# Why should we expect the connection between complex arithmetic and geometry?

I realized that I take it for granted that properties of complex numbers have clear geometric interpretations. Visualizing complex numbers with the help of the complex plane really helps to understand complex arithmetic better and those mysterious properties of holomorphic functions (conformality, Maximum-Modulus Theorem, Argument Principle to name a few) make perfect sense once one knows that complex multiplication is simply a rotation and scaling. But lately I have been asking myself why there should be a connection between complex arithmetic and geometry at all? Of course there is nothing stopping us from interpreting these numbers as points in the plane (after all they are pairs of real numbers) but I am still bewildered by the fact that once we think of them this way, everything else related to complex numbers seems to find a perfect geometrical explanation! For example without a geometrical picture, the only way to understand complex multiplication is the distributive law. But the geometrical interpretation of complex multiplication turns out to be much more elegant and it is almost like it was always meant to be thought of that way. I am really curious to hear your thoughts about this.

• Historically this was not the case. There were different attempts to geometrically visualize complex numbers when they were first being studied. – David K Oct 9 '20 at 17:21
• You may enjoy this video on $e^{i\pi}$ in terms of group theory – spraff Oct 9 '20 at 21:05
• A part form these relations on the surface, Cauchy's integral formula is the Stokes' theorem on the plane (domain of a holomorphic function) and it connect to holomorphic/meromorphic functions with (co)homology. For example, one can think of residue theorem as computing de Rham cohomology of a punctured disk. Analytic continuation and Riemann surfaces as construction of the (universal) covering space etc. – Bumblebee Oct 9 '20 at 21:08
• On the other hand, open unit disk (or any half plane) is a model for the hyperbolic geometry, plane has the Euclidean geometry, and the Riemann sphere (one point cmpactification of the plane) models the spherical geometry. Hence complex plane is naturally very very geometric. This is basically why we see conformality and Mobius transformations. – Bumblebee Oct 9 '20 at 21:19
• In the elementary level, a complex number has Cartesian, polar, exponential, matrix and polynomial representations (I bet you can add more in to this list). Hence blend algebra and geometry together. By the Cauchy completeness of the plane, it has a nice theory of analysis as well. – Bumblebee Oct 9 '20 at 21:28

We can start way earlier to get a geometric interpretation, at the real numbers. Multiplication by a real number is a combination of scaling and mirroring. Multiplying by a a positive number is scaling the real line, multiplying by $$-1$$ is mirroring it at the origin. On an abstract level, a core feature of mirroring is that doing it twice returns the original picture. This gives rise to the interpretation that multiplication by $$-1$$ is a mirroring, since $$(-1)^2=1$$, so multiplying by $$-1$$ twice is the identity.

The complex numbers give rise to a similar interpretation. We can still view multiplication by $$-1$$ as mirroring the plane at the origin, but in a 2d context, we can also see it as a $$180^\circ$$ rotation. They are really the same. But we also get a new element, $$\mathrm i$$. Its basic feature is that $$\mathrm i^2=-1$$, that is, multiplying by $$\mathrm i$$ twice is rotation by $$180^\circ$$. But that's also a core feature of rotation by $$90^\circ$$: rotating by that amount twice is the same as rotating by $$180^\circ$$ once. So that's a good hint that complex multiplication can have something to do with rotations. We just need to find a fitting topology (a scalar product to describe angles, most importantly) which makes multiplication by $$\mathrm i$$ an actual $$90^\circ$$ rotation. And it turns out that the scalar product wrt which $$1$$ and $$\mathrm i$$ form an orthonormal basis does just that. So it's a good idea to choose those as a basis of $$\mathbb C$$ as a real vector space, making them span the coordinate axes. In this picture, multiplication by $$\mathrm i$$ will be guaranteed to be a $$90^\circ$$ rotation. And using some algebra, all other complex multiplications can then be shown to also be rotations and scalings.

