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For a finite dimensional $k$-algebra $A$, each $A^{\otimes n}, n \geq 0$ is a $k$-algebra ($A^0 = k $). Let $T= \prod_{n \geq 0} A^{\otimes n}$. This is a $k$-alegbra with unit $(1,1,\dots)$ and multiplication is component-wise. Let $\Delta^{(n)} : A^{\otimes n} \to T \otimes T$ be the deconcatenation map $$ \Delta^{(n)}(a_1 \otimes \dots \otimes a_n) = \sum_{i=0}^n (a_1 \otimes \dots \otimes a_i) \otimes (a_{i+1} \otimes \dots \otimes a_n ). $$

I want to extend these $\Delta^{(n)}$ to a comultiplication $\Delta : T \to T \otimes T $. This does not seem to work in a straightforward way because if $t = ( t_0, t_1, \dots ) \in T, $ then $\sum_n \Delta^{(n)}(t_n)$ may not be a finite sum of pure tensors in $T \otimes T$ (I have not shown this sum can be infinite, but suspect it can be).

Is there a way to make $T$ into a Hopf algebra so that $\Delta(t) = \Delta^{(n)}(t)$ when $t_i=0$ for $i \ne n$? If not, is there an algebra similar to $T$ where this does work?

Is there a standard way to complete the tensor product and instead get a map $\Delta$ from $T$ to the completion? Does this give rise to a genuine Hopf algebra or some generalization of Hopf alegbras?

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If you did this on the scheme level, you'd instead be taking $\coprod_{n \geq 0} Spec(A)^n$, which is a representing object for the "free monoid generated by Spec(A)". Unfortunately Spec doesn't take infinite products to infinite coproducts, so you will probably need to take a completion with respect to the inverse limit topology. (I assume that the nonexistence of an antipode isn't really the issue you're concerned with.) –  Tyler Lawson Apr 25 '10 at 21:11
    
What are the counit and antipode? –  Charlie Frohman Apr 25 '10 at 21:49
    
Related question: mathoverflow.net/questions/22659/… –  Anton Geraschenko Apr 27 '10 at 1:46

2 Answers 2

up vote 7 down vote accepted

You can indeed complete the tensor product and get a good comultiplication, but it's not strictly speaking a Hopf algebra. An algebraic geometer would call it an affine formal group. If you think of the infinite product $T=\prod_n A^{\otimes n}$ as a pro-object indexed by $\mathbb{N}$ with $T(n) = \prod_{i=0}^n A^{\otimes i}$, then you can tensor $T$ with itself and get a pro-object indexed by $\mathbb{N} \times \mathbb{N}$. Take the limit and that's the completed tensor product you want. Alternatively, you can do it with topologized rings and topologically-complete the tensor product.

For the uncompleted tensor product, there is no comultiplication extending your rule. There is, however, a cofree Hopf algebra (as Anton points out, consult Does the forgetful functor {Hopf Algebras}→{Algebras} have a right adjoint?). That's a sub-Hopf-algebra of the completed Hopf algebra $T$.

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You may want to look at the work by Ron Umble with various coauthors on $A_\infty$-Hopf algebras and bialgebras. I don't recall the details, but I think they are trying to deal with a rather similar situation.

http://arxiv.org/abs/0709.3436

http://arxiv.org/abs/math/0406270

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