# Coalgebraic geometry

Previouslywe suggested that if we think of commutative algebras as secretly being functions on some sort of spaces, we should correspondingly think of cocommutative coalgebras as secretly being distributions on some sort of spaces. In this post we’ll describe what these spaces are in the language of algebraic geometry.

Let $Coalgebraic geometry$ be a cocommutative coalgebra over a commutative ring  $Coalgebraic geometry$ . If we want to make sense of $Coalgebraic geometry$ as defining an algebro-geometric object, it needs to have a functor of points on commutative $Coalgebraic geometry$ -algebras. Here it is:

$Coalgebraic geometry$ .

In words, the functor of points of a cocommutative coalgebra $Coalgebraic geometry$ sends a commutative $Coalgebraic geometry$ -algebra  $Coalgebraic geometry$ to the set $Coalgebraic geometry$ of setlike elements of $Coalgebraic geometry$ . In the rest of this post we’ll work through some examples.

#### Sets

Recall that if $Coalgebraic geometry$ is a set then $Coalgebraic geometry$ is a cocommutative coalgebra with comultiplication coming from the diagonal $Coalgebraic geometry$ . More explicitly, the comultiplication is determined by the condition that $Coalgebraic geometry$ for all $Coalgebraic geometry$ .

The functor of points of this coalgebra sends a commutative $Coalgebraic geometry$ -algebra  $Coalgebraic geometry$ to the set of setlike elements of $Coalgebraic geometry$ , and as we computed before, these are precisely the elements of the form $Coalgebraic geometry$ where

$Coalgebraic geometry$

and $Coalgebraic geometry$ , or equivalently $Coalgebraic geometry$ is a complete orthogonal set of idempotents in $Coalgebraic geometry$ . Together, the $Coalgebraic geometry$ determine a direct product decomposition

$Coalgebraic geometry$

which geometrically corresponds to a decomposition of $Coalgebraic geometry$ into disjoint components $Coalgebraic geometry$ . As mentioned previously, the data of such a decomposition is equivalent to the data of a continuous function from the Pierce spectrum $Coalgebraic geometry$ to $Coalgebraic geometry$ .

In other words, $Coalgebraic geometry$ consists of “locally constant functions from $Coalgebraic geometry$ to $Coalgebraic geometry$ .”

We can also equip $Coalgebraic geometry$ with a group structure, and then $Coalgebraic geometry$ , with the usual Hopf algebra structure, has a functor of points sending a commutative $Coalgebraic geometry$ -algebra $Coalgebraic geometry$ to the group of continuous functions from $Coalgebraic geometry$ to $Coalgebraic geometry$ , with pointwise product.

#### Finite-dimensional algebras

Now we restrict to the case that $Coalgebraic geometry$ is a field.

Let $Coalgebraic geometry$ be a finite-dimensional commutative $Coalgebraic geometry$ -algebra. Then the linear dual $Coalgebraic geometry$ acquires a natural coalgebra structure given by dualizing the algebra structure on $Coalgebraic geometry$ . (We don’t need commutativity to say this.) More explicitly, if $Coalgebraic geometry$ is an element of $Coalgebraic geometry$ , then the comultiplication is

$Coalgebraic geometry$

and the counit is

$Coalgebraic geometry$ .

On the other hand,

$Coalgebraic geometry$ .

We conclude the following.

Lemma:A linear functional  $Coalgebraic geometry$ is setlike if and only if $Coalgebraic geometry$ for all $Coalgebraic geometry$ and $Coalgebraic geometry$ ; in other words, if and only if $Coalgebraic geometry$ is a morphism of $Coalgebraic geometry$ -algebras.

More generally, because $Coalgebraic geometry$ is a finite-dimensional $Coalgebraic geometry$ -vector space, if $Coalgebraic geometry$ is any commutative $Coalgebraic geometry$ -algebra then the natural map

$Coalgebraic geometry$

is an isomorphism. We can check that it’s even an isomorphism of coalgebras, and exactly the same computation as above shows the following.

Corollary:An element of $Coalgebraic geometry$ is setlike if and only if the corresponding element of $Coalgebraic geometry$ is a morphism of $Coalgebraic geometry$ -algebras.

Hence the functor of points of $Coalgebraic geometry$ as a coalgebra is precisely the functor of points of $Coalgebraic geometry$ as an algebra: setlike elements of $Coalgebraic geometry$ correspond to morphisms $Coalgebraic geometry$ of affine schemes over $Coalgebraic geometry$ .

