Talk:Hopf algebra

Bruhat

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)

Examples

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)

I removed the following list of examples from the main page, as I think they are not Hopf algebras as defined in this article. I believe they are possibly examples of "locally compact quantum groups", some sort of topological version of Hopf algebras. AxelBoldt 20:58, 3 Sep 2004 (UTC)

Given a topological group G, we 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 coaction of this Hopf algebra upon noncommutative spaces is as a left (right) comodule. The other Hopf algebra we can construct is the convolution product algebra of distributions over G (i.e. its group ring). This time, the action of this Hopf algebra upon noncommutative spaces is as a left (right) module.
If, in addition, G is a Lie group, it has a Lie algebra g. Its universal enveloping algebra can be turned into a Hopf algebra by εx=0, Δx=x⊗1 1⊗x and Sx=-x for all elements of the Lie algebra. There's an injective homomorphism from this Hopf algebra to the Hopf algebra of convolutions over G such that the image of this homomorphism is the subalgebra generated by the Dirac delta distribution and its derivatives over the identity of G.
Given a Lie superalgebra L, we can turn it into a Hopf algebra as follows: Let A be the unital associative algebra generated by the elements of L and an element g subject to g²=1, gxg=(-1)^xx for pure elements x of L, cx dy (as a Lie superalgebra linear combination)=cx dy (as a linear combination in A) and [x,y]=xy-(-1)^xyyx for pure elements x, y in L. It's not quite the universal enveloping algebra although there is a canonical injective embedding of the universal enveloping algebra within A. Now, let ε(g)=1 and ε(x)=0 and $\Delta g=g\otimes g$ and $\Delta x=x\otimes 1 g^{x}\otimes 1$ for pure elements x in L.

Field?

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)[reply]

The article no longer mentions a field, and gives a general definition. 67.198.37.16 (talk) 16:06, 15 September 2016 (UTC)[reply]

Uniqueness

Is the antipode for the given bialgebra exactly unique or only up to iso? --Anton (talk) 20:39, 1 April 2009 (UTC)[reply]

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)[reply]

renormalization

This topic is shows up in quantum field theory in ways that I don't understand at all (arXiv:hep-th/9912092 from here). Maybe some of that can be useful to the article. 75.57.242.120 (talk) 10:29, 14 March 2011 (UTC)[reply]

Quoting from the abstract:

Renormalization in quantum field theory is a special instance of a general mathematical procedure of extraction of finite values based on the Riemann-Hilbert problem. We shall first show that for any quantum field theory, the combinatorics of Feynman graphs gives rise to a Hopf algebra $\Hc$ which is commutative as an algebra. It is the dual Hopf algebra of the enveloping algebra of a Lie algebra $\ud G$ whose basis is labelled by the one-particle irreducible Feynman graphs.

Alain Connes is awesome! 67.198.37.16 (talk) 16:08, 15 September 2016 (UTC)[reply]

Antipode

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)[reply]

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)[reply]

Tensor Algebra???

In this article, the Tensor Algebra with the coproduct $Deltax=x\otimes 1 1\otimes x$ 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)[reply]

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)[reply]

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)[reply]

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)[reply]

Uniqueness of theta?

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)[reply]