The Stacks project

Lemma 42.56.1. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FEJ. Beware of the difference between the letter 'O' and the digit '0'.