Modeling infinitesimals with 2x2 matrices

[Vaughan Pratt <pratt@cs.Stanford.EDU>]

At some point I'll try to collect my thoughts on Sol Feferman's Thursday
lecture on his alternative to Grothendieck universes, which he objected to
as entailing an infinity of inaccessible cardinals.  (What was Grothendieck's
view of inaccessible cardinals vis a vis his universes?)

During the lecture it struck me that his approach was quite like Robinson's
approach to infinitesimals, in that it constructed lots of models of what
was needed, took the common theory, then constructed a single model from
the many, using techniques of Vaught and others to avoid losing too much
of the common spirit of the many guided by the common theory (not sure if
that captures the idea completely faithfully, but it's something like that).

Thus distracted, I found myself wondering yet again why the d^2 = 0 property
was so difficult for an infinitesimal d.  Having been mulling over the
quaternions lately, it seemed to me there was something of an analogy there,
some property so built into our very psyche that we can't let go of it.
Hamilton finally dropped commutativity, along with any reservations he
might have harbored about vandalizing stone bridges in his own town.

For the quaternions, d^2 = 0 implies d = 0, so this doesn't help.  However the
quaternions have a sibling algebra, just as noncommutative, and of exactly
the same vector space dimension (in fact the only Clifford such, i.e. the
only other real 4D vector space for which ij+ji=0 for all orthogonal vectors
i,j having no real component), that is even better known than the quaternions
(imagine that).

Namely the Clifford algebra of 2x2 real matrices, as a 4D real vector space,
made an algebra with matrix multiplication.

Why not model d as the matrix
0 1
0 0?

This is a perfectly good quantity, adding and scaling just like any
real, e.g. 2d =
0 2
0 0.

And obviously d^2 = 0.

Standard reals x would have the form

x 0
0 x

1+d would therefore be

1 1
0 1

(1+d)^2 then becomes

1 2
0 1

as common sense would indicate.

The determinant of d being 0, one can't divide by it.  But who in their
right mind would want to divide by a quantity infinitesimally close to zero?
Obviously that's going to produce an infinitely large quantity; if you want
to do that, why not just go ahead and divide by zero itself?  As Douglas
Adams pointed out, you may think the store down the road is a fair way away,
but other galaxies are even further away.  To a nematode they're all far away.

On the other hand
1 2
0 1
has a perfectly good reciprocal, namely
1 -2
0 1
again as suggested by common sense.

So the proposal is to base calculus on a field-like object that is a field
in the large, but zero divide errors set in when one gets infinitesimally
close to zero.  Basically what happens with IEEE floating point arithmetic,
but modeled with 2x2 real matrices rather than 64-bit numbers.

Oh, but what about the noncommutativity of 2x2 matrices, might that mess
something up?

Actually no, this two-dimensional algebra consisting of matrices of the form
a b
0 a
is commutative.  So only the zero divisors really close to 0 constitute
any departure at all from the field axioms.

The diagonal element a is the standard real part and the off-diagonal
element b in the upper right gives the infinitesimal displacement.

So we have a real commutative associative algebra of refined numbers,
having a real part and an infinitesimal part, whose only zero divisors are
the infinitesimals.  We don't *have* to think of them as matrices because
we can just write its elements as x+yd by analogy with x+iy, where d is
the above matrix representing the prototypical infinitesimal.  The square
of i is -1, and the square of d is 0.

Moreover x and y in x+yd can be complex.  We then have numbers x+iy+ud+ivd,
which can parsed as either refined complex numbers, namely complex numbers
with refined coefficients x+ud+i(y+vd), or complex refined numbers, namely
refined numbers with complex coefficients x+iy+(u+iv)d.  This is still a
real associative algebra, which through force of habit people will no doubt
want to call a complex commutative associative algebra, but it could just
as legitimately be called a refined associative algebra.

Ok, what about commutative?  Well, the complex numbers are commutative and
the refined numbers are commutative, so how could refining complex numbers
make any difference?

Well, the reason I wrote x+yd rather than x+dy is that, even though the
*natural* thing to do is to make i commute with d, if instead we make
id+di=0, the defining condition for Clifford algebras, then we can fit the
whole thing into 2x2 *real* matrices!

Here I'm using the following 2x2 real matrices for i and d respectively:

(0 -1) (0 1)
(1  0) (0 0)

But now notice that the matrices for 1,i,d,id form a basis for all the
2x2 matrices.  In fact *any* 2x2 matrix [[a,b],[c,d]] can be decomposed as

(d -c) + (a-d b+c)
(c  d)   ( 0   0 )

(I'd appreciate feedback from anyone for whom the above doesn't typeset
readably.)

So to read an arbitary 2x2 real matrix as a refined complex number, take the
bottom row reversed as the complex part and the departure of the top row from
the usual matrix representation of complex numbers as the infinitesimal part,
taking care to get both signs right.

How did I notice this?  Simple.  I knew (i) that id+di=0 would make it a
Clifford algebra, (ii) there are only two 4D Clifford algebras, and (iii) d^2
= 0 -> d = 0 in the quaternions.  This narrows things down to the 2x2 real
matrices, there are no other associative algebras with these properties.
Getting the above decomposition was then just a matter of solving some
trivial linear equations.

This is so simple, and the infinitesimals have been worried at for so long,
that this *has* to be known already.  But then it would really bug me to
have been the last to learn about it -- why wasn't I told, as they say?

Vaughan Pratt