A q-differential on a compact Riemann surface with genus $g\geq 2$ for $q\geq 1$ always has a zero

complex-analysisdivisors-algebraic-geometryriemann-surfaces

While reading about Weierstrass points on a Riemann surface from 'Farkas and Kra' the Wronskian of a basis of q-differentials was introduced .

If $ \phi_1 dz^q,…,\phi_ddz^q$ (where d is the dimension of the space of holomorphic q-differentials) is a basis locally at a point by $z$ in those local coordinates, then the Wronskian is

$\Phi(z)=det \begin{bmatrix}\phi_1(z) dz^q & \phi_2(z) dz^q &…&\phi_d(z)dz^q \\
\phi_1'(z) dz^{q+1} & \phi_2'(z) dz^{q+1} &…&\phi_d'(z)dz^{q+1} \\
. &.&…&.\\
. &.&…&.\\
\phi_1^{(d-1)}(z) dz^{q+d-1} & \phi_2^{(d-1)}(z) dz^{q+d-1} &…&\phi_d^{(d-1)}(z)dz^{q+d-1} \\
\end {bmatrix} $

Now, $P$ given by $z$ in the local coordinate is a q-Weierstrass point iff $\Phi$ has a zero at $z$.

Now,on the question of existence of q-Weierstrass point on every compact Riemann surface of genus $g\geq 2$ and $q \geq1 $ Farkas and Kra state that every $m_q$ differential ($m_q$ being the order of the differential $\Phi$) has a zero. But I don't see why this must be true.

Can somebody please help me clear this doubt?

Best Answer

The degree of a $1$-differential $\omega$ on a Riemann surface of genus $g\geq2$ is at least $2.2-2=2$. Thus , any 1-differential has a zero.

Then The $q$-differential $\omega^q$ (whose local expression where $\omega=f(z)dz$ is given by $f(z)^qdz^q$ and for $q\geq1$ there is no ambiguity in this) has degree $(2g-2)^q\geq2^q>1$. Thus , every $m_q$ differential has zeroes (since all the $q$-differential belong to the $q$-canonical divisor class and their degrees are same).

This fact applied to the Wronskian in the question proves the existence of q-Weierstrass points on compact Riemann surfaces of genus $g$ for $g\geq2$ and $q\geq1$.(on the torus no Weierstrass points exist).

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