zero component of $X$. Then the following diagram is commutative

$$\xymatrix{ \ar[r]&X^{i-3}\ar[d]_{\mathrm{id}}\ar[r]&X^{i-2}\ar[d]... ...0 \\ \ar[r]&X^{i-3}\ar[r]&X^{i-2}\ar[r]&X^{i-1}\ar[r]^{\varphi}&X^i\ar[r]& 0 }$$

The mapping cone of the resulting map is isomorphic to $\iota(H^i(X))[-i]$ in $\mathcal{D}^b(\mathcal{C})$, now using the previous lemma and iterating this construction on the top row of the above diagram we get that $[X]=\sum_{i\in\mathbb{Z}}(-1)^i[\iota (H^i(X))]$. Thus, $\beta\circ \alpha =\mathrm{id}_{K_0(\mathcal{D}^b(\mathcal{C}))}$. $\qedsymbol$