# On the classification of singularities, with an application to non-rotating black holes

In mathematics a singularity is a point at which a mathematical object (e.g., a function) is not defined or behaves badly’ in some way. Singularities can be isolated (e.g., removable singularities, poles and essential singularities) or nonisolated (e.g., branch cuts). For teaching purposes, I want to delve into some of the mathematical aspects of isolated singularities in this note using simple examples involving the complex sine function. I will not consider nonisolated singularities in detail. These are briefly discussed with some examples in this Wikipedia page. I will also briefly look at how singularities arise in the context of black hole physics in a short final section.

Definition: A function $f$ has an isolated singularity at the point $\alpha$ if $f$ is analytic on a punctured open disc $\{z: 0 < |z - \alpha| < r \}$, where $r > 0$, but not at $\alpha$ itself.

Note that a function $f$ is analytic at a point $\alpha$ if it is differentiable on a region containing $\alpha$. Strangely, a function can have a derivative at a point without being analytic there. For example, the function $f(z) = |z|^2$ has a derivative at $z = 0$ but at no other point, as can easily be verified using the Cauchy-Riemann equations. Therefore this function is not analytic at $z = 0$. Also note with regard to the definition of an isolated singularity that the function MUST be analytic on the whole’ of the punctured open disc for the singularity to be defined. For example, despite appearances, the function

$f(z) = \frac{1}{\sqrt{z}}$

does not have a singularity at $z = 0$ because it is impossible to define a punctured open disc centred at $0$ on which $f(z)$ is analytic (the function $z \rightarrow \sqrt{z}$ is discontinuous everywhere on the negative real axis, so $f(z)$ fails to be analytic there).

I find it appealing that all three types of isolated singularity (removable, poles and essential singularities) can be illustrated by using members of the following family of functions:

$f(z) = \frac{\sin(z^m)}{z^n}$

where $m, n \in \mathbb{N}$. For example, if $m = n = 1$ we get

$f_1(z) = \frac{\sin(z)}{z}$

which has a removable singularity at $z = 0$. If $m = 1, n = 3$ we get

$f_2(z) = \frac{\sin(z)}{z^3}$

which has a pole of order $2$ at $z = 0$. Finally, if $m = -1, n = 0$ we get

$f_3(z) = \sin\big( \frac{1}{z} \big)$

which has an essential singularity at $z = 0$. In each of these three cases, the function is not analytic at $z = 0$ but is analytic on a punctured open disc with centre $0$, e.g., $\{z: 0 < |z| < 1\}$ or indeed $\mathbb{C} - \{0\}$ (which can be thought of as a punctured disc with infinite radius). In what follows I will use these three examples to delve into structural definitions of the three types of singularity. I will then explore their classification using Laurent series expansions.

Structural definitions of isolated singularities

Removable singularities

Suppose a function $f$ is analytic on the punctured open disc

$\{z: 0 < |z - \alpha| < r\}$

and has a singularity at $\alpha$. The function $f$ has a removable singularity at $\alpha$ if there is a function $g$ which is analytic at $\alpha$ such that

$f(z) = g(z)$ for $0 < |z - \alpha| < r$

We can see that $g$ extends the analyticity of $f$ to include $\alpha$, so we say that $g$ is an analytic extension of $f$ to the circle

$\{z: |z - \alpha| < r \}$

With removable singularities we always have that $\lim_{z \rightarrow \alpha} f(z)$ exists since

$\lim_{z \rightarrow \alpha} f(z) = g(\alpha)$

(this will not be true for the other types of singularity) and the name of this singularity comes from the fact that we can effectively remove’ the singularity by defining $f(\alpha) = g(\alpha)$.

To apply this to the function

$f_1(z) = \frac{\sin(z)}{z}$

we first observe that the Maclaurin series expansion of $\sin(z)$ is

$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots$ for $z \in \mathbb{C}$

Therefore we can write

$f_1(z) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots$ for $z \in \mathbb{C} - \{0\}$

If we then set

$g(z) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots$ for $z \in \mathbb{C}$

we see that $g(z)$ extends the analyticity of $f_1(z)$ to include $z = 0$. We also see that

$\lim_{z \rightarrow 0} f_1(z) = g(0)$

Therefore $f_1(z)$ has a removable singularity at $z = 0$.

