Category Archives: Complex analysis

Möbius transformations and cross-ratios

A Möbius transformation is a map f:\mathbb C_\infty\to\mathbb C_\infty of the form

\displaystyle f(z)=\frac{az+b}{cz+d},\quad a,b,c,d\in\mathbb C,\quad ad-bc\neq 0

where \mathbb C_\infty:=\mathbb C\cup\{\infty\}is the extended complex plane and the ‘point at infinity’ \infty is defined so that

  1. if c\neq 0 then f(\infty)=a/c and f(-d/c)=\infty;
  2.  if c=0 then f(\infty)=\infty.

The following video gives a very illuminating illustration. The sphere in the video is called the Riemann sphere, which in a sense ‘wraps up’ the extended complex plane into a sphere. Each point on the sphere corresponds to a unique point on the plane (i.e. there is a bijection between points on the extended plane and points on the sphere), with the ‘light source’ being the point at infinity. This bijective correspondence is the main reason for including the point at infinity.

According to the video any Möbius transformation can be generated by the four basic ones: translations, dilations, rotations and inversions:

  1. Translation: f(z)=z+b, b\in\mathbb C
  2. Dilation: f(z)=az, a\in\mathbb R
  3. Rotation: f(z)=e^{i\theta}z, \theta\in[0,2\pi]
  4. Inversion: f(z)=1/z

Exercise. Show that any Möbius transformation is a composition of these four operations.

The Möbius transformations in fact form a group \mathcal M under composition which acts on \mathbb C_\infty. Moreover, we have a surjective homomorphism

\displaystyle\begin{matrix}\hbox{GL}_2(\mathbb C)\to\mathcal M,\quad\left(\begin{matrix} a & b\\ c & d\end{matrix}\right)\mapsto\displaystyle\frac{az+b}{cz+d}\end{matrix}.

Möbius transformations exhibit very interesting properties, some of which are:

Proposition 1. Given distinct z_1,z_2,z_3\in\mathbb C_\infty, there is a unique Möbius map f\in\mathcal M such that

\displaystyle f:\left(\begin{matrix} z_1\\ z_2\\z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)

Proof. It is not difficult to work out that the unique f is given by

\displaystyle f(z)=\frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1}. \square

Proposition 2. The action of \mathcal M on \mathbb C_\infty is sharply triply transitive: if z_1,z_2,z_3\in\mathbb C_\infty are distinct and w_1,w_2,w_3\in\mathbb C_\infty are distinct, then there eixsts a unique f\in\mathcal M such that f(z_i)=w_i for i=1,2,3.

Proof. By proposition 1, there is a unique g\in\mathcal M such that

\displaystyle g:\left(\begin{matrix} z_1\\ z_2\\ z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)

and a unique h\in\mathcal M such that

\displaystyle h:\left(\begin{matrix} w_1\\ w_2\\ w_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right).

Then f=h^{-1}\circ g is the unique map satisfying the required property. \square 

The cross-ratio [z_1,z_2,z_3,z_4] of four distinct points z_1,z_2,z_3,z_4\in\mathbb C_\infty is defined to be the unique \lambda\in\mathbb C_\infty such that if f\in\mathcal M is the unique map satisfying

\displaystyle f:\left(\begin{matrix} z_1\\ z_2\\z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)

then f(z_4)=\lambda, i.e.

\displaystyle [z_1,z_2,z_3,z_4]=\frac{z_1-z_4}{z_3-z_4}\frac{z_2-z_3}{z_2-z_1}.      (*)

One nice thing about cross-ratios is that they are preserved by Möbius transformations.

Proposition 3. If f\in\mathcal M, then [z_1,z_2,z_3,z_4]=[f(z_1),f(z_2),f(z_3),f(z_4)].

Proof. Let g\in\mathcal M such that

\displaystyle g:\left(\begin{matrix} z_1\\ z_2\\ z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right).

