An answer to a John Baez puzzle (I hope : )

Newsgroups: sci.physics.research
Subject: An answer to a John Baez puzzle (I hope : )
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Sender: Doug Sweetser<sweetser@alum.mit.edu>
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John Baez posted the following puzzle back in October
(repeated here to bring back the memories : )

...
>In one world you start with a vector space V with a nondegenerate
>metric g satisfying
>
>g(v,w) = g(w,v)
>
>and cook up the Clifford algebra with
>
>vw + wv = hbar g(v,w)
>
>which when Planck's constant hbar goes to zero, becomes the
>exterior algebra. Then you do all sorts of fermionic stuff.
>
>In the other world you start with a vector space V with a
>symplectic structure, a nondegenerate bilinear omega satisfying
>
>omega(v,w) = -omega(w,v)
>
>and cook of the Weyl algebra with
>
>vw - wv = hbar omega(v,w)
>
>which when hbar goes to zero becomes the symmetric algebra.
>Then you do all sorts of bosonic stuff.
>
>I went on and on about all the wonderful things that happen...
>and then I posed the puzzle: what sort of vector space has
>both a nondegenerate metric AND a symplectic structure? Such
>a thing would let you study both bosons and fermions, in a
>unified way. It would obviously be very important.
>
>
Good puzzle, precisely stated, but I can't answer it. I can
solve a closely related puzzle:

Q*: What sort of field has both a nondegenerate metric AND a
symplectic structure?

A*: The field of quaternions has both properties.

We start with an infinite set of quaternions, all of the form

t + x I + y J + z K

Let v and w be some arbitrary subset of this collection,
equal in their number of elements (for convenience only, let
that number be one so I don't have to deal with sums in
ASCII : )

Define g (where * here means the conjugate or the
transpose in the real matrix representation) as:

g(v, w) = 1/2 (v* w + w* v)

= 1/2 (vt - vx I - vy J - vz K)(wt + wx I + wy J + wz K)
+ (wt - wx I - wy J - wz K)(vt + vx I + vy J + vz K)

= vt wt + vx wx + vy wy + vz wz

This is a standard dot product, but applied to the field of
quaternions. If v = w, the Euclidean norm is the result.
From its definition, g satisfies

g(v, w) = g(w, v)

(A more lengthy analysis that the g defined here does behave
like a nondegenerate metric - the triangle inequality, the
Schwarz inequality, etc., - can be found in the quaternion
quantum mechanics section of my home page, URL at end).

Half the puzzle complete...

Hopefully, g and omega would be simply related, so make the
simplest proposal. Define omega as

omega(v, w) = 1/2 (v* w - w* v)

= 1/2 (vt - vx I - vy J - vz K)(wt + wx I + wy J + wz K)
- (wt - wx I - wy J - wz K)(vt + vx I + vy J + vz K)

= ((vt wx - vx wt) - (vy wz - vz wy)) I

((vt wy - vy wt) - (vz wx - vx wz)) J

((vt wz - vz wt) - (vx wy - vy wx)) K

It should be clear from the definition that

omega(v, w) = - omega(w, v)

Omega looks like it is worth puzzling over! It contains the
negative of a curl and an anti-symmetric product of time and
space. No wonder the math behind bosons and fermions is both
strange and wonderful!

A note on nomenclature: I call g the "Euclidean inner
product" because it gives rise to a Euclidean norm. I refer
to omega as the "Euclidean outer product" because it is the
anti- symmetric partner of the Euclidean inner product.
Together they add up to the quaternion product v* w. There
are two other products of interest, the Grassman inner
product, 1/2(vw + wv), which contains the invariant interval
of special relativity, and the Grassman outer product which
is also known as the cross product. Together the Grassman
inner and outer products add up to v w.

A question remains: did I stretch the rules too much to solve
the puzzle? I can still use all of the tools of calculus
because quaternions are a mathematical field. Since the
Euclidean inner and outer products are quaternions, they
should still be part of this mix. We'll let John decide if
Q* was a valid substitution, and the rest of us if A* might be
important.


Doug Sweetser
Putting quaternions to work since April, 1996

http://world.std.com/~sweetser