I am trying to to prove Proposition 3.9 from Miranda's book "Algebraic Curves and Riemann Surfaces" which he gives without proof. Overall the proof mainly involves a use of the implicit function theorem and does not seem difficult, but I am running into a couple technical problems that I would like to get help with streamlining. Also, I would be satisfied if a response provided a reference where I could find a proof of this result.

Miranda defines a smooth complete intersection curve as the space of solutions to $n-1$ homogeneous polynomials $F_1,...,F_{n-1}$ (in $n+1$ variables) in $\mathbb{P}^n$ such that the $n-1\times n+1$ Jacobian matrix has full rank on every point of the variety. Here's a statement of Miranda's Proposition 3.8: " A smooth complete intersection curve in $\mathbb{P}^n$ is a compact Riemann surface. Moreover, at every point of $X$ one can take as a local coordinate a ratio $x_i/x_j$ of the homogeneous coordinates."

I know how to prove compactness and only want to show $X$ is a Riemann surface. This can be done by using the implicit function theorem to show that the curve is a graph in two coordinates $x_i$ and $x_j$ in a neighborhood of every point, and then defining a chart to be given by $x_i/x_j$. I believe is easy to check that the transition maps between these charts are holomorphisms, but am having a hard time choosing the open sets on which the charts are defined and proving that the charts on these sets are homeomorphisms. This is what I have so far.

Let $p=(x_1:...:x_{n+1})\in X.$ There must be some nonzero coordinate which we may suppose WLOG is the first, so $x_1\neq 0.$ Then if $F=(F_1,...,F_{n-1}),$ Euler's formula on homogeneous polynomials says that $$ \sum x_i\frac{\partial F}{\partial x_i}= (d_1\cdot F_1(x_1,...,x_{n+1}),...,d_{n-1}\cdot F_{n-1}(x_1,...,x_{n+1})) =0. $$ Since $x_1\neq 0$ it follows that $\frac{\partial F}{\partial x_1}$ at $p$ can be represented as a linear combination of the $\frac{\partial F}{\partial x_i}$ for $i\neq 1$. That is, the first column of the Jacobian be be represented in terms of the other columns. Since there are $n$ remaining columns and only $n-1$ rows in the Jacobian, another column can be similarly represented as a linear combination of the others, lets say its $\frac{\partial F}{\partial x_2}.$ Then since the original Jacobian had full rank, the remaining matrix $ \begin{bmatrix} \frac{\partial F_1}{\partial x_3} &\dots &\frac{\partial F_1}{\partial x_{n+1}}\\ \vdots & \dots & \vdots\\ \frac{F_{n-1}}{\partial x_3}& \dots &\frac{\partial F_{n-1}}{\partial x_{n+1}} \end{bmatrix} $ has nonzero determinant. Then by the implicit function theorem there exists a neighborhood $U\subseteq \mathbb{C}^2$ of $(x_1,x_2)$, a neighborhood $V\subseteq \mathbb{C}^{n-1}$ and a unique holomorphic function $g:U\to V$ such that $$\{(x,y)\in U\times V:F(x,y)=0\} =\{(x, g(x)):x\in U\}. $$

We know that $x_1\neq 0.$ Because we want to define a chart map by $x_2/x_1,$ shrink $U$ to a smaller neighborhood of $(x_1,x_2)$ for which the first coordinate is never zero. Then if we use $\overline{U\times V}$ to represent the set of equivalence classes in $\mathbb{P}^n$ represented in $U\times V,$ we can define a chart map $\phi:\overline{U\times V}\to \mathbb{C}$ so that $\phi(x_1:x_2:...:x_{n+1})=x_2/x_1.$ Clearly $\phi$ is continuous. I need to show $\phi(\overline{U\times V})$ is open and $\phi$ is a homeomorphism.

