Augustin-Louis Cauchy
1. Complex Analysis in Higher Dimensions
Complex analysis is suspiciously good. You get contour integration, residues, analytic continuation, conformal mapping, and a theorem saying a function differentiable once is differentiable forever. Nothing in real analysis behaves like that! So the obvious question is whether any of it survives going up a dimension. This post is the first half of finding out. We'll start with a way of looking at holomorphic functions that makes them look like physics, then try the obvious generalization and see why it doesn’t work, then work out what we should have been generalizing instead. I didn't know a lot of the math at the end of this when I started, so we'll build it as we go.
1.1 Polya vector fields
Take a holomorphic \(f(z) = u(x,y) + iv(x,y)\). There's a nice trick, due to Polya, of turning it into a vector field on the plane by taking the conjugate and reading off components:
The minus sign looks arbitrary, but it comes from the conjugation. Why? Recall the Cauchy-Riemann equations,
and now just compute the divergence and the curl of \(\mathbf{P}\). In two dimensions the curl is the single scalar \(\partial_x P_y - \partial_y P_x\):
Both vanish, and each one is exactly one of the Cauchy-Riemann equations. So holomorphic means divergence-free and curl-free. But a planar field with no divergence and no curl is a vacuum electrostatic field, no charge and no current anywhere. Every holomorphic function is an electrostatic field, and every source-free planar field is a holomorphic function! Now we also get something cool when we integrate. Writing \(dz = dx + i dy\) and multiplying out,
The real part is \(\mathbf{P} \cdot d\mathbf{r}\), and the imaginary part is \(\mathbf{P} \cdot \mathbf{n} ds\) with the outward normal \(\mathbf{n} ds = (dy, -dx)\). So
which makes Cauchy's theorem almost embarrassing. If \(f\) is holomorphic inside \(C\) there are no sources and no vortices in there, so nothing flows out and nothing goes around, and both integrals are zero. Poles are where the sources and vortices live! For me, this is super strange but very clean. It is VERY cool to see poles as sources, and makes Cauchy’s theorem geometrically speaking, much more intuitive.
1.2 Two fields you already know
Ok, time for actual examples. Take \(f(z) = 1/z\). Rationalising, \(1/z = (x - iy)/r^2\) with \(r^2 = x^2 + y^2\), so \(u = x/r^2\) and \(v = -y/r^2\), and
Pointing radially outward, falling off like \(1/r\). That's the field of an infinite line charge. Its residue is \(1\), so \(\oint f dz = 2\pi i\), which is zero circulation and \(2\pi\) of flux. That's Gauss's law. Now take \(f(z) = i/z = (y + ix)/r^2\), so \(u = y/r^2\) and \(v = x/r^2\):
Same \(1/r\) falloff, but perpendicular to \(\hat{\mathbf{r}}\) everywhere. It circulates. That's the magnetic field of an infinite line current. Its residue is \(i\), so \(\oint f dz = -2\pi\), which is all circulation and zero flux. Ampere's law! Same pole, rotated by ninety degrees in the complex plane, and electrostatics becomes magnetostatics. Which is a strong hint that this might work!
1.3 Why quaternions don't work
So we want to go up a dimension. The obvious candidate is the quaternions, since \(\mathbb{H} \cong \mathbb{R}^4\) and they're the next division algebra along. Write \(q = x_0 + x_1 i + x_2 j + x_3 k\) with
and define differentiability the way you'd expect, by demanding the limit of a difference quotient exists. Since multiplication doesn't commute we have to pick a side, so let's say
Now push \(h\) along each axis in turn. With \(h = t\) real, \(t\) commutes with everything and we get \(\partial_0 f = f'\). With \(h = ti\) we get \(f'(q)(ti) = t f'(q) i\), so \(\partial_1 f = f' i\), and likewise for the other two:
These are the quaternionic Cauchy-Riemann equations. They look pretty reasonable I think. BUT, watch what happens when we test them against the one thing every \(C^2\) function has to satisfy, which is that mixed partials commute! Differentiate the second equation with respect to \(x_2\), and use the third along the way:
Now do it the other way round, differentiating the third with respect to \(x_1\):
Mixed partials commute, so the left-hand sides are equal, and therefore
But \(ij = k\) and \(ji = -k\). So \(-(\partial_0^2 f) k = (\partial_0^2 f) k\), and multiplying through on the right by \(k^{-1}\) leaves
That's already a big problem, and it gets worse. Since \(\partial_0 f' = \partial_0^2 f = 0\), differentiating \(\partial_1 f = f' i\) with respect to \(x_0\) gives \(\partial_0 \partial_1 f = (\partial_0 f')i = 0\), and swapping the order gives \(\partial_1 \partial_0 f = \partial_1 f'\). So \(\partial_1 f' = 0\), and the same argument kills \(\partial_2 f'\) and \(\partial_3 f'\). The derivative is constant. Call it \(b\), integrate the four equations, and
That's it. That's every quaternion-differentiable function. Affine maps, nothing else. Boo! Anticommutativity made a boring, empty theory.