Then [z_1,z_2,z_3,z_4]=g(z_4). Likewise, if h\in\mathcal M satisfies

\displaystyle h:\left(\begin{matrix} f(z_1)\\ f(z_2)\\ f(z_3)\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right)

then h\circ f=g by proposition 1, so [f(z_1),f(z_2),f(z_3),f(z_4)]=h(f(z_4)) =g(z_4)=[z_1,z_2,z_3,z_4]. \square

From (*) we observe that some permutations of 1,2,3,4 leave the value of the cross-ratio [z_1,z_2,z_3,z_4] unaltered, e.g. (1\ 3)(2\ 4)\in S_4 is one such. What about the others?

Let S_4 act on the indices of the cross-ratio [z_1,z_2,z_3,z_4]. The permutations \sigma\in S_4 that fix [z_1,z_2,z_3,z_4] form the stabiliser subgroup of S_4 of this action. Using transitivity and invariance (propositions 2 and 3), the orbit of [z_1,z_2,z_3,z_4] is just the different assignments of the values 0,1,\infty to z_1,z_2,z_3; i.e. the distinct cross-ratios that we get by permuting the indices are just


[z_2,z_3,z_1,z_4], [z_3,z_1,z_2,z_4], [z_3,z_2,z_1,z_4].

Let f\in\mathcal M such that

\displaystyle f:\left(\begin{matrix} z_1\\ z_2\\z_3\end{matrix}\right)\mapsto\left(\begin{matrix} 0\\ 1\\ \infty\end{matrix}\right).

Then writing \lambda=f(z_4)=[z_1,z_2,z_3,z_4] shows that the values of the above cross-ratios are (not necessarily in this order—too lazy to work out the precise order)

\displaystyle \lambda,\frac 1\lambda, 1-\lambda,\frac{1}{1-\lambda},\frac{\lambda}{\lambda-1},\frac{\lambda-1}{\lambda}

and they (as functions of \lambda) form the subgroup of \mathcal M that fixes the set \{0,1,\infty\}. This group is isomorphic to S_3.

So there are in fact four permutations \sigma\in S_4 such that


and they form a subgroup of S_4 that is isomorphic to the Klein four-group

V_4=\{e, (1\ 2)(3\ 4), (1\ 3)(2\ 4), (1\ 4)(2\ 3)\}.


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Filed under Algebra, Complex analysis

Trivial zeros of the Riemann zeta function

The Riemann hypothesis is perhaps the most famous unsolved problem in mathematics. It says that all non-trivial zeros of the Riemann zeta function have real part 1/2. What is the Riemann zeta function? A common misconception is that it is defined as

\displaystyle\zeta (z)=\sum_{k=1}^\infty k^{-z}=\frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+\dots,\quad z\in\mathbb C.

This is false! Because the series on the right only converges for \Re(z)>1. Also, it is a well-established fact that the trivial zeros of the Riemann zeta function occur at the negative even numbers -2,-4,-6\dots. If we substitute z=-2 in the above equation we get 1^2+2^2+3^2+\dots which is far from being zero. So even though these are called ‘trivial’ zeros, it seems highly non-trivial why these are actually zeros. What’s going on?

The problem lies in the above definition. This equation does NOT define the Riemann zeta function, which is a meromorphic function (see below) on the whole complex plane. The above definition only works for \Re(z)>1. Then what about the other values of z? Before proceeding further, let us familiarise ourselves with some terminology from the theory of complex analysis.

