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Questions about algebraic structures known as quantum groups, and their categories of representations. Quasitriangular Hopf algebras and their Drinfel'd twists, triangular Hopf algebras, $C^\star$ quantum groups, h-adic quantum groups, various semisimplified categories at roots of unity which are called "quantum groups", bicrossproduct quantum groups, and quantum groups coming from braided tensor categories.
6
votes
Cartier-Kostant-Milnor-Moore theorem
The case of irreducible, cocommutative Hopf algebras, over a field with $char(k)> 0$, is discussed in Sweedler's textbook on Hopf algebras, Ch.$XIII$, sect. $13.2$. (See prop. $13.2.2$, $13.2.3$).
…
6
votes
Cocommutativity, comultiplication and coalgebra maps
Given a (coassociative and counital) coalgebra $(C,\Delta,\varepsilon)$, over a field $k$, we can form the tensor product coalgebra $(C\otimes C,\Delta_{C\otimes C},\varepsilon_{C\otimes C})$ through: …
4
votes
1
answer
664
views
Primitive elements in group hopf algebras over fields of non-zero characteristic
An element $x$ of a Hopf algebra $H$, is called a primitive element if $\Delta(x)=1\otimes x+x\otimes 1$. The set of primitive elements of $H$ is denoted $P(H)$. It can be shown that:
"If $H$ is a $ …
2
votes
Accepted
What are the primitive elements in a polynomial Hopf algebra with primitive indeterminates?
No, in general the claim is not true:
To see why, consider a field $k$ of characteristic $p$ and take the polynomial hopf algebra $k[x]$ (in a single variable). Then $x$ is primitive and so is $x^p$ …
6
votes
Accepted
When is this map of Hopf algebras Surjective?
Some thoughts, regarding question (a):
In the case of a pointed, cocommutative hopf algebra $H$ over a field $k$ of characteristic $0$, by the Cartier-Konstant-Milnor-Moore theorem (see: Classifica …
4
votes
Kazhdan-Lusztig equivalence for Lie super-algebras
In: Kazhdan-Lustzig polynomials and character formulae for the Lie superalgebra $gl(m|n)$, J. Amer. Math. Soc. 16 (2002), 185–231, J. Brundan develops a conjecture on the characters for the irreducibl …
6
votes
Coalgebras(or quantum groups) which admit a linear operator satisfying certain functional eq...
About your first question:
Since you are asking for an example, take any group hopf algebra $k\mathbb{G}$, pick some subset $S\subset \mathbb{G}$ and denote $kS$ the linear subspace of $k\mathbb{G}$ …
8
votes
Accepted
Inner automorphisms of Hopf algebras
I am not sure if the following is the kind of answer you are expecting, but take the (left) adjoint action $(ad_l h)\triangleright k=\sum h_1 kS(h_2)$ of a hopf algebra $H$ on itself.
(It is known tha …
2
votes
q-difference equations and quantum mechanics
Regarding the first part of the question:
Have there ever been actual uses of q-calculus and quantum groups to computing or understanding solutions of Schrodinger equations, or functions actually …
4
votes
2
answers
651
views
Cocommutativity, comultiplication and coalgebra maps
Given a coalgebra $(C,\Delta,\varepsilon)$, over a field, the following is a well-known property:
the comultiplication $\Delta:C\to C\otimes C$ is a coalgebra map if and only if $C$ is cocommutat …
12
votes
What is quantum algebra?
I think that a modern realistic perception of the term "quantum algebra" has to be understood in its historical context, that is, the algebraic/geometric methods, originating from the study of the qua …
1
vote
Accepted
quantum affine $gl_2$
I guess you mean the following presentation in terms of generators and relations:
The excerpt is from:
Evaluation modules for quantum toroidal ${\mathfrak{gl}}_n$ algebras, arXiv:1709.01592v4 [math. …
1
vote
Commutative and Cocommutative Quantum Groups
If the definition of a finite quantum group, you use, is a pair $(A,\Phi)$ of a finite dimensional $C^*$-algebra $A$, with a comultiplication $\Phi$, such that $(A,\Phi)$ is a Hopf $*$-algebra, then t …
3
votes
On a dual of Kaplansky's $2^{nd}$ conjecture: admissible algebras?
A couple of examples of non-admissible algebras:
$\bullet$ It is known (see here, p.222) that a semisimple Hopf algebra is also separable (as an algebra). Consequently, a semisimple but non-separabl …
4
votes
Can one define quantized universal enveloping algebras in a basis-free way?
I do not know the answer in general. But towards the end of the OP you say:
"Even a construction that still involves generators and relations but avoids choosing a basis of the root system would …