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## Coalgebraic geometry

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

$\displaystyle D(-) : \text{CAlg}(k) \ni R \mapsto |D \otimes_k R| \in \text{Set}$.

In words, the functor of points of a cocommutative coalgebra $D$ sends a commutative $k$-algebra $R$ to the set $|D \otimes_k R|$ of setlike elements of $D \otimes_k R$. In the rest of this post we’ll work through some examples.

Sets

Recall that if $X$ is a set then $k[X]$ is a cocommutative coalgebra with comultiplication coming from the diagonal $\Delta : X \to X \times X$. More explicitly, the comultiplication is determined by the condition that $\Delta(x) = x \otimes x$ for all $x \in X$.

The functor of points of this coalgebra sends a commutative $k$-algebra $R$ to the set of setlike elements of $R[X]$, and as we computed before, these are precisely the elements of the form $\sum_x r_x x$ where

$\displaystyle r_x r_y = \begin{cases} r_x & \text{ if } x = y \\ 0 & \text{ otherwise} \end{cases}$

and $\displaystyle \sum r_x = 1$, or equivalently $r_x$ is a complete orthogonal set of idempotents in $R$. Together, the $r_x$ determine a direct product decomposition

$\displaystyle R \cong \prod_x R r_x \cong \prod_x R/(r_x)$

which geometrically corresponds to a decomposition of $\text{Spec } R$ into disjoint components $\text{Spec } R r_x \cong \text{Spec } R/(r_x)$. As mentioned previously, the data of such a decomposition is equivalent to the data of a continuous function from the Pierce spectrum $\pi_0(\text{Spec } R)$ to $X$.

In other words, $|R[X]|$ consists of “locally constant functions from $\text{Spec } R$ to $X$.”

We can also equip $X$ with a group structure, and then $k[X]$, with the usual Hopf algebra structure, has a functor of points sending a commutative $k$-algebra $R$ to the group of continuous functions from $\pi_0(\text{Spec } R)$ to $X$, with pointwise product.

Finite-dimensional algebras

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

Let $A$ be a finite-dimensional commutative $k$-algebra. Then the linear dual $A^{\ast}$ acquires a natural coalgebra structure given by dualizing the algebra structure on $A$. (We don’t need commutativity to say this.) More explicitly, if $f : A \to k$ is an element of $A^{\ast}$, then the comultiplication is

$\displaystyle \Delta(f) = \left( a \otimes b \mapsto f(ab) \right) \in (A \otimes A)^{\ast} \cong A^{\ast} \otimes A^{\ast}$

and the counit is

$\varepsilon(f) = f(1)$.

On the other hand,

$\displaystyle f \otimes f = \left( a \otimes b \mapsto f(a) f(b) \right) \in A^{\ast} \otimes A^{\ast}$.

We conclude the following.

Lemma: A linear functional $f : A \to k$ is setlike if and only if $f(ab) = f(a) f(b)$ for all $a, b \in A$ and $f(1) = 1$; in other words, if and only if $f$ is a morphism of $k$-algebras.

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

$\displaystyle A^{\ast} \otimes_k R \to \text{Hom}_k(A, R)$

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 $A^{\ast} \otimes_k R$ is setlike if and only if the corresponding element of $\text{Hom}_k(A, R)$ is a morphism of $k$-algebras.

Hence the functor of points of $A^{\ast}$ as a coalgebra is precisely the functor of points of $A$ as an algebra: setlike elements of $A^{\ast} \otimes_k R$ correspond to morphisms $\text{Spec } R \to \text{Spec } A$ of affine schemes over $k$.

The dual map $A \to A^{\ast}$ induces an equivalence of categories between finite-dimensional commutative algebras and finite-dimensional cocommutative coalgebras over $k$, so we can learn something about the latter by learning something about the former. Every finite-dimensional commutative algebra over a field $k$ 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 $A/J(A) \cong A/N(A)$ is a finite-dimensional commutative semisimple $k$-algebra, hence factors as a finite product of finite field extensions of $k$.

Hence, up to taking finite extensions, $\text{Spec } A$ looks like a finite set of points together with some “nilpotent fuzz.” $A$ looks like functions on this and $A^{\ast}$ looks like distributions; both are equally sensitive to the “nilpotent fuzz,” as we saw previously in the special case of primitive elements.

Infinite-dimensional algebras

Again let $k$ be a field. Let $A$ be a commutative $k$-algebra, not necessarily finite-dimensional. Then it is no longer true that we can put a coalgebra structure on $A^{\ast}$: when we try dualizing the multiplication, the map $A^{\ast} \otimes A^{\ast} \to (A \otimes A)^{\ast}$ goes in the wrong direction to get a comultiplication.

