2
$\begingroup$

Notations : $R$ is a commutative ring with unity. $P(R)$ is the category of finitely generated projective $R-$ modules, $Ch^{b}(P(R))$ is the the category of bounded chain complexes on $P(R)$ and $C^q(P(R))$ is the category of bounded exact chain-complexes on $P(R)$.

Each of the above mentioned categories are exact categories. If I define the weak equivalence class as the isomorphism classes then $K_0$ of each of the categories will be the quotient of the free group generated by the isomorphism classes of the elements $[C]$ where $C \in ob\;\mathcal{C} .$ ($\mathcal{C}$ being any of the above categories). If we replace $R$ by a field $\mathbb{F}$ then my question is will the inclusion functor $i : C^q(P(\mathbb{F})) \longrightarrow Ch^{b}(P(\mathbb{F}))$ induce an injective group homomorphism from $$K_0C^q(P(\mathbb{F})) \longrightarrow K_0Ch^{b}(P(\mathbb{F}))?$$ My attempt :

Proposition : For an exact category $\mathcal{C}$ if $[A_1] = [A_2]$ in $K_0(\mathcal{A})$ then there are short exact sequences $0 \rightarrow C' \rightarrow C_1 \rightarrow C'' \rightarrow 0$ and $0 \rightarrow C' \rightarrow C_2 \rightarrow C'' \rightarrow 0$ such that $A_1 \oplus C_1 \cong A_2 \oplus C_2$.

So what I was attempting to show that the kernel of the induced map is trivial, now any typical element in $K_0C^q(P(\mathbb{F}))$ is either $[F.,d]$ or $[F.,d] -[F'.,d']$; ($d,d'$ being the differential). If $[F.,d] \longmapsto 0$ then $F.$ as a complex is itself $0$ according to the proposition (because here the SES of the proposition will split). Thus no problem here

For the second case what I have figured out so far is that if $[F. d] = [F'.,d']$ in $K_0Ch^{b}(P(\mathbb{F}))$ then for each $n$ I have $F_n = F'_n$. (Because there will be a splitting in the SES of the proposition) but then I am unable to proceed further, if you could please point me to the right direction I will be grateful.

$\endgroup$
2
$\begingroup$

Yes, it's injective. Both $K_0$ groups are free with infinite countable basis. The basis of the first one is formed by the chain complexes $$\cdots\rightarrow0\rightarrow\mathbb{F}\rightarrow\mathbb{F}\rightarrow0\rightarrow\cdots$$ concentrated in two consecutive degrees (the middle arrow is the identity). The basis of the second one is the union of this set with the set formed by the complexes $$\cdots\rightarrow0\rightarrow\mathbb{F}\rightarrow0\rightarrow\cdots$$ concentrated in a single dimension. The morphism between them is given by the inclusion of the first basis into the second one.

EDIT: since you didn't specify the exact structure but you took the trivial weak equivalences, I also assumed you consider the trivial exact structure. If you take the exact structure given by short exact sequences instead, then the source is the same since all bounded exact complexes are projective. In the target, each of the first generators is the sum of two of the second kind, the obvious ones, so a basis of the target is given by the second set. The morphism is still injective since it looks like $$\bigoplus_{-\infty}^{+\infty}\mathbb{Z}\rightarrow \bigoplus_{-\infty}^{+\infty}\mathbb{Z}\colon e_i\mapsto e_{i+1}-e_i.$$

$\endgroup$
3
  • $\begingroup$ Right sir, I understand now, and by trivial exact structure you mean exact sequence as in $R$ modules right? Thank you for your answer. $\endgroup$
    – Dibya
    Feb 23 at 17:43
  • $\begingroup$ Also then this will be true for any $R$ as well right for which $P(R)$ is $Free(R)$. $\endgroup$
    – Dibya
    Feb 23 at 17:45
  • 1
    $\begingroup$ The trivial exact structure has just split short exact sequences. This is not valid for rings other than fields, in general. $\endgroup$ Feb 24 at 0:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.