## Representations of Bialgebras

What’s so great about bialgebras? Their categories of representations are monoidal!

Let’s say we have two algebra representations and . These are morphisms in the category of -algebras, and so of course we can take their tensor product . But this is *not* a representation of the same algebra. It’s a representation of the tensor square of the algebra:

Ah, but if we have a way to send to (an algebra homomorphism, that is), then we can compose it with this tensor product to get a representation of . And that’s exactly what the comultiplication does for us. We abuse notation slightly and write:

where the homomorphism of this representation is the comultiplication followed by the tensor product of the two homomorphisms, followed by the equivalence of algebras.

Notice here that the underlying vector space of the tensor product of two representations is the tensor product of their underlying vector spaces . That is, if we think (as many approaches to representation theory do) of the vector space as fundamental and the homomorphism as extra structure, then this is saying we can put the structure of a representation on the tensor product of the vector spaces.

Which leads us to the next consideration. For the tensor product to be a monoidal structure we need an associator. And the underlying linear map on vector spaces must clearly be the old associator for . We just need to verify that it commutes with the action of .

So let’s consider three representations , , and . Given an algebra element and vectors , , and , we have the action

On the other hand, if we associate the other way we have the action

Where we have used the Sweedler notation to write out the comultiplications of . But now we can use the coassociativity of the comultiplication — along with the fact that, as algebra homomorphisms, the representations are linear maps — to show that the associator on intertwines these actions, and thus acts as an associator for the category of representations of as well.

We also need a unit object, and similar considerations to those above tell us it should be based on the vector space unit object. That is, we need a homomorphism . But linear maps from the base field to itself (considered as a one-dimensional vector space) are just multiplications by field elements! That is, the algebra is just the field itself, and we need a homomorphism . This is precisely what the counit provides! I’ll leave it to you to verify that the left and right unit maps from vector spaces intertwine the relevant representations.

[...] of Hopf Algebras I We’ve seen that the category of representations of a bialgebra is monoidal. What do we get for Hopf algebras? What does an antipode buy us? Duals! At least when we restrict [...]

Pingback by Representations of Hopf Algebras I « The Unapologetic Mathematician | November 12, 2008 |

[...] we showed that the category of representations of a Hopf algebra has duals. This is on top of our earlier result that the category of representations of any bialgebra is monoidal. Let’s look at this a [...]

Pingback by The Category of Representations of a Hopf Algebra « The Unapologetic Mathematician | November 18, 2008 |

[...] buy us? It turns out that the category of representations of a cocommutative bialgebra is not only monoidal, but it’s also symmetric! Indeed, given representations and , we have the tensor product [...]

Pingback by Cocommutativity « The Unapologetic Mathematician | November 19, 2008 |

[...] have a group . This gives us a cocommutative Hopf algebra. Thus the category of representations of is monoidal — symmetric, even — and has duals. Let’s consider these structures a bit more [...]

Pingback by The Category of Representations of a Group « The Unapologetic Mathematician | November 21, 2008 |

[...] is even easier than tensor products were, and we don’t even need to be a bialgebra. An element of is just a pair with and . [...]

Pingback by Direct Sums of Representations « The Unapologetic Mathematician | December 9, 2008 |