I am having difficulty understanding the Riemann surface for the inverse of the Gaussian integral $\int_0^z e^{\zeta^2}d\zeta$ in the complex plane.

Let $g(z)=\int_0^z e^{\zeta^2}d\zeta$, so $g'$ is never zero, and so is locally 1-1. Also, it seems that $g:\mathbb{C}\rightarrow\mathbb{C}$ must be surjective, by the following argument. Since $g$ is odd, if its range omits $w$ it must also omit $-w$, so by the little Picard theorem it can omit only 0, but it doesn't omit 0.

So $g$ is a nonconstant locally 1-1 entire function mapping $\mathbb{C}$ onto $\mathbb{C}$. Now, it seems not hard to prove that the Riemann surface of the inverse of $g$ has a single sheet and covers $\mathbb{C}$, and so gives a global inverse. (I will give more details below.) But then $g$ is a holomorphic homeomorphism, and the only such maps from $\mathbb{C}$ to itself are the affine maps $z\mapsto az+b$. (Note that $\lim_{z\rightarrow\infty}g(z)=\infty$ by a compactness argument, so setting $g(\infty)=\infty$ gives a meromorphic function by the Riemann removable singularities theorem.)

Here is the argument about the existence of the global inverse. Points on the Riemann surface correspond to function elements (power series), so say $g(z_0)=w_0$ and say $h_1$ and $h_2$ are two local inverses to $g$ both centered at $w_0$ and converging in a small disk about it. Since $g(h_1(w))=w=g(h_2(w))$ for all $w$ in the disk and $g$ is locally 1-1, it follows that $h_1\equiv h_2$ in the disk and so define the same function element. So every $w$ has at most one point above it in the Riemann surface. However, since $g$ is onto and an open map, there is a holomorphic local inverse around each $w$, and so every point has a point above it in the surface. (If you prefer germs to function elements, you can make pretty much the same argument.)

Where is the flaw? Can someone provide a reference discussing this surface?

Edit [Slaps forehead] I see the problem. Let $R$ be the Riemann surface. It is true that as a covering of the $w$-plane (domain of $g$), $R$ is single-sheeted. But as a surface mapping onto the $z$-plane (range of $g$), it is not, although it's a locally 1-1 holomorphic map onto the $z$-plane. So it can't be a covering of the $z$-plane.

It's an interesting surface, and if someone has a reference giving more info about it, I'd appreciate it.


It is a very nice surface but the map is not 1-1. At infinity the map has an essential singularity so every value except possibly one is taken infinitely many times (Picard's Great Theorem). The symmetry argument you give shows every value is taken.

The Riemann surface argument (for the 1-1 correspondance) only works if any bounded path in the image, starting say at the origin, $w=\gamma(t)$, $0\leq t \leq 1$, $\gamma(0)=0$ may be lifted to a bounded path in the source. The lifted path is the solution of the ode: $$ \frac{dz}{dt}=\frac{d\gamma(t)}{dt} \frac{1}{g'(z(t))}, \ \ z(0)=0.$$ Any simply connected neighborhood $U$ of the origin in the image then lifts to a simply connected nghb $V$ of the origin in the source and $g:V\rightarrow U$ is indeed 1-1. The problem here is that the path from
e.g. $0$ to $i\sqrt{\pi}/2$ lifts to a path going to infinity and there is no unique lift beyond, so a 1-1 correspondence is no longer guaranteed (and fails here).

  • $\begingroup$ But it is locally 1-1, yes? The Riemann surface argument seems to show that any surjective locally 1-1 holomorphic map is injective, however this appears to be a counter-example. So there's something wrong with either the Riemann surface argument, or one of the other claims. It differ crucially from the exponential map, because exp is not surjective (unless the argument using little Picard is flawed). $\endgroup$ – Michael Weiss Aug 18 '16 at 22:36
  • $\begingroup$ By "surjective", I meant surjective from $\mathbb{C}$ to $\mathbb{C}$. The simple connectedness of $\mathbb{C}$ is also crucial, of course. $\endgroup$ – Michael Weiss Aug 18 '16 at 22:44
  • $\begingroup$ Your little Picard seems convincing so the map should be surjective (and numerically it seems to be the case). But the 'lifting' argument does not work. If you take circles of radius $r$ centered at the origin and you use local 1-1 to lift it by the covering map, then it works for small circles but when it becomes of radius $\sqrt{\pi/2}$ then the lift will explode (parts going off to infinity) so there is a discontinuity in the lift. But this just gives a hint of something going wrong. I'll see if I can find a more rigorous proof. $\endgroup$ – H. H. Rugh Aug 18 '16 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.