Subject: categories: Does duality categorify?

From: "John Baez" <[email-masked]>
Date: Thu, 27 Apr 2006 20:00:13 -0700 (PDT)

Vaughn Pratt asks:

> Does duality categorify?  If so, how?  If not, why not?

Of course this is a huge question, but the answer is:
surely it must!

My own favorite duality is between compact Hausdorff spaces
and commutative C*-algebras, because my advisor was Irving
Segal.  Elements of such a C*-algebra can be thought of as
continuous functions on its "spectrum", which is a compact
Hausdorff space.  This map from the algebra to function on
its spectrum is called the "Gelfand transform"; you can think
of the Fourier transform as a special case.

If you categorify this you get something called the
Doplicher-Robert theorem, which is a duality between certain
"compact groupoids" and certain "symmetric monoidal C*-categories".
I tried to explain this here:

http://front.math.ucdavis.edu/q-alg/9609018

However, if I really wanted to categorify dualities in general,
I'd start with less complicated examples.

> For example distributive lattices categorify to distributive algebras.
> The dual of a distributive lattice is a partially ordered Stone space.
> What does categorification do here?

Unfortunately I don't know what a distributive algebra is
or in what sense it's a categorified distributive lattice.

If I had to think about this, I'd probably start with something
I understand ever so slightly better, like the duality between
finite posets and finite distributive lattices.

Best,
jb

PS - there's some nice feedback on our questions about the
algebraic closure of the rationals here:

http://groups.google.com/group/sci.math.research/browse_thread/thread/61e30c22ee6c27b6/f647aee0763a349c?&hl=en#f647aee0763a349c

Briefly, while the existence of an algebraic closure of Q
can be shown without choice, it uniqueness-up-to-isomorphism
seems to require choice.  Also, while arithmetic operations
in Qbar are computable, they seem to present interesting challenges.
Quoting David Madore:

 The usual manner is to represent a real element of Qbar by

 * its minimal polynomial over Q (or perhaps, some polynomial, not
 necessarily minimal, but probably at least separable, of which it is a
 root),

 * an interval which isolates the root from all other roots (or the
 number of the root in the usual order on the reals).

 Basically the trick is that sums and products can be computed by
 universal rules (if P1 and P2 are polynomials over Q, there is a
 polynomial, which can be given universally in function of the
 coefficients of P1 and P2, whose roots are the sums of roots of P1 and
 P2, and ditto for the product), and roots can always be isolated using
 Sturm-Liouville (in other words, you can narrow the interval as much
 as you want since Sturm-Liouville lets you count the number of roots
 in any given interval).

 This is for real algebraics; for the full Qbar, you just represent a
 complex number by its real and imaginary parts (both of which are
 algebraic if the complex is algebraic).

 Actually programming this is *unbelievably* painful.  As for the
 algorithmic complexity, I think it's not that bad, in the sense that
 if x and y have small height (for any reasonable definition of
 "height") then computing x+y can be done in a reasonable time, but
 there's a catch: the height of x+y grows considerably larger than that
 of x or y, so any actual computation can become terribly difficult.
 (The same problem happens for rationals: computing r+s where r and s
 are rationals is polynomial in the height of r and s, but try
 computing something like 1/2+1/3+1/5+1/7+1/11+1/13+1/17...)