# Rank of coherent sheaf on complex manifolds

Let $$X$$ be a complex smooth connected manifold of dimension $$n$$ and let $$\mathcal{O}_{X,x}$$ be the ring of germs of holomorphic functions at $$x$$. Since the stalk of $$\mathcal{O}_X$$ at $$x$$ does not depend in which open subset is contained the point, then it can be identified with the ring of convergent power series in $$n$$ variables. Let $$K(x)$$ be the field of fractions of $$\mathcal{O}_{X,x}$$, i.e. it is the field of germs of meromorphic functions at $$x$$.

Definition: Let $$\mathcal{F}$$ be a sheaf over $$X$$, then we say that $$\mathcal{F}$$ is coherent if for each $$x\in X$$ there exist a neighborhood $$U$$ of $$x$$ and an exact sequence $$\begin{equation*}\mathcal{O}_U^{\oplus q}\rightarrow\mathcal{O}_U^{\oplus p}\rightarrow\mathcal{F}_{|U}\rightarrow 0\end{equation*}$$for some $$p,q>0$$ where $$\mathcal{O}_U$$ is the restriction to $$U$$ of the sheaf of holomorphic functions on $$X$$. In particular the sheaf $$\mathcal{F}$$ is locally finitely presented.

Definition: Let $$\mathcal{F}$$ be a coherent sheaf on $$X$$, since the stalk $$\mathcal{F}_x$$ is a finitely-generated $$\mathcal{O}_{X,x}$$-module then we can define the rank of $$\mathcal{F}$$ to be $$\begin{equation*}\text{Rank}(\mathcal{F}):=\text{dim}_{K(x)}\bigg(\mathcal{F}_x\bigotimes_{\mathcal{O}_{X,x}}K(x)\bigg) \end{equation*}$$

## Question 1

Since $$X$$ is assumed to be smooth and connected, is the rank of a coherent sheaf $$\mathcal{F}$$ constant and equal to $$r$$ say?

Now suppose that $$\mathcal{F}$$ is coherent and torsion-free (i.e. $$\mathcal{F}_x$$ is a torsion-free $$\mathcal{O}_{X,x}$$-module for each $$x\in X$$), then it can be proved that there exist a closed complex manifold $$Y\subset X$$ of codimension at least $$2$$ sucht that $$\mathcal{F}|_{X\setminus Y}$$ is locally-free and in particular has finite constant rank (since over this open subset is a holomorphic vector bundle, am I right?).

## Question 2

Adding the hypothesis of being torsion-free, what can we say about the rank over $$Y$$? Could it jump?

• Well, no you can still take a skyscraper sheaf of stalk $\mathbb{C}$ somewhere. This will be coherent.
– Ahr
Jan 25, 2021 at 9:31

To your first question: No, you can not in general assume that the rank of a coherent sheaf on a connected manifold is constant. In fact, vector bundles on connected manifolds are precisely the coherent sheaves with constant rank.

A little more involved example would be the cokernel sheaf of the following morphism. Let $$B\subset \mathbb{C}^n$$ be the unit ball and let $$\phi\colon \mathcal{O}_B \to \mathcal{O}_B^n$$ be given by $$f\mapsto \left(z_1f,...,z_nf\right)$$. The cokernel is coherent but not a vector bundle, since the cokernel is not free at $$0$$.

Explicitly, if it were a vector bundle then $$\text{coker}\left(\phi\right)\cong \mathcal{O}_B^{n-1}$$, since every holomorphic bundle over $$B$$ is trivial. And then the sequence

$$0 \to \mathcal{O}_B \to \mathcal{O}_B^{n} \to \mathcal{O}_B^{n-1} \to 0$$

would be spilt exact, meaning there would exist $$g \colon \mathcal{O}_B^{n} \to \mathcal{O}_B$$ such that $$g\circ f = \text{id}$$, which is clearly not possible.

