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 being monogenic: My original claim was that for monogenic functions, but today, I miscalculated the partials for an example that led me to believe that 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 and , respectively, to be a single sum ranging from instead of a term for and , respectively, plus a sum for . Later, when there were two parameters in the sum, I claimed that implies that ; this, of course, is false, since is identified with 1 so that . 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 , an equivalent representation of which is where and where represent the scalar and vector parts of , respectively. In particular, then, if we consider to be the derivative of , it follows that precisely when and . With regards to the scalar part of , this implies that

.

In particular, then, the vector , and so each component must be zero. Hence, for .

If we then turn our attention to the vector part of , we see that , i.e. that

.

Note that the two sums in sum to zero precisely when each sum itself is equal to zero due to the linear independence of the basis elements , . In particular, then, the second sum in 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 cited there should be matrices instead of . Recall also that the first equation in the system of equations shown in the original entry – the equation – is achieved by combining the equation of “mixed partials” of the form from the second sum in with the fact that 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 -module theory. Until next time….

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