If $B$ is a bialgebra in the category $\tt{Vect}$ of vector spaces (over $\mathbb C$, for example) then $B$ can't have two different antipodes.
Is this true for bialgebras in an arbitrary symmetric monoidal category?
$\begingroup$
$\endgroup$
11
asked Aug 4, 2014 at 7:08
-
1$\begingroup$ This is really obvious if you look at the standard proof of the uniqueness of the antipode. The antipode is defined as the $*$-inverse of the identity in the convolution algebra. The convolution algebra is a honest algebra, not an algebra-over-a-category; so inverses are unique if they exist. $\endgroup$– darij grinbergAug 4, 2014 at 18:05
-
1$\begingroup$ @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. $\endgroup$– Sergei AkbarovAug 5, 2014 at 1:15
-
1$\begingroup$ Oh -- but it is very easy to prove the associativity of convolution without Sweedler's notation, just using the axioms of a coalgebra and of an algebra. $\endgroup$– darij grinbergAug 5, 2014 at 1:18
-
$\begingroup$ I did not see such a proof. Do you mean this trick with string diagrams? $\endgroup$– Sergei AkbarovAug 5, 2014 at 1:20
-
$\begingroup$ Denoting the multiplication of the algebra by $m$ and the comultiplication of the coalgebra by $\Delta$, we have ... $\endgroup$– darij grinbergAug 5, 2014 at 1:29
|
Show 6 more comments
2 Answers
$\begingroup$
$\endgroup$
7
Yes, as the following string diagram proof shows.
answered Aug 4, 2014 at 7:53
-
$\begingroup$ Evan, I don't understand these pictures. Do you know a text where they are explained? $\endgroup$ Aug 4, 2014 at 7:59
-
$\begingroup$ Yes, I guessed. However, a text with explanations would be highly appreciated. $\endgroup$ Aug 4, 2014 at 10:01
-
3$\begingroup$ @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? :) $\endgroup$ Aug 4, 2014 at 14:53
-
2$\begingroup$ @SergeiAkbarov A good start would be the seminal paper by Joyal and Street, The Geometry of Tensor Calculus (in Adv. Math.). The technique is scattered all across the literature, so much so that there is an embarrassment of riches, but the nLab page referenced in Dmitri's answer might also be a good place to look. $\endgroup$– Todd Trimble ♦Aug 4, 2014 at 19:30
-
2$\begingroup$ @SergeiAkbarov I learned this diagrammatic technique from papers by Shahn Majid, I think he described it quite well in his book "Foundations of quantum group theory". But I understand what you mean, and I agree, especially since there are so many variations. One should take the time to explain the symbols if one uses diagrammatic notation. $\endgroup$– UwFAug 4, 2014 at 19:40
$\begingroup$
$\endgroup$
0
Nice answer by Evan Jenkins. (Here are references for string diagrams: http://ncatlab.org/nlab/show/string+diagram#details)
The bottom line is that being a Hopf algebra is a property of a bialgebra rather than an extra structure. A Hopf algebra is usaully defined as a bialgebra with an antipode. Then the "propertiness" means that the antipode is unique. This indeed holds in any braided monoidal category.
However, there is another definition of a Hopf algebra which emphasizes the "propertiness". It is based on the generalization of the following simple group-theoretic fact: A monoid is a group if and only if the maps $$(a, b) \mapsto (ab, b), \qquad (a, b) \mapsto (a, ab)$$ are invertbile. Indeed, the inverses are given by $$(a, b) \mapsto (ab^{-1}, b), \qquad (a, b) \mapsto (a, a^{-1}b).$$
Generalizing this to bialgebras (within any braided monoidal category), a Hopf algebra is a bialgebra $B$ for which the maps $$B\otimes B \xrightarrow{1\otimes d} B\otimes B\otimes B \xrightarrow{m \otimes 1} B\otimes B$$ $$B\otimes B \xrightarrow{d\otimes 1} B\otimes B\otimes B \xrightarrow{1\otimes m} B\otimes B$$ are invertible. These maps are called left and right fusion maps.
This definition is equivalent to the definition via an antipode. The anipode can be recovered as $$B \xrightarrow{1\otimes \eta} B\otimes B \xrightarrow{l^{-1}} B\otimes B \xrightarrow{\epsilon\otimes 1} B$$
where $l$ is the left fusion map. Or it can be recovered by a similar composite involving the inverse of the right fusion map.
answered Aug 4, 2014 at 17:01