
12.10 K-groups

Definition 12.10.1. Let $\mathcal{A}$ be an abelian category. We denote $K_0(\mathcal{A})$ the zeroth $K$-group of $\mathcal{A}$. It is the abelian group constructed as follows. Take the free abelian group on the objects on $\mathcal{A}$ and for every short exact sequence $0 \to A \to B \to C \to 0$ impose the relation $[B] - [A] - [C] = 0$.

Another way to say this is that there is a presentation

$\bigoplus _{A \to B \to C\text{ ses}} \mathbf{Z}[A \to B \to C] \longrightarrow \bigoplus _{A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})} \mathbf{Z}[A] \longrightarrow K_0(\mathcal{A}) \longrightarrow 0$

with $[A \to B \to C] \mapsto [B] - [A] - [C]$ of $K_0(\mathcal{A})$. The short exact sequence $0 \to 0 \to 0 \to 0 \to 0$ leads to the relation $[0] = 0$ in $K_0(\mathcal{A})$. There are no set-theoretical issues as all of our categories are “small” if not mentioned otherwise. Some examples of $K$-groups for categories of modules over rings where computed in Algebra, Section 10.54.

Lemma 12.10.2. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor between abelian categories. Then $F$ induces a homomorphism of $K$-groups $K_0(F) : K_0(\mathcal{A}) \to K_0(\mathcal{B})$ by simply setting $K_0(F)([A]) = [F(A)]$.

Proof. Proves itself. $\square$

Suppose we are given an object $M$ of an abelian category $\mathcal{A}$ and a complex of the form

12.10.2.1
$$\label{homology-equation-cyclic-complex} \xymatrix{ \ldots \ar[r] & M \ar[r]^\varphi & M \ar[r]^\psi & M \ar[r]^\varphi & M \ar[r] & \ldots }$$

In this situation we define

$H^0(M, \varphi , \psi ) = \mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(\varphi ) , \quad \text{and}\quad H^1(M, \varphi , \psi ) = \mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\psi ).$

Lemma 12.10.3. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory and set $\mathcal{B} = \mathcal{A}/\mathcal{C}$.

1. The exact functors $\mathcal{C} \to \mathcal{A}$ and $\mathcal{A} \to \mathcal{B}$ induce an exact sequence

$K_0(\mathcal{C}) \to K_0(\mathcal{A}) \to K_0(\mathcal{B}) \to 0$

of $K$-groups, and

2. the kernel of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$ is equal to the collection of elements of the form

$[H^0(M, \varphi , \psi )] - [H^1(M, \varphi , \psi )]$

where $(M, \varphi , \psi )$ is a complex as in (12.10.2.1) with the property that it becomes exact in $\mathcal{B}$; in other words that $H^0(M, \varphi , \psi )$ and $H^1(M, \varphi , \psi )$ are objects of $\mathcal{C}$.

Proof. We omit the proof of (1). The proof of (2) is in a sense completely combinatorial. First we remark that any class of the type $[H^0(M, \varphi , \psi )] - [H^1(M, \varphi , \psi )]$ is zero in $K_0(\mathcal{A})$ by the following calculation

\begin{align*} 0 & = [M] - [M] \\ & = [\mathop{\mathrm{Ker}}(\varphi )] + [\mathop{\mathrm{Im}}(\varphi )] - [\mathop{\mathrm{Ker}}(\psi )] - [\mathop{\mathrm{Im}}(\psi )] \\ & = [\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\psi )] - [\mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(\varphi )] \\ & = [H^1(M, \varphi , \psi )] - [H^0(M, \varphi , \psi )] \end{align*}

as desired. Hence it suffices to show that any element in the kernel of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$ is of this form.