1.4 What Cauchy-Riemann actually is
Okay. So the difference quotient was the wrong thing to generalize. Which raises the question of what we should have generalized instead, and to answer that we need to ask what the Cauchy-Riemann equations really are, structurally, rather than where they came from. Let’s define
Then \(f\) is holomorphic exactly when \(\bar{\partial} f = 0\), which you can check by expanding: \((\partial_x + i\partial_y)(u + iv) = (\partial_x u - \partial_y v) + i(\partial_y u + \partial_x v)\), and setting both parts to zero gives back the Cauchy-Riemann equations. But now look at what happens if you multiply the two operators together:
Cauchy-Riemann is a first-order operator that factors the Laplacian. This is actually why holomorphic functions are automatically harmonic. So that's what we generalise. Not "the difference quotient converges" but "there's a first-order operator whose square is the Laplacian."
1.5 Building the operator
Let's just demand it and see what we're forced into. In \(n\) dimensions, look for an operator
where the \(e_\mu\) are constant coefficients of some kind, currently unknown, and require
Square it out. Since the coefficients are constant they pass through the derivatives, so
Now split that sum into the terms where \(\mu = \nu\) and the terms where they differ. For the off-diagonal terms, the pair \((\mu,\nu)\) and the pair \((\nu,\mu)\) both appear, and \(\partial_\mu \partial_\nu = \partial_\nu \partial_\mu\), so we can collect them:
We want this to be \(\sum_\mu \partial_\mu^2\) and nothing else. Matching term by term, there is no choice at all:
The coefficients have to square to one and anticommute with each other. No real or complex numbers do that, so whatever the \(e_\mu\) are, they aren't numbers. Pauli and Dirac faced similar problems in physics, so we should not be discouraged. There is always a solution! It turns out they're the generators of what's called a Clifford algebra, and we actually didn’t define it but it was forced. But what’s cool is that noncommutativity destroyed the quaternion attempt, but here it's the requirement. Just to check in two dimensions, where we know the answer. We need \(e_1, e_2\) with \(e_1^2 = e_2^2 = 1\) and \(e_1 e_2 = -e_2 e_1\). Consider the product \(e_1 e_2\) and square it:
So \(e_1 e_2\) squares to \(-1\). It behaves exactly like \(i\), which is a slightly weird thing to have come out of an anticommutation rule. And if we multiply our operator by \(e_1\) on the left,
which, writing \(I = e_1e_2\), is \(\partial_x + I \partial_y\). That's \(\bar{\partial}\). So in two dimensions the construction gives us back the Cauchy-Riemann operator, which is a good sign that we generalized the right thing. Functions satisfying \(Df = 0\) are called monogenic, and since \(D^2 = \nabla^2\), every monogenic function is harmonic componentwise, exactly as holomorphic functions are. That's the higher-dimensional replacement for holomorphic. One last thing. In three dimensions the algebra has eight independent elements, and the ones built from an even number of generators, namely \(1\), \(e_1e_2\), \(e_2e_3\), \(e_3e_1\), each square to \(-1\) and multiply among themselves exactly like \(i\), \(j\), \(k\). The quaternions are sitting inside the Clifford algebra as its even part. So, they were never really the wrong objects!
1.6 Where this goes
Ok, now we have a candidate. There's a first-order operator generalising Cauchy-Riemann, it factors the Laplacian, it reproduces the classical theory in two dimensions, and the functions it annihilates are harmonic. It turns out there's also a Cauchy theorem, an integral formula and a residue theory, which is a lot more than the quaternion attempt gave us. The part I want to get to next is what happens when you run the same construction in Minkowski space instead. The demand becomes \(D^2 = \Box\) rather than \(D^2 = \nabla^2\), which flips some signs in the anticommutation relations, and the operator you get is \(\nabla = \gamma^\mu \partial_\mu\). Then you bundle the electric and magnetic fields into a single object \(F\) and every one of Maxwell's equations turns into
so that in vacuum the electromagnetic field is a monogenic function of spacetime. Maxwell's equations are the Cauchy-Riemann equations of \(\mathbb{R}^{1,3}\), which is the four-dimensional version of the thing we noticed right at the top, that holomorphic functions are electrostatic fields! But that's the next post.