Complex analysis is one of the most beautiful branches of mathematics. It deals with functions f:\mathbb C\to \mathbb C in the complex plane that are differentiable in some open subset D\subseteq \mathbb C. (Open sets are like open intervals; open intervals are line segments excluding the endpoints, open sets in the complex plane can be thought of as discs excluding their boundaries.) Some basic definitions are:

  • f:D\to\mathbb C is said to be analytic (or holomorphic) at z_0 if it is differentiable in a neighbourhood of z_0, i.e. at all points z such that |z-z_0|<\varepsilon for some \varepsilon >0.
  • f above is called entire if it is analytic at every z\in\mathbb C.
  • Every analytic function has a Laurent series, which is like a Taylor series, but negative powers are allowed.
  • If \displaystyle f(z)=\sum_{k=-n}^\infty a_k(z-z_0)^k is a Laurent series in some neighbourhood of z_0 then (z-z_0)^nf(z) is analytic, and z_0 is called a pole of f of order n. The coefficient a_{-1} of (z-z_0)^{-1} is the residue of f at z_0, which we shall denote as \text{res}(f,z_0).
  • If f:D\to\mathbb C is analytic except for a set of isolated poles, it is called meromorphic.

Unlike real analysis, amazing things happen when analysis is done in the complex plane. Some important results in complex analysis are:

\displaystyle\int_\gamma f(z)dz=0.

\displaystyle f(w)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-w}dz.

  • Cauchy’s residue theorem: If A is the set of poles of f inside the closed curve \gamma and f is analytic inside \gamma except for these poles, then

\displaystyle \int_\gamma f(z)dz=2\pi i\sum_{a\in A}\text{res}(f,a),

The identity theorem gives a useful method of extending a function analytically. Suppose D'\subseteq D\subseteq\mathbb C are open sets and f:D'\to\mathbb C and g:D\to\mathbb C are analytic. If f=g on D', and we were to define f on D preserving analyticity, then the identity theorem says f=g on D. g in this case is called the (unique) analytic continuation of f to D. Now let’s go back to the Riemann zeta function. It’s proper definition is the following:

\zeta(z)=\begin{cases}&\displaystyle\sum_{k=1}^\infty k^{-z},\;\Re(z)>1,\\ &\text{analytic continuation elsewhere.}\end{cases}

Recall the gamma function

\displaystyle\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}dt,\; \Re(z)>0.

It can be shown using integration by parts that \Gamma(z+1)=z\Gamma(z). The function \Gamma(z+1) is analytic for \Re(z)>-1, so this equation provides the analytic continuation of \Gamma(z) to \Re(z)>-1. By repeating this procedure we obtain the analytic continuation of \Gamma to the whole complex plane (except for the non-positive integers, which are in fact the poles of \Gamma).

It is an exercise to show that

\displaystyle \zeta(z)=\frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t-1} dt,\; \Re(z)>1.

It is another exercise to show that

\displaystyle \frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t-1} dt=\frac{\Gamma(1-z)}{2\pi i}\int_{\gamma}\frac{t^{z-1}}{e^{-t}-1}dt

where \gamma is the Hankel contour:


Now the right side of the above equation is analytic for \Re(z)<1. Hence this equation provides the analytic continuation of \zeta(z) to \mathbb C\backslash\{1\} (in fact, there is a pole at z=1.)

The Bernoulli numbers B_n are defined by the equation

\displaystyle \frac{1}{e^t-1}=\sum_{m=0}^\infty B_m\frac{t^{m-1}}{m!},

and B_0=1, B_1=1/2, B_{2m+1}=0 for m=1,2,\dots. Substituting this into the equation above, setting z=-n and using \Gamma(n+1)=n! we obtain

\displaystyle \zeta(-n)=\frac{n!}{2\pi i}\int_{\gamma}\frac{t^{-n-1}}{e^{-t}-1}dt =\frac{n!}{2\pi i}\sum_{m=0}^\infty (-1)^{m-1}\frac{B_m}{m!}\int_\gamma t^{m-n-2}dt

By Cauchy’s residue theorem,

\displaystyle\int_\gamma t^{m-n-2} dt=\begin{cases} &2\pi i\;\text{ if }m=n+1\\ & 0 \; \text{ otherwise}.\end{cases}

Thus \displaystyle\zeta(-n)=(-1)^n\frac{B_{n+1}}{n+1}, which is zero if n>1 is even.

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Filed under Complex analysis