Differential forms

Question

I failed to understand the definition of holomorphic $1$-form on Riemann surfaces. Can one explain it here? I saw two definitions in the books of Miranda and Farkas-Kra.

Definition 1.: Suppose that $w_1=f(z)dz$ is a holomorphic $1$-form in the coordinate $z$, defined on an open set $V_1$. Also suppose that $w_2=g(w)dw$ be a holomorphic $1$-form in the coordinate $w$, defined on an open set $V_2$. Let $T$ be a holomorphic mapping from open set $V_2$ to $V_1$. We say that $w_1$ transforms to $w_2$ under $T$ if

(1) $g(w)=f(T(w))T'(w)$.

Let $X$ be a Riemann surface. A holomorphic $1$-form on $X$ is a collection of holomorphic $1$-forms $\{w_{\phi}\}$, one for each chart $\phi\colon U\rightarrow V $ in the co-ordinate of the target $V$, such that if two charts $\phi_i\colon U_i\rightarrow V_i$ (for $i=1,2$) have overlapping domains then the associated $1$-form $w_{\phi_1}$ transforms to $w_{\phi_2}$ under the change of coordinate mapping $T=\phi_1 \circ \phi_{2}^{-1}$.

Question 1: Can one explain the meaning of (1)? Can we say anything about it in terms of commutative diagram?

---

Definition 2. Let $M$ be a Riemann surface. A $1$-form $w$ on $M$ is an (ordered) assignment of two continuous functions $f$ and $g$ to each local coordinate $z(=x+iy)$ on $M$ such that

(2) $ fdx + gdy$

is invariant under coordinate changes; that is, if $\tilde{z}$ is another local coordinate on $M$ and the domain of $\tilde{z}$ intersects non-trivially with the domain of $z$, and if $w$ assigns the functions $\tilde{f},\tilde{g}$ to $\tilde{z}$, then

(3) $ \tilde{f}(\tilde{z})=\frac{\partial x}{\partial \tilde{x}} f(z(\tilde{z})) + \frac{\partial y}{\partial \tilde{x}} g(z(\tilde{z})) $

(4) $ \tilde{g}(\tilde{z})=\frac{\partial x}{\partial \tilde{y}} f(z(\tilde{z})) + \frac{\partial y}{\partial \tilde{y}} g(z(\tilde{z})) $

Question 2 What is the meaning of these equations? How did they come (or how we imposed these conditions for the definition of holomorphic $1$-form? Can we say anything about them in terms of commutative diagrams.

Answer 1

I will assume that you know how to handle smooth differential forms on smooth real manifolds, this is a necessary prerequisite for understanding complex and holomorphic differential forms on complex manifolds.

(I won't mention commutative diagrams because I don't know how to draw these here, because I don't know which ones would be useful and because I'm confident that you'll come up with the ones you find useful yourself after you have understood the definition in the most simple setting.)

Suggestion: Let's look at the most simple situation, $\mathbb{C}$ as both a one dimensional complex manifold and as a two dimensional smooth real manifold $\mathbb{R}^2$.

On $\mathbb{R}^2$ we have a global chart, the cartesian coordinates, with coordinates denoted by $(x, y)$. In every point we have a basis of the tangential space $(\partial_x, \partial_y)$ and the dual basis of differential forms in the cotangential space $(d x, d y)$. Every (smooth) differential form can therefore be written (in this chart) as $$ d w = f_1(x, y) d x + g_1(x, y) d y $$ where we say that $d w$ is smooth if $f$ and $g$ are.

Now we can make a coordinate change in $\mathbb{C} = \mathbb{R}^2$ to the new coordinates $$ z = x + i y $$ and $$ \bar{z} = x - i y $$ This is an example of a biholomorphic mapping of an open set $V_2 = \mathbb{C}$ to $V_1 = \mathbb{C}$, which is therefore a change of coordinates on $\mathbb{C}$ (both as a real smooth and as a complex manifold). Now we can express every differential form $w$ again in the new coordinates: $$ d w = f_2(z, \bar{z}) d z + g_2(z, \bar{z}) d \bar{z} $$ The $f_2, g_2$ are related to $f_1, g_1$ (how?).

