# Complete intersection curve is a Riemann surface - Miranda Prop. 3.9

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!

• 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. Commented Aug 30, 2021 at 23:15
• 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. Commented Aug 30, 2021 at 23:34

## 1 Answer

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 $$U_k$$.

• Do you have a reference for the holomorphic implicit mapping theorem? A textbook is okay. Commented Jun 9, 2021 at 16:10
• I don't have a textbook, but this question and answer may help : math.stackexchange.com/questions/2901576/… Commented Jun 9, 2021 at 16:16
• 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. Commented Jun 9, 2021 at 16:33
• Technically the implicit mapping theorem is valid for $\mathbb{C}^n$ or $\mathbb{C}^{n+1}.$ How do you translate this to $\mathbb{CP}^n$? Commented Jun 9, 2021 at 16:40
• 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. Commented Jun 9, 2021 at 16:56