Lemma 42.55.1. Let $X$ be a scheme. There is a ring map

$\psi ^2 : K_0(\textit{Vect}(X)) \longrightarrow K_0(\textit{Vect}(X))$

which sends $[\mathcal{L}]$ to $[\mathcal{L}^{\otimes 2}]$ when $\mathcal{L}$ is invertible and is compatible with pullbacks.

Proof. Let $X$ be a scheme. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module. We will consider the element

$\psi ^2(\mathcal{E}) = [\text{Sym}^2(\mathcal{E})] - [\wedge ^2(\mathcal{E})]$

of $K_0(\textit{Vect}(X))$.

Let $X$ be a scheme and consider a short exact sequence

$0 \to \mathcal{E} \to \mathcal{F} \to \mathcal{G} \to 0$

of finite locally free $\mathcal{O}_ X$-modules. Let us think of this as a filtration on $\mathcal{F}$ with $2$ steps. The induced filtration on $\text{Sym}^2(\mathcal{F})$ has $3$ steps with graded pieces $\text{Sym}^2(\mathcal{E})$, $\mathcal{E} \otimes \mathcal{F}$, and $\text{Sym}^2(\mathcal{G})$. Hence

$[\text{Sym}^2(\mathcal{F})] = [\text{Sym}^2(\mathcal{E})] + [\mathcal{E} \otimes \mathcal{F}] + [\text{Sym}^2(\mathcal{G})]$

In exactly the same manner one shows that

$[\wedge ^2(\mathcal{F})] = [\wedge ^2(\mathcal{E})] + [\mathcal{E} \otimes \mathcal{F}] + [\wedge ^2(\mathcal{G})]$

Thus we see that $\psi ^2(\mathcal{F}) = \psi ^2(\mathcal{E}) + \psi ^2(\mathcal{G})$. We conclude that we obtain a well defined additive map $\psi ^2 : K_0(\textit{Vect}(X)) \to K_0(\textit{Vect}(X))$.

It is clear that this map commutes with pullbacks.

We still have to show that $\psi ^2$ is a ring map. Let $X$ be a scheme and let $\mathcal{E}$ and $\mathcal{F}$ be finite locally free $\mathcal{O}_ X$-modules. Observe that there is a short exact sequence

$0 \to \wedge ^2(\mathcal{E}) \otimes \wedge ^2(\mathcal{F}) \to \text{Sym}^2(\mathcal{E} \otimes \mathcal{F}) \to \text{Sym}^2(\mathcal{E}) \otimes \text{Sym}^2(\mathcal{F}) \to 0$

where the first map sends $(e \wedge e') \otimes (f \wedge f')$ to $(e \otimes f)(e' \otimes f') - (e' \otimes f)(e \otimes f')$ and the second map sends $(e \otimes f) (e' \otimes f')$ to $ee' \otimes ff'$. Similarly, there is a short exact sequence

$0 \to \text{Sym}^2(\mathcal{E}) \otimes \wedge ^2(\mathcal{F}) \to \wedge ^2(\mathcal{E} \otimes \mathcal{F}) \to \wedge ^2(\mathcal{E}) \otimes \text{Sym}^2(\mathcal{F}) \to 0$

where the first map sends $e e' \otimes f \wedge f'$ to $(e \otimes f) \wedge (e' \otimes f') + (e' \otimes f) \wedge (e \otimes f')$ and the second map sends $(e \otimes f) \wedge (e' \otimes f')$ to $(e \wedge e') \otimes (f f')$. As above this proves the map $\psi ^2$ is multiplicative. Since it is clear that $\psi ^2(1) = 1$ this concludes the proof. $\square$

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