Any element $x$ in $K_0(\mathcal{C})$ can be represented as the difference $x = [P] - [Q]$ of two objects of $\mathcal{C}$ (fun exercise). Suppose that this element maps to zero in $K_0(\mathcal{A})$. This means that there exist

1. a finite set $I = I^{+} \amalg I^{-}$,

2. for each $i \in I$ a short exact sequence

$0 \to A_ i \to B_ i \to C_ i \to 0$

in the abelian category $\mathcal{A}$

such that

$[P] - [Q] = \sum \nolimits _{i \in I^{+}} ([B_ i] - [A_ i] - [C_ i]) - \sum \nolimits _{i \in I^{-}} ([B_ i] - [A_ i] - [C_ i])$

in the free abelian group on the objects of $\mathcal{A}$. We can rewrite this as

$[P] + \sum \nolimits _{i \in I^{+}} ([A_ i] + [C_ i]) + \sum \nolimits _{i \in I^{-}} [B_ i] = [Q] + \sum \nolimits _{i \in I^{-}} ([A_ i] + [C_ i]) + \sum \nolimits _{i \in I^{+}} [B_ i].$

Since the right and left hand side should contain the same objects of $\mathcal{A}$ counted with multiplicity, this means there should be a bijection $\tau$ between the terms which occur above. Set

$T^{+} = \{ p\} \ \amalg \ \{ a, c\} \times I^{+}\ \amalg \ \{ b\} \times I^{-}$

and

$T^{-} = \{ q\} \ \amalg \ \{ a, c\} \times I^{-}\ \amalg \ \{ b\} \times I^{+}.$

Set $T = T^{+} \amalg T^{-} = \{ p, q\} \amalg \{ a, b, c\} \times I$. For $t \in T$ define

$O(t) = \left\{ \begin{matrix} P & \text{if} & t = p \\ Q & \text{if} & t = q \\ A_ i & \text{if} & t = (a, i) \\ B_ i & \text{if} & t = (b, i) \\ C_ i & \text{if} & t = (c, i) \end{matrix} \right.$

Hence we can view $\tau : T^{+} \to T^{-}$ as a bijection such that $O(t) = O(\tau (t))$ for all $t \in T^{+}$. Let $t^{-}_0 = \tau (p)$ and let $t^{+}_0 \in T^{+}$ be the unique element such that $\tau (t^{+}_0) = q$. Consider the object

$M^{+} = \bigoplus \nolimits _{t \in T^{+}} O(t)$

By using $\tau$ we see that it is equal to the object

$M^{-} = \bigoplus \nolimits _{t \in T^{-}} O(t)$

Consider the map

$\varphi : M^{+} \longrightarrow M^{-}$

which on the summand $O(t) = A_ i$ corresponding to $t = (a, i)$, $i \in I^{+}$ uses the map $A_ i \to B_ i$ into the summand $O((b, i)) = B_ i$ of $M^{-}$ and on the summand $O(t) = B_ i$ corresponding to $(b, i)$, $i \in I^{-}$ uses the map $B_ i \to C_ i$ into the summand $O((c, i)) = C_ i$ of $M^{-}$. The map is zero on the summands corresponding to $p$ and $(c, i)$, $i \in I^{+}$. Similarly, consider the map

$\psi : M^{-} \longrightarrow M^{+}$

which on the summand $O(t) = A_ i$ corresponding to $t = (a, i)$, $i \in I^{-}$ uses the map $A_ i \to B_ i$ into the summand $O((b, i)) = B_ i$ of $M^{+}$ and on the summand $O(t) = B_ i$ corresponding to $(b, i)$, $i \in I^{+}$ uses the map $B_ i \to C_ i$ into the summand $O((c, i)) = C_ i$ of $M^{+}$. The map is zero on the summands corresponding to $q$ and $(c, i)$, $i \in I^{-}$.

Note that the kernel of $\varphi$ is equal to the direct sum of the summand $P$ and the summands $O((c, i)) = C_ i$, $i \in I^{+}$ and the subobjects $A_ i$ inside the summands $O((b, i)) = B_ i$, $i \in I^{-}$. The image of $\psi$ is equal to the direct sum of the summands $O((c, i)) = C_ i$, $i \in I^{+}$ and the subobjects $A_ i$ inside the summands $O((b, i)) = B_ i$, $i \in I^{-}$. In other words we see that

$P \cong \mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Im}}(\psi ).$

In exactly the same way we see that

$Q \cong \mathop{\mathrm{Ker}}(\psi )/\mathop{\mathrm{Im}}(\varphi ).$

Since as we remarked above the existence of the bijection $\tau$ shows that $M^{+} = M^{-}$ we see that the lemma follows. $\square$

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