# What do poles of differentials on a curve mean?

Let C be a (smooth proper) curve. I think of an element f of the (algebraic) function field of C as, literally, a map from C to P1. Poles of f literally mean the fiber over infinty, and div f makes good sense to me.

How should I think of a pole of a differential? I.e., let K(C) be the function field of C and define Omega(C) to be the set of symbols df for f in K(C), modulo the usual relations. Let w in Omega(C). I understand quite well the yoga of defining and manipulating div w; my question is -- how, exactly, should I think of poles of w?

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Omega(C) is not just stuff of the form df. – David Hansen Apr 19 '11 at 5:31

This is not unlike Francois' answer.

First, let's look at the naive description of the order of a zero/pole of a differential $\omega$ on $C$ at some point $p$: write $\omega = f(z)\;dz$ in terms of some local coordinate $z$ near $p$, and then define $\operatorname{ord}_p(\omega) = \operatorname{ord}_p(f)$. Explicitly, this says two things:

• Every differential is proportional to the differential of a local coordinate, locally;
• The differential of a local coordinate is regular where it is defined.

In order for that to make sense, you have to check that for any other local coordinate $w$, the ratio (derivative) $dw/dz$ is regular at $p$. Of course, some precise algebraic computation is necessary, but intuitively, this is just the statement that both $z$ and $w$ have "slope 1" at $p$, so are equal to order one.

The fact that you have to choose a local coordinate is what is troubling you (it also troubles me); it comes about because there is no impartial basis for comparison, like there is with rational functions, which you can just compare to the function 1. The way around this, which also frees you from coordinate choices, is to talk about the entire sheaf of differentials rather than individual differentials.

Let's define $\Omega_C$ to be the sheaf of regular (Kähler) differentials on $C$, as defined in any basic algebraic geometry book. You give me $\omega$, a rational section of $\Omega_C$, or in other words, an element of $\Omega_C(U)$ for some open set $U$, and we want to find its divisor. Here is how we restate the first part of the above computations:

• The fact that $\omega$ can be expressed locally in terms of rational functions means that $\Omega_C$ is a line bundle (which is is; the trivializations are choices of local coordinate).

What about the second part? Let's continue: a section of $\Omega_C(U)$ is the same thing as a map $\phi \colon \mathcal{O}_C|_U \to \Omega_C|_U$, sending the rational function 1 to the differential $\omega$. Suppose for the sake of argument that $\omega$ had only zeros but no poles; then around a point $p \in C \setminus U$, it would look like $z^n\;dz$, choosing a local coordinate, and therefore, the image of $\phi$ would look like the ideal $(z^n) \subset \mathcal{O}_{C,p}$. More intrinsically, the cokernel of $\phi$ would have length $n$ at $p$. Thus, the convention that $dz$ is regular means that:

• When $\phi$ is an inclusion of sheaves, the order of vanishing of $\omega$ at $p$ is the length of $\operatorname{coker}(\phi)$ at $p$.

Of course, $\omega$ has poles, since you said $C$ is proper. Thus, $\phi$ does not even extend to an inclusion of sheaves. However, we want to think of a pole of something as being like a zero of the inverse, and we know how to find zeros. Suppose that we extend $\phi$ as much as possible, so that its zeros lie in $U$, form the divisor of zeros:

$$D_Z = \operatorname{div}(\omega)_Z = \sum_p \ell(\operatorname{coker}(\phi)|_p)p$$

and replace $\phi$ by its induced map $\phi \colon \mathcal{O}_C(-D_Z)|_{U \setminus D_Z} \to \Omega_C|_{U \setminus D_Z}$. Then this new $\phi$ is an isomorphism, and we can invert it; the poles of $\phi$ are by definition the zeros of $\phi^{-1}$. The divisor of poles $D_P = \operatorname{div}(\omega)_P$, defined as the divisor of zeros of $\phi^{-1}$, is disjoint from $D_Z$, because the new $\phi$ already is an isomorphism on $D_Z$, so that, after twisting by $-D_P$, $\phi^{-1}$ extends to all of $C$ (you should convince yourself, by playing with DVR's, that it really does). Then

$$\operatorname{div}(\omega) = \operatorname{div}(\omega)_Z - \operatorname{div}(\omega)_P$$

is the canonically-defined divisor of $\omega$. The short definition of this divisor is therefore:

• $\operatorname{div}(\omega)$ is the unique divisor $D$ such that the induced map $\phi \colon \mathcal{O}_C(-D)|_U \to \Omega_C|_U$ extends to an isomorphism of line bundles.

This is what the divisor corresponding to a line bundle usually means. Note that none of this is particular to differential forms, but allows you to define the zeros and poles of any rational section of any line bundle.

