I realise that this question might be rather basic but however I was unable to find the answer in any textbook nor manage to figure out the answer. The question is the following: given two Hopf algebras $H_1,H_2$ is there a canonical way to turn the direct sum $H_1 \oplus H_2$ into a Hopf algebra? I have a problem already at the level of bialgebras: the counit is always a algebra morphism (linear and multiplicative) but the set of all linear multiplicative maps $\omega: H_1 \oplus H_2 \to \mathbb{C}$ is a sum of such sets corresponding to $H_1$ nad $H_2$. So I don't see how to define (in a canonical way) the counit map $\varepsilon_{H_1 \oplus H_2}$.
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3 Answers
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The answer is "no". In the commutative case, this would be asking for the disjoint union of two group schemes to be a group scheme. Even more concretely (say, working with finite dimensional commutative Hopf algebras over $\mathbb{C}$), there is no natural way to make the disjoint union of two finite groups into a finite group. The algebras of functions on the groups would be the Hopf algebras in question. Your difficulty with the counit amounts to the problem of making a canonical pointed set from the disjoint union of two pointed sets.
answered Dec 23, 2013 at 1:13
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3$\begingroup$ Nice question and nice answer. Maybe groupoids and Hopf algebroids can come in to solve the problem? $\endgroup$ Commented Dec 23, 2013 at 8:36
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4$\begingroup$ @FernandoMuro That is an excellent suggestion. If we expand our scope to Hopf algebroids, we get a well-behaved direct sum, and in the commutative case, this yields the disjoint union of groupoids of affine schemes. $\endgroup$– S. Carnahan ♦Commented Dec 23, 2013 at 9:50
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The answer to your question appears to be given in Comm. Math. Phys.(1999) 38(2), 519-524 where it's mentioned that because $(H_1\oplus H_2)\otimes (H_1\oplus H_2)\neq (H_1\otimes H_1)\oplus (H_2\otimes H_2)$ we cannot in general construct a Hopf algebra on $H_1\oplus H_2$ directly from the Hopf algebra structures on $H_1$ and $H_2$. However, they do define a separate direct sum of Hopf algebras which can be regarded as a Hopf algebra.
answered Dec 23, 2013 at 1:12
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This is a positive version of the answer "no".
As already mentioned before, the disjoint union of groups is not a group, BUT it is a grupoid!
In a similar way, the direct sum of two Hopf algebras is not a Hopf algebra, but a weak Hopf algebra. There is no problem for multiplication and comultiplication, the problem is, as you saw, the counit, but for weak Hopf algebras is the counit axiom that is precisely different. You can check in
https://arxiv.org/pdf/2008.00606.pdf UNIVERSAL QUANTUM SEMIGROUPOIDS, HONGDI HUANG, CHELSEA WALTON, ELIZABETH WICKS, AND ROBERT WON
the definition of weak Hopf (Definition 2.1), Lemma 2.24 is precisely your question.
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$\begingroup$ Great answer! Now I have a better idea of what weak Hopf algebras are for. $\endgroup$ Commented Sep 1, 2020 at 23:18