The dual map $Coalgebraic geometry$ induces an equivalence of categories between finite-dimensional commutative algebras and finite-dimensional cocommutative coalgebras over $Coalgebraic geometry$ , so we can learn something about the latter by learning something about the former. Every finite-dimensional commutative algebra over a field $Coalgebraic geometry$ is in particular  Artinian , and so factors as a finite product of Artinian local rings. The nilradical of such a ring coincides with its Jacobson radical, and the quotient $Coalgebraic geometry$ is a finite-dimensional commutative semisimple $Coalgebraic geometry$ -algebra, hence factors as a finite product of finite field extensions of $Coalgebraic geometry$ .

Hence, up to taking finite extensions, $Coalgebraic geometry$ looks like a finite set of points together with some “nilpotent fuzz.” $Coalgebraic geometry$ looks like functions on this and $Coalgebraic geometry$ looks like distributions; both are equally sensitive to the “nilpotent fuzz,” as we saw previouslyin the special case of primitive elements.

#### Infinite-dimensional algebras

Again let $Coalgebraic geometry$ be a field. Let $Coalgebraic geometry$ be a commutative $Coalgebraic geometry$ -algebra, not necessarily finite-dimensional. Then it is no longer true that we can put a coalgebra structure on $Coalgebraic geometry$ : when we try dualizing the multiplication, the map $Coalgebraic geometry$ goes in the wrong direction to get a comultiplication.

Intuitively, the problem is that because we’re using the algebraic tensor product $Coalgebraic geometry$ to define coalgebras, the comultiplication $Coalgebraic geometry$ can only output a sum of finitely many tensors, and so has trouble dealing with distributions that are not “compactly supported.”

However, it is possible to rescue this construction as follows. If $Coalgebraic geometry$ is a commutative $Coalgebraic geometry$ -algebra, define its  finite dual

$Coalgebraic geometry$

to consist of all linear functionals $Coalgebraic geometry$ factoring through a finite quotient $Coalgebraic geometry$ of $Coalgebraic geometry$ (as a $Coalgebraic geometry$ -algebra). Geometrically, these are the distributions with “finite support,” and they do in fact have a comultiplication, as follows. If $Coalgebraic geometry$ factors through a finite quotient $Coalgebraic geometry$ , then

$Coalgebraic geometry$

factors through

$Coalgebraic geometry$

and the quotient map $Coalgebraic geometry$ dualizes to a map $Coalgebraic geometry$ , giving us an element of $Coalgebraic geometry$ coming from $Coalgebraic geometry$ , and hence giving an element of $Coalgebraic geometry$ . This is our comultiplication. The counit is $Coalgebraic geometry$ as usual; this poses no problems.

The result we showed in the finite-dimensional case above shows the following here.

Theorem:Let $Coalgebraic geometry$ be a commutative $Coalgebraic geometry$ -algebra and let $Coalgebraic geometry$ be its finite dual. Then the setlike elements of $Coalgebraic geometry$ can naturally be identified with the $Coalgebraic geometry$ -algebra homomorphisms  $Coalgebraic geometry$ which factor through a finite quotient of $Coalgebraic geometry$ .

Geometrically, this says that the functor of points of $Coalgebraic geometry$ sends an affine scheme $Coalgebraic geometry$ to maps from $Coalgebraic geometry$ to the spectrum of the profinite completion

$Coalgebraic geometry$

of $Coalgebraic geometry$ . In other words, $Coalgebraic geometry$ itself is the coalgebra of distributions on the profinite completion.

Example. Let $Coalgebraic geometry$ , so that $Coalgebraic geometry$ is the affine line. The distributions on the affine line with finite support, or equivalently the profinite completion of $Coalgebraic geometry$ , can be very explicitly classified. By the Chinese remainder theorem, the finite quotients of $Coalgebraic geometry$ take the form

$Coalgebraic geometry$

where the $Coalgebraic geometry$ are irreducible polynomials over $Coalgebraic geometry$ . This is a finite product, hence a finite direct sum, of vector spaces, and so any linear functional on it breaks up as a direct sum of linear functionals on each piece, so we can restrict attention to linear functionals on $Coalgebraic geometry$ (distributions “supported on $Coalgebraic geometry$ “) without loss of generality.

