# GIBBERISH: what Arthur C Clarke told us

Mitchell Porter (mitch@thehub.com.au)
Sat, 18 Oct 1997 00:27:30 +1000 (EST)

Michael Butler wrote

> But Arthur C. Clarke has already told us that the locus of the
> contravariant tensor has noncommutative divergence in the region of the
> transfinite singularity!

As an exercise, I thought I'd try to find a meaning for the
purportedly meaningless part of that sentence.

Working backwards, the first stumbling block is "transfinite
singularity". This is presumably a singularity in some function.
The question is: in what sense is it "transfinite"?
The function might be transfinite-valued at this point,
or the singularity might be located in a transfinite region
of the function's domain (the set of values for which the
function is defined). Consider, for example, the function x^2.
It is defined for transfinite ordinals like omega just as
much as it is for finite numbers. (For those who haven't
run across it, omega is the first number that comes after
all the finite numbers 1, 2, 3,... After omega come
omega+1, omega+2, ... 2*omega, 2*omega+1, ..., "and so on".)
An example of a function with a transfinite singularity,
in the second sense, is f(x) = 1/(x-omega), which is
singular at x=omega.

The next stumbling block is "noncommutative divergence".
Commutativity is the property that a.b = b.a, something
that's true when "." is multiplication and a and b are
integers, but not true when a and b are rotations of
neighboring Rubik's cube faces and "." means "followed by";
it's one of the basic properties by which one might seek
to characterize an abstract algebra. Divergence can
mean the opposite of convergence, but it also refers
to a quantity in vector mechanics, and this seems the
most sensible interpretation. Let's assume that we're
interested in the divergence of some field in this
second sense. Then what is it that the divergence does not
commute with? Perhaps the divergence takes values in a
noncommutative algebra in this region? But I find it hard to
see why the divergence wouldn't then take values from that
same algebra everywhere else.

So the main problem here is to give a nontrivial meaning to
the notion of "noncommutative divergence". One way out
is to assume that by "noncommutative" we mean not commuting
with a *particular element* of the algebra of divergence values.
Since we are apparently dealing with tensor quantities
this is no great mystery - most matrices do not commute.

The final stumbling block is that it is the *locus* which
is said to have a "divergence". A locus is a region, so
we might suppose that when we talk about the divergence
of the locus, we are integrating divergence across the
locus. But what do we mean by *the* locus of the
contravariant tensor? The expression "the locus of X"
normally would mean "the region in which X can be
found". But then we would be saying "the divergence
of the region is noncommutative in a particular region",
which does not make obvious sense. Therefore we are
probably using the term "locus" as a shorthand for
some already introduced expression, for example
"locus of points for which ... has the same value
as it does at the current point".

So the complete translation of our original statement
might run:

"But Arthur C. Clarke has already told us that the
integral of the divergence of the contravariant tensor
(over a constant-valued neighborhood of the point we
are considering), in the region of the transfinite
singularity, does not commute with e (where e is
some privileged element of the divergence algebra)!"

Constructing a function for which this is true,
and a context in which it makes sense to point all
this out, is left as an exercise for the reader.

-mitch
http://www.thehub.com.au/~mitch