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## Coalgebras of distributions

Mathematicians are very fond of thinking about algebras. In particular, it’s common to think of commutative algebras as consisting of functions of some sort on spaces of some sort.

Less commonly, mathematicians sometimes think about coalgebras. In general it seems that mathematicians find these harder to think about, although it’s sometimes unavoidable, e.g. when discussing Hopf algebras. The goal of this post is to describe how to begin thinking about cocommutative coalgebras as consisting of distributions of some sort on spaces of some sort.

Functions vs. distributions

Distributions are typically defined as being duals (spaces of continuous linear functionals) to topological vector spaces of functions. Loosely speaking, a distribution is something you can integrate a class of functions against; it’s a kind of generalized measure.

For example, the dual of the space $C(K)$ of continuous functions on a compact Hausdorff space $K$ (with the sup norm topology) is a space of (signed) Radon measures on $K$. A class of examples closer to the examples we’ll be considering, although it involves more technicalities than we’ll need, is the dual of the space $C^{\infty}(M)$ of smooth functions on a smooth manifold $M$ (with the Fréchet topology), which can be thought of as “distributions with compact support” on $M$.

The simplest examples of distributions are the Dirac delta distributions, definable in great generality: as linear functionals on spaces of functions they are precisely the evaluation functionals

$\displaystyle \delta_x : f \mapsto f(x)$.

When we take duals to spaces of smooth functions, as opposed to continuous functions, we get more interesting distributions “supported at a point” given by taking derivatives. For example, on $M = \mathbb{R}$, at every point $x \in \mathbb{R}$ there are linear functionals on $C^{\infty}(\mathbb{R})$ given by

$\displaystyle \delta_x^{(k)} : f \mapsto (-1)^k f^{(k)}(x)$.

These distributions are named using derivative notation because they are the distributional derivatives of $\delta_x$.

The two most important things to keep in mind about the difference between functions and distributions is the following.

1. Functions pull back, while distributions push forward.
2. Functions form commutative algebras, while distributions form cocommutative coalgebras.

These points are closely related: the multiplication on functions resp. the comultiplication on functions, can be described using pullback resp. pushforward along the diagonal map

$\displaystyle \Delta : X \to X \times X$.

Namely, because we can multiply functions on $X$ by functions on $Y$ to get functions on $X \times Y$, for any reasonable notion of functions $F(-)$ we get a dual map

$\displaystyle F(X) \otimes F(X) \to F(X \times X) \xrightarrow{F(\Delta)} F(X)$

giving the multiplication on functions.

The situation for distributions is similar but less straightforward: if $D(-)$ is any reasonable notion of distributions we get a map

$\displaystyle D(X) \xrightarrow{D(\Delta)} D(X \times X)$

To get a comultiplication from this we’d like for there to be an isomorphism, or at least a map, from $D(X \times X)$ to $D(X) \otimes D(X)$. Unfortunately, the map that exists usually goes in the other direction, and usually will not be an isomorphism unless $\otimes$ is some kind of completed tensor product.

Nevertheless, in some examples, and/or with the right modified notion of tensor product, the required maps do exist and we do get a comultiplication on distributions.

In addition to comultiplication, coalgebras also need a counit. In the case of distributions on spaces this counit comes from pushing forward along the unique map $X \to \bullet$, getting a map

$\varepsilon : D(X) \to D(\bullet)$

which, if we think of distributions as generalized measures, computes the “total measure” of a measure.

The diagonal

The appearance of the diagonal map above can be put into a more abstract context. Recall that in any category $C$ with finite products, every object is canonically a cocommutative comonoid in a unique way, via the diagonal map

$\displaystyle \Delta : c \mapsto c \times c$.

A typical example for us will be $C = \text{Set}$, and in general we’ll want to think of $C$ as a category of “spaces.” We can get both commutative monoids and cocommutative comonoids out of diagonal maps as follows.

If $F : C^{op} \to V$ is a contravariant functor out of $C$ (describing a notion of “functions”) to a symmetric monoidal category $V$ (typically something like $V = \text{Vect}$) which is lax symmetric monoidal in the sense that it is equipped with natural transformations

$\displaystyle F(c) \otimes F(d) \to F(c \times d)$

compatible with symmetries (plus some unit stuff), then pulling back along the diagonal endows each $F(c)$ with the structure of a commutative monoid in $V$.

Example. If $C = \text{Set}, V = \text{Vect}$, then we can take $F(c)$ to consist of all functions $c \to k$, where $k$ is the underlying field. If $C = \text{FinSet}$, then $F$ is even symmetric monoidal in the sense that the natural transformations above are isomorphisms.

