# What is a coalgebra intuitively?

How to think about coalgebras? Are there geometric interpretations of coalgebras? If I think of algebras and modules as spaces and vectorbundles, what are coalgebras and comodules? What basic examples of coalgebras should one keep in mind?

Anything that helps to think about coalgebras without headache is welcome ;)

It seems like there are two basic sources of examples:

1.

The basic structure you have on a space (set, scheme...) is the diagonal morphism $\Delta :X \to X\times X$. Functions on spaces are contravariant, which is why functions on a space form an algebra: $f.g = \Delta ^\ast (f \boxtimes g)$. We also have the morphism $\pi : X \to pt$, and the unit in this algebra is $\pi ^\ast 1$.

If we chose some covariant linearization of our space (like measures, topological chains...) then this space would be a coalgebra, with comultiplication given by $\Delta_\ast$, and counit $\pi_\ast$. So, for example we have the coalgebra $C_\ast (X)$ of (say, singular) chains on a topological space.

Such coalgebras are naturally cocommutative.

In nice cases, we have a pushforward and a pullback (e.g. functions on a finite set, equipped with a measure), and the algebra and coalgebra structures together form a Frobenius algebra (the inner product is $\pi _\ast \Delta ^\ast$.

2.

If our space $X$ is a group (or maybe just a monoid), then we have a multiplication map $m: X\times X \to X$, and then $m^\ast$ equips the space of functions (or, e.g. cochains) with the structure of a colagebra. If the multiplication on $X$ has a unit $e: pt \to X$, then $e^\ast$ is the counit of this algebra.

This coalgebra will not be cocommutative in general (unless $(X,m)$ is commutative).

If we also remember the ordinary multiplication structure (coming from $\Delta ^\ast$), then these two structures toegether form a Hopf algebra (also using the inversion $i:X \to X$ in the group).

Julian's example is of this second kind.

All the examples I know, are morally either of type 1 or type 2, but sometimes you need to have a very broad definition of "function"or "measure"!

• I think the Davids would approve of this answer. – Mike Skirvin Sep 27 '11 at 14:53
• For source 1 you need your covariant linearization to be at least lax monoidal, right? – Qiaochu Yuan Sep 27 '11 at 18:39
• @Qiaochu: Yeah, I guess so – Sam Gunningham Sep 27 '11 at 20:00
• I have just a comment about comodules in your pictures. In type 2, they arise for example from actions of monoids on spaces. In type 1 the arise for example from coactions in spaces. But it turns out that coactions in spaces is just a complicated expression for a map of spaces! – Jan Weidner Jan 31 '12 at 13:35

Coalgebras appear naturally in combinatorics as describing ways one can decompose objects into other objects of the same type. For example, the coalgebra structure on $k[x]$ given by

$$x^n \mapsto \sum_{k=0}^n {n \choose k} x^k \otimes x^{n-k}$$

describes the ways in which one can decompose a set into two subsets. (Note that the convolution product of linear functionals $k[x] \to k$ can be identified with the product of exponential generating functions.)

As a more complicated example, there is a coalgebra describing the ways in which one can decompose a connected region of $\mathbb{R}^2$ tiled by finitely many squares into two such regions. There are endless variations on this construction.

I learned this point of view from Gian-Carlo Rota's Coalgebras and bialgebras in combinatorics, which I currently cannot find an online copy of...

• This type of answer applies very generally to coalgebras of endofunctors. A binary tree for example can be described as a set $X$ equipped with a function $X \to 1 + X^2$, which sends a point to its pair of children (or to the singleton of $1$ if their are no children). Needless to say, these sorts of parsing operation occurs all over combinatorics. – Todd Trimble Sep 27 '11 at 16:57

The elements of coalgebras can often be thought of as having a distribution-like character, as can be seen from various examples. For example, the cofree coalgebra cogenerated by a single element (let us say over a field $k$ of characteristic zero) can be realized as a localization of $k[x]$ at the prime ideal $(x)$, which we can think of as sitting inside a space of formal power series

$$k[x]_{(x)} \hookrightarrow k[[x]]$$

by expanding inverses of elements prime to $x$ in geometric series. A way to think about the formal series $\sum_n a_n x^n$ that so arise is that $x^n$ is a symbol for a derivative of a Dirac functional in the sense of distributions, i.e., we think of a formal series as a formal sum

$$\sum_n a_n \frac{\delta_{0}^{(n)}}{n!}$$

where $\delta_0 = eval_0: k[x] \to k$ is the Dirac functional at $0$. The comultiplication can be read off from the adjoint rule

$$\langle \Delta(m), f \otimes g\rangle = \langle m, f \cdot g\rangle$$

where $f, g \in k[x]$ are polynomial "functions". I wrote up a calculational exposition of this point of view on the cofree coalgebra at the $n$-Category Café here.

