Let $X$ be a compact connected Riemann surface. Let $\{h_1(z),\ldots,h_g(z)\}$ be a base of the space of holomorphic forms on $X$ in a local map centered in $p\in X$. I’d like to know how to show that $p$ is a Weierstrass point iff the Wronskian
$$W(z):=\begin{vmatrix}h_1(z) & h_2(z) & \ldots & h_g(z) \\h_1’(z) & h_2’(z) & \ldots & h_g’(z) \\ \vdots & \vdots & & \vdots \\h_1^{(g-1)}(z) & h_2^{(g-1)}(z) & \ldots & h_g^{(g-1)}(z)\end{vmatrix}$$
verifies $W(0)=0$.
I honestly don’t really know how to tackle this problem. The only thing I could think of is trying to show that if $W(0)=0$ then all of the vectors $\{h_1^{(i)}(0), h_2^{(i)}(0), \ldots, h_g^{(i)}(z)\}$ (for $0\leq i \leq g-1$) lie in a $(g-1)$-dimensional subspace of $\mathbb{C}^g$, hoping that this could somehow be used to conclude that the functions $\{h_i\}$ are linearly dependent? Or in other words, that if $\{h_1(z),\ldots,h_g(z)\}$ are linearly independent holomorphic functions defined in a neighborhood of $z=0$, then $W(0) \neq 0$.
But the problem seems complicated to me and I don’t exactly see how this connects with Weierstrass points (if that’s of any use, I understand nearly everything on the Wikipedia article about them). Any help would be appreciated.
Edit: I’m using the following definition of Weierstrass point: $p\in X$ is a Weierstrass point iff $\dim(L(g(p))) >1$.
I claim that we can organize the $h_i$ so that $\def\ord{\operatorname{ord}}\ord_p(h_i) < \ord_p(h_{i+1})$ for all $i$. First, pick $h_1$ to be a member of the $h_i$ so that $\ord_p(h_i)$ is minimal. Now for each $h_i$ with $\ord_p(h_1)=\ord_p(h_i)$, we can replace $h_i$ by $h_i+ch_1$ for some constant $c$ so that we still have a basis and $\ord_p(h_1)<\ord_p(h_i+ch_i)$: just cancel the lowest degree term in the local power series expansion. Repeating this process, we have the claim.
Now we can reduce the vanishing of the Wronskian to the existence of a Weierstrass point. The vector $(h_i(z),h_i'(z),\cdots,h_i^{(g-1)}(z))$ consists of $\ord_p(h_i)$ zeroes followed by a nonzero entry followed by arbitrary entries, so the Wronskian matrix is lower-triangular and thus the determinant is the product of the diagonal entries. Further, if there's a zero on the diagonal, then all entries on the diagonal which are below and to the right of that zero must be zero as well, so the determinant of the Wronskian matrix is zero iff $h_g$ is of order at least $g$ at $p$. Such a function will demonstrate that we have $l(gp)>1$: by Riemann-Roch, $l(gp)-l(K-gp)=1$, and $\ord_p(h_g)\geq g$ implies $l(K-gp)\geq 0$, so $l(gp)>1$ and we're done.