Let $D$ be the open unit disk in the complex plane and $U_1,U_2,\,\ldots\,,U_n$ be an open cover of the puntured disk $D^*= D\setminus\{0\}$. Suppose on each open set $U_j$ there is an injective holomorphic function $f_j$ such that $df_j=df_k$ on every intersection $U_j\cap U_k$.

Question: Is it true that the differentials $df_j$ glue together to a meromorphic 1-form on $D$?

Remark: If the residue is zero then it is true (with help of Picard's theorem).

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    $\begingroup$ Maybe you want to assume that the $U_j$ are connected; otherwise it's not true even in the residue zero case. $\endgroup$ Apr 16, 2011 at 4:02
  • $\begingroup$ No, Tom Goodwillie, I don't see any necessity to suppose the U_j connected. But if you prefer you can suppose them connected. $\endgroup$
    – MathOMan
    Apr 16, 2011 at 8:02
  • $\begingroup$ FYI: The question is copy-pasted from the Wikipedia article on the Picard theorem, see the 4th note at en.wikipedia.org/wiki/Picard_theorem $\endgroup$ Apr 19, 2011 at 23:31
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    $\begingroup$ @ Gunnar: yes the the question is the same as the conjecture in the Wikipedia article you mention. There is a good reason for that: the conjecture is due to MathOMan ! This is not apparent from the article because in the bibliography MathOMan appears under his real name . (I'm divulging no secret since MathOMan says so himself in his answer below) $\endgroup$ Apr 23, 2011 at 15:58
  • $\begingroup$ @Georges: Thank you for clearing that up. $\endgroup$ May 1, 2011 at 13:31

3 Answers 3


If the $U_i$'s are assumed to be connected (as in Tom Goodwillie's comment) the answer is yes. There are counterexamples if the $U_i$ are not connected.

The hypothesis implies that the differentials $df_j$ patch together to define a differential form $\omega$ on $D \setminus 0$ that is never 0 (on $D \setminus 0$). Integration of $\omega$ defines a covering space of $D \setminus 0$: that is, we can choose a basepoint, say $* = 1/2$, elements of the covering space $Y$ consist of points $z \in D \setminus 0$ together with a value that can be obtained by integrating $\omega$ along some path in $D \setminus 0$ from $*$ to $z$.

If $\omega$ is the differential of an injective function on each of finitely many connected $U_j$'s (following Tom Goodwilli's comment), then each $U_j$ can be lifted to the covering space $Y$. This implies that the branched cover has finitely many sheets. Since $\pi_1(D \setminus 0)$ is abelian, $Y$ an $n$-fold cyclic cover isomorphic to $z \mapsto z^n$. In these coordinates, pullback of $\omega$ to $Y$ can be integrated to give a function $g$ from $Y$ to the Riemann sphere such that satisfies $g(\zeta y) = g(\zeta) + C$ for some constant $C$, where $\zeta$ is a primitive $n$th root of unity. But then $n*C = 0$, so $C = 0$, so $g$ comes from a function on $D \setminus 0$; since it is finite-to-one near 0, by Picard's theorem this is a removable singularity (as a map to the Riemann sphere), it extends to a meromorphic function on $D$, and its differential is therefore meromorphic.

If the $U_j$'s are not assumed connected, the covering space $Y$ still exists, but it need not have finitely many sheets. In the infinite-sheeted case, the covering space is the universal cover of $D \setminus 0$ isomorphic to $z \mapsto \exp(z)$ from the right halfplane $Re(z) < 0$ to $D \setminus 0$. The integral of the pullback of $\omega$ to the right halfplane is a function that has the form $g(z) = a z + g_0(z)$, where $g_0(z)$ is the pullback of a function $f_0$ defined on $D \setminus 0$. (The linear term takes care of the integral of $\omega$ on a loop around the origin). The function $f_0$ must be an immersion (locally univalent function) from $D \setminus 0$ to $\mathbb C$. Such functions can be rather wild, for example, $f_0(z) = exp(1/z)$. The integral of $\omega$ itself could then be expressed as the multi-function $f(z) = z^a f_0(z)$, where $a$ is any complex number.