This is where I get a bit shaky. I know I can define an inverse to $\phi$ to look something like $\psi(z)=(1:z:g(1,z))$ but this may not work since we may not have $(1,z)\in U$ and so $g$ may not be well-defined there. Modifying this idea, we could define $\psi(z)=(x_1,zx_1,g(z_1,zx_1))$ and this will be well-defined on some neighborhood of $x_2/x_1$ since $(x_1,x_1\cdot \frac{x_2}{x_1})\in U.$ But then I think we would need to shrink the domain of the chart $\phi$ for $\phi$ to be inverse to $\psi.$ Do we need to some version of the open mapping theorem to show that the image of $\phi$ is open, or will this follow directly from the continuity of an inverse? I'm not sure about the best way to manage the details at this point of the proof.

I really appreciate any help. Thanks!

  • $\begingroup$ I posted what I think is a fairly complete answer to this question as Proposition 2 on my blog at tobysmathblog.wordpress.com/2021/06/12/riemann-surfaces. It may be overcomplicated, but I've found it surprisingly difficult to write a technically correct solution to this question. $\endgroup$
    – subrosar
    Commented Aug 30, 2021 at 23:15
  • $\begingroup$ Also, I think Pene Papin's answer is correct but there are some details missing. I think he is defining $\Psi_k:U_k\to \mathbb{C^{n-1}}$ by $\Psi_k([1:z_1:...:z_n])=(F_1,...,F_{n-1})(1,z_1,...,z_n).$ To prove $\Psi_k$ is a submersion I think one needs to use Euler's formula like I did above to prove the Jacobian of $\Psi_k$ is still a submersion, since this is the Jacobian of $(F_1,...,F_{n-1})$ with the first column deleted. $\endgroup$
    – subrosar
    Commented Aug 30, 2021 at 23:34

1 Answer 1


If I well understood your question, you may want to give local coordinates for $X$. In fact, I think one may proceed like this. Let $U_k:=\{[z_0:\ldots: z_n]\in\mathbb{CP}^n: z_k\neq 0\}$ be the standard affine charts. Since the question is local, one only needs to give local coordinates on $U_k\cap X$ for each $k$. The tuple $(F_1,\ldots, F_{n-1})$ gives a map $$\Psi_k: U_k\to\mathbb C^{n-1}.$$ The fact that $F_1,\ldots, F_{n-1}$ gives a smooth complete intersection just says that this map $\Psi_k$ is submersive over $0\in\mathbb C^{n-1}$ (i.e. the differential of $\Psi_k$ is surjective over $0\in\mathbb C^{n-1}$). Hence, you can use holomorphic implicit mapping theorem to give local coordinates for $\Psi_k^{-1}(0)$. Don't forget $\Psi_k^{-1}(0)$ is nothing but your $X$ intersecting with $U_k$.

  • $\begingroup$ Do you have a reference for the holomorphic implicit mapping theorem? A textbook is okay. $\endgroup$
    – subrosar
    Commented Jun 9, 2021 at 16:10
  • 1
    $\begingroup$ I don't have a textbook, but this question and answer may help : math.stackexchange.com/questions/2901576/… $\endgroup$ Commented Jun 9, 2021 at 16:16
  • $\begingroup$ Great! Since they only state the one variable case, I also found a statement of the multivariable case in "Holomorphic Functions of Several Variables" by Kaup and Kaup. $\endgroup$
    – subrosar
    Commented Jun 9, 2021 at 16:33
  • $\begingroup$ Technically the implicit mapping theorem is valid for $\mathbb{C}^n$ or $\mathbb{C}^{n+1}.$ How do you translate this to $\mathbb{CP}^n$? $\endgroup$
    – subrosar
    Commented Jun 9, 2021 at 16:40
  • 1
    $\begingroup$ I took an affine cover $\{U_k\}$ of $\mathbb{CP}^n$ where $U_k\cong \mathbb C^n$ and utilise the implicit mapping for each of these $U_k$. Best regards. $\endgroup$ Commented Jun 9, 2021 at 16:56

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