Zero in the Grothedieck group of the derived category

Question

I have a problem. I was wondering whether there is a precise answer to the following question.

Let $\mathcal{A}$ be an abelian category and $\mathcal{D}^b(\mathcal{A})$ its bounded derived category. Then we have the usual definition of Grothendieck group $\mathcal{K}_0(\mathcal{D}^b(\mathcal{A}))$ by modulo $[B] =[A]+[C] $ for every distinguished triangles $A \rightarrow B \rightarrow C \rightarrow A[1]$.

$\bf My$ $\bf Question:$ What are zeros $[X] = 0$ in $\mathcal{K}_0(\mathcal{D}^b(\mathcal{A}))$ for $X \in \mathcal{D}^b(\mathcal{A})$? Namely, how can we characterize those complexes who are zero in $\mathcal{K}_0(\mathcal{D}^b(\mathcal{A}))$?

Thanks very much!

Answer 1

The problem of understanding elements in $\mathcal{K}_0(\mathcal{D}^b(\mathcal{A}))$ can be somewhat reduced to that of understanding elements in $\mathcal{K}_0(\mathcal{A})$ via the isomorphism $$\mathcal{K}_0(\mathcal{D}^b(\mathcal{A})) \to \mathcal{K}_0(\mathcal{A}): [X] \mapsto \sum_{i=-\infty}^{\infty}(-1)^i [H^i(X)].$$

Thus if you know the cohomology of an element $X$ of the derived category, and if you understand $\mathcal{K}_0(\mathcal{A})$ well enough, then you can decide if $[X]=0$.