bio | website | mathnet.ru/php/… |
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location | ||
age | ||
visits | member for | 3 years |
seen | 14 hours ago | |
stats | profile views | 1,476 |
Aug 12 |
asked | Is the ideal of functions vanishing at a set complementable in $C(X)$? |
Aug 12 |
revised |
Identities that connect antipode with multiplication and comultiplication
added 35 characters in body |
Aug 12 |
comment |
Identities that connect antipode with multiplication and comultiplication
As far as I understand, Sweedler considers Hopf algebras in the category of vector spaces. Is the same true for all (braded?) monoidal categories? |
Aug 12 |
revised |
Identities that connect antipode with multiplication and comultiplication
deleted 3 characters in body |
Aug 12 |
asked | Identities that connect antipode with multiplication and comultiplication |
Aug 5 |
accepted | Is antipode unique for bialgebras in arbitrary monoidal categories? |
Aug 5 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
They are visual, that's important. It will take me some time to look at the explanations. Thank you, Darij! |
Aug 5 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
Yes, I expected something like this from the very beginning. But, Darij, it's not correct to write $f\otimes g\otimes h$. You should write either $(f\otimes g)\otimes h$ or $f\otimes (g\otimes h)$, and this is a supplementary headache here. If you don't mind, I think, I'll accept the answer by Evan after getting accustomed to his style. :) |
Aug 5 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
I did not see such a proof. Do you mean this trick with string diagrams? |
Aug 5 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
@darij grinberg: This becomes obvious only if the assocoativity of convolution in $Hom(B^c,B^a)$ is already proved. This fact is not obvious -- in the book by Dascalescu, Nastasescu and Raianu, for example, they use the Sweedler notations for proving it. But the Sweedler notations, I would say, is a risky way in arbitrary monoidal categories. Chari and Pressley mention this fact, but without references. These were two sources for me. I saw before the pictures like those that Evan Jenkins gave in his answer, but I did not understand them and I did not know that they can be used here. |
Aug 5 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
Dima, thank you! |
Aug 5 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
Todd and UwF, thank you! |
Aug 4 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
@UwF: I guessed also. :) But this style of explanation - "I drop a hint, you guess!" - is a bit annoying when you need a reference for a paper. Are there texts where this technique is described more or less intelligible? :) |
Aug 4 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
Yes, I guessed. However, a text with explanations would be highly appreciated. |
Aug 4 |
comment |
Is antipode unique for bialgebras in arbitrary monoidal categories?
Evan, I don't understand these pictures. Do you know a text where they are explained? |
Aug 4 |
revised |
Is antipode unique for bialgebras in arbitrary monoidal categories?
added 3 characters in body |
Aug 4 |
asked | Is antipode unique for bialgebras in arbitrary monoidal categories? |
Jul 2 |
awarded | Curious |
Jun 5 |
comment |
Characterization of $l_p$ up to a linear isometry
Ah, I see... You mean that the unit ball (of the norm) must be a polytope, and in addition it must have $2n$ facets. Is this true? Anyway, this looks too complicated... Is there a way to describe the situation without the additional assumption that the inut ball is a polytope? |
Jun 5 |
comment |
Characterization of $l_p$ up to a linear isometry
Mark, if we take this definition of facet, then for an absorbing bounded balanced convex set $B$ in ${\mathbb R}^3$ having 6 facets is not equivalent to being a unit ball for a space isometrically isomorphic to $\ell_\infty^3$. You can take $B=\{x\in {\mathbb R}^3: \ \max_i|x_i|\le 1\quad \&\quad x_1^2+x_2^2+x_3^2\le 2\}$ (the intersection of a cube with the edge 2 and the usual ball of radius $\sqrt{2}$) -- this set has 6 facets but it is not a cube. |