# Catching Up

So I managed to go nine months without ever looking at this thing. That’s unfortunate. It is, however, an indication of how the past two semesters have been for me, time-wise.

Without moving backwards, I’ll leave this for an update:

• My GPA is pretty not-terrible.
• I managed to exempt all of my qualifying exams by scoring A- or higher in the classes needed.
• I’ve gotten to know some professors who seem not to hate me.
• I’ve become a better mathematician.

All of those are important, of course, but the last one is the one that matters most.

When I started this thing, I’d meant it to be a place for sharing lots of general math stuff. Now that I (hopefully) have more time, I’m expecting more sharing to happen. Just to get things off on the right foot, here’s part of a little something I worked on last semester.

Here, we’re considering functions $f:\mathbb{H}\to\mathbb{H}$ which are quaternion-valued functions of a quaternion variable. The function $f$ is said to be (Feuter) Regular at $q\in\mathbb{H}$ provided $\overline{\partial}_\ell f(q)=0$, where

$\overline{\partial}_\ell=\dfrac{\partial}{\partial t}+i\dfrac{\partial}{\partial x}+j\dfrac{\partial}{\partial y}+k\dfrac{\partial}{\partial z}$

The Cauchy-Fueter Integral Formula
If $f$ is regular at every point of the positively-oriented parallelepiped $K$ and $q_0$ is a point in the interior $K^\circ$ of $K$, then

$f(q_0)=\dfrac{1}{2\pi^2}\displaystyle\int_{\partial K} \dfrac{(q-q_0)^{-1}} {|q-q_0|^2} \,Dq\,f(q).$

Proof. The proof given largely follows the outline given by Sudbury. For an equivalent statement translated from the language of differential forms into the framework of vector fields and their integrals, see the proof on page 12 of Deavours.

First, let $K$ be a 4-parallelepiped in $\mathbb{H}$ and let $q_0$ be an element of its interior $K^\circ$. Define the function $g$ so that

$g(q)=-\partial_r\left(\dfrac{1}{|q-q_0|^2}\right)=\dfrac{(q-q_0)^{-1}}{|q-q_0|^2},$

where

$\partial_r=\dfrac{\partial}{\partial t}-\dfrac{\partial}{\partial x}i-\dfrac{\partial}{\partial y}j-\dfrac{\partial}{\partial z}k.$

Clearly, then, $g$ is real differentiable at every point $q\neq q_0$ in $K$. Moreover, a simple computation verifies that $\overline{\partial}_r g=0$ except at $q=q_0$. From other results (see Sudbury), it follows that $d(g\,Dq\,f)=0$ for all $q_0\neq q\in K^\circ$ by regularity of $f$.

Next, we use the dissection method from Goursat’s theorem. In particular, consider replacing $K$ by a smaller parallelepiped $\hat{K}$ for which $q_0\in\hat{K}^\circ\subset K$. Because $f$ has a removable singularity at $q=q_0$, it follows that $f$ is continuous at $q_0$, whereby it follows that choosing $\hat{K}$ small enough allows approximation of $f(q)$ by $f(q_0)$. Finally, let $S$ be a 3-sphere with center $q_0$ and Euclidean volume element $dS$ on which

$Dq=\dfrac{(q-q_0)}{|q-q_0|}dS.$

Such substitution is possible by way of a change of variables in the definition of $Dq$. This change of variables results in the following:

$\begin{array}{rcl} \displaystyle\int_{\partial K}\dfrac{(q-q_0)^{-1}}{|q-q_0|^2}Dq\,f(q) & = & \displaystyle\int_{\partial K}g\,Dq\,f(q) \\[1.5em] & = & \displaystyle\int_S g\,\dfrac{(q-q_0)}{|q-q_0|}dSf(q_0) \\[1.5em] & = &f(q_0)\displaystyle\int_S\dfrac{dS}{|q-q_0|^3}\\[1.5em] & = & f(q_0)\left(2\pi^2\right)\end{array}$

Dividing both sides by $2\pi^2$ yields the exact statement of the result. $\,\square$

The above is part of a (rather long) paper I finished last semester that’s in the infancy of a future as a potential pre-print/publication. I’m not going to link to it until I’ve done a bit more with it. Worth noting, too, is that this is the first time I’ve ever used WordPress’s incarnation of LaTeX, meaning there are many kinks to work out and much tinkering to do. I’m hoping that I can fit this blog into my plans for the rest of the semester, though, so maybe that will force me to do more with it.