Maybe, this is a problem of linear algebra. But I do not know how to calculate it. Let $E$ be a vector bundle of rank $2$ over an algebraic surface. If $H=S^{2n}E\bigotimes (\operatorname{det} E)^{\bigotimes -n}$, then $\operatorname{det} H$ is trivial, why?



How to calculate the determinant bundle
Take a local basis $e\_1,e\_2$ for $E$, and then get a local basis $E\_0=e\_1^{2n}, E\_1=e\_1^{2n-1}e\_2, \dots, E\_{2n}=e\_2^{2n}$ for $S^{2n}E$. If we scale $e\_1$ by $a\_1$ and $e\_2$ by $a\_2$, we scale $E\_j$ by $a\_1^{2n-j}a\_2^j$. So we scale $E\_0 \wedge \dots \wedge E\_{2n}$ by $a\_1^p a\_2^p$ where $p=0+1+2+\dots+2n=(2n+1)n$. But in $S^{2n}E \otimes \det E^{\otimes -n}$, we replace each $E\_j$ by something with an extra $a\_1^{-n}a\_2^{-n}$, say $E'\_j = E\_j \otimes (e\_1 \wedge e\_2)^{-\otimes n}$ scales by $a\_1^{n-j} a\_2^{j-n}$. So $E\_0' \wedge \dots \wedge E\_{2n}'$ scales by $a\_1^p a\_2^p$ where $p=n+(n-1)+\dots+(-n+1)+(-n)=0$. So $S^{2n}E \otimes \det E^{\otimes -n}$ has local section $E\_0' \wedge \dots \wedge E\_{2n}'$ invariant under these rescalings. It is easy to see that it is also invariant under adding multiples of $e\_2$ to $e\_1$ and vice versa, because of the wedge products, so that gives us invariance of our section under all linear transformations of $e\_1$, $e\_2$ with coefficients locally defined holomorphic functions. But then any two such sections must agree on overlaps, since the sections of $E$ will agree up to linear transformations.