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In the beginning, it is good to gain intuition by considering the case where $C$ is a (compact) Riemann surface. Then you can think of a differential on $C$ as a meromorphic section of some holomorphic line bundle on $C$.

More precisely, if you consider the trivial holomorphic line bundle $\mathcal{O}_C$ (which you can think of as the space $C \times \mathbf{C}$ over $C$, then its holomorphic sections are just the holomorphic functions on (open sets of) $C$, while its meromorphic sections are the meromorphic functions on $C$.

In the same way, there is a holomorphic line bundle $\Omega^1_C$ over $C$ (called the canonical bundle). The fiber of $\Omega^1_C$ over $p \in C$ is the cotangent space of $C$ at $p$ (this is a $\mathbf{C}$-line). Then the holomorphic sections of $\Omega^1_C$ are just holomorphic differential forms on $C$ (usually denoted by $\Omega^1(C)$), while the meromorphic sections of $\Omega^1_C$ are the differentials you consider.

More generally, given any holomorphic line bundle $\mathcal{L}$ on $C$, you can consider its space of holomorphic/meromorphic sections. For every meromorphic section $s$ of $\mathcal{L}$, you can make sense of $s$ being holomorphic at a given point $p \in C$. Moreover, you can define $\operatorname{ord}_p(s) \in \mathbf{Z}$. Finally you can define $\operatorname{div}(s)$, which is a divisor on $C$ and is well-defined up to the principal divisors.

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I think of them pretty simply as differential forms with zeros in the denominator of the "coefficient function" upon choosing a local uniformizing parameter (which is really just the definition).

Edit: The following isn't quite right, as pointed out in the comments. See below for attempt at fixing it.

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However, if you want something more akin to your "map to $\mathbb{P}^1$" description, you might try something like this: to the invertible sheaf $\Omega$ you can associate a projective bundle $\mathbb{P}(\Omega)$ equipped with a map $\pi:\mathbb{P}(\Omega)\to C$ whose fibers are the projectivizations of the fibers of $\Omega$. Then a meromorphic differential form should correspond to a section $s$ to $\pi$ and the poles of the form are the points where $s(x)=\infty$.

I can't say that I've ever seen exactly this written down, but it seems quite reasonable...

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Perhaps one should consider the projectivization of the bundle $\Omega\oplus\mathcal{O}$ instead. Clearly if I take a regular differential form $\omega$, it gives rise to a section of $\mathbb{P}(\Omega\oplus \mathcal{O})$ by considering the image of $\omega\oplus 1$ in the projectivization.

Arguing locally, it seems that if I take a meromorphic differential form of the form $t^{-n}udt$ where $u$ is a local unit, I can associate to it the image of $udt\oplus t^n$ in the projectivization. Now glue over the curve to associate a section of this projective bundle to your chosen meromorphic differential form. If everything glues without incident, it seems that the resulting section should have the desired property that the poles are the points mapping to $\infty$ in the fiber, just by construction.

Hopefully this makes more sense than my first attempt :)

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bold P of a line bundle is just isomorphic to C? – mdeland Apr 19 '11 at 18:33
Oops. I guess my comment isn't at all right as stated... I still wonder if there is a $\mathbb{P}^1$ bundle over $C$ with the property I mentioned. It clearly isn't $\mathbb{P}(\Omega)$, which as you point out is just $C$! Let me think... – Ramsey Apr 19 '11 at 18:59
Perhaps you want the following: consider $\Omega$ as a $\mathbb{G}_m$-torsor (the punctured line bundle) and allow $\mathbb{G}_m$ to act on $\mathbb{P}^1$ by $u \cdot [x : y] = [ux : y]$. Take the associated $\mathbb{P}^1$-bundle. – Ryan Reich Apr 19 '11 at 23:56

The right way to think about this is to look at the whole set of zeroes and poles of the meromorphic 1-form $\omega$, in other words to consider the divisor

$\textrm{div}(\omega)=\sum_{p} \textrm{ord}_p(\omega) \cdot p$.

Every divisor of this type is called a canonical divisor. Two canonical divisors are always linearly equivalent and their degree is $2g-2$, where $g$ is the genus of the curve. The linear system of all canonical divisors is denoted by $|K|$, and it is the basic tool for the study of algebraic curves.

Now let $D$ be an effective divisor on $C$; then the linear space of meromorphic 1-forms having poles at most in the set $D$ is given by the cohomology group $H^0(K+D)$ or, using Serre duality, by $H^1(-D)$.

This is pretty basic stuff and you should find it in any textbook dealing with algebraic curves or Riemann surfaces (see for instance Miranda's book).

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