From Coalgebras to Algebras via Lax Monoidal Duality

Duality Is Lax Monoidal: Why Coalgebras Dualize to AlgebrasThe canonical map $V^*\otimes W^*\to (V\otimes W)^*$CoherenceLax monoidal functors preserve monoid objectsA coalgebra is a monoid in the opposite categoryWhy does the same argument not dualize an algebra?What fails in infinite dimensions?Finite dimensions: lax becomes strong

 

Duality Is Lax Monoidal: Why Coalgebras Dualize to Algebras

There is a familiar asymmetry in linear algebra.

If $C$$C$ is a coalgebra over a field $k$$k$, then its full linear dual

$$C^{\ast }={\mathrm{Hom}}_k(C,k)$$

is naturally an algebra. But if $A$$A$ is an algebra, its full dual $A^{\ast }$$A^*$ is not, in general, naturally a coalgebra.

In finite dimensions the asymmetry disappears: finite-dimensional algebras and coalgebras dualize into one another.

At first sight this looks like a technical issue caused by infinite-dimensional vector spaces. There is, however, a clean categorical explanation:

$$(-{)}^{\ast }:{\mathbf{Vect}}_k^{op}⟶{\mathbf{Vect}}_k$$

is naturally a lax monoidal functor, and it becomes strong monoidal after restricting to finite-dimensional vector spaces.

Since lax monoidal functors preserve monoid objects, the dual of a coalgebra is automatically an algebra. The failure in the opposite direction is precisely the failure of the lax monoidal structure map to be invertible.

This is what we will prove.

The canonical map $V^{\ast }\otimes W^{\ast }\to (V\otimes W{)}^{\ast }$$V^*\otimes W^*\to (V\otimes W)^*$

Let $V$$V$ and $W$$W$ be vector spaces over $k$$k$.

There is a canonical linear map

$${\mu }_{V,W}:V^{\ast }\otimes W^{\ast }⟶(V\otimes W{)}^{\ast }$$

defined on pure tensors by

$${\mu }_{V,W}(f\otimes g)(v\otimes w)=f(v)g(w).$$

Equivalently, a finite sum

$$\sum _{i=1}^n{f}_i\otimes g_i$$

is sent to the bilinear form

$$(v,w)⟼\sum _{i=1}^n{f}_i(v)g_i(w).$$

There is also a canonical unit map

$${\mu }_0:k⟶k^{\ast }$$

sending $\lambda \in k$$\lambda\in k$ to the linear functional

$$x⟼\lambda x.$$

Under the usual identification $k^{\ast }\cong k$$k^*\cong k$, this is simply the identity.

These maps are natural in $V$$V$ and $W$$W$.

Now regard linear duality as the covariant functor

$$D=(-{)}^{\ast }:{\mathbf{Vect}}_k^{op}⟶{\mathbf{Vect}}_k.$$

The maps above have exactly the form required for a lax monoidal structure:

$$D(V)\otimes D(W)⟶D(V\otimes W),$$

together with

$$k⟶D(k).$$

So it remains only to check coherence.

Coherence

Take

$$f\in U^{\ast },g\in V^{\ast },h\in W^{\ast }.$$

There are two ways to use the maps $\mu$$\mu$ to send

$$f\otimes g\otimes h$$

to an element of

$$(U\otimes V\otimes W{)}^{\ast }.$$

Evaluated on

$$u\otimes v\otimes w,$$

both give

$$f(u)g(v)h(w).$$

Thus the associativity coherence diagram commutes.

Similarly, the unit maps give

$$1\cdot f(v)=f(v)=f(v)\cdot 1,$$

so the left and right unit coherence diagrams commute as well.

Hence

$$\boxed{{\displaystyle (-{)}^{\ast }:{\mathbf{Vect}}_k^{op}\to {\mathbf{Vect}}_k\text{ is lax monoidal}.}}$$

In fact, it is lax symmetric monoidal: the canonical map is compatible with the symmetry

$$V\otimes W\cong W\otimes V.$$

Lax monoidal functors preserve monoid objects

The relevance of this observation comes from a general fact.

