Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:13:05Z http://mathoverflow.net/feeds/question/22536 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22536/hopf-algebra-structure-on-prod-n-a-otimes-n-for-an-algebra-a Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$ Jonah Blasiak 2010-04-25T20:52:29Z 2010-06-07T09:52:35Z <p>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 $$\Delta^{(n)}(a_1 \otimes \dots \otimes a_n) = \sum_{i=0}^n (a_1 \otimes \dots \otimes a_i) \otimes (a_{i+1} \otimes \dots \otimes a_n ).$$</p> <p>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). </p> <blockquote> <p>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?</p> <p>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?</p> </blockquote> http://mathoverflow.net/questions/22536/hopf-algebra-structure-on-prod-n-a-otimes-n-for-an-algebra-a/22541#22541 Answer by Mikael Vejdemo-Johansson for Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$ Mikael Vejdemo-Johansson 2010-04-25T22:03:37Z 2010-04-25T22:03:37Z <p>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.</p> <p><a href="http://arxiv.org/abs/0709.3436" rel="nofollow">http://arxiv.org/abs/0709.3436</a></p> <p><a href="http://arxiv.org/abs/math/0406270" rel="nofollow">http://arxiv.org/abs/math/0406270</a></p> http://mathoverflow.net/questions/22536/hopf-algebra-structure-on-prod-n-a-otimes-n-for-an-algebra-a/27335#27335 Answer by Tilman for Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$ Tilman 2010-06-07T09:52:35Z 2010-06-07T09:52:35Z <p>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.</p> <p>For the uncompleted tensor product, there is no comultiplication extending your rule. There is, however, a cofree Hopf algebra (as Anton points out, consult <a href="http://mathoverflow.net/questions/22659/does-the-forgetful-functor-hopf-algebrasalgebras-have-an-adjoint" rel="nofollow">http://mathoverflow.net/questions/22659/does-the-forgetful-functor-hopf-algebrasalgebras-have-an-adjoint</a>). That's a sub-Hopf-algebra of the completed Hopf algebra $T$.</p>