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Why smooth projective algebraic curves over complex numbers are compact and orientable? I want to use it to show algebraic smooth curves are topologically identical to toruses with different genuses.

(projective algebraic curves are zeros of a homogenous polynomial of three variables with complex coefficients in CP2. We will call a projective algebraic curve smooth in point p if its gradient is not zero at point p.)

Toby Mak
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Ahmadreza Khazaeipoul
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    You should include your definition of "smooth plane algebraic curve" -- some people might assume you're talking about a scheme, when I think you're talking about a manifold. – diracdeltafunk Aug 30 '20 at 03:12
  • algebraic curves are zeros of a polynomial of two variables with complex coefficients. We call an algebraic curve smooth in point p if its gradient is not zero at point p. – Ahmadreza Khazaeipoul Aug 30 '20 at 03:14
  • Thank you! Just to make sure: you really mean the set of zeros in $\mathbb{C}\mathbb{P}^2$ of a homogeneous polynomial in three variables, right? If you mean the set of zeros in $\mathbb{C}^2$ of a polynomial of two variables, then these are not necessarily compact – e.g. the set of zeros of $(x,y) \mapsto x-y : \mathbb{C}^2 \to \mathbb{C}$ (which has nonzero gradient everywhere) is homeomorphic to $\mathbb{C}$. – diracdeltafunk Aug 30 '20 at 03:20
  • Yes, you are right. I will edit it. – Ahmadreza Khazaeipoul Aug 30 '20 at 03:21

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  1. Show that each smooth plane algebraic curve is closed. Potential approach: consider the restriction of the polynomial map $\mathbb{C}^3\setminus\{0\} \to \mathbb{C}$. Compose with the quotient map $\mathbb{C} \to \mathbb{C}/\mathbb{C}^\times$. Show this descends to $\mathbb{CP}^2$. Since $\mathbb{CP}^2$ is compact, this implies that each smooth plane algebraic curve is compact.

  2. Show that each smooth plane algebraic curve is a $1$-dimensional complex manifold. Potential approach: work in local coordinates on $\mathbb{CP}^2$ to show that each smooth plane algebraic curve is locally homeomorphic to the zero set in $\mathbb{C}^2$ of a smooth complex polynomial of two variables. Then show that these zero sets are always $1$-dimensional complex manifolds.

  3. We conclude that each smooth plane algebraic curve is a compact $1$-dimensional complex manifold. Every complex manifold is orientable as a real manifold (compare the Jacobians of the real and complex transition maps to see that the real Jacobians must have positive determinant -- in other words, holomorphic maps preserve orientation), so we are done.

diracdeltafunk
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