Let

$$F:(\mathcal{C},\otimes ,{\mathbb{1}}_{\mathcal{C}})⟶(\mathcal{D},\otimes ,{\mathbb{1}}_{\mathcal{D}})$$

be a lax monoidal functor, with structure maps

$${\varphi }_{X,Y}:F(X)\otimes F(Y)⟶F(X\otimes Y)$$

and

$${\varphi }_0:{\mathbb{1}}_{\mathcal{D}}⟶F({\mathbb{1}}_{\mathcal{C}}).$$

Suppose $M$$M$ is a monoid object in $\mathcal{C}$$\mathcal C$, with multiplication

$$m:M\otimes M⟶M$$

and unit

$$u:{\mathbb{1}}_{\mathcal{C}}⟶M.$$

Then $F(M)$$F(M)$ becomes a monoid object in $\mathcal{D}$$\mathcal D$ by defining

$$F(M)\otimes F(M)\stackrel{{\varphi }_{M,M}}{\to }F(M\otimes M)\stackrel{F(m)}{\to }F(M)$$

as its multiplication, and

$${\mathbb{1}}_{\mathcal{D}}\stackrel{{\varphi }_0}{\to }F({\mathbb{1}}_{\mathcal{C}})\stackrel{F(u)}{\to }F(M)$$

as its unit.

The monoid axioms follow exactly from the coherence axioms for the lax monoidal structure together with the monoid axioms for $M$$M$.

Thus

$$\boxed{{\displaystyle \text{lax monoidal functors preserve monoid objects}.}}$$

Notice that no inverse to ${\varphi }_{X,Y}$$\phi_{X,Y}$ is needed.

This directionality is the important point.

A coalgebra is a monoid in the opposite category

Let

$$C=(C,\mathrm{\Delta },\epsilon )$$

be a coalgebra in ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$.

Thus we have

$$\mathrm{\Delta }:C⟶C\otimes C$$

and

$$\epsilon :C⟶k.$$

Passing to the opposite category reverses the arrows. Hence in ${\mathbf{Vect}}_k^{op}$$\mathbf{Vect}_k^{op}$ they become

$$C\otimes C⟶C$$

and

$$k⟶C.$$

The coassociativity and counit axioms become precisely the associativity and unit axioms.

Therefore

$$\boxed{{\displaystyle \text{a coalgebra in }{\mathbf{Vect}}_k\text{ is a monoid object in }{\mathbf{Vect}}_k^{op}.}}$$

We may now apply the lax monoidal functor

$$(-{)}^{\ast }:{\mathbf{Vect}}_k^{op}⟶{\mathbf{Vect}}_k.$$

Since lax monoidal functors preserve monoids, $C^{\ast }$$C^*$ automatically becomes a monoid object in ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$, hence an algebra.

Its multiplication is the composite

$$C^{\ast }\otimes C^{\ast }\stackrel{{\mu }_{C,C}}{\to }(C\otimes C{)}^{\ast }\stackrel{{\mathrm{\Delta }}^{\ast }}{\to }{C}^{\ast }.$$

Thus, for $f,g\in C^{\ast }$$f,g\in C^*$,

$$(fg)(c)=(f\otimes g)(\mathrm{\Delta }(c)).$$

Writing

$$\mathrm{\Delta }(c)=\sum c_{(1)}\otimes c_{(2)}$$

in Sweedler notation gives

$$(fg)(c)=\sum f(c_{(1)})g(c_{(2)}).$$

This is exactly the usual convolution multiplication on $C^{\ast }$$C^*$.

The unit is obtained from the counit

$$\epsilon :C\to k.$$

Dualizing gives

$${\epsilon }^{\ast }:k^{\ast }⟶C^{\ast },$$

and using $k\cong k^{\ast }$$k\cong k^*$ we obtain

$$1⟼\epsilon .$$

So the unit of the dual algebra is precisely the counit of the original coalgebra.

The usual algebra structure on $C^{\ast }$$C^*$ is therefore not an additional construction that happens to work.

It is forced by monoidality:

$$\boxed{{\displaystyle \text{convolution on }{C}^{\ast }\text{ is the monoid structure transported by the lax monoidal dual functor}.}}$$

Why does the same argument not dualize an algebra?

Now let

$$A=(A,m,u)$$

be an algebra.

Its multiplication is

$$m:A\otimes A⟶A.$$

Dualizing gives

$$m^{\ast }:A^{\ast }⟶(A\otimes A{)}^{\ast }.$$

But to make $A^{\ast }$$A^*$ into a coalgebra, we would need a comultiplication

$$A^{\ast }⟶A^{\ast }\otimes A^{\ast }.$$

So we would like to continue with a map

$$(A\otimes A{)}^{\ast }⟶A^{\ast }\otimes A^{\ast }.$$

The lax monoidal structure gives exactly the opposite direction:

$$A^{\ast }\otimes A^{\ast }⟶(A\otimes A{)}^{\ast }.$$

Thus lax monoidality alone cannot produce a coalgebra structure on $A^{\ast }$$A^*$.

Categorically, there is a simple reason.

An algebra in ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$ becomes a comonoid in ${\mathbf{Vect}}_k^{op}$$\mathbf{Vect}_k^{op}$, but lax monoidal functors preserve monoids, not comonoids.