Motivation for the definition of "holomorphic differential form":

The differential $d f$ of a smooth function $f$ is a real smooth differential form. Similarily, one would like to define "holomorphic differential form" in a way such that the differential $d f$ of a holomorphic function $f$ is a holomorphic differential form. Since $f$ is holomorphic iff it depends on the coordinate $z$ only, one defines a holomorphic differential form to be a differential form that does depend on the coordinate $z$ only, that is $d w$ is holomorphic iff it can be written as $$ d w = g(z) d z $$ with a holomorphic function $g(z)$. This definition makes explicitly use of the global coordinate chart of $\mathbb{C}$ that we defined above. So, in order to show that "holomorphic differential form" is well defined on $\mathbb{C}$ seen as a complex manifold, we need to show that coordinate changes on $\mathbb{C}$ do not map holomorphic differential forms to non-holomorphic or vice versa.

A coordinate change in our example would be any biholomorphic map of an open subset of $\mathbb{C}$ to another open subset of $\mathbb{C}$. Suggestion: Pick an easy example, write $d w$ in real coordinates and do a coordinate transform.

After that, have another look at the two definitions you cited.

Answer 2

The easiest way to understand differential 1-forms on a Riemann surface is to recall single variable calculus. For example, suppose $F$ is a function of one variable $x$. Then, its differential is $$dF = F'(x) \, dx$$ and now suppose $x$ is itself a function of another variable $y$, say $x = T(y)$. Then, $$dx = T'(y) \, dy$$ and so $$dF = F'(T(y)) T'(y) \, dy$$ If we write $f = F'$ and $\omega = dF$ we get $$\omega = f(x) \, dx = f(T(y)) T'(y) \, dy$$ which looks just like your first equation. This is the basic idea which motivates the abstract construction of differential 1-forms.

I don't think these equations are easily expressed in terms of commutative diagrams, as they are based on arbitrary choices of charts. An intrinsic definition of holomorphic differential 1-forms would be something along the lines of "holomorphic sections of the cotangent bundle". From a sheaf point of view, we're saying that the $\mathbb{C}$-space of holomorphic differential 1-forms defined on $U$, an open subset of the Riemann surface $S$, $U$ biholomorphic to $V$, an open subset of the complex plane $\mathbb{C}$, is isomorphic to the $\mathbb{C}$-space of holomorphic functions on $V$, and the equations in your post define the restriction maps of the sheaf.

Answer 3

a 1-form $w = f(z)dz$ is something to be thought of as a derivative : around any point (except finitely many) $z_0$, it locally has a unique primitive function $g(z)$ such that $g(z_0) = 0$ and $dg(z) = w$. We understand what $dg(z) = f(z)dz$ means when we are in an open of $\mathbb{C}$. The "transformation" formula is simply saying what the derivative of the transformed function is if we know the original derivative.

For example, if you pick the riemann sphere, with its two charts $\phi_1 : u \in \mathbb{C} \to U_1, \phi_2 : v \in \mathbb{C} \to U_2$, such that $\phi_i^{-1}(U_1 \cap U_2) = \mathbb{C}^*$ and $\phi_1^{-1}(z)\phi_2^{-1}(z) = uv = 1$ for all $z \in U_1 \cap U_2$.

on $U_1$ you have the 1-form $du$. For any $u_0 \in \mathbb{C}$ the local solution to $df = du$ that is $0$ at $u_0$ is $g_1(u) = u-u_0$. Now if $u_0 \neq 0$, you can change the chart : you get $v_0 = 1/u_0$, and $g_2(v) = g_1 \circ \phi_1^{-1} \circ \phi_2 (v) = g_1(u = 1/v) = 1/v - u_0 = 1/v - 1/v_0$. The derivative of $g_2$ is $dg_2 = -dv/v^2$. Thus, the 1-form $du$ has changed into $-dv/v^2$ under the transformation $T = \phi_1^{-1} \circ \phi_2 $