## 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 15.118. By More on Algebra, Remark 15.118.1 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. $\square$

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)”.

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