In the simplest case, $Coalgebraic geometry$ is a linear polynomial $Coalgebraic geometry$ . Then the linear dual of $Coalgebraic geometry$ has a basis consisting of taking each of the first $Coalgebraic geometry$ terms of the Taylor series expansion of a polynomial in $Coalgebraic geometry$ centered at $Coalgebraic geometry$ : these are (up to the issue of dividing by various factors if $Coalgebraic geometry$ has positive characteristic) the derivatives of the Dirac delta at $Coalgebraic geometry$ .

In the general case we can understand what’s happening using Galois descent. After passing to a suitable field extension $Coalgebraic geometry$ of $Coalgebraic geometry$ , namely the splitting field of $Coalgebraic geometry$ , the quotient $Coalgebraic geometry$ breaks up further into linear factors. In the case that $Coalgebraic geometry$ is Galois, linear functionals on $Coalgebraic geometry$ can be interpreted as $Coalgebraic geometry$ -invariant distributions on $Coalgebraic geometry$ . Geometrically we should think of a finite set of “fuzzy” points acted on by the Galois group; examples of Galois-invariant distributions on this include the sum of Dirac deltas at each point, or the sum of derivatives of Dirac deltas at each point. If $Coalgebraic geometry$ isn’t Galois (meaning that $Coalgebraic geometry$ is inseparable), there is actually extra “fuzziness” that could be hidden over $Coalgebraic geometry$ and only becomes visible over $Coalgebraic geometry$ .

Subexample. Let $Coalgebraic geometry$ and consider the quotient $Coalgebraic geometry$ of $Coalgebraic geometry$ . After passing to the Galois extension $Coalgebraic geometry$ , this quotient becomes $Coalgebraic geometry$ , and it’s clear that the dual space has a natural basis given by two Dirac deltas, one at $Coalgebraic geometry$ and one at $Coalgebraic geometry$ . The corresponding linear functionals $Coalgebraic geometry$ are just evaluation at these two points.

Unfortunately, these Dirac deltas don’t directly make sense over $Coalgebraic geometry$ . Instead, there are two Galois-invariant linear combinations that do: we can take

$Coalgebraic geometry$

which, up to a factor of $Coalgebraic geometry$ , takes the real part of $Coalgebraic geometry$ , and

$Coalgebraic geometry$

which, again up to a factor of $Coalgebraic geometry$ , takes the imaginary part.

#### Cartier duality

We mostly restricted to the case of a field $Coalgebraic geometry$ above because over a field duality behaves in the following very nice way.

Theorem:The functor $Coalgebraic geometry$ is a contravariant equivalence of symmetric monoidal categories between the symmetric monoidal category $Coalgebraic geometry$ of finite-dimensional $Coalgebraic geometry$ -vector spaces and itself.

Because this equivalence is symmetric monoidal, it induces various further equivalences.

Corollary:The functor $Coalgebraic geometry$ is a contravariant equivalence of categories between finite-dimensional $Coalgebraic geometry$ -algebras and finite-dimensional $Coalgebraic geometry$ -coalgebras, and between finite-dimensional commutative $Coalgebraic geometry$ -algebras and finite-dimensional cocommutative $Coalgebraic geometry$ -coalgebras.

These remain symmetric monoidal equivalences if we equip everything with the usual tensor product (which for commutative algebras is the coproduct and for cocommutative coalgebras is the product, so in this case we get that the equivalence is symmetric monoidal for free). We can even ask for both an algebra and a coalgebra structure at once, which gives us this.

Corollary ( Cartier duality ): The functor $Coalgebraic geometry$ is a contravariant equivalence of categories between finite-dimensional commutative and cocommutative Hopf algebras over $Coalgebraic geometry$ and itself.

Finite-dimensional commutative and cocommutative Hopf algebras over $Coalgebraic geometry$ are the analogues of finite abelian groups in the world of algebraic geometry over $Coalgebraic geometry$ : more precisely, they are finite (in the sense that they are Spec of a finite-dimensional algebra) commutative (because “abelian” means something else in algebraic geometry) group schemes (meaning Spec of a commutative Hopf algebra).