Dually, if $D : C \to V$ is a covariant functor out of $C$ (describing a notion of “distributions”) to a symmetric monoidal category $V$ which is oplax symmetric monoidal in the sense that it is equipped with natural transformations

$\displaystyle D(c \times d) \to D(c) \otimes D(d)$

compatible with symmetries (plus unit stuff as above), then pushing forward along the diagonal endows each $D(c)$ with the structure of a cocommutative comonoid in $V$.

Example. If $C = \text{Set}, V = \text{Vect}$, then we can take $D(c)$ to consist of the free $k$-vector space $k[c]$ on $c$, where $k$ is the underlying field. Without any finiteness hypotheses, this is even symmetric monoidal.

Sets as coalgebras

Let’s slightly generalize the construction above. Let $k$ be a commutative ring (in fact we could take a commutative semiring here). Then we have a free $k$-module functor from sets to $k$-modules. The above construction shows that this functor can be regarded as taking values in cocommutative coalgebras over $k$, so in fact we have a functor

$\displaystyle \text{Set} \ni X \mapsto k[X] \in \text{Coalg}(k)$.

At this point it will be convenient to introduce the following definition.

Definition: An element $d$ of a coalgebra $(D, \Delta, \varepsilon)$ (where $\Delta : D \to D \otimes D$ is the comultiplication and $\varepsilon : D \to k$ is the counit) is setlike if $\Delta d = d \otimes d$ and $\varepsilon(d) = 1$. If $D$ is a coalgebra, we’ll write $|D|$ for its set of setlike elements.

(The more common term is grouplike, but that term is really only appropriate to the case of Hopf algebras, since in that case the setlike elements form a group. Here the setlike elements only form a set.)

Now we can describe $k[X]$, as a coalgebra, as being freely generated by setlike elements. Thinking in terms of distributions, setlike elements correspond to Dirac distributions, and so it’s reasonable to think of them as the “points” of a coalgebra, or more precisely of a hypothetical space on which the coalgebra is distributions.

Proposition: The functor $X \mapsto k[X]$ from sets to coalgebras above has a right adjoint sending a coalgebra $D$ to its set $|D|$ of setlike elements.

Proof. We want to show that if $X$ is a set and $D$ is a coalgebra, we have a natural bijection

$\displaystyle \text{Hom}(k[X], D) \cong \text{Hom}(X, |D|)$.

But this is clear from the observation that $k[X]$ is a free $R$-module on setlike elements, from which it follows that a coalgebra homomorphism $R[X] \to D$ is uniquely and freely determined by what it does to each element $x \in X$. These elements must be sent to some setlike element of $D$ and can be sent to any such element. $\Box$

In praticular, the functor $D \mapsto |D|$ is represented by the coalgebra $k[\bullet]$ (of “distributions on a point”).

Lemma: Suppose $k$ has no nontrivial idempotents (that is, it is a connected ring). Then the setlike elements of $k[X]$ are precisely the elements $X \in k$: that is, the unit $X \mapsto |R[X]|$ of the above adjunction is an isomorphism.

Proof. Suppose $\sum k_x x \in k[X]$ is a setlike element. Then

$\displaystyle \Delta \left( \sum_x k_x x \right) = \sum_x k_x x \otimes x$

must be equal to

$\displaystyle \left( \sum_x k_x x \right) \otimes \left( \sum_x k_x x \right) = \sum_{x, y} k_x k_y x \otimes y$

which happens if and only if $k_x k_y = k_x$ if $y = x$ and $0$ otherwise. The counit condition is

$\displaystyle \varepsilon \left( \sum_x k_x x \right) = \sum k_x = 1$.

Altogether, the condition that $\sum k_x x$ is primitive is precisely the condition that the elements $k_x$ are a complete set of orthogonal idempotents in $R$. Since $R$ has no nontrivial idempotents by assumption, each $k_x$ is equal to either $0$ or $1$. Since they are orthogonal (meaning $k_x k_y = 0$ if $x \neq y$), at most one of them is equal to $1$. And since they sum to $1$, exactly one of them is equal to $1$. Hence our setlike element is some $x \in X$. $\Box$

The correct statement without the hypothesis that $k$ is connected, which is not hard to extract from the above argument, is that the setlike elements of $k[X]$ in general correspond to functions from the set of connected components of $\text{Spec } k$ to $X$ with finite image, or equivalently to continuous functions from the Pierce spectrum $\pi_0(\text{Spec } k)$ to $X$.

