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).

Different possible branch cuts for the principal argument, principal logarithm and principal square root functions

For some work I was doing with a student, I was trying to find different ways of proving the familiar result that the complex square root function $f(z) = \sqrt{z}$ is discontinuous everywhere on the negative real axis. As I was working on alternative proofs it became very clear to me how sensitive’ all the proofs were to the particular definition of the principal argument I was using, namely that the principal argument $\theta = \text{Arg}z$ is the unique argument of $z$ satisfying $-\pi < \theta \leq \pi$. In a sense, this definition manufactures’ the discontinuity of the complex square root function on the negative real axis, because the principal argument function itself is discontinuous here: the principal argument of a sequence of points approaching the negative real axis from above will tend to $\pi$, whereas the principal argument of a sequence approaching the same point on the negative real axis from below will tend to $-\pi$. I realised that all the proofs I was coming up with were exploiting this discontinuity of the principal argument function. However, this particular choice of principal argument function is completely arbitrary. An alternative could be to say that the principal argument of $z$ is the unique argument satisfying $0 \leq \theta < 2\pi$ which we can call $\text{Arg}_{2\pi} z$. The effect of this choice of principal argument function is to make the complex square root function discontinuous everywhere on the positive real axis! It turns out that we can choose an infinite number of different lines to be lines of discontinuity for the complex square root function, simply by choosing different definitions of the principal argument function. The same applies to the complex logarithm function. In this note I want to record some of my thoughts about this.

The reason for having to specify principal argument functions in the first place is that we need to make complex functions of complex variables single-valued rather than multiple-valued, to make them well-behaved with regard to operations like differentiation. Specifying a principal argument function in order to make a particular complex function single-valued is called choosing a branch of the function. If we specify the principal argument function to be $f(z) = \text{Arg} z$ where $-\pi < \text{Arg} z \leq \pi$ then we define the principal branch of the logarithm function to be

$\text{Log} z = \text{log}_e |z| + i \text{Arg} z$

for $z \in \mathbb{C} - \{0\}$, and the principal branch of the square root function to be

$z^{\frac{1}{2}} = \text{exp}\big(\frac{1}{2} \text{Log} z \big)$

for $z \in \mathbb{C}$ with $z \neq 0$.

If we define the functions $\text{Log} z$ and $z^{\frac{1}{2}}$ in this way they will be single-valued, but the cost of doing this is that they will not be continuous on the whole of the complex plane (essentially because of the discontinuity of the principal argument function, which both functions inherit’). They will be discontinuous everywhere on the negative real axis. The negative real axis is known as a branch cut for these functions. Using this terminology, what I want to explore in this short note is the fact that different choices of branch for these functions will result in different branch cuts for them.

To begin with, let’s formally prove the discontinuity of the principal argument function $f(z) = \text{Arg} z$, $z \neq 0$, and then see how this discontinuity is inherited’ by the principal logarithm and square root functions. For the purposes of the proof we can consider the sequence of points

$z_n = |\alpha| \text{e}^{(-\pi + 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x < 0 \}$

Clearly, as $n \rightarrow \infty$, we have $z_n \rightarrow -|\alpha| = \alpha$. However,

$f(z_n) = \text{Arg} \big( |\alpha| \text{e}^{(-\pi + 1/n)i}\big)$

$= -\pi + \frac{1}{n}$

$\rightarrow -\pi$

whereas

$f(\alpha) = \text{Arg}\big(|\alpha| \text{e}^{\pi i} \big) = \pi$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal argument function is discontinuous at all points on the negative real axis.

Now consider how the following proof of the discontinuity of $f(z) = z^{\frac{1}{2}}$ on the negative real axis depends crucially on the discontinuity of $\text{Arg} z$. We again consider the sequence of points

$z_n = |\alpha| \text{e}^{(-\pi + 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x < 0 \}$

so that $z_n \rightarrow -|\alpha| = \alpha$. However,

$f(z_n) = z_n^{\frac{1}{2}} = \text{exp}\big(\frac{1}{2} \text{Log} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |z_n| + \frac{1}{2} i \text{Arg} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |\alpha| + \frac{1}{2} i (- \pi + \frac{1}{n}) \big)$

$\rightarrow |\alpha|^{\frac{1}{2}} \text{e}^{-i \pi /2} = - i |\alpha|^{\frac{1}{2}}$

whereas

$f(\alpha) = \big( |\alpha| \text{e}^{i \pi}\big)^{\frac{1}{2}}$

$= |\alpha|^{\frac{1}{2}} \text{e}^{i \pi/2} = i |\alpha|^{\frac{1}{2}}$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal square root function is discontinuous at all points on the negative real axis.

