## 45.13 Exterior powers and K-groups

We do the minimal amount of work to define the lambda operators. Let $X$ be a scheme. Recall that $\textit{Vect}(X)$ denotes the category of finite locally free $\mathcal{O}_ X$-modules. Moreover, recall that we have constructed a zeroth $K$-group $K_0(\textit{Vect}(X))$ associated to this category in Derived Categories of Schemes, Section 36.38. Finally, $K_0(\textit{Vect}(X))$ is a ring, see Derived Categories of Schemes, Remark 36.38.6.

Lemma 45.13.1. Let $X$ be a scheme. There are maps

$\lambda ^ r : K_0(\textit{Vect}(X)) \longrightarrow K_0(\textit{Vect}(X))$

which sends $[\mathcal{E}]$ to $[\wedge ^ r(\mathcal{E})]$ when $\mathcal{E}$ is a finite locally free $\mathcal{O}_ X$-module and which are compatible with pullbacks.

Proof. Consider the ring $R = K_0(\textit{Vect}(X))[[t]]$ where $t$ is a variable. For a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ we set

$c(\mathcal{E}) = \sum \nolimits _{i = 0}^\infty [\wedge ^ i(\mathcal{E})] t^ i$

in $R$. We claim that given a short exact sequence

$0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$

of finite locally free $\mathcal{O}_ X$-modules we have $c(\mathcal{E}) = c(\mathcal{E}') c(\mathcal{E}'')$. The claim implies that $c$ extends to a map

$c : K_0(\textit{Vect}(X)) \longrightarrow R$

which converts addition in $K_0(\textit{Vect}(X))$ to multiplication in $R$. Writing $c(\alpha ) = \sum \lambda ^ i(\alpha ) t^ i$ we obtain the desired operators $\lambda ^ i$.

To see the claim, we consider the short exact sequence as a filtration on $\mathcal{E}$ with $2$ steps. We obtain an induced filtration on $\wedge ^ r(\mathcal{E})$ with $r + 1$ steps and subquotients

$\wedge ^ r(\mathcal{E}'), \wedge ^{r - 1}(\mathcal{E}') \otimes \mathcal{E}'', \wedge ^{r - 2}(\mathcal{E}') \otimes \wedge ^2(\mathcal{E}''), \ldots \wedge ^ r(\mathcal{E}'')$

Thus we see that $[\wedge ^ r(\mathcal{E})]$ is equal to

$\sum \nolimits _{i = 0}^ r [\wedge ^{r - i}(\mathcal{E}')] [\wedge ^ i(\mathcal{E}'')]$

and the result follows easily from this and elementary algebra. $\square$

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