How does one show that slope stability of a vector bundle is an open condition with respect to the polarisation?

Question

I would like a source for the following result, which I expect to be true (probably well known):

Let $X$ be a complex projective variety, $L$ an ample line bundle and $E$ a slope stable vector bundle with respect to $L$. Then for all line bundles $H$, there exists an $\epsilon>0$ such that $E$ is stable with respect to $L+\epsilon H$.

For some fixed subsheaf $F$, that the stability condition is open is clear. However since a priori one would need to check infinitely many subsheaves, it is not immediately obvious that such an $\epsilon$ exists. I would be interested also to know of a reference for the corresponding analytic question regarding Hermite-Einstein metrics.

Answer 1

I don't know a reference either but you can proceed as follows...

In the vector space of numerical $\mathbb{R}$-divisors, $N^1(X)_{\mathbb{R}}$, being ample is an open condition. Take an open ball $B \subset N^1(X)_{\mathbb{R}}$ satisfying $L \in B$. Look at the map from $N^1(X)_{\mathbb{R}}$ to numerical $\mathbb{R}$-curves $N_1(X)_{\mathbb{R}}$ by taking dim($X$)-1 powers. This map is an open map on the ample cone (Hard-Lefschetz?) so $Image(B)$ will be some neighborhood of $L^{d-1}$. In this neighborhood take images of $\mathbb{Q}$-ample divisors S = {$A_1^{d-1},...,A_m^{d-1}$} such that the cone over the convex hull of S, ($\mathbb{R}^{\ge 0}\cdot S$) contains $L^{d-1}$. Each of these $A_i$ will have a corresponding "maximal destabilizing subsheaf" $\mathcal{F}_i \subset \mathcal{E}$ [see e.g. Huybrechts pg 17]. Most importantly for our purposes, if $\mu_{A_i}(\mathcal{F})$ is the slope of $\mathcal{F}$ with respect to $A_i$ then for all subsheaves $\mathcal{F} \subset E$ we know $\mu_{A_i}(E) - \mu_{A_i}(\mathcal{F}) \ge \mu_{A_i}(E) - \mu_{A_i}(\mathcal{F}_i) =: b_i$.

Now, we can define the slope of a torsion free sheaf with respect to any $\mathbb{R}$-curve class $\lbrack C \rbrack$:

$\mu_{\lbrack C \rbrack} (\mathcal{F}) := \frac{\lbrack C \rbrack \cdot c_1(\mathcal{F})}{rank(\mathcal{F})}$.

The reason this is valuable is because fixing $\mathcal{F}$, $\mu_{\lbrack - \rbrack} (\mathcal{F})$ is a linear function in $\lbrack C \rbrack$, as is $\mu_{\lbrack - \rbrack} (E) - \mu_{\lbrack - \rbrack} (\mathcal{F})$. At each of the $\lbrack A_i^{d-1} \rbrack$ we can bound all the linear functions $\mu_{\lbrack - \rbrack} (E) - \mu_{\lbrack - \rbrack} (\mathcal{F})$ (where $\mathcal{F}$ is a subsheaf of $E$) from below by $b_i$. Moreover, we can bound all $\mu_{\lbrack - \rbrack} (E) - \mu_{\lbrack - \rbrack} (\mathcal{F})$ at $L^{d-1}$ from below by $\frac{1}{r(r-1)}$ (where $r$ is the rank of $E$). So the linear functions $\mu_{\lbrack - \rbrack} (E) - \mu_{\lbrack - \rbrack} (\mathcal{F})$ can be bounded below along the segment connecting $L^{d-1}$ to $A_i^{d-1}$ by the unique line $l_{below,i}$ which is $\frac{1}{r(r-1)}$ at $L^{d-1}$ and $b_i$ at $A_i^{d-1}$.

Now, the point is that if $\mu_{\lbrack M^{d-1} \rbrack} (E) - \mu_{\lbrack M^{d-1} \rbrack} (\mathcal{F})$ is positive for all $\mathcal{F} \subset E$ then $E$ is stable with respect to $M$. In the direction of $A_i^{d-1}$ from $L^{d-1}$ we know $l_{below,i}$ is positive for at least a short time - and thus all the $\mu_{\lbrack - \rbrack} (E) - \mu_{\lbrack - \rbrack} (\mathcal{F})$ are also positive. So there is some $\epsilon_i$ so that $E$ is stable with respect to $\lbrack C \rbrack$ if $\lbrack C \rbrack$ is within $\epsilon_i$ of $L^{d-1}$ in the direction of $A_i^{d-1}$.

Finally, I claim that (fixing $E$) the curve classes for which $E$ is slope semistable is a convex cone, and if $\lbrack C \rbrack$ is in the interior of the convex cone then $E$ is stable with respect to $\lbrack C \rbrack$. The convex cone of the zeroes of the $l_{below,i}$ is a neighborhood of curve classes for which $E$ is stable. Taking the preimage under the d-1 powers map we get an open neighborhood of $L$ where $E$ is slope stable.