Corollary: Let $k$ have no nontrivial idempotents. Then the functor $X \mapsto k[X]$ is an equivalence of categories from sets to cocommutative coalgebras over $k$ which are free on setlike elements.

In other words, as a slogan, sets are coalgebras of Dirac deltas.

Proof. We showed that the unit of the adjunction between sets and coalgebras is an isomorphism on sets. In general, an adjunction restricts to an equivalence of categories between the subcategories on which the unit resp. the counit of the adjunction are isomorphisms. So it remains to determine for which coalgebras the counit of the adjunction is an isomorphism. Explicitly, the counit is the natural map

$\displaystyle k[|D|] \to D$

from the free $k$-module on the setlike elements of a coalgebra $D$ to $D$. If this is an isomorphism, then $D$ must in particular be free on some setlike elements. Conversely, if $D = k[X]$ is free on setlike elements, then the lemma above shows that $|D| \cong X$ naturally, so that $k[|D|] \cong k[X] \to D$ is an isomorphism. $\Box$

This equivalence induces an equivalence between groups and cocommutative Hopf algebras over $k$ which are free (as modules) on setlike (here “grouplike”) elements.

Beyond Dirac deltas

We’ve said a lot about setlike elements of coalgebras, or equivalently about Dirac delta distributions. But coalgebras have lots of other kinds of elements in general. For example, if $\mathfrak{g}$ is a Lie algebra, its universal enveloping algebra $U(\mathfrak{g})$ has a natural comultiplication given by extending

$\displaystyle \Delta X = 1 \otimes X + X \otimes 1$

where $X \in \mathfrak{g}$; that is, each $X$ is primitive. In a geometric story about distributions, where do the primitives?

The first observation is that in an arbitrary coalgebra there isn’t an element called $1$, so coalgebras don’t have a notion of primitive element. What makes the element $1 \in U(\mathfrak{g})$ special is that it is in fact the unique setlike element: it satisfies $\Delta 1 = 1 \otimes 1$ and is the only element of $U(\mathfrak{g})$ with this property. So whatever primitivity means, geometrically it has something to do with a fixed setlike element, or in distributional terms with a fixed Dirac delta.

Definition: Let $s$ be a setlike element of a coalgebra $D$. An element $d \in D$ is primitive with respect to $s$ if

$\displaystyle \Delta d = s \otimes d + d \otimes s$

and $\varepsilon (d) = 0$.

We can get a big hint about what this definition means by going back to the example of distributions coming from taking the dual of the space of smooth functions $C^{\infty}(\mathbb{R})$. Consider the distribution

$\displaystyle - \delta_r' : C^{\infty}(\mathbb{R}) \ni f \mapsto f'(r) \in \mathbb{R}$.

How does comultiplication act on this distribution? To answer that question we need to see what this distribution does to a product $f(x) g(x)$ of functions (since this describes the action of the distribution on at least a dense subspace of the pullback of functions along the diagonal map $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$). The answer, using the product rule, is that

$\displaystyle fg \mapsto f(r) g'(r) + f'(r) g(r)$.

This gives that

$\displaystyle \Delta (-\delta_r') = \delta_r \otimes (-\delta_r') + (-\delta_r') \otimes \delta_r$

and tells us that primitivity is a reflection of the Leibniz rule for derivations: saying that an element $d$ is primitive with respect to a setlike element $s$ means that if $s$ is a “point,” or more precisely a Dirac delta at a point, then $d$ is a “directional derivative” in a tangent direction at that point. Similarly, computing the pushforward to a point means differentiating constant functions (which are the functions pulled back from a point), which gives zero.

More formally, we can say the following.

Theorem: Let $s$ be a setlike element of a cocommutative coalgebra $D$ over $k$, and let $d$ be an arbitrary element. Then $d$ is primitive with respect to $s$ iff $d + \epsilon s$ is a setlike element of $D[\epsilon]/\epsilon^2$.

Proof. Computation. $\Box$

Intuitively, $s$ is primitive with respect to $d$ iff both $d$ and $d + \epsilon s$ are “points,” where the $\epsilon$ indicates that they are “infinitesimally close” points.

The fact that $U(\mathfrak{g})$, as a Hopf algebra, is generated by primitive elements can be interpreted geometrically as saying that it corresponds to distributions “supported at a point.” In fact it is possible to describe $U(\mathfrak{g})$ as distributions supported at the identity on a Lie group $G$ with Lie algebra $\mathfrak{g}$.