oh. so then your objects are actually free algebras for the monad that adds two new vectors to a space... does this lead anywhere? let the objects of *V* be sets (of base vectors). a morphism from A to B is a linear operator from R^A to R^B (an AxB-"matrix"). the monad *V*--->*V* maps A |--> A+2, where 2 = {0,1}. it extends the linear operators so that the adjoined constants are preserved. let *K* be the kleisli category for (-)+2. the wedge functor V : *K* x *K* ---> *K* maps <A,B> |---> A + {m} + B. given two *K*-arrows, ie linear operators A--f--> 0+B+1 and C--g-->0+D+1, we need to define A+m+C ---fVg---> 0+B+m+D+1. conjoin the A -->1 minor of f and the C-->0 minor of g to get the m-component ("column") of f V g. now what might be the final coalgebra of WX = XVX? if all of the above happens over the category *S* of sets and functions, instead of *V* of sets and linear operators --- then the final coalgebra is (0,1). remember that this is just the generators, so when you really make the final coalgebra by adding 2, you get [0,1]. so this is just the freyd construction. the coalgebraic structure takes each number to one of its binary representations. when we work over *V*, and someone gives me a coalgebra X ---> WX in *K*, ie a linear operator X -----> 0+X+m+X+1, this coalgebra structure unfolds each x from X into a tree of real numbers. there are exactly 2|X|+3 branches coming out of each node that is not a leaf; 3 out of those are leaves, and each of the remaining 2|X| nodes has 2|X|+3 branches coming out of it. the final coalgebra Z -----> WZ would have to consist of such trees as well, of width 2|Z|+3. each tree induced by x from X would have to correspond to a unique Z-tuple of real numbers. hmm. it seems clear that Z must be wider than any X. but size is a bad reason for something not to exist. what if we have two universes? can we span all 2|X|+3-trees by the vectors from some base Z? g'night, -- dusko Tom Leinster wrote:
Dusko Pavlovic wrote:
doesnt the 0-dim vector space carry the final (terminal) coalgebra? (it has just one vector, so there is not much choice for v_0 and v_1.)
Nope - v_0 and v_1 have to be linearly independent. (This is like the condition in Peter's result that the two distinguished points of the set be distinct, without which the result degenerates; and as in Peter's result, it can be regarded as a kind of flatness condition.)
Incidentally, I was probably wrong to suspect that the answer is something to do with measure, as the question is posed over an arbitrary field. I now suspect that there's a similar question whose answer has to do with measure, but I won't attempt any further speculation here.
Tom
Here's a question belonging to (10), to which I don't know the answer. Let C be the category whose objects are triples (V, v_0, v_1) where V is a vector space and v_0 and v_1 are linearly independent vectors in V, and whose maps preserve linear structure and the `basepoints'. There's a `wedge' functor C x C --> C defined by
(V, v_0, v_1) wedge (W, w_0, w_1) = ( (V + W)/~, v_0, w_1)
where (V + W)/~ is the direct sum with v_1 identified with w_0. (So dim(V wedge W) = dim V + dim W - 1.) There's then an endofunctor G of C given by self-wedging. Question: what, if any, is the terminal G-coalgebra?
(I suspect the answer is something to do with measure/integration - again see Peter's previous postings - but really have no idea.)