I just went through a proof of the counting of Weierstrass points on a Riemann surface (References: Reyssat, Quelques aspects des surfaces de Riemann and Farkas & Kra, Riemann Surfaces) that says that the total number of Weierstrass points on Riemann surface, taking into account their "weight" is $(g-1)g(g+1)$, where $g$ is the genus.
A key point of this proof is the introduction of the Wronskian of a basis of the space $\Omega^1(X) = H^0(X,\Omega^1_X)$ of holomorphic differentials. More precisely: if your base is $\omega_1, \ldots, \omega_g$ and that in the coordinate $z$ they are written $\omega_j = f_j dz$, you consider the Wronskian W_z = \begin{vmatrix} f_1 & f_2 & \ldots & f_g \\ f'_1 & f'_2 & \ldots & f'_g \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(g-1)} & f_2^{(g-1)} & \ldots & f_g^{(g-1)} & \end{vmatrix} Then, a simple but unenlightening computation proves that, through a change of coordinates $z \mapsto w$, the Wronskian is transformed by multiplication by $\left(\frac{dz}{dw}\right)^q$ where $q=\frac{g(g+1)}2$. So that means that the Wronskian of differential forms exists as a q-differential (and moreover, because I considered a basis of $\Omega^1(X)$, the Wronskian is canonically defined, up to a constant, which is very important for the said proof but, I think, irrelevant for my question). After these quite indigestible prolegomena, here is my question:
Is there a "higher-level" way of seeing that the Wronskian of a bunch of forms exists as a $q$-differential, without doing any explicit boring computation on the expressions in charts?
Of course, every enlightening comment on all this stuff will be greatly appreciated.