What is a most elementary Coalgebra?

Let's start first with the (de-categorified) definition.

If $k$ is your favorite field, then a $(k)$-coalgebra is

• A $k$-vector space $A$
• A $k$-linear map $\Delta : A \to A \otimes A$ (called "comultiplication")
• A $k$-linear map $\epsilon : A \to K$ (called the "counit")

so that

• $(\Delta \otimes \text{id}_A) \circ \Delta = (\text{id}_A \otimes \Delta) \circ \Delta$ (the "coassociativity" axiom)
• $(\epsilon \otimes \text{id}_A) \circ \Delta = \text{id}_A = (\text{id}_A \otimes \epsilon) \circ \Delta$ (the "coidentity" axiom)

Here, as usual $\otimes$ is the tensor product, and $\text{id}_A$ is the identity map on $A$.

Ok, this is a weird looking structure. Why should we care about it? Well recall the usual definition of a $(k)$-algebra. It's

• A $k$-vector space $A$
• A $k$-bilinear map $m : A \times A \to A$ (called "multiplication")
• A $k$-linear map $1 : k \to A$ (called the "unit")

so that

• $m \circ (m \times \text{id}_A) = m \circ (\text{id}_A \times m)$ (the "associativity" axiom)
• $m \circ (\text{id}_A \times 1) = \text{id}_A = m \circ (1 \times \text{id}_A)$ (the "identity" axiom)

Now if you stare at these requirements for a while, you'll find that they're perfectly backwards. Where algebras have a map $A \times A \to A$, coalgebras have a map $A \to A \otimes A$. Where associativity says that the two obvious ways to multiply three things $A \times A \times A \to A$ are actually the same, coassociativity says that the two obvious ways to comultiply into three things $A \to A \otimes A \otimes A$ are the same.

This is where the "co" prefix comes from. We use "co" in category theory to indicate that we are doing something "with the arrows reversed". Since these axioms come from taking the axioms of an algebra and "turning the arrows around", we call the resulting structure a "co-algebra".

If you're thinking "why would anyone care about these things?", you're not alone. A priori the definition doesn't look particularly natural, and honestly I don't know of any reasons to care about just coalgebras (at least at time of writing). But I do know why we might care about Hopf Algebras. A hopf algebra is a vector space that is both an algebra and a coalgebra in a way that the two structures are "compatible" (this is mediated by an "antipode" $s$). Hopf algebras have use in representation theory, physics, and combinatorics, and you can find some reasons to care about them (and, by extension, coalgebras) here. There are also some discussions here about examples of coalgebras (and, by extension, why we might care).

Here's probably the simplest example of a coalgebra:

Let $A = k^2$, with basis elements $x$ and $y$. Then let

• $\Delta(x) = x \otimes x$ and $\Delta(y) = y \otimes y$
• $\epsilon(x) = \epsilon(y) = 1$

where we extend these maps to all of $A$ by linearity.

Then notice

$$ x \overset{\Delta}{\longmapsto} x \otimes x \overset{\Delta \otimes \text{id}_A}{\longmapsto} (x \otimes x) \otimes x $$

and

$$ x \overset{\Delta}{\longmapsto} x \otimes x \overset{\text{id}_A \otimes \Delta}{\longmapsto} x \otimes (x \otimes x) $$

are actually the same map (likewise when applied to $y$). This is the coassociativity axiom. I'll leave it as an instructive exercise to check the coidentity axiom.

As a slightly less trivial example, consider $k[x]$, which has a basis $\{x^0, x^1, x^2, x^3, \ldots\}$. Then we get a coalgebra structure by considering

• $\Delta(x^n) = \sum_k x^k \otimes x^{n-k}$
• $\epsilon(x^0) = 1$
• $\epsilon(x^n) = 0$ for $n > 1$

again, extended linearly.

Now it's a bit more tedious to check the axioms, but it's doable if you persevere.

This coalgebra apparently has applications in combinatorics, though I admit I don't know much about this myself. For more you might look into Joni and Rota's Coalgebras and Bialgebras in Combinatorics, available here, say.

I hope this helps ^_^

The simplest examples arise from free modules with chosen bases (over a commutative ring; for a non-commutative ring you have to use bimodules to get the tensor product to work).

Explicitly, the tensor product $M\otimes M$ of a a free $R$-module $M$ with chosen basis $X\subseteq M$ has basis $\{x_1\otimes x_2:x_1,x_2\in X\}\cong X\times X$. It turns out that the choice of basis $X$ for $M$ gives a coalgebra structure with comultiplication the linear map $M\to M\otimes M$ generated by $x\mapsto x\otimes x$ and that has a counit the linear map $M\to R$ generated by $x\mapsto 1$. More explcitly, the comultiplication is given by $\sum_ir_ix_i\mapsto\sum r_ix_i\otimes x_i$, and the counit by $\sum_i r_ix_i\mapsto\sum_ir_i$.

The next simplest examples arise from free modules of finite rank (i.e. admitting a finite basis) with an algebra structure. Namely, given an $R$-algebra $M$ with multiplication $m\colon M\otimes M\to M$ and unit $e\colon R\to M$, these correspond to an $R$-linear map $m^\vee\colon M^\vee\to(M\otimes M)^\vee$ and an $R$-linear map $e^\vee\colon M^\vee\to R$, where $M^\vee=\{f\colon M\to R\colon f$ is $R$-linear$\}$. In the case $M$ is free of finite rank, we have an isomorphism $(M\otimes M)^\vee\cong M^\vee\otimes M^\vee$, so a (unital) algebra structure on $M$ induces a coalgebra (with counit) structure on $M^\vee$, the $R$-module dual to $M$.