Suppose $f_1,\dots,f_g$ are holomorphic functons on a domain $U\subset\mathbb{C}$. By the Wronskian determinant $f_1,\dots,f_g$ one means the determinant of the matrix of derivatives $f_k^{(m)},$ where $0\leq m\leq g-1, 1\leq k\leq g,$ i.e., \begin{equation} W(f_1,\dotsc,f_g) := \det \begin{pmatrix} f_1 & f_2 & \dots & f_g \\ f_1' & f_2' & \dots & f_g' \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(g-1)} & f_2^{(g-1)} & \dots & f_g^{(g-1)} \end{pmatrix} \end{equation} One can show that if the functions $f_1,\dots,f_g$ are linearly independent over $\mathbb{C}$, then the Wronskian isn't identically zero. Now suppose $X$ is a compact Riemann surface of genus $g\geq 1$ and $\omega_1,\dots,\omega_g$ is a basis of $\Omega(X).$ For any coordinate neighborhood $(U,z)$ we can define a holomorphic function $W_z(\omega_1,\dots,\omega_g)$ on $U$ as follows. The $1$-form $\omega_k$ may be written $\omega_k=f_k dz$ on $U$. Set $W_z(\omega_1,\dots,\omega_g)=W(f_1,\dots,f_g),$ And under change of coordinates it changes in the following way:

Suppose $(U,z)$ and $(\tilde{U},\tilde{z})$ are two neighborhoods on $X$. Then on $U\cap \tilde{U}$ we get $W_z(\omega_1,\dots,\omega_g)=\big(\frac{d\tilde{z}}{dz}\big)^N W_{\tilde{z}}(\omega_1,\dots,\omega_g),$ where $N=\frac{g(g+1)}{2}.$ It seems that this relation suggests that the Wronskain is a line bundle. But again the definition is dependent on the choice of basis of $\Omega(X)$ and if $\tilde{\omega}_1,\dots,\tilde{\omega}_g$ is another basis of $\Omega(X),$ then $\exists$ constants $c_{jk}\in\mathbb{C}$ with det$(c_{jk})=:c\neq 0$ such that $\omega_j=\sum_k c_{jk}\tilde{\omega}_k.$ Then $W_z(\omega_1,\dots,\omega_g)=c W_z(\tilde{\omega}_1,\dots,\tilde{\omega}_g).$

Is it really a line bundle on the Riemann surface? If not how else can we visualize it ?