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Sep
10 |
comment |
Does the fundamental theorem of calculus require continuity of the function being integrated?
For a piecewise continuous function $f$, the statement probably changes to $F'=f$ a.e. |
Sep
3 |
revised |
Free $k[x_1, \dots, x_n]^{S_n}$-module?
fixed typo |
Sep
3 |
comment |
What's the relation between half-twists and star structures on Hopf algebras?
Let us continue this discussion in chat. |
Sep
3 |
comment |
What's the relation between half-twists and star structures on Hopf algebras?
The bar involution is an algebra automorphism defined by $\overline{E}=E$, $\overline{F}=F$ and $\overline{q}=q^{-1}$. The relation $[E,F]=(K^2-K^{-2})/(q-q^{-1})$ implies that $\overline{K}=K^{-1}$ since $\overline{[E,F]}=[E,F]$. I guess if ``the star is antilinear'' means take complex conjugates of coefficients, then, for $q$ a root of unity, this would be compatible. |
Sep
3 |
comment |
What's the relation between half-twists and star structures on Hopf algebras?
Maybe I am misunderstanding what you mean by $U_qSL(2)_{\mathbb{C}}$. Is this the quantum group over $\mathbb{C}(q)$, or $U_qSL(2)\otimes_{\mathbb{Q}(q)}\mathbb{C}$ where $\mathbb{C}$ is regarded as a $\mathbb{Q}(q)$ module with $q$ acting as a scalar? |
Sep
3 |
comment |
What's the relation between half-twists and star structures on Hopf algebras?
Are you sure both send $K\mapsto K$? My first thought was that they differ by the bar involution, but then one should be $K\mapsto K^{-1}$. |
Sep
2 |
answered | Free $k[x_1, \dots, x_n]^{S_n}$-module? |
Sep
1 |
revised |
Constructing a simple $A$-module
another typo |
Sep
1 |
revised |
Constructing a simple $A$-module
fixed typo |
Sep
1 |
answered | Constructing a simple $A$-module |
Aug
27 |
comment |
Generalized identities of (soluble) groups
I don't understand your definition. If $a_1,\ldots,a_n\in G$, what does $x^{a_1}\cdots x^{a_n}$, $x\in G$ mean? |
Jul
14 |
comment |
What are the “tensor-closed” object of the BGG category $\mathcal{O}$ of a semisimple Lie algebra $\mathfrak{g}$?
No, I was mistaken. I think one can probably show that the tensor product of a Verma module with an infinite dimensional module is not finitely generated. |
Jul
14 |
comment |
Decomposition of quadratic polynomials inti irreducible representations of affine group over a finit field
@Thomas $V$ is not closed under addition. It is just a set. |
Jul
14 |
comment |
What are the “tensor-closed” object of the BGG category $\mathcal{O}$ of a semisimple Lie algebra $\mathfrak{g}$?
Have you looked at sl_2 yet? I think that will give a negative answer to your question. |
Jul
13 |
comment |
Decomposition of quadratic polynomials inti irreducible representations of affine group over a finit field
That is a good point. |
Jul
13 |
answered | Decomposition of quadratic polynomials inti irreducible representations of affine group over a finit field |
Jul
8 |
comment |
SO(3) transformation that produces a reflection
I believe $H(u)=u-(2u\cdot v^T)v$ is what you meant. This is a reflection. The product of two reflections is a rotation. |
Jul
7 |
comment |
Irreducible unitary representations of semidirect groups of a discrete abelian group by a discrete group
@YemonChoi Frobenius reciprocity seems to be okay. The proof I know is fine for the situation. The only issue might be the equality between $\dim H_\Gamma$ and the hom space, but given we have an invariant inner product, that seems fine too. |
Jul
7 |
comment |
Irreducible unitary representations of semidirect groups of a discrete abelian group by a discrete group
@YemonChoi I think it's okay. Isn't Fronenius reciprocity really a statement about adjointness of tensor product and hom. The proof I know is for modules over some rings S<R, though I don't recall the exact hypotheses. I will certainly double check. |
Jul
7 |
revised |
Irreducible unitary representations of semidirect groups of a discrete abelian group by a discrete group
deleted 6 characters in body |