All Questions
Tagged with hopf-algebras noncommutative-geometry
15 questions
6
votes
0
answers
349
views
Quantum Hilbert's fifth problem
Hilbert's fifth problem inquires whether every locally Euclidean group is necessarily a Lie group. Von Neumann demonstrated that this is indeed true for the compact case.
The definition of a quantum ...
4
votes
1
answer
266
views
Hopf "algebroid" structure of a groupoid convolution algebra?
This question is already posted in math.stackexchange, but didn't receive any answer. I'm not sure if this question fits in here, but surely someone in here can guide me to the correct answer.
To make ...
6
votes
1
answer
338
views
Invertible elements of the Hopf algebra quantum $SU(2)$
Let $SU_q(2)$ be the (polynomial) Hopf algebra introduced by Woronocicz called the quantum special unitary group. For details see
https://en.wikipedia.org/wiki/Compact_quantum_group
(Note that on the ...
6
votes
0
answers
200
views
What is a quantum analogue of the fact that the second fundamental group of every Lie group is trivial?
What is an appropriate version of the following fact in terms of Hopf algebras and quantum groups:
"For every connected Lie group $G$ the second fundamental group $\pi_2(G)$ is trivial?"
Is there ...
3
votes
1
answer
170
views
Reduced compact quantum group and left and right multiplication
Let $(A,\Delta)$ be a compact quantum group in the sense of Woronowicz, and let $A_0$ be its dense Hopf subalgebra. We can construct from the Haar state $h:A \to \mathbb{C}$ an inner product
$$
\...
2
votes
0
answers
60
views
Integrals in noncommutative graded algebras which are not necessarily Hopf
Let $\mathbf{k}$ be a field. Let $A$ be a finite dimensional $\mathbb{Z}_{\geq 0}$-graded $\mathbf{k}$-algebra such that $A^0=\mathbf{k}1$. Let $m$ be the maximal non-negative integer such that $A^m\...
3
votes
1
answer
176
views
Noncommutative Leray - Hirsch theorem in the context of noncommutative principal bundles
In the literature, are there some researchs on non commutative analogy of Leray-Hirsch theorem in the context of non commutative Principal bundles?
15
votes
1
answer
657
views
Is every finite quantum group a quantum symmetry group?
This post is basically a quantum extension of Is every finite group a group of “symmetries”?
Here finite quantum group means finite dimensional Hopf ${\rm C}^{\star}$-algebra.
Frucht's theorem ...
2
votes
0
answers
85
views
Relative version of Hopf cyclic cohomology
In this paper Connes and Moscovici introduced the Hopf algebra of transverse differential operators in order to compute the index formula for the diffeomorphism invariant geometry. They developed the ...
6
votes
1
answer
157
views
The Quantum Group ${\cal O}_q(SL(n))$, for $q>1$
For the quantum group ${\cal O}_q(SL(n))$, $q\in \mathbb{R}$, I have read, without a proof, that for $p>1$, there exists a $q\in (0,1)$ such that
$$
{\cal O}_p(SL(n)) \simeq {\cal O}_q(SL(n)).
$$
...
8
votes
1
answer
222
views
Hopf Galois extensions and conditional expectations for C* algebras
Suppose that $H$ is a Hopf algebra with normalised invariant integral (appropriate side) $\int:H\to \mathbb{C}$. The $H$ right comodule algebra $P$ is a Hopf Galois extension, so the canonical map $P\...
11
votes
1
answer
556
views
Generators of the Odd Dimensional Quantum Spheres
As is well-known, the $(2N-1)$-quantum sphere $S^{2N-1}_q$ is defined to be the invariant subalgebra of $SU_q(N)$ under the coaction $\Delta_R = (id \otimes \pi) \circ \Delta$, where $\Delta$ is the ...
2
votes
2
answers
397
views
Finding the Universal Ideal of a (Covariant) Differential Calculus
Let $(\Omega,d)$ be a differential calculus over an algebra $A$. It is easy to show that $\Omega$ is always equal to a quotient of $\Omega_u(A)$, the universal calculus over $A$, by some ideal $N$ of $...
13
votes
6
answers
2k
views
Hopf algebras arising as Group Algebras
Every commutative $C^*$-algebra is isomorphic to the set of continuous functions, that vanish at infinity, of a locally compact Hausdorff space. Every commutative finite dimensional Hopf algebra is ...
38
votes
6
answers
4k
views
Why Drinfel'd-Jimbo-type quantum groups?
Hopf algebras are pretty easy to motivate, as a not-necessarily-commutative generalization of the ring of functions on an algebraic group (and there are many other ways in which they come up). I like ...