To transport a comonoid structure, one would need structure maps in the reverse direction,

$$F(X\otimes Y)⟶F(X)\otimes F(Y).$$

In other words, one needs an oplax monoidal structure, or, most conveniently, a strong monoidal structure whose structural maps can be inverted.

This is precisely what fails for unrestricted linear duality.

What fails in infinite dimensions?

The canonical map

$${\mu }_{V,W}:V^{\ast }\otimes W^{\ast }⟶(V\otimes W{)}^{\ast }$$

is always injective, but it need not be surjective.

The image consists of bilinear forms that can be written as finite sums

$$B(v,w)=\sum _{i=1}^n{f}_i(v)g_i(w).$$

Equivalently, if we regard a bilinear form as a linear map

$$\tilde{B}:V⟶W^{\ast },$$

then every element in the image of $V^{\ast }\otimes W^{\ast }$$V^*\otimes W^*$ gives a map of finite rank.

Indeed,

$$\tilde{B}(v)=\sum _{i=1}^n{f}_i(v)g_i$$

has image contained in

$$\mathrm{span}\{g_1,\dots ,g_n\}.$$

Infinite-dimensional spaces admit bilinear forms of infinite rank, so not every element of $(V\otimes W{)}^{\ast }$$(V\otimes W)^*$ comes from $V^{\ast }\otimes W^{\ast }$$V^*\otimes W^*$.

For example, let $V$$V$ be an infinite-dimensional vector space with basis

$$\{e_i{\}}_{i\in I}.$$

Define a bilinear form by

$$B(e_i,e_j)={\delta }_{ij}.$$

The associated map

$$V⟶V^{\ast }$$

sends $e_i$$e_i$ to the corresponding coordinate functional $e_i^{\ast }$$e_i^*$, so it has infinite-dimensional image.

Hence $B$$B$ cannot lie in the image of

$$V^{\ast }\otimes V^{\ast }⟶(V\otimes V{)}^{\ast }.$$

Thus in general

$$V^{\ast }\otimes W^{\ast }⊊(V\otimes W{)}^{\ast }.$$

This is the concrete linear-algebraic obstruction behind the categorical asymmetry.

Finite dimensions: lax becomes strong

Now restrict to finite-dimensional vector spaces.

If $V$$V$ and $W$$W$ are finite-dimensional, then the canonical map

$$V^{\ast }\otimes W^{\ast }⟶(V\otimes W{)}^{\ast }$$

is an isomorphism.

For example, if

$$\{v_i\}$$

and

$$\{w_j\}$$

are bases of $V$$V$ and $W$$W$, with dual bases

$$\{v_i^{\ast }\}\text{and}\{w_j^{\ast }\},$$

then

$$\{v_i^{\ast }\otimes w_j^{\ast }\}$$

maps to the dual basis of

$$\{v_i\otimes w_j\}.$$

Hence

$$\boxed{{\displaystyle (-{)}^{\ast }:{\mathbf{Vect}}_{k,\mathrm{fd}}^{op}⟶{\mathbf{Vect}}_{k,\mathrm{fd}}\text{ is strong monoidal}.}}$$

In fact it is strong symmetric monoidal.

Now the structure map can be inverted:

$$(V\otimes W{)}^{\ast }\stackrel{\sim }{\to }{V}^{\ast }\otimes W^{\ast }.$$

Therefore strong monoidal duality can transport both monoid and comonoid structures.

For a finite-dimensional algebra $A$$A$, dualizing its multiplication gives

$$m^{\ast }:A^{\ast }⟶(A\otimes A{)}^{\ast },$$

and then using the inverse monoidal structure gives

$$A^{\ast }\stackrel{m^{\ast }}{\to }(A\otimes A{)}^{\ast }\stackrel{{\mu }_{A,A}^{-1}}{\to }{A}^{\ast }\otimes A^{\ast }.$$

This is the comultiplication on $A^{\ast }$$A^*$.

Likewise the unit

$$u:k⟶A$$

dualizes to

$$u^{\ast }:A^{\ast }⟶k^{\ast }\cong k,$$

which becomes the counit.

Thus every finite-dimensional algebra has a dual coalgebra.

Conversely, every coalgebra, finite-dimensional or not, has a dual algebra.

So the familiar asymmetry can be summarized as

$$\begin{array}{c}\boxed{{\displaystyle \begin{array}{c}\text{coalgebra }C\\ \Downarrow \text{always}\\ \text{algebra }{C}^{\ast }\end{array}}}\end{array}$$

whereas

$$\begin{array}{c}\boxed{{\displaystyle \begin{array}{c}\text{algebra }A\\ \Downarrow \text{automatically only in finite dimensions}\\ \text{coalgebra }{A}^{\ast }.\end{array}}}\end{array}$$

The distinction is exactly the distinction between lax and strong monoidality.