The Unapologetic Mathematician

Mathematics for the interested outsider


In yesterday’s post I used the group algebra \mathbb{F}[G] of a group G as an example of a coalgebra. In fact, more is true.

A bialgebra is a vector space A equipped with both the structure of an algebra and the structure of a coalgebra, and that these two structures are “compatible” in a certain sense. The traditional definitions usually consist in laying out the algebra maps and relations, then the coalgebra maps and relations. Then they state that the algebra structure preserves the coalgebra structure, and that the coalgebra structure preserves the algebra structure, and they note that really you only need to require one of these last two conditions because they turn out to be equivalent.

In fact, our perspective allows this equivalence to come to the fore. The algebra structure makes the bialgebra a monoid object in the category of vector space over \mathbb{F}. Then a compatible coalgebra structure makes it a comonoid object in the category of algebras over \mathbb{F}. Or in the other order, we have a monoid object in the category of comonoid objects in the category of vector spaces over \mathbb{F}. And these describe essentially the same things because internalizations commute!

Okay, let’s be explicit about what we mean by “compatibility”. This just means that — on the one side — the coalgebra maps are not just linear maps between the underlying vector spaces, but actually are algebra homomorphisms. On the other side, it means that the algebra maps are actually coalgebra homomorphisms.

Multiplication and comultiplication being compatible actually mean the same thing. Take two algebra elements and multiply them, then comultiply the result. Alternatively, comultiply each of them, and the multiply corresponding factors of the result. We should get the same answer whether we multiply or comultiply first. That is: \Delta\circ\mu=(\mu\otimes\mu)\circ(1_A\otimes\tau_{A,A}\otimes1_A)\circ(\Delta\otimes\Delta), where \tau is the twist map, exchanging two factors.

Let’s check this condition for the group algebra \mathbb{F}[G]:

\begin{aligned}\left[\mu\otimes\mu\right]\left(\left[1_A\otimes\tau_{A,A}\otimes1_A\right]\left(\left[\Delta\otimes\Delta\right](e_g\otimes e_h)\right)\right)=\\\left[\mu\otimes\mu\right]\left(\left[1_A\otimes\tau_{A,A}\otimes1_A\right](e_g\otimes e_g\otimes e_h\otimes e_h)\right)=\\\left[\mu\otimes\mu\right](e_g\otimes e_h\otimes e_g\otimes e_h)=e_{gh}\otimes e_{gh}=\\\Delta(e_{gh})=\Delta\left(\mu(e_g\otimes e_h)\right)\end{aligned}

Similarly, if we multiply two algebra elements and then take the counit, it should be the same as the product (in \mathbb{F}) of the counits of the elements. Dually, the product of two copies of the algebra unit should be the algebra unit again, and the counit of the algebra unit should be the unit in \mathbb{F}. It’s straightforward to verify that these hold for \mathbb{F}[G].

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November 6, 2008 - Posted by | Algebra, Category theory


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