Intuitively, the problem is that because we’re using the algebraic tensor product $\otimes_k$ to define coalgebras, the comultiplication $D \to D \otimes D$ 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 $A$ is a commutative $k$-algebra, define its finite dual

$A^{!} = \{ f : A \to k \mid f(I) = 0, I \text{ an ideal }, \dim A/I < \infty \}$

to consist of all linear functionals $A \to k$ factoring through a finite quotient $A/I$ of $A$ (as a $k$-algebra). Geometrically, these are the distributions with “finite support,” and they do in fact have a comultiplication, as follows. If $f \in A^{\ast}$ factors through a finite quotient $A/I$, then

$\Delta(f) = \left( v \otimes w \mapsto f(vw) \right) \in (A \otimes A)^{\ast})$

factors through

$\displaystyle (A/I \otimes A/I)^{\ast} \cong (A/I)^{\ast} \otimes (A/I)^{\ast}$

and the quotient map $A \to A/I$ dualizes to a map $(A/I)^{\ast} \to A^{\ast}$, giving us an element of $A^{\ast} \otimes A^{\ast}$ coming from $(A/I)^{\ast} \otimes (A/I)^{\ast}$, and hence giving an element of $A^{!} \otimes A^{!}$. This is our comultiplication. The counit is $\varepsilon f = f(1)$ as usual; this poses no problems.

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

Theorem: Let $A$ be a commutative $k$-algebra and let $A^{!}$ be its finite dual. Then the setlike elements of $A^{!} \otimes_k R$ can naturally be identified with the $k$-algebra homomorphisms $A \to R$ which factor through a finite quotient of $A$.

Geometrically, this says that the functor of points of $A^{!}$ sends an affine scheme $\text{Spec } R$ to maps from $\text{Spec } R$ to the spectrum of the profinite completion

$\displaystyle \widehat{A} = \lim_{I, \dim A/I < \infty} A/I$

of $A$. In other words, $A^{!}$ itself is the coalgebra of distributions on the profinite completion.

Example. Let $A = k[x]$, so that $\text{Spec } A = \mathbb{A}^1$ is the affine line. The distributions on the affine line with finite support, or equivalently the profinite completion of $k[x]$, can be very explicitly classified. By the Chinese remainder theorem, the finite quotients of $k[x]$ take the form

$\displaystyle \prod_i k[x]/f_i(x)^{m_i}$

where the $f_i(x)$ are irreducible polynomials over $k$. 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 $k[x]/f_i(x)^{m_i}$ (distributions “supported on $\text{Spec } k[x]/f_i(x)$“) without loss of generality.

In the simplest case, $f_i(x)$ is a linear polynomial $x - a_i$. Then the linear dual of $k[x]/(x - a_i)^{m_i}$ has a basis consisting of taking each of the first $m_i$ terms of the Taylor series expansion of a polynomial in $k[x]$ centered at $a_i$: these are (up to the issue of dividing by various factors if $k$ has positive characteristic) the derivatives of the Dirac delta at $a_i$.

In the general case we can understand what’s happening using Galois descent. After passing to a suitable field extension $L$ of $k$, namely the splitting field of $f_i(x)$, the quotient $k[x]/f_i(x)^{m_i}$ breaks up further into linear factors. In the case that $L$ is Galois, linear functionals on $k[x]/f_i(x)^{m_i}$ can be interpreted as $\text{Gal}(L/k)$-invariant distributions on $\text{Spec } L[x]/f_i(x)^{m_i}$. 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 $L$ isn’t Galois (meaning that $f_i(x)$ is inseparable), there is actually extra “fuzziness” that could be hidden over $k$ and only becomes visible over $L$.

Subexample. Let $k = \mathbb{R}$ and consider the quotient $\mathbb{R}[x]/(x^2 + 1)$ of $\mathbb{R}[x]$. After passing to the Galois extension $L = \mathbb{C}$, this quotient becomes $\mathbb{C}[x]/(x + i)(x - i) \cong \mathbb{C} \times \mathbb{C}$, and it’s clear that the dual space has a natural basis given by two Dirac deltas, one at $x = i$ and one at $x = -i$. The corresponding linear functionals $\mathbb{C}[x] \to \mathbb{C}$ are just evaluation at these two points.

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

$\displaystyle \delta_i + \delta_{-i} : \mathbb{R}[x] \ni f \mapsto f(i) + f(-i) \in \mathbb{R}$

which, up to a factor of $2$, takes the real part of $f(i)$, and

$\displaystyle \frac{\delta_i - \delta_{-i}}{i} : \mathbb{R}[x] \ni f \mapsto \frac{f(i) - f(-i)}{i} \in \mathbb{R}$

which, again up to a factor of $2$, takes the imaginary part.

