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?
Best Answer
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 $M$.