Holomorphic 1-forms on a smooth affine plane curve

Question

I'm looking at Miranda's Algebraic Curves and Riemann Surfaces. In particular, Chapter IV.I, question C : Let $X$ be a smooth affine plane curve defined by $f(u,v) =0$. Show that $du$ and $dv$ define holomorphic 1-forms on $X$, as do $p(u,v)du$ and $p(u,v)dv$ for any polynomial $p(u,v)$. Show that if $r(u,v)$ is any rational function, then $r(u,v)du$ and $r(u,v)dv$ are meromorphic 1-forms on $X$. Show that $(\partial f/\partial u)du = -(\partial f/\partial v)dv$ as holomorphic 1-forms on $X$.

I understand what a holomorphic 1-form is, but I'm not sure how I'm supposed to show that these particular examples are 1-forms on $X$. The reason for this mainly is that I'm unsure of when something fails to be a holomorphic 1-form.

Answer 1

Think about what you do when you define a holomorphic form over an open set $U\subseteq \mathbf{C}$: you choose a coordinate $z$ and then you decide to consider all the expressions $$f(z)\,\mathrm dz$$ for $f$ holomorphic over $U$. If you want to transport this notion to a general Riemann surface, you are faced with the problem that no unique coordinate can be given, but you can count on some charts. So what you do is to allow collections of homolorphic forms defined over the elements of an atlas, with some nice compatibility properties.

That's what you need to prove in your case: you need to recognise that $u$ and $v$ are two allowed coordinates in your affine curve $X$. This is true because in the affine case projecting to variables gives rise to charts.

I think you can handle the other parts once you have figured out this.

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If you would like to warm up with elementary and down-to-earth methods in algebraic curve, there is a nice book by Egbert Brieskorn, called Plane algebraic curves. It carries out some classical topics which usually are covered in more abstract ways and gives a very interesting insight of the "simple" algebraic geometry.

Answer 2

Hey I am also working through that problem set in Dr. Miranda's book!

I think the solution with those first four problems (IV.1.A-D) is to check what happens to the given forms under transition maps using the formula given by $g(w) = f(T(w))T'(w)$ (a chain rule-looking thing) in the beginning of the section. So in any nonempty intersection of domains of two charts on a Riemann surface, the collection of 1-forms corresponding to a given $f(z)dz$ (where $z$ is the coordinate of one of the charts) should be compatible, i.e. $g(w) = f(T(w))T'(w)$ should be the holomorphic function of the 1-form $g(w)dw$ when you transition from $f(z)dz$. (Here, $w$ is the coordinate of the second chart.)

If you're only given a 1-form $f(z)dz$ on one chart, then compute what the other 1-forms should be (via the formula and transition maps) to define a 1-form on the entire Riemann surface (by which I really mean a collection of 1-forms, one for each chart, which are compatible in the sense that Dr. Miranda defines). The only condition you need to note, then, is that transitioning gives another $holomorphic$ 1-form. I just started on this exercise set last night after finishing my first-read through of almost the entire book (trying to get that connection between Picard group avatars down over the break), and it really helped me to go back to the sections on holomorphic maps on the specific Riemann surfaces and the charts of the specific Riemann surfaces (to get the explicit transition maps that Dr. Miranda derives/gives).

In your particular problem, the transition maps are either the identity or (as it turns out) a holomorphic function $g(z)$ (or $g(w)$ or $g(u)$, depending on what variables you're using). Also, you may note that $p(u,g(u))$ is certainly holomorphic for a holomorphic function $g$ and a polynomial $p(u,v)$. A similar comment might be made about $r(u,g(u))$ for a rational function $r(u,v)$.

Hope that helps! (Also, kudos for working on this material as an undergraduate! I'm a first year graduate student, and it's still difficult material for me. This book does have great exposition, though, and Dr. Miranda really meant for it to be accessible.)

