In the thread Why do we categorize all other (iso.) singularities as "essential"?, here is one of the questions that was asked:

Do we not care about essential singularities to classify them further?

The accepted answer for this question addresses this as follows:

So all essential singularities have some things in common, but on the other hand this should not lead us to believe that they are all the same. What they have in common is complicated behaviour, but they can be complicated in very different ways! Indeed, different transcendental entire functions (those that have an essential singularity at infinity; i.e. are not polynomials) can vary very much with respect to their behavior near infinity. Just for example, for some such functions, such as $z\mapsto e^z$, there exist curves tending to infinity on which the function is bounded, while for others this is not the case.

I see that essential singularities can be very different, but perhaps there can still be a classification. Here is how I would state my question:

If we say that functions having an isolated essential singularity at $0$ are equivalent when they differ by multiplication by a non-vanishing holomorphic function, is there a nice or fundamental set of representatives of this equivalence relation?

More formally:

Let $\mathcal{S}$ (for "singularity") be the set of germs of functions at $0$ that are holomorphic except for an isolated essential singularity at $0$; in other words, let $\mathcal{S}$ be the quotient of $$\left\{(f,U)\,\middle\vert\, \begin{array}{c} U\subseteq\mathbb{C}\text{ is a neighborhood of }0,\\ f\colon U\to \widehat{\mathbb{C}}\text{ is holomorphic on }U\setminus\{0\},\\ f\text{ has an essential singularity at }0\end{array}\right\}$$ by the usual equivalence relation for germs.

Let $\mathcal{I}$ (for "invertible") be the set of germs of holomorphic functions at $0$ that are non-vanishing in a neighborhood of $0$; thus, for any $[f]\in \mathcal{I}$, we also have $[1/f]\in\mathcal{I}$. Multiplication of germs at a point is well-defined, so we have a map $\mathcal{I}\times\mathcal{S}\to\mathcal{S}$; it is in fact a group action.

Is there a nice or fundamental set of representatives for the orbits of this action?

Very naively, I might guess that a set of representatives is given by $e^{1/f}$ for holomorphic functions $f$ vanishing at $0$, together with any function obtainable from these by repeated composition with $e^z$ (thus, for example, $e^{e^{1/z}}$), but I don't know how I would go about determining whether this is correct.

P.S. I'm pretty weak with complex analysis; explanations assuming little background knowledge would be very welcome.

  • $\begingroup$ I don't have an answer, but your conjectured classification isn't true. A typical function with an essential singularity will have infinitely many zeroes in any neighborhood of the singularity. This makes it impossible to describe the function as the exponential of anything. $\endgroup$
    – Jim Belk
    May 14, 2013 at 2:06

1 Answer 1


The problem with the quotient by multiplication by invertible functions is that it barely makes a dent in the size of the space. The quotient remains too big to admit any explicit set of representatives (in contrast to the case of poles, which get represented by $\{z^{-n}:n\ge 1\}$). For example, a function can have any number of asymptotic values, to which it tends along various complicated curves terminating at the essential singularity. This structure remains untouched by the quotient. On the other hand, taking the quotient we lose the structure of vector space.

A better way to take a quotient of singular germs is additively: up to addition of a holomorphic function in a neighborhood of $0$. Then the orbits have a nice set of representatives: namely, functions holomorphic in $\widehat{\mathbb C}\setminus \{0\}$ and vanishing at $\infty$. Indeed, any function holomorphic in $\{z:0<|z|<r\}$ can be expanded into Laurent series $\sum_{n\in\mathbb Z}c_n z^n$; we get a representative by keeping only the terms with $n<0$.

Having taken the quotient in the preceding paragraph, we can just as well replace $z$ by $1/z$, thus putting equivalence classes of essential singularities in bijection with transcendental entire functions vanishing at $0$. The "vanishing at $0$" is more of a distraction, of which we can get rid by dividing by $z$. So, the question becomes: how to classify transcendental entire functions? Given the existence of several monographs written on the subject of entire functions, we can't have much hope for an exhaustive structure theorem.

Although the Weierstrass factorization theorem allows us to write down $f$ as a certain product, this product involves $e^g$, where $g$ is just another entire function. The factorization works much better when $f$ has finite order because $g$ becomes a polynomial then (Hadamard factorization). In this way, we get a classification of entire functions of finite order, and - returning to the original formulation - of essential singularities at $0$ such that $$\limsup_{r\to 0}\frac{1}{\log (1/r)}\log\log \sup_{|z|\ge r} |f(z)|<\infty \tag1$$ Every such singularity is essentially described by a polynomial and a sequence of complex numbers converging to $0$.


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