To your second question: Even if your sheaf is torsion-free, the rank can still jump. One difference is that the singular set of a coherent sheaf has codimension at least $$1$$ and the singular set of a torsion-free coherent sheaf has codimension at least $$2$$. So for example, a torsion-free coherent sheaf on a connected complex $$1$$-dimensional manifold is locally free and thus has constant rank.

In general the rank of a coherent sheaf $$F$$ is upper-semicontinuous, i.e. for every $$p\in M$$ there exists an open subset $$U$$ such that $$\forall q \in U \colon \text{rank}_q\left(F\right)\leq \text{rank}_p\left(F\right)$$.

Edit: I realized your second question was not addressed by my answer, since you asked about the rank of torsion-free sheaf over its singular set. For this note that pullback preserves coherence and fiber rank is invariant under pullback. Hence again one can at most expect there to be a codimension $$1$$ subset in $$Y$$ such that $$i^*_YF$$ has constant rank away from this set. Here $$i_Y$$ denotes the inclusion of $$Y$$ in $$M$$.

Question: "Now suppose that $$F$$ is coherent and torsion-free (i.e. $$F_x$$ is a torsion-free $$O_{X,x}$$-module for each $$x∈X$$), then it can be proved that there exist a closed complex manifold $$Y⊂X$$ of codimension at least 2 sucht that $$F|X∖Y$$ is locally-free and in particular has finite constant rank (since over this open subset is a holomorphic vector bundle, am I right?)."

Answer: If $$X^a\subseteq \mathbb{P}^n_{\mathbb{C}}$$ is a complex projective manifold and $$E^a$$ is a coherent $$\mathcal{O}_{X^a}$$-module, it follows $$X^a:=X$$ is algebraic and to $$E^a$$ corresponds a coherent $$\mathcal{O}_X$$-module $$E$$ with the property that there is an open sub variety $$U\subseteq X$$ where the restriction $$E_U$$ is a locally trivial $$\mathcal{O}_U$$-module of finite rank $$e$$. This implies the same result for $$E^a$$ I believe: The $$\mathcal{O}_{X^a}$$-module $$E^a$$ when restricted to $$U$$ is locally trivial of finite rank $$e$$. I believe the functor $$F$$ (and its "inverse" $$G$$) that associates the coherent $$\mathcal{O}_X$$-module $$E$$ to the $$\mathcal{O}_{X^a}$$-module $$E^a$$ respects restrictions and isomorphisms. Hence when $$E_U \cong \mathcal{O}_U^e$$ it follows

$$E^a_U \cong G(E_U) \cong G(\mathcal{O}_U^e)\cong \mathcal{O}_{U^a}^e.$$

You must check the original GeoAlg-paper where this correspondence is proved to exist.

There is the "generic freeness" result in Matsumura (page 185, \$24):

Theorem 1. Let $$A$$ be any noetherian ring and let $$M$$ be a fintely generated $$A$$-module. It follows there is an element $$a\in A$$ such that $$M_a$$ is a free $$A_a$$-module of finite rank.

Corollary. Let $$Y \subseteq \mathbb{P}^n_{\mathbb{C}}$$ be any projective sub scheme/variety and let $$\mathcal{E}$$ be a coherent $$\mathcal{O}_Y$$-module. It follows there is an open set $$U \subseteq Y$$ with $$\mathcal{E}_U$$ a locally trivial $$\mathcal{O}_U$$-module of finite rank.

Proof. Choose any open affine subset $$U:=Spec(A) \subseteq Y$$ and let $$\mathcal{E}(U):=E$$. It follows from Theorem 1 there is an open subset $$D(a) \subseteq U$$ where $$E_a \cong \mathcal{E}(D(a))$$ is a finite rank trivial $$\mathcal{O}_{D(a)}$$-module. Hence there is an open subscheme $$D(a) \subseteq U \subseteq Y$$ where $$\mathcal{E}_U$$ is locally trivial of finite rank. QED.

Hence it seems the result holds in greater generality if your complex manifold/space is projective.