Let $X$ be a complex surface, $K$ the canonical divisor, $F$ a rank 2 vector bundle on $X$, $n$ a positive integer, $S^nF$ the $n$-th symmetric product of $F$, $P$ a rational divisor, and $a$ a rational number such that $naP$ is integral. In the proof of Proposition 2.3 of this paper (link: https://mathscinet.ams.org/mathscinet-getitem?mr=1272710), there is the following statement:
$h^2(S^nF(-naP))=h^0(S^nF(naP+K-n\cdot \det(F)))$ by duality.
It seems duality means Serre duality (https://en.wikipedia.org/wiki/Serre_duality#Algebraic_theorem), but I can't see how does the equality hold. Serre duality is $h^i(E)=h^{2-i}(K\otimes E^*)$, so applying in this situation, we have $K\otimes (S^nF(-naP))^*=K\otimes (S^nF\otimes (-naP))^*=K\otimes (S^nF)^* \otimes (naP)=K\otimes S^n(F^*)\otimes (naP)$. But how do we have $h^0(K\otimes S^n(F^*)\otimes (naP))=h^0(S^nF(naP+K-n\cdot \det(F)))$?
The missing detail is the isomorphism $$ F^* \cong F\otimes\det(F)^{-1} $$ which holds true because the rank of $F$ is 2 and hence there is a non-degenerate pairing $F \otimes F \to \det(F)$, and which implies after taking $n$-th symmetric powers $$ S^nF^* \cong S^nF \otimes \det(F)^{-n}. $$