# The Unapologetic Mathematician

## Coalgebras

Okay, back to business. We’re about to need a little more algebraic structure floating around. This is something that’s always present, but many approaches don’t explicitly mention it until much later. Since I’m taking a categorical view of things, it’s easier to show what’s really going on right away.

Remember that an $\mathbb{F}$-algebra is a monoid object in the category of vector spaces over $\mathbb{F}$. Dually, an $\mathbb{F}$-coalgebra is a comonoid object in the category of vector spaces over $\mathbb{F}$. That’s all well and good, but what’s a comonoid object? We’ve mentioned them before, but let’s be more explicit this time around.

Remember that a monoid object was a functor from a certain category we cooked up to mirror the axioms of a monoid. We gave the category objects $M^{\otimes n}$ corresponding to the natural numbers, corresponding to lists of monoid elements. We have a map $\mu:M\otimes M\rightarrow M$ corresponding to multiplication, and a map $\iota:\mathbf{1}\rightarrow M$ picking out the unit in the monoid.

So a comonoid object will be a functor from the dual of this category! That is, we’ve still got all the same objects, but now we have a “comultiplication” arrow $\Delta:C\rightarrow C\otimes C$, and a “counit” arrow $\epsilon:C\rightarrow\mathbf{1}$.

Now, the model category describing monoid objects isn’t just objects and arrows. We also have the relations that make a monoid a monoid: the associative law $\mu\circ(\mu\otimes1_M)=\mu\circ(1_m\otimes\mu)$, and the left and right unit laws $\mu\circ(1_M\otimes\iota)=1_M=\mu\circ(\iota\otimes1_M)$.

Dually, we must have dual relations for comonoid objects. We have a coassociative law $(\Delta\otimes1_M)\circ\Delta=(1_M\otimes\Delta)\circ\Delta$, and left and right counit laws $(1_M\otimes\epsilon)\circ\Delta=1_M=(\epsilon\otimes1_M)\circ\Delta$.

We could write these down in terms of commuting diagrams, but it’s even more instructive to look at “string diagrams” like we did before. This makes the sense of what’s going on all the clearer.

So a coalgebra is a comonoid object in the category of vector spaces over $\mathbb{F}$. That is, it’s an $\mathbb{F}$ vector space $C$, equipped with a linear comultiplication $\Delta$ and a linear counit $\epsilon$, which satisfy the coassociative and counit laws. I’ll admit that this seems an extremely quirky structure to discuss, so an example is in order. The one we care most about right now is the group algebra. Yes, it turns out to also be a coalgebra!

To really wrap our heads around it, let’s start with a finite group $G$. Then we get a finite-dimensional vector space $\mathbb{F}[G]$, with a basis $e_g$ indexed by elements of $G$. Let’s forget, for the moment, that we have a multiplication and a unit. Instead, we define the comultiplication by $\Delta(e_g)=e_g\otimes e_g$ for each basis element. We also define the counit by $\epsilon(e_g)=1$ for each element $g\in G$. Both of these maps extend by linearity.

Now, let’s check the coassociative property. It suffices to check it on basis elements, because the extensions by linearity have to agree. In this case we have

$\begin{aligned}\left[\Delta\otimes1_{\mathbb{F}[G]}\right]\left(\Delta(e_g)\right)=\left[\Delta\otimes1_{\mathbb{F}[G]}\right](e_g\otimes e_g)=e_g\otimes e_g\otimes e_g\\=\left[1_{\mathbb{F}[G]}\otimes\Delta\right](e_g\otimes e_g)=\left[1_{\mathbb{F}[G]}\otimes\Delta\right]\left(\Delta(e_g)\right)\end{aligned}$

Similarly, we can check the right counit law:

$\begin{aligned}\left[1_{\mathbb{F}[G]}\otimes\epsilon\right]\left(\Delta(e_g)\right)=\left[1_{\mathbb{F}[G]}\otimes\epsilon\right](e_g\otimes e_g)=e_g\\=\left[\epsilon\otimes1_{\mathbb{F}[G]}\right](e_g\otimes e_g)=\left[\epsilon\otimes1_{\mathbb{F}[G]}\right]\left(\Delta(e_g)\right)\end{aligned}$

and the left counit law is similar. Thus these maps do indeed describe the structure of a coalgebra.

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Posted by John Armstrong | Algebra, Category theory

## 3 Comments	»

1. [...] In yesterday’s post I used the group algebra of a group as an example of a coalgebra. In fact, more is [...]

Pingback by | November 6, 2008 | Reply

2. [...] As we work with coalgebras, we’ll need a nice way to write out the comultiplication of an element. In the group algebra [...]

Pingback by | November 10, 2008 | Reply

3. [...] this concept, we say that a coalgebra is cocommutative if we can swap the outputs from the comultiplication. That is, if . Similarly, [...]

Pingback by | November 19, 2008 | Reply

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