The Stacks project

17.25 Rank and determinant

Let $(X, \mathcal{O}_ X)$ be a ringed space. Consider the category $\textit{Vect}(X)$ of finite locally free $\mathcal{O}_ X$-modules. This is an exact category (see Injectives, Remark 19.9.6) whose admissible epimorphisms are surjections and whose admissible monomorphisms are kernels of surjections. Moreover, there is a set of isomorphism classes of objects of $\textit{Vect}(X)$ (proof omitted). Thus we can form the zeroth Grothendieck $K$-group $K_0(\textit{Vect}(X))$. Explicitly, in this case $K_0(\textit{Vect}(X))$ is the abelian group generated by $[\mathcal{E}]$ for $\mathcal{E}$ a finite locally free $\mathcal{O}_ X$-module, subject to the relations

\[ [\mathcal{E}'] = [\mathcal{E}] + [\mathcal{E}''] \]

whenever there is a short exact sequence $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ of finite locally free $\mathcal{O}_ X$-modules.

Ranks. Assume all stalks $\mathcal{O}_{X, x}$ are nonzero rings. Given a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$, the rank is a locally constant function

\[ \text{rank}_\mathcal {E} : X \longrightarrow \mathbf{Z}_{\geq 0},\quad x \longmapsto \text{rank}_{\mathcal{O}_{X, x}} \mathcal{E}_ x \]

See Lemma 17.14.4. By definition of locally free modules the function $\text{rank}_\mathcal {E}$ is locally constant. If $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ is a short exact sequence of finite locally free $\mathcal{O}_ X$-modules, then $\text{rank}_\mathcal {E} = \text{rank}_{\mathcal{E}'} + \text{rank}_{\mathcal{E}''}$, Thus the rank defines a homomorphism

\[ K_0(\textit{Vect}(X)) \longrightarrow \text{Map}_{cont}(X, \mathbf{Z}),\quad [\mathcal{E}] \longmapsto \text{rank}_\mathcal {E} \]

Determinants. Given a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ we obtain a disjoint union decomposition

\[ X = X_0 \amalg X_1 \amalg X_2 \amalg \ldots \]

with $X_ i$ open and closed, such that $\mathcal{E}$ is finite locally free of rank $i$ on $X_ i$ (this is exactly the same as saying the $\text{rank}_\mathcal {E}$ is locally constant). In this case we define $\det (\mathcal{E})$ as the invertible sheaf on $X$ which is equal to $\wedge ^ i(\mathcal{E}|_{X_ i})$ on $X_ i$ for all $i \geq 0$. Since the decomposition above is disjoint, there are no glueing conditions to check. By Lemma 17.25.1 below this defines a homomorphism

\[ \det : K_0(\textit{Vect}(X)) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (X),\quad [\mathcal{E}] \longmapsto \det (\mathcal{E}) \]

of abelian groups. The elements of $\mathop{\mathrm{Pic}}\nolimits (X)$ we get in this manner are locally free of rank $1$ (see below the lemma for a generalization).

Lemma 17.25.1. Let $X$ be a ringed space. Let $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ be a short exact sequence of finite locally free $\mathcal{O}_ X$-modules, Then there is a canonical isomorphism

\[ \det (\mathcal{E}') \otimes _{\mathcal{O}_ X}\det (\mathcal{E}'') \longrightarrow \det (\mathcal{E}) \]

of $\mathcal{O}_ X$-modules.

Proof. We can decompose $X$ into disjoint open and closed subsets such that both $\mathcal{E}'$ and $\mathcal{E}''$ have constant rank on them. Thus we reduce to the case where $\mathcal{E}'$ and $\mathcal{E}''$ have constant rank, say $r'$ and $r''$. In this situation we define

\[ \wedge ^{r'}(\mathcal{E}') \otimes _{\mathcal{O}_ X} \wedge ^{r''}(\mathcal{E}'') \longrightarrow \wedge ^{r' + r''}(\mathcal{E}) \]

as follows. Given local sections $s'_1, \ldots , s'_{r'}$ of $\mathcal{E}'$ and local sections $s''_1, \ldots , s''_{r''}$ of $\mathcal{E}''$ we map

\[ s'_1 \wedge \ldots \wedge s'_{r'} \otimes s''_1 \wedge \ldots \wedge s''_{r''} \quad \text{to}\quad s'_1 \wedge \ldots \wedge s'_{r'} \wedge \tilde s''_1 \wedge \ldots \wedge \tilde s''_{r''} \]

where $\tilde s''_ i$ is a local lift of the section $s''_ i$ to a section of $\mathcal{E}$. We omit the details. $\square$

