Consider three functions $f, g$ and $h$ on a smooth curve $X$ over $\mathbb{C}$. I have found the following equality: $$\sum (res(f\frac{dg}{g})\frac{dh}{h}-res(f\frac{dh}{h})\frac{dg}{g})=0.$$

Here the sum is over complex points of $X$, $res$ means residue of the corresponding meromorphic form on $X$. The most important point is that object $$res(f\frac{dg}{g})\frac{dh}{h}$$ should be considered as an element of the module of absolute Kahler differentials $\Omega^1_{\mathbb{C}/\mathbb{Q}}.$

For $f=1$ it gives the Weil reciprocity law, since $$res(\frac{dg}{g})\frac{dh}{h}-res(\frac{dh}{h})\frac{dg}{g}=ord(g)\frac{dh}{h}-ord(h)\frac{dg}{g}=$$ $$=\frac{d\{g,h\}_W}{\{g,h\}_W},$$ where $\{g,h\}_W$ stands for the Weil symbol. So, $$0=\sum (res(\frac{dg}{g})\frac{dh}{h}-res(\frac{dh}{h})\frac{dg}{g})= =\sum\frac{d\{g,h\}_W}{\{g,h\}_W}=\frac{d\prod \{g,h\}_W}{\prod \{g,h\}_W}.$$ From this it follows that $\prod \{g,h\}_W$ is algebraic complex number. Since the product of Weil symbols depends continously on the functions, it should be constant. But for $g=h=1$ it is equal to one. This finishes the proof of the Weil reciprocity law.

Question 1: How one can prove this formula? I have found quite a long proof with Newton polygons, following ideas of Askold Khovansky.

Question 2: Can one interprete it as some kind of a reciprocity law for surfaces?

Question 3: I am almost sure that it is known. I would be very grateful for any reference.

  • $\begingroup$ The sum of interest is a difference between expressions of the form $\sum_P {\rm{res}}_P(\omega)\omega'$ for rational 1-forms $\omega$ and $\omega'$, but such a sum vanishes because it is $\omega'$ times the scalar $\sum_P {\rm{res}}_P(\omega) = 0$. Am I misunderstanding something? $\endgroup$
    – user27920
    Jul 10, 2014 at 0:05
  • $\begingroup$ The argument "by continuity" (I suppose "by holomorphicity") on non-zero functions is a bit hand-wavy, since to make sense of it one should restrict to $g$ and $h$ inside a fixed space such as functions with poles controlled by a divisor, but then the nature of the definition of the Weil symbol has delicate behavior when pole-orders change within that space. So it isn't clear if this really gives a proof of Weil reciprocity just from the residue theorem alone. $\endgroup$
    – user27920
    Jul 10, 2014 at 0:16
  • 1
    $\begingroup$ I think the expressions outside of the residue are supposed to be evaluated at the points, i.e. the sum is $\sum_P \mathrm{res}_P(\omega) \omega'(P)$. This alone does not make sense, but the difference should make sense (like in the Weil reciprocity). This forces $\mathrm{res}(f d\mathrm{log}(h))\mathrm{ord}(g) = \mathrm{res}(f d\mathrm{log}(g))\mathrm{ord}(h)$ at every point. If you take $f=1/x$, $g=x$ and $h = x+x^2$, the expression still doesn't make sense at $x=0$. $\endgroup$ Jul 10, 2014 at 5:56
  • 1
    $\begingroup$ There are higher analogs of Weil reciprocity called Suslin reciprocity laws (due to Milnor for $X=\mathbb{P}^1$, Bass-Tate and Suslin in general). They state that the composite $K^M_n(k(X))\rightarrow \prod_{x\in X} K^M_{n-1}(k(x))\rightarrow K^M_{n-1}(k)$ is trivial ($K^M$ are the Milnor $K$-theory groups). For $n=1$ this is "sum of the residues of a rational function is zero" and for $n=2$ this is Weil reciprocity. You can take the $n=3$ Suslin reciprocity which would involve 3 functions and apply the differential symbol, but it would be some statement in $\Omega^2_{\mathbb{C}}$. $\endgroup$ Jul 10, 2014 at 6:02
  • $\begingroup$ About the argument "by continuity": it is hand-wavy, but I think, one can make sense out of it by adding parameters. $\endgroup$ Jul 10, 2014 at 7:52

1 Answer 1


Look at so-called Parshin reciprocity law, for example Mazin' approach http://www.math.utoronto.ca/mmazin/thesis.pdf

I guess that is exactly what you want.

If I am not mistaken, your version follows from the standard version of Weil reciprocity law for the functions $f/g,f/h$, in the spirit of logarithmic differentials (cf. http://www.math.toronto.edu/askold/osaka.pdf) If not, it is very strange.


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