Talk:Hopf algebra

message: On distributions on a topological group - there is a theory due to Bruhat, where test functions are the Schwarz-Bruhat functions. But I think what is meant here is the appropriate group algebra concept, which can be illustrated by convolution of measures (? appropriate hypotheses). Since the point made is about variance, it shouldn't matter so much which convolution algebra is taken.
Charles Matthews 11:07, 2 Nov 2003 (UTC)
thread_title: Bruhat

message: : can form two different Hopf algebras over it. The first is the algebra of continuous functions from G to K whose product is the pointwise product. ε acting on a function gives its value at the identity and (Δ f)(x,y)=f(xy) for all x and y in G. (Sf)(x)=f(x-1). 
The formula (Δ f)(x,y)=f(xy) does not work formally, since (Δ f) is supposed to be an element of the tensor product of C(G,K) and C(G,K). So apparently some map from C(GxG,K) to the tensor product of C(G,K) and C(G,K) is silently being used. I can see that it works for finite discrete G, but in general I don't know what map to use. Or are we using some "continuous" tensor product here? AxelBoldt 20:22, 10 Aug 2004 (UTC)
thread_title: Examples

message: I always thought one can define a Hopf algebra over an arbitrary integral domain, not neccesary a field. At the very least, Hopf algebras over integral domains are being studied. Elenthel 22:24, 6 June 2006 (UTC)
replies: text: The article no longer mentions a field, and gives a general definition. 67.198.37.16 (talk) 16:06, 15 September 2016 (UTC)
thread_title: Field?

message: Is the antipode for the given bialgebra exactly unique or only up to iso? --Anton (talk) 20:39, 1 April 2009 (UTC)
replies: text: Unique. The set Hom(H,H) is a monoid w.r.t. the convolution product. In this monoid, the antipode S : H --> H is inverse to the identity morphism 1 : H --> H.  —Preceding unsigned comment added by 219.117.195.84 (talk) 02:13, 25 May 2009 (UTC)
thread_title:  Uniqueness 

message: Having only come across Hopf algebras in algebraic topology, the antipode was something new to me from this page. I think it would be helpful to others like me to add the fact that if a bialgebra is graded (with non-negative grades only) and the zero grade part of it is isomorphic to the underlying field, then an antipode always exists, so such a bialgebra can always be made into a Hopf algebra. There is a proof of this at why graded bi algebras have antipodes.
Now, in the algebraic topology case that the bialgebra is the cohomology ring of a path-connected H-space, we also have that for an element h of grade n>0, the grade n component of (Δh) is 1\otimes h + h\otimes 1. Am I right in thinking that this is not in general true for an arbitrary Hopf algebra ?
Rfs2 (talk) 10:37, 21 July 2011 (UTC)
replies: text: In general, Hopf algebra are not graded. In general the structure constants are not so simple, i.e. in general \Delta h =\sum_k h^'_k \otimes h^{}_k where the sum is not a sum over grading, but simply a sum over other elements in the algebra. Its exactly like structure constants in a Lie algebra, but going the other way.  I vaguely recall that there are some Steeenrod-y things that have this kind of more complex structure. 67.198.37.16 (talk) 14:39, 17 September 2016 (UTC)
thread_title:  Antipode 

message: In this article, the Tensor Algebra with the coproduct Delta x = x\otimes 1 + 1 \otimes x is given as an example of a Hopf algebra. However, in the article Tensor algebra, it is mentioned that this coproduct does not turn the Tensor algebra into a Hopf algebra. One of the two articles has to be wrong. Which one is it? --2001:41B8:83F:1004:0:0:FFFE:C0E6 (talk) 11:19, 15 May 2015 (UTC)
replies: text: It seems, from reading tensor algebra, that it depends on a choice of coproduct. With a correct choice, a tensor algebra is a Hopf algebra. -- Taku (talk) 16:09, 15 May 2015 (UTC)
replies: text: I don't see anywhere where it says that, but the article on tensor algebra is seriously messed up. 67.198.37.16 (talk) 04:15, 18 September 2016 (UTC)
replies: text: Oh, I see, it says that in cofree coalgebra. But that's different -- no one is claiming that cofree coalgebras are Hopf algebras (although I guess they could be if all the right definitions are made, it being essentially just the dual to the tensor algebra). In the meanwhile, I am cleaning up the confusion that was in tensor algebra. 67.198.37.16 (talk) 16:35, 18 September 2016 (UTC)
thread_title:  Tensor Algebra??? 

message: I don't understand how theta can be unique in the braided monoidal categories section. Isn't the whole point of a braiding (rather than a symmetry) that theta(a, b, c, d) may not be equal to the inverse of theta(a, c, b, d)? Cyrapas (talk) 19:06, 19 October 2023 (UTC)
thread_title:  Uniqueness of theta?