Now suppose we choose a different branch for the principal logarithm and square root functions, say $\text{Arg}_{2\pi} z$ which as we said earlier satisfies $0 \leq \text{Arg}_{2\pi} z < 2\pi$. The effect of this is to change the branch cut of these functions to the positive real axis! The reason is that the principal argument function will now be discontinuous everywhere on the positive real axis, and this discontinuity will again be inherited’ by the principal logarithm and square root functions.

To prove the discontinuity of the principal argument function $f(z) = \text{Arg}_{2\pi} z$ on the positive real axis we can consider the sequence of points

$z_n = \alpha \text{e}^{(2 \pi - 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x > 0 \}$

We have $z_n \rightarrow \alpha$. However,

$f(z_n) = \text{Arg}_{2\pi} \big(\alpha \text{e}^{(2\pi - 1/n)i}\big)$

$= 2\pi - \frac{1}{n}$

$\rightarrow 2\pi$

whereas

$f(\alpha) = \text{Arg}_{2\pi}(\alpha) = 0$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal argument function is discontinuous at all points on the positive real axis.

We can now again see how the following proof of the discontinuity of $f(z) = z^{\frac{1}{2}}$ on the positive real axis depends crucially on the discontinuity of $\text{Arg}_{2\pi} z$ there. We again consider the sequence of points

$z_n = \alpha \text{e}^{(2\pi - 1/n)i}$

where

$\alpha \in \{ x \in \mathbb{R}: x > 0 \}$

so that $z_n \rightarrow \alpha$. However,

$f(z_n) = z_n^{\frac{1}{2}} = \text{exp}\big(\frac{1}{2} \text{Log} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |z_n| + \frac{1}{2} i \text{Arg}_{2\pi} z_n \big)$

$= \text{exp}\big( \frac{1}{2} \text{log}_e |\alpha| + \frac{1}{2} i (2 \pi - \frac{1}{n}) \big)$

$\rightarrow \alpha^{\frac{1}{2}} \text{e}^{i 2 \pi /2} = - \alpha^{\frac{1}{2}}$

whereas

$f(\alpha) = \alpha^{\frac{1}{2}}$

Therefore $f(z_n) \not \rightarrow f(\alpha)$, so the principal square root function is discontinuous at all points on the positive real axis.

There are infinitely many other branches to choose from. In general, if $\tau$ is any real number, we can define the principal argument function to be $f(z) = \text{Arg}_{\tau} z$ where

$\tau \leq \text{Arg}_{\tau} < \tau + 2\pi$

and this will give rise to a branch cut for the principal logarithm and square root functions consisting of a line emanating from the origin and containing all those points $z$ such that $\text{arg}(z) = \tau$ modulo $2\pi$.

Topological equivalence of the 2-sphere and the extended complex plane (‘the Riemann sphere’)

I was chatting to a friend the other day about how the Riemann sphere arises when the point at infinity is added to the complex plane to give the extended complex plane $\Sigma = \mathbb{C}\bigcup \{\infty\}$. In this short note I want to quickly set out the mathematical details of this, i.e., an explicit homeomorphism which establishes the topological equivalence of the 2-sphere and the extended complex plane, giving rise to the name Riemann sphere for the latter.

