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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.

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2 Answers

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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This is not an answer.

You might (and I really mean you might) find the beginning of section 2.2 and Remark 2.2.9 helpful in http://www.math.leidenuniv.nl/~rdejong/publications/Thesiswebversion.pdf

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