The Wronskian of holomorphic differentials as a q-differential

Solution 1:

I'm afraid this is only an idea for an answer, but you should be able to fill in the details.

A $q$-differential on $X$ is a section of the bundle $q K_X = (\Omega^1_X)^{\otimes q}$. We want to see that the Wronskian is a section of some $q K_X$. Now, the Wronskian is the determinant of the $(g-1)^{th}$ Taylor polynomials of sections of $K_X$. These sections are most naturally interpreted as sections of the $(g-1)^{th}$ jet bundle $J^{g-1}K_X$ of $K_X$, so the Wronskian is a section of the line bundle $\det J^{g-1}K_X$. We want to see that this bundle is a multiple of $K_X$.

Here Chern classes should come to our resque. One can see that if $L$ is a line bundle then the Chern classes of $J^kL$ are the same as the Chern classes of $$ L \otimes \bigl( 1 \oplus \Omega^1_X \oplus \ldots \oplus Sym^k\Omega^1_X \bigr). $$ In our case $Sym^k \Omega^1_X = (\Omega^1_X)^{\otimes k} = k K_X$ since $\Omega^1_X$ is a line bundle. Plugging in $L = K_X$ should show us exactly which multiple of $K_X$ the bundle $\det J^{g-1}K_X$ is, and thus give the $q$ we're looking for.

Unfortunately I don't get the right multiple $q$ here; I get $q = 1 + \binom{g}{2}$. I'm sure this is due to a stupid error I've made somewhere along the way, one that I trust you'll correct without trouble. In any case I think this is a good candidate for the "high-level" approach to the Wronskian you wanted.