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Trying to see the proof of embedding the Jacobian of a Compact Riemann Surface $X$ using Theta functions. So, using the Theta divisor we have the corresponding line bundle say $L$, we want to prove that the map $\iota_{L^3}:J(X)\rightarrow \mathbb{C}\mathbb{P}^N$ is an embedding. We have dim$H^0(J(X),\mathcal{O}(L))=1,$ the holomomorphic section induced by the Theta function. I am stuck at the immersion part.

To prove that $\iota_{L^3}$ is an immersion, we want to prove that if $J(X)=\mathbb{C}^n/\Lambda,$ and $\pi:\mathbb{C}^n\rightarrow J(X)$ be the quotient map, then given $z\in\mathbb{C}^n,$ we need to prove that $d\iota_{L^3}$ is injective at all $\pi(z).$ Now suppose $\theta_0,\dots,\theta_N$ are the functions whose corresponding sections form a basis of holomorphic sections of $L^3.$ Say we want to check the immersion condition on a chart where the first coordinate of $\mathbb{C}\mathbb{P}^N$ is nonzero. So, w.r.t the charts this means we need to check that the function $\mathbb{C}^n\rightarrow \mathbb{C}^N,z\mapsto (\frac{\theta_1}{\theta_0},\dots,\frac{\theta_N}{\theta_0})$ has full rank Jacobian at all points and similarly on other charts. The book (Griffiths and Harris) says that this condition is equivalent to say that the matrix \begin{bmatrix} \theta_0(z)&\dots&\theta_N(z)\\ \frac{\partial\theta_0}{\partial z_1}(z)&\dots&\frac{\partial\theta_N}{\partial z_1}(z)\\ \vdots& &\vdots\\ \frac{\partial\theta_0}{\partial z_n}(z)&\dots&\frac{\partial\theta_N}{\partial z_n}(z) \end{bmatrix} has rank $n+1$. I can't really understand this equivalence condition. Can someone please help me?

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    $\begingroup$ Hint: $\theta _0^{n+1}d\left(\dfrac{\theta _1}{\theta _0}\right)\wedge\ldots \wedge d\left(\dfrac{\theta _n}{\theta _0}\right)=$ $\sum_i (-1)^i \theta _id\theta_0\wedge\ldots \wedge \widehat{d\theta _i}\wedge\ldots \wedge d\theta _n$, where the $\widehat{d\theta _i}$ means that this term is deleted. Note that this is a general fact about maps into projective space, nothing particular to do with Jacobians. $\endgroup$
    – abx
    Commented Apr 7, 2020 at 13:04
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    $\begingroup$ Just apply both sides to $\dfrac{\partial }{\partial z_1}\wedge \ldots \wedge \dfrac{\partial }{\partial z_n}$, this gives the equality of the Griffiths-Harris determinant with the Jacobian determinant of your map. $\endgroup$
    – abx
    Commented Apr 7, 2020 at 18:24

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