The final piece of this problem will require you to use the Implicit Function Theorem, stated (in its 2-D or whatever correct small dimension) form in the section on smooth affine plane curves (and really the reason that we can make Riemann surfaces of these objects in the first place).

Answer 3

To answer your questions you should know the following: $X$ has a holomorphic atlas (F. Kirwan, "Complex Algebraic Curves", Cambridge University Press 1992, Proposition 5.27), and definitions of holomorphic/meromorphic differential forms (Chapter 6, Definitions 6.4, 6.5 from the book by Kirwan).

Fix a point $p$ of $X$. Let $t$, $t(p) = 0$ be a local coordinate (a chart) defined in an open neighborhood of $p$ in $X$. Then there exist two holomorphic (near $0$) functions $u(t)$, $v(t)$ which define the inverse chart. Therefore, we have the following.

1. $du$, $dv$ are holomorphic near $p$ on $X$ since $u'(t)dt$, $v'(t)dt$ are holomorphic near $0$.
2. $r(u, v)du$ and $r(u, v)dv$ are meromorphic since $r(u(t), v(t))u'(t)dt$ and $r(u(t), v(t))v'(t)dt$ are meromorphic near $0$.
3. You will get the desired equality by differentiating the identity $f(u(t), v(t)) = 0$.

The book "Plane Algebraic Curves" by E. Brieskorn and H. Knorrer (pp. 628-631) may be also helpful.

Answer 4

It seems to me that no one has given a complete answer so far.

1. Preliminaries.
Let $X=\{(u,v)\ \vert \ f(u,v)=0\}$ be a smooth affine plane curve. Consider the subsequent open subsets of $X$ $$X_1:=\{(u_0,v_0)\in X \ \vert \ \dfrac{\partial f}{\partial v}(u_0,v_0)\neq 0\}$$ $$X_2:=\{(u_0,v_0)\in X \ \vert \ \dfrac{\partial f}{\partial u}(u_0,v_0)\neq 0\}.$$ Then the following two projections define an atlas of $X$ $$u\colon X_1\rightarrow Y_1\subset \mathbb{C}; (u_0,v_0)\mapsto u_0$$ $$v\colon X_2\rightarrow Y_2\subset \mathbb{C}; (u_0,v_0)\mapsto v_0.$$ Here $Y_1$ and $Y_2$ denote the image of the projections $u$ and $v$, respectively. By the Implicit Function Theorem for Holomorphic Functions, both charts $u$ and $v$ admit holomorphic two-sided inverses $g\colon Y_1\rightarrow X_1$ and $h\colon Y_2\rightarrow X_2$, respectively.

2. $du$ and $dv$ define holomorphic $1$- forms.
We only consider the $1$-form specified by the single formula $du$. The other case is treated analogously.

In Miranda's terminology, we have to show that $du$ transforms uniquely to a holomorphic $1$-form under the change of coordinate mapping $T(v)=(v\circ g)(v)$. Now, $du$ transforms uniquely to $T'(v)dv$. Since $T(v)$ is holomorphic, so is $T'(v)$. The claim follows.

3. $p(u,v)du$ and $p(u,v)dv$ define holomorphic $1$-forms.
Analogously to above, we only consider the $1$-form specified by the single formula $p(u,v)du$. Now, under the change of coordinate mapping $T(v)=(v\circ g)(v)$ the single formula $p(u,v)du$ transforms uniquely to $p(T(v),v)\cdot T'(v)dv$. Since both $T(v)$ and $p(u,v)$ are holomorphic, so is $p(T(v),v)\cdot T'(v)$. The claim follows.

4. $r(u,v)du$ and $r(u,v)dv$ define meromorphic $1$-forms.
This proof proceeds in complete analogy to the cases 2. and 3. It boils down to the observation that the map $r(T(v),v)\cdot T'(v)$ is meromorphic for any rational function $r(u,v)$.

5. $(\partial f/\partial u)du = -(\partial f/\partial v)dv$ as holomorphic 1-forms on $X$.
This is proven in Andrew D. Hwang's answer here.