Free-Forgetful Adjunction for Vect

6 Dec 1997

      I've done my best on this, for two weeks. There comes a point at which
one calls for help, because one isn't seeing it.

A full-rank set of vectors in a vector space defines a basis, obviously.
And given an n-dimensional  Euclidean manifold, with specified origin,
an arbitrary set of n points, "in general position" (i.e. not contained
in any proper subspace), defines a basis for it.

In other words, a choice of simplex defines a choice of frame.

Now, MacLane sets this up as an example of the "free-forgetful" adjoint
relationship. Postulate functors V, U between the two categories:

                V ->
        Set(s)          Vect(or Spaces)
                <- U

and domains I in Set, and V in Vect.

The adjunction is:

        V I   ->  P             U P  <-  I

        UV I  ->  U P           VU P <-  VI
        |      /                |      /
        ^     /                 v     /
        |u   /                  |n   /
        |   /                   |   /
          I                       P

The interpretation is:

        V : Set -> Vect

takes a set I to V I, the span of I over some field. (Set-to-set arrows
go to linear maps.)

        U : Vect -> Set

is the forgetful functor, taking each vector space to "its underlying set";
which as far as one can tell, is the vector space considered as the total
set of linear combinations of the basis elements, with all arrows mapped
to nowhere-land.

The adjoint isomorphism is:

        [ V I  -> P ]   ~~   [ I -> U P ]
           frame                simplex

That makes sense, as far as it goes. Now consider the unit and counit.

        u: id(Sets)  ->  U V

        => u(I): I ->  UV(I)

        is the natural transformation from any set I to the elements
        of its span; which set is isomorphic to R^|I|, and can be read
        as the flat space in which I resides. That makes sense if I is
        a set of points, but is meaningless if I is, say, the set {1,2,3}.

        n : V U -> id(Vect)

        =>  n(P) : VU(P) -> P

        is the natural transformation from the span of the set of elements
        of a vector space, to the space itself. But the set of elements
        is infinite. VU(P) would have to be a vector space with an infinite
        basis.

My perplexities:

(1) As above, the unit and counit do meaningless things.

(2) The vector spaces defined by V are free on I. It's meaningless to
ask whether the elements of I are in general position; they are assumed
to be. But in a geometric interpretation, one wants to introduce the
arrows to the Boolean domain:

        C2 : P x P -> B : condition for two points to coincide

        C3: P x P x P -> B: " " 3 points to be collinear

        C4: PxPxPxP -> B:   " " 4 points to be coplanar

        .... and so on up to |I|.

There has to be a sensible place to plug these into Vect, and I expected
the adjoint construction to be it.

(3) Otherwise, what is the damn thing good for? By comparison, the
Diagonal-Product and Product-Exponential units and counits are the
most useful things about those constructions; and the Free-Forgetful
adjunction for monoids gives us the Kleene closure. But the Free-
Forgetful adjunction for Vect almost seems vacuous.

An alternative way to see the problem is, to say that Set is not the
right category to be looking for adjoints in. Geometrically, we want
Vect to be related not to spaces, but to "linear structures" on spaces,
in the sense that a manifold atlas genuinely describes a differentiable
structure. Maybe we want something like:

               V ->
        Flat            Vect
               <- U

where Flat (= linear structure) is equipped with the incidence arrows
C2, C3,.., C|I|, and admits only  m : Flat -> Flat which are U-images
of the exterior products in Vect, and preserve the C's.

This is related to the question: is the exterior product (i.e. determinant)
the unit of some adjunction?

Any help really appreciated.

Jonathan Burns