A general heuristic here is that it's hard to multiply distributions, but one can often "comultiply" them by this adjoint rule.

Another example is given by homology theory. Here the adjoint pairing (say if we are thinking in terms of De Rham theory, where cocycles are given by smooth functions) is given by integration of an $n$-form over an $n$-chain:

$$\langle c, \omega \rangle = \int_c \omega$$

so here we are thinking of $n$-chains as acting on $n$-forms much as currents. In this way, the homology coalgebra also has a distributional character (with the same sort of interpretation of comultiplication).

• By the way, I don't know why my computer didn't notify me of Sam's answer that appeared before mine, which has a similar thematic character. – Todd Trimble Sep 27 '11 at 15:07

One extra way in which you can view coalgebras is as (a special kind of) functors from algebras to algebras.

Let us fix a coalgebra $C$. For each algebra $A$ we can consider the vector space $F_C(A)=\hom(C,A)$ of all linear maps $C\to A$. This is an algebra with respect to the convolution product, so that for all $f$, $g\in F_C(A)$ the product is the function $f\star g:C\to A$ given by $$f\star g = \mu_A\circ f\otimes g\circ\Delta_C;$$ here $\mu_A:A\otimes A\to A$ is the multiplication in $A$, and $\Delta_C:C\to C\otimes C$ is the comultiplication in $C$. It is easy to turn $F_C$ into a functor.

For example, if $M^n$ the the coalgebra of $n\times n$ comatrices, then $F_{M^n}(A)=M_n(A)$ is the algebra of $n\times n$ matrices with entries in $A$. &c.

• Of course, you can turn this around and view algebras as functors from coalgebras to algebras---in case you also want some 'intuition' about algebras :) – Mariano Suárez-Álvarez Sep 27 '11 at 17:51
• Maybe someone can tell us exactly which functors $\mathrm{Alg}\to\mathrm{Alg}$ we get from coalgebras? – Mariano Suárez-Álvarez Sep 27 '11 at 18:07
• @Mariano: Every continuous endofunctor of algebras is representable (SAFT) if we compose it with the forgetful functor, and the representing object comes equipped with a coalgebra structure. – Martin Brandenburg Oct 1 '11 at 10:03

a pretty general kind of coalgebra is the path coalgebra of a quiver (directed graph).

• Ah thanks, I was not aware of this example! A path gets mapped to the sum of all ways to decompose it into two pieces, right? Does this coalgebra structure play an important role somewhere? – Jan Weidner Jul 23 '12 at 7:49

The following does not answer your question, but gives a particular example in a special case:

The commutative Hopf algebras have a geometric interpretation, they are the coordinate rings of affine group schemes via Yoneda Lemma. Via this interpretation comodules for the coordinate ring are just modules for the group scheme.

Examples of Hopf algebras (I'm not that familiar with coalgebras, that are not Hopf algebras) that I think one should keep in mind:

• group algebras
• enveloping algebras of Lie algebras
• restricted enveloping algebras
• quantized enveloping algebras
• small quantum groups
• Comatrix coalgebras are a nice example of coalgebras which are not Hopf algebras. – Mariano Suárez-Álvarez Sep 27 '11 at 15:48
• (they don't have one-dimensional comodules, so they can't be Hopf) – Mariano Suárez-Álvarez Sep 27 '11 at 15:52

This is also coming more from the Hopf point of view than the strictly coalgebra point of view.

If you think of algebras heuristically as functions, then a Hopf algebra heuristically can be thought of as the function algebra of a group. A representation of the group $G$ on a vector space $V$ can be thought of as a map $$G \times V \to V.$$ Dualizing gives (roughly) a map $$V^* \to V^* \otimes \mathcal{F}(G),$$ where $\mathcal{F}(G)$ is some appropriate Hopf algebra of functions on $G$. In other words a comodule can be thought of as a representation of the group structure underlying the coalgebra.

If you have a space $V$ which is both a module and a comodule for $\mathcal{F}(G)$ with some compatibility, then $V$ is the space of sections of some homogeneous vector bundle over $G$ (the Hopf Module Theorem). Functions on $G$ act by pointwise multiplication on sections, and $G$ also acts on sections by translation on the base.

• Coalgebras would be more like function algebras of semigroups, right? (there need not be an antipode) – Yemon Choi Sep 27 '11 at 20:00
• Yes, that's correct - a semigroup is all that's needed for a coalgebra structure. – MTS Sep 27 '11 at 20:52