Poles of order k, k > 0

Suppose a function $f$ is analytic on the punctured open disc

$\{z: 0 < |z - \alpha| < r\}$

and has a singularity at $\alpha$. The function $f$ has a pole of order $k$ at $\alpha$ if there is a function $g$, analytic at $\alpha$ with $g(\alpha) \neq 0$, such that

$f(z) = \frac{g(z)}{(z - \alpha)^k}$ for $0 < |z - \alpha| < r$

With poles of order $k$ we always have that

$f(z) \rightarrow \infty$ as $z \rightarrow \alpha$

(which distinguishes them from removable singularities)

and

$\lim_{z \rightarrow \alpha} (z - \alpha)^k f(z)$

exists and is nonzero (since $\lim_{z \rightarrow \alpha} (z - \alpha)^k f(z) = g(\alpha) \neq 0$).

To apply this to the function

$f_2(z) = \frac{\sin(z)}{z^3}$

we first observe that

$f_2(z) = \frac{\sin(z)/z}{z^2} = \frac{g(z)}{z^2}$ for $z \in \mathbb{C} - \{0\}$

where $g$ is the function

$g(z) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots$ for $z \in \mathbb{C}$

Since $g(0) = 1 > 0$, we see that $f_2(z)$ behaves like $\frac{1}{z^2}$ near $z = 0$ and

$f_2(z) \rightarrow \infty$ as $z \rightarrow 0$

so the singularity at $z = 0$ is not removable. We also see that

$\lim_{z \rightarrow 0} z ^2 f_2(z) = g(0) = 1$

Therefore the function $f_2(z)$ has a pole of order $2$ at $z = 0$.

Essential singularities

Suppose a function $f$ is analytic on the punctured open disc

$\{z: 0 < |z - \alpha| < r\}$

and has a singularity at $\alpha$. The function $f$ has an essential singularity at $\alpha$ if the singularity is neither removable nor a pole. Such a singularity cannot be removed in any way, including by mutiplying by any $(z - \alpha)^k$, hence the name.

With essential singularities we have that

$\lim_{z \rightarrow \alpha} f(z)$

does not exist, and $f(z)$ does not tend to infinity as $z \rightarrow \alpha$.

To apply this to the function

$f_3(z) = \sin\big( \frac{1}{z}\big)$

we observe that if we restrict the function to the real axis and consider a sequence of points

$z_n = \frac{2}{(2n + 1) \pi}$

then we have that $z_n \rightarrow 0$ whereas

$f_3(z_n) = \sin\big(\frac{(2n + 1) \pi}{2}\big) = (-1)^n$

Therefore

$\lim_{z \rightarrow 0} f_3(z)$

does not exist, so the singularity is not removable, but it is also the case that

$\lim_{z \rightarrow 0} f_3(z) \not \rightarrow \infty$

so the singularity is not a pole. Since it is neither a removable singularity nor a pole, it must be an essential singularity.

Classification of isolated singularities using Laurent series

By Laurent’s Theorem, a function $f$ which is analytic on an open annulus

$A = \{z: 0 \leq r_1 < |z - \alpha| < r_2 \leq \infty \}$

(shown in the diagram) can be represented as an extended power series of the form

$f(z) = \sum_{n = -\infty}^{\infty} a_n(z - \alpha)^n$

$= \cdots + \frac{a_{-2}}{(z - \alpha)^2} + \frac{a_{-1}}{(z - \alpha)} + a_0 + a_1 (z - \alpha) + a_2 (z - \alpha)^2 + \cdots$

for $z \in A$, which converges at all points in the annulus. It is an extended’ power series because it involves negative powers of $(z - \alpha)$. (The part of the power series involving negative powers is often referred to as the singular part. The part involving non-negative powers is referred to as the analytic part). This extended power series representation is the Laurent series about $\alpha$ for the function $f$ on the annulus $A$. Laurent series are also often used in the case when $A$ is a punctured open disc, in which case we refer to the series as the Laurent series about $\alpha$ for the function $f$.

The Laurent series representation of a function on an annulus $A$ is unique. We can often use simple procedures, such as finding ordinary Maclaurin or Taylor series expansions, to obtain an extended power series and we can feel safe in the knowledge that the power series thus obtained must be the Laurent series.

