The n-Category Café

Hi,

The proof of the fact that Mat(N) is the PROP for bicommutative bialgebras can be found in the draft version of my paper where I construct the “PRO” corresponding to a category of games and strategies interpreting a fragment of first-order propositional logic (comments are of course very welcome!). I think that the credits and coffees for this result should go to Albert Burroni and Yves Lafont who introduced the general methodology for constructing such PROs. As far as I know, the notions of polygraph and canonical form used to prove the result were introduced by Albert in Higher dimensional word problem with application to equational logic and deeply studied and refined by Yves in Towards an Algebraic Theory of Boolean Circuits. I haven’t checked the details, but this method should extend to the case of Mat(Z) without major difficulties (I can give a notion of canonical form for this PROP or more details if anyone is interested).

I am intrigued by this result.

I also have a question. Greg Kuperberg has shown that the category of cobordisms between surfaces is the PROP for involutory Hopf algebras; see

• Involutory Hopf algebras and 3-manifold invariants Internat. J. Math. 2 (1991), no. 1, 41–66, arXiv:math/9201301 .

Then we should see Mat(Z) as a quotient of this symmetric category. This might even give a strategy for proving the result on Mat(Z).

I don’t claim to deserve any coffee for that, but I don’t think all bicommutative Hopf bialgebras are involutive${}^{(1)}$, are they?

If not, then the theorem should probably be stated for involutive bicommutative Hopf algebras, right?

(1) From wikipedia, a Hopf bialgebra is involutive when its antipode $S$ is such that $S^2 = \mathit{id}$.

Hello !

I have a question/comment (it might be stupid but …) about John’s first coffee challenge.

1) Denote by C the (Z-linear) PROP of bicommutative bialgebras. As a free Z-module, the space C(2,2) is generated by three elements (identity, permutation and product followed by coproduct).

I might have misunderstood the challenge itself. But it seems to me that there is no chance for the existence of an isomorphsim C –> Mat(Z).

2) To my opinion a more reasonable statement should be that there is an isomorphism between the PROP Mat(Z) and the PROP of commutative (or cocommutative) bialgebras.
Let me explain why.

a) Denote by BiAlg the (Z-linear) PROP of bialgebras. Enriquez and Etingof (see Proposition 6.2 of math/0306212) found a Z-linear basis of BiAlg(n,m) (see also example 59 in Markl’s review math/0601129).

The main idea is that using compatibility between product and coproduct you can write any composition of many of them in a unique way with all products on the right and all coproducts on the left.

As far as I understand the basis can be described as follows :
a Z-linear basis of BiAlg(n,m) is given by triples (s,u,v) where
* s is a permutation of [k].
* u=((u_1,p_1),…,(u_k,p_k)), where (u_1,…,u_k) is a k-partition of n and each p_i is a permutation of [u_i].
* v=((v_1,q_1),…,(v_k,q_k)), where (v_1,…,v_k) is a k-partition of m and each q_i is a permutation of [v_k].

b) for comBiAlg, the PROP of commutative bialgeras, the basis is the same but without the p_i’s.

c) Now I think I can construct a map from such triples (the ones defining the basis of comBiAlg) to Mat(Z) that is compatible with the PROP-structure.

c1) Let us first forget about q_i’s (and thus construct a map from C to Mat(Z)).
- to each u_i associate the line-matrix of lenght u_i with all entries being 1’s. Then construct a matrix
U:=diag(u_1,…u_k)
which will have the following form
111000000
000110000
000001111
if u_1=3,u_2=2 and u_3=4 .
- to each v_j associate the column-matrix of lenght v_j with all entries being 1’s.
Construct a matrix V similarly to the above construction for U.
- then multiply VSU, where S is the permutation matrix for s.

c2) for non-trivial q_i’s one has to take a different column-matrix v_i with entries being 1’s and 0’s (with a least one 1).
It can be constructed a s follows :
- consider the permutation matrix of q_i.
- if the element on the top-left of this matrix is 1 (resp. 0), then the first entry of v_i is 1 (resp. 0).
- remove the first line and first column of the permutation matrix.
- do the same with the new matrix you obtain to find the second entry of v_i
- etc …

The map constructed in c1) is NOT an isomorphism , because of 1).

I did not check if the map c2) is, but it seems more reasonable. Also in the case c2) one has to check that it is a morphism of PROPs (in the case c1) it is obvious).

This comment is very much longer than what I thought first.
Maybe I am very far from the orignial question, and I hope I did not write too many meaningless things.

Damien