Cartier duality

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

Theorem: The functor $V \mapsto V^{\ast}$ is a contravariant equivalence of symmetric monoidal categories between the symmetric monoidal category $\text{FinVect}$ of finite-dimensional $k$-vector spaces and itself.

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

Corollary: The functor $A \mapsto A^{\ast}$ is a contravariant equivalence of categories between finite-dimensional $k$-algebras and finite-dimensional $k$-coalgebras, and between finite-dimensional commutative $k$-algebras and finite-dimensional cocommutative $k$-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 $H \mapsto H^{\ast}$ is a contravariant equivalence of categories between finite-dimensional commutative and cocommutative Hopf algebras over $k$ and itself.

Finite-dimensional commutative and cocommutative Hopf algebras over $k$ are the analogues of finite abelian groups in the world of algebraic geometry over $k$: 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 $G$ is a finite abelian group, and $k[G]$ is its group algebra, regarded as a Hopf algebra in the usual way (so cocommutative for general reasons, and commutative because $G$ is abelian). Then the Cartier dual of $k[G]$ is the function algebra $k^G$, regarded as a Hopf algebra in the usual way (commutative for general reasons, and cocommutative because $G$ is abelian).

Subexample. If $G = \mathbb{Z}/n\mathbb{Z}$ is the cyclic group of order $n$, then $\text{Spec } k[\mathbb{Z}/n\mathbb{Z}]$, as a group scheme, has functor of points

$\displaystyle \text{CAlg}(k) \ni R \mapsto \{ r \in R : r^n = 1 \} \in \text{Ab}$

sending a commutative $k$-algebra $R$ to the group of $n^{th}$ roots of unity in $R$. This group scheme has its own name in algebraic geometry: it’s called $\mu_n$. On the other hand, its Cartier dual $\text{Spec } k^{\mathbb{Z}/n\mathbb{Z}}$ is the “constant” group scheme with value $\mathbb{Z}/n\mathbb{Z}$: it has functor of points

$\displaystyle \text{CAlg}(k) \ni R \mapsto \{ f : \pi_0(\text{Spec } R) \to \mathbb{Z}/n\mathbb{Z} \} \in \text{Ab}$

sending a commutative $k$-algebra $R$ to, as above, the group of locally constant functions from $\text{Spec } R$ to $\mathbb{Z}/n\mathbb{Z}$. This is the same functor of points we get if we think about $k[\mathbb{Z}/n\mathbb{Z}]$ as a coalgebra, and its name is just $\mathbb{Z}/n\mathbb{Z}$.

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

$\displaystyle \text{CAlg}(k) \ni R \mapsto |H \otimes_k R| \in \text{Ab}$

sends a commutative $k$-algebra $R$ to the group (really a group now, since we are in a Hopf algebra) of setlike elements of $H \otimes_k R$.

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 $k[\bullet]$ 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

$\displaystyle k[x, x^{-1}]$

where the comultiplication is $\Delta(x) = x \otimes x$, the counit is $\varepsilon (x) = 1$, and the antipode is $S(x) = x^{-1}$. This Hopf algebra is commutative, and thinking of it as a group scheme, it is a very famous one, the multiplicative group scheme $\mathbb{G}_m$, whose functor of points

$\displaystyle \mathbb{G}_m : \text{CAlg}(k) \ni R \mapsto R^{\times} \in \text{Ab}$

sends a commutative $k$-algebra to its group of units. Morphisms $k[x, x^{-1}] \to H$ of Hopf algebras correspond to setlike elements of $H$, and if $H$ is commutative these in addition correspond to morphisms $\text{Spec } H \to \mathbb{G}_m$ 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 $1$-dimensional representation in the world of group schemes.

Cartier duality can then be interpreted as follows: if $\text{Spec } H$ is a finite commutative group scheme, then “characters of $\text{Spec } H$” forms another finite commutative group scheme, whose functor of points

$\displaystyle \text{CAlg}(k) \ni R \mapsto \{ f : \text{Spec } H_R \to \mathbb{G}_m \} \in \text{Ab}$

sends a commutative $k$-algebra $R$ to the group (under pointwise multiplication) of characters of the base change $\text{Spec } H_R = \text{Spec } H \otimes_k R$. But we saw earlier that this is nothing more than the set of setlike elements of $H \otimes_k R$, or equivalently the set of homomorphisms $H^{\ast} \to R$,  and so this is precisely the functor of points of the Cartier dual $H^{\ast}$ 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 $V$ is just $V$ 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 $A \to \mathbb{C}^{\times}$, and which can be interpreted as Cartier duality for constant group schemes over $\mathbb{C}$.