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.


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.


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