# J.-P. Serre: Duality of regular differentials on singular curves

I already asked this on math.stackexchange.com, but didn't get any responses. I hope it is appropriate here.

Let $$X'$$ be an irreducible singular algebraic curve over an algebraically closed field $$k$$, let $$X \to X'$$ be its normalization, and consider a singular point $$Q \in X'$$. Let $$K = Q(X)$$ be the function field of $$X$$ and $$X'$$.

Let $$\mathcal{O}_Q' = \mathcal{O}_{X', Q}$$ be the stalk of the structure sheaf of $$X'$$ at $$Q$$, and let $$\mathcal{O}_Q = \bigcap_{P \mapsto Q} \mathcal{O}_P$$ be its normalization. Here $$\mathcal{O}_P$$ is the stalk of the structure sheaf of $$X$$ at $$P \in X$$, and the intersection is over all points mapping to $$Q$$.

In Algebraic Groups and Class Fields by J.-P. Serre, chapter IV §3, Serre introduces the module $$\underline{\Omega}_Q'$$ of regular differentials at $$Q$$. A differential $$\omega \in D_k(K)$$ is called regular, iff $$\begin{equation}\sum_{P \mapsto Q} \operatorname{Res}_P(f \omega) = 0 \quad \text{for all} \ f\in \mathcal{O}_Q'.\end{equation}$$

Similarly to $$\mathcal{O}_Q$$, Serre defines $$\underline{\Omega}_Q = \bigcap_{P \mapsto Q} \Omega_P.$$ Since every differential $$\omega \in \underline{\Omega}_Q$$ has no poles at any point $$P \mapsto Q$$, clearly $$\operatorname{Res}_P(f \omega) = 0$$ for $$f \in \mathcal{O}_Q'$$, so that $$\underline{\Omega}_Q \subset \underline{\Omega}_Q'$$.

Now to my question: The mapping \begin{align} \mathcal{O}_Q / \mathcal{O}_Q' \times \underline{\Omega}_Q' / \underline{\Omega}_Q & \to k \\ (f, \omega) & \mapsto \sum_{P \mapsto Q} \operatorname{Res}_P(f \omega) \end{align} is clearly bilinear and well-defined. Serre claims, that it is a perfect pairing, but I don't know why. I think we have to show two things:

1. If $$f \in \mathcal{O}_Q$$, with the property that for each $$\omega \in \underline{\Omega}_Q'$$, one has $$\sum_P \operatorname{Res}_P(f \omega) = 0$$, then in fact $$f \in \mathcal{O}_Q'$$.
2. If $$\omega \in \underline{\Omega}_Q'$$, such that for each $$f \in \mathcal{O}_Q$$, one has $$\sum \operatorname{Res}_P(f \omega) = 0$$, then $$\omega \in \underline{\Omega}_Q$$, i.e. $$\omega$$ is regular at every $$P \mapsto Q$$.

Any help would be appreciated :)

• Most of this should be in Tate's paper Residues of differentials on curves''. Oct 21, 2020 at 13:37
• @Meow As far as I can tell from skimming the paper, Tate only considers regular curves. The phenomena discussed here happen exactly when $X'$ is not regular. Oct 21, 2020 at 14:03
• Whoops, sorry! I didn't read the question properly. Oct 22, 2020 at 12:00

In the algebraic setting, I solved the second case: Given $$\omega \in \underline \Omega_Q' \setminus \underline \Omega_Q$$, we want to find an $$f \in \mathcal O_Q$$, such that $$\sum_P \operatorname{Res}_P(f \omega) \neq 0$$. The basic idea is that, as $$\omega \notin \underline\Omega_Q$$, there exists a point $$P_1 \mapsto Q$$ such that $$\omega$$ has a pole of order $$n > 0$$ at $$P_1$$. So if $$t$$ is a local parameter at $$P_1$$, then $$\operatorname{Res}_{P_1}(t^{n-1} \omega) \neq 0$$. However it is not clear if $$t \in \mathcal O_Q$$, and we also have to control the residues at the other points.

To solve this, let $$n_P$$ be the order of the pole of $$\omega$$ at $$P \mapsto Q$$, which is possibly $$0$$ if $$\omega$$ does not have a pole at $$P$$. Then define a divisor $$D = \left(\sum_{P\to Q} n_P [P]\right) - [P_1].$$ If we find a meromorphic function $$f$$ whose divisor of zeroes is exactly $$D$$, then $$\sum \operatorname{Res}_P(f\omega) = \operatorname{Res}_{P_1}(f \omega) \neq 0.\tag{1}$$ Pick any additional point $$R \in X$$, which does not map to $$Q$$. Then by Riemann-Roch, if $$N \gg 0$$, there exists a function $$f \in H^0(\mathcal O(N[R] - D)) \setminus H^0(\mathcal O(N[R] - D - [P_1])),$$ which is exactly an $$f$$ such that (1) holds.

