# Dual bundle of a stable vector bundle.

Given a vector bundle $E \to X$ over a complex curve $X$, let its rank and degree be $r$ and $d$ respectively. Then the slope of $E$ is defined to be

$\mu(E):= \frac{d}{r}.$

$E$ is defined to be stable if $\mu(U) < \mu(E)$ for every non-trivial proper subbundle $U$ of $E$.

I'm trying to decide whether the following is true: $E$ is stable if and only if its dual bundle $E^*$ is stable.

If would like the following to hold: $\mu(U) < E \iff \mu(E/U) > \mu(E)$ because then given a subbundle $U$ I can dualize the ses

$0 \to U \to E \to E/U \to 0$

to give ses

$0 \to (E/U)^* \to E^* \to U^* \to 0$,

from which it follows that

$E^*$ stable $\implies \mu((E/U)^*) < \mu(E^*) \implies \mu((E/U))>\mu(E) \implies \mu(U) < \mu(E).$

However, this does not seem like it would hold in general, but I can't think of a counter example (i.e stable $E$ with $E^*$ not stable).

Thanks.

This follows from the fact that stability can also be checked with respect to quotients of $E$, that is $E$ is stable if and only if for every quotient bundle $E\rightarrow Q$ we have $\mu(Q)>\mu(E)$.
This follows from the fact that both degree and rank are additive with respect to exact sequences. More explicitly, for any short exact sequence $$0 \to E \to F \to G \to 0$$ one has $$\deg F = \deg E + \deg G$$ and $$\operatorname{rk} F = \operatorname{rk}E + \operatorname{rk} G$$. Thus everything can be deduced to the following fact about rational numbers: $$\frac{d_1}{r_1}<\frac{d_1+d_2}{r_1+r_2}$$ if and only if $$\frac{d_1+d_2}{r_1+r_2}<\frac{d_2}{r_2}$$.
Indeed let $$d_1$$ (resp. $$d_2$$ be the degree of $$U$$ (resp. $$E/U$$) and $$r_1$$ (resp. $$r_2$$) its rank.
By definition $$\mu(U) < \mu (E)$$ means $$\frac{d_1}{r_1} < \frac{d_1+d_2}{r_1+r_2}$$ which can be re-written as $$d_1r_1+d_1r_2 canceling $$d_1r_1$$ we get $$d_1r_2.
By the same considerations the condition $$\mu(E)<\mu(E/U)$$ is equivalent to $$d_1r_2+d_2r_2 which leads to the same $$d_1r_2.