Claim for any complex number $a$ and any locally univalent function $f_0: D \setminus 0 \to \mathbb C$ there is a finite cover $U_i$ of $D \setminus 0$ (where the $U_i$ are not connected) such that on each $U_i$ $\omega$ is the differential of a univalent function $f$. Proof of Claim: Cover $D \setminus 0$ by countably many open sets such that the integral of $\omega$ in each set is univalent. Every cover of a 2-manifold has a refinement that is locally at most 3-to-1, so its nerve is a 2-complex. (This is one characterization of topological dimension). It has a further refinement $U$, corresponding to the barycentric subdivision, by a covering that can be partitioned into three parts $U = A \cup B \cup C$ where the elements of $A$ are disjoint, the elemetns of $B$ are disjoint, and the elements of $C$ are disjoint. We can integrate $\omega$ on each of the elements of $A$, $B$ and $C$, and then add suitable constants to make the images in each of $A$, $B$ and $C$ disjoint.

In summary: we've found three holomorphic functions $f_A$, $f_B$ and $f_C$ defined on the union of $A$, the union of $B$ and the union of $C$ whose differentials all equal $\omega$, but $\omega$ has an essential singularity at the origin.

  • $\begingroup$ This is a beautiful answer! $\endgroup$ Apr 16, 2011 at 23:23
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    $\begingroup$ I don't understand the second sentence in the third paragraph. Why finitely many sheets? $\endgroup$ Apr 19, 2011 at 16:50
  • $\begingroup$ @Tom Goodwillie: this was a lapse. I'll redo things. $\endgroup$ Apr 19, 2011 at 17:06

Thanks for the answer! I apologize for answering with a delay (am on biking trip in Corsica). Yes, Tom Goodwillie is right about the necessity of the connectiveness condition. Bill Thurston's answer, however, does not convince me. I will explain why:

Call $\omega$ the holomophic form defined on the punctured unit disk. There are two cases:

  1. The residue is zero. Then there is a holomorphic function on the unit disk such that $df=\omega.$ Thus $df=df_j$ for every j. Using the connectedness of $U_j$ one can add a constant to $f_j$ such that $f=f_j$ on $U_j$. Then one concludes easily with Picard that f is meromorphic on the unit disk.

  2. The residu $a$ is not zero. Then integrating $\omega$ yields a cover of infinite order because each turn around the origin adds the number $2\pi a$. So here I do not agree with Bill Thurston who says the order is finite. What actually happens is that the primitive of $\omega$ is of the form: holomorphic function on the punctured disk + $a\times$ logarithm.

Now here comes my explanation of Bill Thurston's mistake when he says that the covering space is of finite order. The best way to define the Riemann R(g) surface of an analytic germ g is to say that R(g) is the connected component of g in the total space $|\mathcal{O}|$ of the sheaf of holomorphic functions. There are two natural projections defined on R(g). The first projection sends each germ to its centerpoint ("projection on the variable plane"). The second projection sends each germ to the value the germ takes in its centerpoint ("projection on the value plane"). It is the first projection that we are interested in. But when arguing that his covering space is finite above the punctured unit disk, Bill Thurston actually uses the functions $f_j$ as map, thus taking the second projection instead of the first.

  • $\begingroup$ You're right, as Tom Goodwillie also commented to my post, it doesn't follow that the cover is finite to one. I'll redo things. $\endgroup$ Apr 19, 2011 at 17:06
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    $\begingroup$ To G. Magnusson: Yes, it is true that I stated this "conjecture" a dozen years ago in an article at Inst. Fourier (the same year that I left maths). I didn't know that meanwhile it got on Wikipedia! Last friday, Georges Elencwajg, a former professor of mine told me about Mathoverflow and so I posted this forgotten question here. And shame on me that I didn't see by myself the "connectedness condition" pointed out by Tom Goodwillie and Bill Thuirston! Now I will add it in the Wikipedia version. I have some naïve idea about Riemann surfaces but I lack the technique to tackle such a problem. $\endgroup$
    – MathOMan
    Apr 21, 2011 at 21:49
  • $\begingroup$ @MathOMan: Oh, I'm sorry, I didn't realize it was your conjecture. To be honest I thought some guy was posting problems from Wikipedia. My bad. $\endgroup$ May 1, 2011 at 13:28

Here is another conjecture that seems to me equivalent to my initial question. (It is a kind of "ugly" version above the value plane, the initial question above the variable plane, the disk, being the "nice" point of view.)

Let $f : E\to \mathbb{C}$ be a connected étale space and $t:E\to E$ an isomorphism ("translation") such that $f\circ t=f+2i\pi a$ where $a$ is a non-zero complex number. Then $\mathbb{Z}$ acts on $E$ via the "translations" $t^n$. The 1-form $df$ clearly induces a 1-form on the quotient space $E/\mathbb{Z}$ which we still denote by $df$. Conjecture: if the quotient space is biholomorphic to the punctured disk then the pullback of $df$ does not have an essential singularity at the center of the disk (and its residue at the center is exactly $a$).


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