I think I can also show this in the analytic setting. The argument may possibly work in the algebraic setting if one considers completions, but I don't feel too familiar with completions, and my own interest in this is actually analytic.

Let $$X'_1, \dotsc, X'_r$$ be the irreducible components of $$X'$$ at $$Q$$. Then $$X$$ is the disjoint union of the normalizations $$X_i \to X_i'$$. Since we work in the analytic setting and the question is local in $$X'$$ we can assume that each $$X_i$$ contains exactly one point $$P_i \mapsto Q$$. Algebraically this means $$\mathcal O_Q \cong \prod_i \mathcal O_{P_i}$$. Since the $$\mathcal O_{P_i}$$ are regular rings, we can choose a local coordinate $$\mathcal O_{P_i} \cong \mathbb C\{x_i\}$$.

1. Consider an element $$f \in \mathcal O_Q \setminus \mathcal O_Q'$$. We have to find a differential form $$\omega \in \underline{\Omega}_Q'$$ such that $$\sum_P \operatorname{Res}_P(f \omega) \neq 0$$.

1.1. Suppose $$f(P_i) \neq f(P_j)$$ for some $$i$$ and $$j$$, consider the differential form $$\omega$$ which vanishes at all $$P_k$$ for $$i \neq k \neq j$$, and has simple poles at $$P_i$$ and $$P_j$$ with residues $$\operatorname{Res}_{P_i} = - \operatorname{Res}_{P_j} = 1.$$ Since each $$g \in \mathcal O_Q'$$ can be developed locally at $$P_i$$ in a power series, $$\operatorname{Res}_{P_i}(g \omega) = g(Q)$$, and similarly $$\operatorname{Res}_{P_j}(g \omega) = - g(Q)$$. Hence $$\omega \in \underline{\Omega}_Q'$$, but $$\sum_{P \mapsto Q} \operatorname{Res}_P(f \omega) = f(P_i) - f(P_j) \neq 0.$$

1.2. If $$f(P_i) = f(P_j)$$ for all $$i,j$$, one irreducible component of $$X'$$ has to be singular.

Actually, that statement is not correct. As an example take $$X'$$ to be the union of three lines in $$\mathbb C^2$$, intersecting in zero. The normalisation map is given by \begin{align*} \mathbb C\{x,y\}/(x(x-y)y) & \to \mathbb C\{x\} \times \mathbb C\{y\} \times \mathbb C\{z\} \\ x & \mapsto (x,0,z/2) \\ y & \mapsto (0,y,z/2). \end{align*} The function $$f = (x,y,0)$$ takes the same value in all three points over the singularity, but does not come from the left: Any function $$g$$ which restricts to $$x$$ on $$\{y = 0\}$$ and to $$y$$ on $$\{x = 0\}$$ is of the form $$g = x + y + xy h$$, so that $$g$$ has a non-zero linear term in $$z$$.

It is sufficient to show the existence of $$\omega$$ on the irreducible component, so we might suppose $$X'$$ itself is irreducible. This means we have an inclusion $$\mathcal O_Q' \subset \mathbb C\{x\}$$. Since the quotient $$0 \to \mathcal O_Q' \to \mathbb C\{x\} \to \Bbb C^\delta \to 0$$ is finite-dimensional one has $$x^k \in \mathcal O_Q'$$ for some $$k > \delta$$, and so $$\mathcal O_Q'$$ is given by the vanishing of $$\delta$$ linear equations on the coefficients $$a_1, \dotsc, a_{k-1}$$ of a power series $$\sum_n a_n x^n$$. Let $$l = \gamma_1 a_1 + \dotsb + \gamma_{k-1} a_{k-1}$$ be one of those linear equations with $$l(f) \neq 0$$. Then define $$\omega = \left( \frac{\gamma_1x^{k-2} + \dotsb + \gamma_{k-2} x + \gamma_{k-1}}{x^{k}} \right) dx$$ such that for each power series $$g \in \mathbb C\{x\}$$ we have $$\operatorname{Res}_0(g \omega) = l(g)$$. Thus $$\omega \in \underline{\Omega}_Q'$$ and $$\operatorname{Res}_0(f \omega) \neq 0$$.

2. Suppose $$\omega \in \underline{\Omega}'_Q$$, but $$\omega \notin \underline{\Omega}_Q$$. Then $$\omega$$ has a pole at some $$P_i$$, and we can write $$\omega = h(x_i) dx_i$$ for some Laurentseries $$h(x_i) = \sum_{k \geq -n} h_k x^k_i, \quad h_{-n} \neq 0.$$ Thus $$\operatorname{Res}_{P_i}(x_i^{n-1} \omega) = h_{-n} \neq 0$$. So if we define $$f = (0, \dotsc, 0, x_i^{n-1}, 0, \dotsc, 0) \in \prod_j \mathcal O_{P_j}$$ then $$\sum_j \operatorname{Res}_{P_j}(f \omega) = h_{-n} \neq 0$$.