Example. Suppose $Coalgebraic geometry$ is a finite abelian group, and $Coalgebraic geometry$ is its group algebra, regarded as a Hopf algebra in the usual way (so cocommutative for general reasons, and commutative because $Coalgebraic geometry$ is abelian). Then the Cartier dual of $Coalgebraic geometry$ is the function algebra $Coalgebraic geometry$ , regarded as a Hopf algebra in the usual way (commutative for general reasons, and cocommutative because $Coalgebraic geometry$ is abelian).

Subexample. If $Coalgebraic geometry$ is the cyclic group of order $Coalgebraic geometry$ , then $Coalgebraic geometry$ , as a group scheme, has functor of points

$Coalgebraic geometry$

sending a commutative $Coalgebraic geometry$ -algebra $Coalgebraic geometry$ to the group of $Coalgebraic geometry$ roots of unity in $Coalgebraic geometry$ . This group scheme has its own name in algebraic geometry: it’s called $Coalgebraic geometry$ . On the other hand, its Cartier dual $Coalgebraic geometry$ is the “constant” group scheme with value $Coalgebraic geometry$ : it has functor of points

$Coalgebraic geometry$

sending a commutative $Coalgebraic geometry$ -algebra $Coalgebraic geometry$ to, as above, the group of locally constant functions from $Coalgebraic geometry$ to $Coalgebraic geometry$ . This is the same functor of points we get if we think about $Coalgebraic geometry$ as a coalgebra, and its name is just $Coalgebraic geometry$ .

Cartier duality can be described as switching between two possible functors of points for a finite-dimensional commutative and cocommutative Hopf algebra $Coalgebraic geometry$ as above: one based on thinking of $Coalgebraic geometry$ as a group object in finite schemes, and one based on thinking of $Coalgebraic geometry$ itself as a group object in finite-dimensional cocommutative coalgebras. In the second description, the functor of points

$Coalgebraic geometry$

sends a commutative $Coalgebraic geometry$ -algebra $Coalgebraic geometry$ to the group (really a group now, since we are in a Hopf algebra) of setlike elements of $Coalgebraic geometry$ .

As it turns out, it’s possible to give a description of what this functor is doing without explicitly thinking about coalgebras or Cartier duality. Namely, we saw above that the coalgebra $Coalgebraic geometry$ of distributions on a point represents the setlike elements functor on coalgebras. We can ask what represents the setlike elements functor on Hopf algebras, and it’s not hard to see that the answer is the Hopf algebra whose underlying algebra is

$Coalgebraic geometry$

where the comultiplication is $Coalgebraic geometry$ , the counit is $Coalgebraic geometry$ , and the antipode is $Coalgebraic geometry$ . This Hopf algebra is commutative, and thinking of it as a group scheme, it is a very famous one, the  multiplicative group scheme $Coalgebraic geometry$ , whose functor of points

$Coalgebraic geometry$

sends a commutative $Coalgebraic geometry$ -algebra to its group of units. Morphisms $Coalgebraic geometry$ of Hopf algebras correspond to setlike elements of $Coalgebraic geometry$ , and if $Coalgebraic geometry$ is commutative these in addition correspond to morphisms $Coalgebraic geometry$ of affine group schemes. A morphism from an affine group scheme to the multiplicative group is called a  character : it is the correct notion of a $Coalgebraic geometry$ -dimensional representation in the world of group schemes.

Cartier duality can then be interpreted as follows: if $Coalgebraic geometry$ is a finite commutative group scheme, then “characters of $Coalgebraic geometry$ ” forms another finite commutative group scheme, whose functor of points

$Coalgebraic geometry$

sends a commutative $Coalgebraic geometry$ -algebra $Coalgebraic geometry$ to the group (under pointwise multiplication) of characters of the base change $Coalgebraic geometry$ . But we saw earlier that this is nothing more than the set of setlike elements of $Coalgebraic geometry$ , or equivalently the set of homomorphisms $Coalgebraic geometry$ ,  and so this is precisely the functor of points of the Cartier dual $Coalgebraic geometry$ as previously defined.

Once Cartier duality is described in terms of characters, it seems a little more suprising: since the dual of the dual of a finite-dimensional vector space $Coalgebraic geometry$ is just $Coalgebraic geometry$ again, we conclude that taking characters of the characters of a finite commutative group scheme gets us the same group scheme again. This should be compared to  Pontryagin duality for finite abelian groups, which says the same thing, where “characters” means homomorphisms $Coalgebraic geometry$ , and which can be interpreted as Cartier duality for constant group schemes over $Coalgebraic geometry$ .