User mathoman - MathOverflow

Meromorphic 1-form and Picard's theorem

2011-04-15T23:36:53Z

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).

Answer by MathOMan for Basic question about branch points on Riemann surfaces

2011-04-23T17:11:27Z

Another very concrete answer but without any formula:

I suppose you know how to construct "with papers, scissors and glue" a Riemann surface with a single branch point of order 2 (for example the surface of the function $\sqrt z$). Now take the Riemann surface of the logarithm. It has a countable infinite number of sheets. On each sheet you can add "with papers, scissors and glue" a branch point of order two, and this at any place you wish except above the origin. In this way you construct, for any given countable set $A\subset\mathbb C$ a Riemann surface $f : X \to \mathbb C$ which has a branch point above every point of A.

Moreover you see in the same way, that you can prescribe any (finite or infinite) order to each branch point (just glue more sheets, as you would do for $\sqrt[n]z$ or $\ln$); and you can also prescribe the number of branch points you want to have above each point of A (above each point of A you may want to have a countable number of distinct branch points).

Special linear group of a quotient

2011-04-21T22:02:28Z

Let J be a non-trivial ideal of a commutative ring A. The canonical map from A to the quotient A/J induces a homomorphism $\varphi : SL_n(A) \to SL_n(A/J)$. In general $\varphi$ is not surjective (for an example you can look here, comment 7 octobre 2010).

Question: Is there an example of non-surjectiveness using only basics known to an undergrad?

Answer by MathOMan for Basic question about branch points on Riemann surfaces

2011-04-21T18:54:50Z

Here is another example using only algebraic and not transcendental functions. Take three distinct complex numbers $z_1,z_2,z_3$ and let $X$ be the Riemann surface obtained by integrating the form $\sqrt{(z-z_1)(z-z_2)(z-z_3)}dz$. The map $g: X \to\mathbb C$ which sends each germ $\varphi_z\in X$ to $z$ is a covering with infinitely many sheets above the $z$-plane. (The ramification points are exactly above $z_1,z_2,z_3$, and there are infinitely many; each one is of order 2. This is so because near $z_k$ the form is like $(z-z_k)^{1/2}dz$ and therefore the primitive is like $(z-z_k)^{3/2}+cte$.) Now consider the holomorphic map $f: X\to\mathbb C$ which sends each germ $\varphi_z\in X$ to $\varphi_z(z)$. The branch points of $f$ are exactly the same as the branch points of $g$, but whilst they are of order 2 for $g$ they are of order 3 for $f$. (To see why this is so resolve the equation $\zeta=(z-z_k)^{3/2}+cte$ in $z$.) Now comes the important point: For nearly all choices of $z_1,z_2,z_3$ the periods $a_1,a_2,a_3$ of $ydz$ on the elliptic curve $y^2=(z-z_1)(z-z_2)(z-z_3)$ are such that the lattice $\mathbb{Z}a_1+\mathbb{Z}a_2+\mathbb{Z}a_3$ is dense in the plane. This means that the set of the images of the branch points of $f$ is dense in the plane.

In this way you can construct many other examples...

Answer by MathOMan for Meromorphic 1-form and Picard's theorem

2011-04-19T19:08:07Z

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 $a$).

Answer by MathOMan for Meromorphic 1-form and Picard's theorem

2011-04-18T20:33:34Z

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:

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.
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.

Comment by MathOMan

2011-05-09T18:33:30Z

This is the second question of Arnold's Trivium. You can find the solutions to nearly all those questions (except for 27, 41, 51, 58, 68, 69, 70, 73, 74) here: mathoman.com/index.php/…

Comment by MathOMan

2011-04-22T14:57:32Z

My example must be modified in order to work: instead of taking three distinct points take five (thus replacing the elliptic curve by a hyperelliptic one to assure that there are really three periods for the form ydx).

Comment by MathOMan

2011-04-21T21:49:04Z

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.

Comment by MathOMan

2011-04-16T08:02:27Z

No, Tom Goodwillie, I don't see any necessity to suppose the U_j connected. But if you prefer you can suppose them connected.