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Questions tagged [hopf-algebras]

A Hopf algebra is a vector space $H$ over a field $k$ endowed with an associative product $\times:H\otimes_k H\to H$ and a coassociative coproduct $\Delta:H\to H\otimes_k H$ which is a morphism of algebras. Unit $1:k\to H$, counit $\epsilon:H\to k$ and antipode $S:H\to H$ are also required. Such a structure exists on the group algebra $k G$ of a finite group $G$.

5 votes
0 answers
37 views

Let $k$ be a field, say $k=\mathbb C$. If $(A,\Delta_A,\varepsilon_A)$ is a coalgebra and $(B,m_B,1_B)$ is an algebra, then $\mathrm{Hom}_k(A,B)$ is an algebra under the convolution product $$ (f*g)(...
Zhiyuan Wang's user avatar
2 votes
0 answers
50 views

Let $A$ be an algebra and $H$ be a Hopf algebra. We say that $H$ is a Hopf symmetry of $A$ if $A$ is an $H$-module algebra, i.e., if $H$ has an action on $A$ that respects the multiplication and unit ...
Zhiyuan Wang's user avatar
2 votes
0 answers
53 views

Let $\mathcal{C}$ be a unitary fusion category, and let $Z(\mathcal{C})$ be its Drinfeld center. Suppose $Z(\mathcal{C})$ has a fiber functor $F: Z(\mathcal{C})\to \mathrm{Vec}$. Question: must $\...
Zhiyuan Wang's user avatar
6 votes
1 answer
144 views

I've often encountered (even on this forum) the claim that if $A$ and $B$ are Hopf algebras together with a Hopf pairing $A \otimes B \rightarrow \mathbb{k}$, then the Drinfeld quantum double $D = A \...
Sasha's user avatar
  • 101
2 votes
1 answer
78 views

Take a Hopf algebra $H$ and a subalgebra $A$, then look at the Hopf dual $H^{\circ}$ of $H$ and the dual coalgebra $A^{\circ}$ of the algebra $A$. We have a map $\rho:H^{\circ} \to A^{\circ}$ given ...
Jacques Holstein's user avatar
11 votes
2 answers
219 views

Consider the quantum group $U_q(\mathfrak{sl}_2)$ for generic $q$, which is a $\mathbb{C}$-algebra. There are two "completions" of this quantum group that one can consider to accommodate a ...
Minkowski's user avatar
  • 937
5 votes
1 answer
258 views

I'm trying to get a better understanding of seemingly different concepts of quantum groups, especially in the case of the $SL(2)$ group. (1) Drinfeld-Jimbo deformation I'm following "Quantum ...
MintChocolate's user avatar
1 vote
0 answers
78 views

Let $H$ be a Hopf algebra endowed with a $*$-structure, making it a Hopf $*$-algebra. Take a finite-dimensional $H$-comodule $(V,\Delta_V)$ and choose a basis $e_i$. In terms of this basis we have ...
Mili Fishta's user avatar
0 votes
0 answers
118 views

I was reading this post: Is it possible to construct a formal group law from a Lie group without choosing coordinates?. My question is how the correspondence between Lie groups and formal groups can ...
LefevresL's user avatar
8 votes
0 answers
330 views

Up to what order are finite-dimensional Hopf $\mathbb{C}$-algebras classified? Is there a table of this classification available somewhere?
Mili Fishta's user avatar
1 vote
0 answers
90 views

Given a Lie algebra $\mathfrak{g}$ which has a universal enveloping algebra $U(\mathfrak{g})$ and $\mathfrak{g}^{*}$ is a linear poisson structure. Is there a canonical linear poisson structure on $U(...
LefevresL's user avatar
1 vote
2 answers
471 views

I am intrested in the concrete case of a Lie algebra $\mathfrak{g}$ which has a universal enveloping algebra $U(\mathfrak{g})$ and $\mathfrak{g}^{*}$ is a linear poisson structure. Is the Sweedler ...
LefevresL's user avatar
2 votes
0 answers
45 views

A Hopf algebra $H$ is called quasitriangular if it has an invertible $R$-matrix. In this case, the category of left $H$-modules becomes a braided monoidal category. I known that Sweedler’s 4-...
Pedro Meireles's user avatar
1 vote
0 answers
140 views

$\newcommand\hash{\mathbin\#}$I am reading the paper "E. Kirkman, J. Kuzmanovich, and J. J. Zhang. “Gorenstein Subrings of Invariants under Hopf Algebra Actions”. In: Journal of Algebra 322 (2009)...
StAKmod's user avatar
  • 151
4 votes
0 answers
103 views

Consider $\mathcal{M}_A$ the monoidal category of right $A$-modules, for some Hopf algebra $A$, where the tensor product of two $A$-modules is defined in the usual way via the coproduct of $A$. In the ...
Mili Fishta's user avatar

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