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 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.
\[ \lambda ^ r : K_0(\textit{Vect}(X)) \longrightarrow K_0(\textit{Vect}(X)) \]
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
in $R$. We claim that given a short exact sequence
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
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
Thus we see that $[\wedge ^ r(\mathcal{E})]$ is equal to
and the result follows easily from this and elementary algebra. $\square$
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