# Realization

Today was grocery day in the Stover household, which means we basically spent the day driving around, picking up amazing savings due to my wife’s couponing and essentially getting nothing else done whatsoever.

Fortunately, I was able to squeeze in about 30 minutes of math while sitting in the local Target’s snack bar / Starbucks area. In particular, I took some time to read a bit further into my professor’s paper on M-conformal Cliffordian functions, and in so doing, I came to a realization.

The last time I wrote here about that paper, I sketched a small proof of an elementary claim that probably required no proof. As a result of that entry, today has been as complete roller coaster for me.

First, I thought I’d misquoted the definition of a function $f:\Omega\to\mathcal{A}_n$ being monogenic: My original claim was that $D_nf=0$ for monogenic functions, but today, I miscalculated the partials for an example that led me to believe that $\overline{D_n}f=0$ was actually the criteria. That’s not correct at all.

Now, I realize what the criteria really is, but at the same time I realize that one small detail of that proof was incorrect. In particular, I combined the summed expressions for $f$ and $D_n$, respectively, to be a single sum ranging from $l=0,\ldots, n$ instead of a term for $u_0$ and $\partial/\partial x_0$, respectively, plus a sum for $l=1,\ldots,n$. Later, when there were two parameters $l,m=0,\ldots,n$ in the sum, I claimed that $l=m$ implies that $e_l^2=-1$; this, of course, is false, since $e_0$ is identified with 1 so that $e_0^2=1$. On the other hand, with a proper bit of rigor, the proof is still essentially correct.

Here’s why:

Here’s one way to think about the problem. Let $f(x)=u_0(x)+\sum_{l=1}^n u_l(x)e_l$, an equivalent representation of which is $f=\mathbf{sc}(f)+\mathbf{vec}(f)$ where $\mathbf{sc}(f)=u_0(x)$ and where $\mathbf{vec}(f)=f(x)-u_0(x)$ represent the scalar and vector parts of $f$, respectively. In particular, then, if we consider $D_nf$ to be the derivative of $f$, it follows that $D_nf=0$ precisely when $D_n[\mathbf{sc}(f)]=0$ and $D_n[\mathbf{vec}(f)]=0$. With regards to the scalar part of $f$, this implies that

$D_n(u_0(x))=0\implies\displaystyle\frac{\partial u_0}{\partial x_0}+\sum_{l=1}^n\frac{\partial u_0}{\partial x_l}=0$.

In particular, then, the vector $\left(\partial u_0/\partial x_0,\cdots,\partial u_0/\partial x_n\right)^T=0$, and so each component must be zero. Hence, $\partial u_0/\partial u_k=0$ for $k=0,\ldots,n$.

If we then turn our attention to the vector part of $f$, we see that $D_n[\mathbf{vec}(f)]=0$, i.e. that

$\begin{array}{rcl}0 & = & \displaystyle\left(\frac{\partial}{\partial x_0}+\sum_{l=1}^n e_l\frac{\partial}{\partial x_l} \right)\circ\left(\sum_{k=1}^n u_k(x)e_k\right) \\[2em] & = & \displaystyle\sum_{k=1}^n \frac{\partial u_k}{\partial x_0}e_k + \sum_{k,l=1}^n e_l\,e_k\frac{\partial u_k}{\partial x_l}\,\,\,\,\,\,\,\,\,\,(1)\end{array}$.

Note that the two sums in $(1)$ sum to zero precisely when each sum itself is equal to zero due to the linear independence of the basis elements $e_l$, $l=0,1,\ldots,n$. In particular, then, the second sum in $(1)$ equals zero and is precisely the sum I used for the matrix analogy in my original solution. Among the necessary corrections is to note that the matrices $M,M'$ cited there should be $n\times n$ matrices instead of $(n+1)\times(n+1)$. Recall also that the first equation in the system of equations shown in the original entry – the equation $\sum_{l=0}^n \partial u_l/\partial x_l=0$ – is achieved by combining the equation of “mixed partials” of the form $\sum_{k=l=1}^n \partial u_l/\partial x_l = 0$ from the second sum in $(1)$ with the fact that $\partial u_0/\partial x_0 = 0$ from above.

whew

This, sirs and madames, is what happens when one doesn’t protect against carelessness. I need to weed that out of my repertoire and fast. Blah.

Anyway, I’m gonna try to learn some Algebraic Geometry and maybe apply that to some $D$-module theory. Until next time….