The 2-sphere in $\mathbb{R}^3$, namely

$S^2 = \{(x_1, x_2, x_3) \in \mathbb{R}^3| x_1^2 + x_2^2 + x_3^2 = 1\}$

is visualised as sitting in a coordinate system with its centre at the origin, and the complex plane $\mathbb{C}$ is identified with the plane $x_3 = 0$ by identifying $z = x + iy$ ($x, y \in \mathbb{R}$) with $(x, y, 0)$ for all $z \in \mathbb{C}$. The point $N = (0, 0, 1)$ is identified as the ‘north pole’ of $S^2$, and stereographic projections from $N$ then give rise to a bijective map between $S^2\backslash \{N\}$ and $\mathbb{C}$ of the form

$\pi: S^2\backslash \{N\} \rightarrow \mathbb{C}$

$\pi: Q \mapsto P$

such that the points $N$, $Q$, and $P$ are collinear (but note that we are excluding $N$ from the domain of the bijection at this stage). The situation is illustrated in the following diagram.

To show that $\pi$ is in fact a homeomorphism (i.e., a continuous bijection from $S^2\backslash \{N\}$ to $\mathbb{C}$ whose inverse is also continuous), let $P = (x, y, 0)$ where $z = x + iy \in \mathbb{C}$, and let

$Q = (x_1, x_2, x_3) \in S^2 \backslash \{N\}$

Since $P$, $Q$, and $N$ are collinear, there is some constant $t$ such that

$P = N + t(Q - N)$

Therefore considering the coordinates separately in this equation we have

$\frac{x}{x_1} = t$      $\frac{y}{x_2} = t$      $\frac{1}{1 - x_3} = t$

and so

$x = \frac{x_1}{1-x_3}$    $y = \frac{x_2}{1-x_3}$

Therefore we have

$z = x + iy = \frac{x_1 + ix_2}{1 - x_3}$

and this is the map $\pi: Q \mapsto P$.

To get the inverse map $\pi^{-1}: P \mapsto Q$, we observe that

$x^2 + y^2 + 1 = \frac{x_1^2}{(1-x_3)^2} + \frac{x_2^2}{(1-x_3)^2} + \frac{(1-x_3)^2}{(1-x_3)^2}$

$= \frac{1 - x_3^2}{(1-x_3)^2} + \frac{1 - 2x_3 + x_3^2}{(1-x_3)^2}$   (using $x_1^2 + x_2^2 + x_3^2 = 1$)

$= \frac{2 - 2x_3}{(1-x_3)^2}$

$= \frac{2}{1-x_3}$

Therefore $\pi^{-1}: P \mapsto Q$ is given by

$x_1 = x(1-x_3) = \frac{2x}{x^2 + y^2 + 1}$

$x_2 = y(1-x_3) = \frac{2y}{x^2 + y^2 + 1}$

$1 - x_3 = \frac{2}{x^2 + y^2 + 1}$ $\Rightarrow$ $x_3 = \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1}$

These expressions show that both $\pi$ and $\pi^{-1}$ are continuous, so $\pi$ is a homeomorphism between $S^2\backslash \{N\}$ and $\mathbb{C}$, i.e., $S^2\backslash \{N\}$ and $\mathbb{C}$ are topologically equivalent.

To complete the picture we need to extend the homeomorphism $\pi$ to include the point $N$. We do this by defining

$\pi(N) = \infty$

where $\infty$, known as the point at infinity, is the distinguishing feature that makes the geometry of the present context non-Euclidean (it can be viewed as the point in this geometry where lines which start out parallel eventually meet, something which is impossible in Euclidean geometry). With the addition of the point at infinity into the picture we get the full homeomorphism

$\pi: S^2 \rightarrow \Sigma$

between the 2-sphere and the extended complex plane. This explains why the extended complex plane $\Sigma$ is referred to as the Riemann sphere. It is because the extended complex plane is homeomorphic (i.e., topologically equivalent) to the 2-sphere in $\mathbb{R}^3$.

Intuitively, points $Q$ on the 2-sphere $S^2$ which are close to the north pole $N$ correspond under $\pi$ to complex numbers $P = z$ with large magnitude $|z|$, i.e., to complex numbers which are ‘closer to infinity’ in a sense. Similarly, points $Q\prime$ on the 2-sphere which are close to the south pole $S = (0, 0, -1)$ correspond to complex numbers $z\prime$ with small magnitude $|z\prime|$. Points on the equator of the 2-sphere, which intersects the plane $x_3 = 0$, correspond to the unit circle $|z| = 1$ in the complex plane. The situation is illustrated in the following diagram.