Laurent series expansions can be used to classify singularities by virtue of the following result: If a function $f$ has a singularity at $\alpha$ and if its Laurent series expansion about $\alpha$ is

$f(z) = \sum_{n = -\infty}^{\infty} a_n(z - \alpha)^n$

then

(a) $f$ has a removable singularity at $\alpha$ iff $a_n = 0$ for all $n < 0$;

(b) $f$ has a pole of order $k$ at $\alpha$ iff $a_n = 0$ for all $n < -k$ and $a_{-k} \neq 0$;

(c) $f$ has an essential singularity at $\alpha$ iff $a_n \neq 0$ for infinitely many $n < 0$.

To apply this to our three examples, observe that the function

$f_1(z) = \frac{\sin(z)}{z}$

has a singularity at $0$ and its Laurent series expansion about $0$ is

$\frac{\sin(z)}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \frac{z^6}{7!} + \cdots$

for $z \in \mathbb{C} - \{0\}$. This has no non-zero coefficients in its singular part (i.e., it only has an analytic part) so the singularity is a removable one.

The function

$f_2(z) = \frac{\sin(z)}{z^3}$

has a singularity at $0$ and its Laurent series expansion about $0$ is

$\frac{\sin(z)}{z^3} = \frac{1}{z^2} - \frac{1}{3!} + \frac{z^2}{5!} - \cdots$

for $z \in \mathbb{C} - \{0\}$. This has $a_n = 0$ for all $n < -2$ and $a_{-2} \neq 0$, so the singularity in this case is a pole of order $2$.

Finally, the function

$f_3(z) = \sin\big( \frac{1}{z} \big)$

has a singularity at $0$ and its Laurent series expansion about $0$ is

$\sin \big(\frac{1}{z} \big) = \frac{1}{z} - \frac{1}{3! z^3} + \frac{1}{5! z^5} - \cdots$

for $z \in \mathbb{C} - \{0\}$. This has $a_n \neq 0$ for infinitely many $n < 0$ so the singularity here is an essential singularity.

Singularities in Schwarzschild black holes

One often hears about singularities in the context of black hole physics and I wanted to quickly look at singularities in the particular case of non-rotating black holes. A detailed investigation of the various singularities that appear in exact solutions of Einstein’s field equations was conducted in the 1960s and 1970s by Penrose, Hawking, Geroch and others. See, e.g., this paper by Penrose and Hawking. There is now a vast literature on this topic. The following discussion is just my own quick look at how the ideas might arise.

The spacetime of a non-rotating spherical black hole is usually analysed using the Schwarzschild solution of the Einstein field equations for an isolated spherical mass $m$. In spherical coordinates this is the metric

$\Delta \tau = \bigg[ \big(1 - \frac{k}{r}\big) (\Delta t)^2 - \frac{1}{c^2} \bigg\{\frac{(\Delta r)^2}{\big(1 - \frac{k}{r}\big)} + r^2(\Delta \theta)^2 + r^2 \sin^2 \theta (\Delta \phi)^2\bigg\} \bigg]^{1/2}$

where

$k = \frac{2mG}{c^2}$ and $m$ is the mass of the spherically symmetric static object exterior to which the Schwarzschild metric applies. If we consider only radial motion (i.e., world lines for which $\Delta \theta = \Delta \phi = 0$) the Schwarzschild metric simplifies to

$(\Delta \tau)^2 = \big(1 - \frac{k}{r}\big) (\Delta t)^2 - \frac{1}{c^2}\frac{(\Delta r)^2}{\big(1 - \frac{k}{r}\big)}$

We can see that the $\Delta r$ term in the metric becomes infinite at $r = k$ so there is apparently a singularity here. However, this singularity is `removable’ by re-expressing the metric in a new set of coordinates, $r$ and $t^{\prime}$, known as the Eddington-Finkelstein coordinates. The transformed metric has the form

$(\Delta \tau)^2 = \big(1 - \frac{k}{r}\big) (\Delta t^{\prime})^2 - \frac{2k \Delta t^{\prime} \Delta r}{cr} - \frac{(\Delta r)^2}{c^2}\big(1 + \frac{k}{r}\big)$

which does not behave badly at $r = k$. In general relativity, this type of removable singularity is known as a coordinate singularity. Another example is the apparent singularity at the $90^{\circ}$ latitude in spherical coordinates, which disappears when a different coordinate system is used.

Since the term $\big(1 - \frac{k}{r} \big)$ in the Schwarzschild metric becomes infinite at $r = 0$, it appears that we also have a singularity at this point. This is not a removable singularity and can in fact be recognised in terms of the earlier discussion above as a pole of order 1 (also called a simple pole).