• Indeed, this is Viewpoint 2 in this post regarding real multiplication. – user21820 Oct 10 '20 at 14:00

From the point of view of group theory there is a deep reason : the group of similitudes ( ratio-of-lengths preserving maps) of a (Euclidean) plane is isomorphic to the group of affine (or anti-affine) transformation of a complex line $$(z\to az+b$$ or $$z\to a \bar z+b$$). This (exceptional) isomorphism enables us to do geometry by using complex numbers.

This is even more clear if we go to the projective line (the Riemann Sphere). The group of projective transformations of a projective line $$PGL(2,C)$$ is isomorphic to Möbius group of conformal maps of a sphere $$PSO(3,1)$$.

• I would not yet call that a reason but just a deeper level of the phenomenon. The reason would be why the group of similitudes is isomorphic to the group of affine transformations. – Paul Sinclair Oct 9 '20 at 19:28
• From this point of view you can argue that it is all just shades of the exceptional Lie isomorphisms. Forexample, the projective Quaternionic line has transformation group PSL(2,H) isomorphic to the mobius group PSO(5,1). Picking a neutral signature Hermitian form on H^2 then gives you "SU*(2)" and PSO(4,1). Thus we have the conformal geometry of S^4 and S^3 – Callum Oct 11 '20 at 0:23
• If you accept that group=geometry, you accept that if two geometries are the same, this is because the two groups are the same... – Thomas Oct 11 '20 at 7:42

You can arrive at complex arithmetic from geometric intuition if you start with transformations of the plane.

It is well known that matrices which preserve angles (i.e. map shapes to similar shapes) and orientation are of the form $$cR(\theta)$$, where $$c$$ is a positive number and $$R(\theta)$$ is a rotation matrix. That is, $$cR(\theta) = c\pmatrix{\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta}.$$

Since $$c$$ and $$\theta$$ are arbitrary, these are all matrices of the form $$\pmatrix{a & -b \\ b & a}$$ for $$a, b \in \mathbb R$$ (except for the null matrix).

Now, after adjoining the null matrix, this set becomes a vector space of dimension two, closed under matrix multiplication, and where all non-null elements have a multiplicative inverse ($$c^{-1} R(-\theta)$$).

The interesting part is that we can choose a basis like this: $$\pmatrix{a & -b \\ b & a} = a \pmatrix{1 & 0 \\ 0 & 1} + b \pmatrix{0 & -1 \\ 1 & 0} = a I + b J,$$ where $$I$$ is the identity matrix and $$J=\pmatrix{0 & -1 \\ 1 & 0}$$ is a matrix which, under matrix multiplication, has the property $$J^2=-I$$. That is, it is in some sense the "square root" of $$-I$$. It also represents rotation in 90º (like the complex $$i$$ does). Indeed: $$J = R(\pi/2)$$ and, as expected, $$J e_1 = e_2$$ and $$J e_2 = -e_1$$.

Moreover, if you work out the product rule, it is exactly the one that arises in complex numbers:

$$\pmatrix{a & -b \\ b & a} \cdot \pmatrix{c & -d \\ d & c} \\ = \pmatrix{ac-bd & -(ad+bc) \\ ad+bc & ac-bd} \\ = (ac-bd)I + (ad+bc)J.$$

Furthermore, we can define subtraction, division, and all arithmetic operations for them in a way parallel to how they are defined for complex numbers.

Finally, add to this that the subspace generated by $$I$$ is an algebraic copy of $$\mathbb R$$, so you can view the full space as an extension of $$\mathbb R$$.

### To sum up

• Angle and orientation preserving linear transformations carry great geometric meaning (similarity).
• They form a two-dimensional space, which you can think of as an algebraically compatible extension of $$\mathbb R$$.
• They have two components, one in the direction of the identity/unit and one in the direction of a $$\pi/2$$ rotation.
• This space can be constructed from $$\mathbb R$$ just by adjoining an outside element $$J$$ such that $$J^2$$ is minus the identity (and extending the usual algebraic rules).
• This is essentially the same recipe as the one for constructing the complex numbers.