Let $(X, \mathcal{O}_ X)$ be a ringed space. Instead of looking at finite locally free $\mathcal{O}_ X$-modules we can look at those $\mathcal{O}_ X$-modules $\mathcal{F}$ which are locally on $X$ a direct summand of a finite free $\mathcal{O}_ X$-module. This is the same thing as asking $\mathcal{F}$ to be a flat $\mathcal{O}_ X$-module of finite presentation, see Lemma 17.18.3. If all the stalks $\mathcal{O}_{X, x}$ are local, then such a module $\mathcal{F}$ is finite locally free, see Lemma 17.14.6. In general however this will not be the case; for example $X$ could be a point and $\Gamma (X, \mathcal{O}_ X)$ could be the product $A \times B$ of two nonzero rings and $\mathcal{F}$ could correspond to $A \times 0$. Thus for such a module the rank function is undefined. However, it turns out we can still define $\det (\mathcal{F})$ and this will be an invertible $\mathcal{O}_ X$-module in the sense of Definition 17.24.1 (not necessarily locally free of rank $1$). Our construction will agree with the one above in the case that $\mathcal{F}$ is finite locally free. We urge the reader to skip the rest of this section.

Lemma 17.25.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a flat and finitely presented $\mathcal{O}_ X$-module. Denote

\[ \det (\mathcal{F}) \subset \wedge ^*_{\mathcal{O}_ X}(\mathcal{F}) \]

the annihilator of $\mathcal{F} \subset \wedge ^*_{\mathcal{O}_ X}(\mathcal{F})$. Then $\det (\mathcal{F})$ is an invertible $\mathcal{O}_ X$-module.

Proof. To prove this we may work locally on $X$. Hence we may assume $\mathcal{F}$ is a direct summand of a finite free module, see Lemma 17.18.3. Say $\mathcal{F} \oplus \mathcal{G} = \mathcal{O}_ X^{\oplus n}$. Set $R = \mathcal{O}_ X(X)$. Then we see $\mathcal{F}(X) \oplus \mathcal{G}(X) = R^{\oplus n}$ and correspondingly $\mathcal{F}(U) \oplus \mathcal{G}(U) = \mathcal{O}_ X(U)^{\oplus n}$ for all opens $U \subset X$. We conclude that $\mathcal{F} = \mathcal{F}_ M$ as in Lemma 17.10.5 with $M = \mathcal{F}(X)$ a finite projective $R$-module. In other words, we have $\mathcal{F}(U) = M \otimes _ R \mathcal{O}_ X(U)$. This implies that $\det (M) \otimes _ R \mathcal{O}_ X(U) = \det (\mathcal{F}(U))$ for all open $U \subset X$ with $\det $ as in More on Algebra, Section 15.116. By More on Algebra, Remark 15.116.1 we see that

\[ \det (M) \otimes _ R \mathcal{O}_ X(U) = \det (\mathcal{F}(U)) \subset \wedge ^*_{\mathcal{O}_ X(U)}(\mathcal{F}(U)) \]

is the annihilator of $\mathcal{F}(U)$. We conclude that $\det (\mathcal{F})$ as defined in the statement of the lemma is equal to $\mathcal{F}_{\det (M)}$. Some details omitted; one has to be careful as annihilators cannot be defined as the sheafification of taking annihilators on sections over opens. Thus $\det (\mathcal{F})$ is the pullback of an invertible module and we conclude. $\square$


Comments (2)

Comment #6476 by Peng DU on

In "Ranks.", need to change “Thus the rank defines a homomorphism K0(Vec(X))⟶Mapcont(X,Z)” to “thus the rank defines a homomorphism K0(Vect(X))⟶Mapcont(X,Z)”.


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 0B37. Beware of the difference between the letter 'O' and the digit '0'.