Enriched Convolution from the Lax Monoidality of Hom

Enriched Hom Is Lax Monoidal: The Categorical Origin of ConvolutionEnriched categoriesMonoidal $\mathcal V$-categoriesThe enriched hom functorThe enriched hom functor is lax monoidalNaturality of the laxatorAssociativity coherenceUnit coherenceComonoids and monoidsThe self-enriched caseExample: convolution algebrasExample: the dual algebra of a coalgebraExample: convolution of equivariant mapsExample: dg convolutionThe general picture

 

Enriched Hom Is Lax Monoidal: The Categorical Origin of Convolution

In the previous discussion, we saw that lax monoidal functors preserve monoid objects. This explained why the linear dual of a coalgebra is naturally an algebra.

There is a more general construction behind that phenomenon.

Suppose $\mathcal{C}$$\mathcal C$ is a monoidal category enriched over a symmetric monoidal category $\mathcal{V}$$\mathcal V$. Then the morphisms from an object $C$$C$ to an object $A$$A$ do not merely form a set: they form an object

$$\mathcal{C}(C,A)\in \mathcal{V}.$$

If $C$$C$ is a comonoid object and $A$$A$ is a monoid object in $\mathcal{C}$$\mathcal C$, we will prove that

$$\mathcal{C}(C,A)$$

is naturally a monoid object in $\mathcal{V}$$\mathcal V$.

The mechanism is exactly the same as before:

$$\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}$$

is a lax monoidal enriched functor, while $(C,A)$$(C,A)$ is a monoid object in

$${\mathcal{C}}^{op}⊠\mathcal{C}.$$

Convolution is simply the monoid structure transported by this lax monoidal functor.

Enriched categories

Let

$$(\mathcal{V},\otimes ,\mathbb{I})$$

be a symmetric monoidal category.

A $\mathcal{V}$$\mathcal V$-category $\mathcal{C}$$\mathcal C$ consists of a collection of objects together with, for every pair $X,Y$$X,Y$, a hom-object

$$\mathcal{C}(X,Y)\in \mathcal{V}.$$

Composition is itself a morphism in $\mathcal{V}$$\mathcal V$:

$$\circ :\mathcal{C}(Y,Z)\otimes \mathcal{C}(X,Y)⟶\mathcal{C}(X,Z),$$

and the identity of an object $X$$X$ is represented by a morphism

$$j_X:\mathbb{I}⟶\mathcal{C}(X,X).$$

These satisfy enriched versions of associativity and the unit laws.

When

$$\mathcal{V}=\mathbf{Set},$$

this is just an ordinary category.

When

$$\mathcal{V}={\mathbf{Vect}}_k,$$

the hom-objects are vector spaces and composition is bilinear, so a ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$-category is precisely a $k$$k$-linear category.

When

$$\mathcal{V}={\mathbf{Ch}}_k,$$

the hom-objects are chain complexes, giving dg-categories.

Thus enrichment replaces hom-sets by hom-objects carrying whatever structure is present in $\mathcal{V}$$\mathcal V$.

Monoidal $\mathcal{V}$$\mathcal V$-categories

Now suppose $\mathcal{C}$$\mathcal C$ is also monoidal.

For convolution, it is not enough that $\mathcal{C}$$\mathcal C$ be both monoidal and enriched independently: the tensor product must itself respect the enrichment.

To formulate this, recall that if $\mathcal{C}$$\mathcal C$ and $\mathcal{D}$$\mathcal D$ are $\mathcal{V}$$\mathcal V$-categories, their tensor product

$$\mathcal{C}⊠\mathcal{D}$$

has objects $(X,Y)$$(X,Y)$ and hom-objects

$$(\mathcal{C}⊠\mathcal{D})((X,Y),(X^{\prime },Y^{\prime }))=\mathcal{C}(X,X^{\prime })\otimes \mathcal{D}(Y,Y^{\prime }).$$

The symmetry of $\mathcal{V}$$\mathcal V$ allows the factors to pass through one another when defining composition.

A monoidal $\mathcal{V}$$\mathcal V$-category is a $\mathcal{V}$$\mathcal V$-category $\mathcal{C}$$\mathcal C$ whose tensor product

$$\otimes :\mathcal{C}⊠\mathcal{C}⟶\mathcal{C}$$

is a $\mathcal{V}$$\mathcal V$-functor, together with the usual associator and unitors, now required to be $\mathcal{V}$$\mathcal V$-natural.

The fact that $\otimes$$\otimes$ is a $\mathcal{V}$$\mathcal V$-functor means, in particular, that for any four objects there is a canonical morphism

$${\tau }_{X,X^{\prime };Y,Y^{\prime }}:\mathcal{C}(X,X^{\prime })\otimes \mathcal{C}(Y,Y^{\prime })⟶\mathcal{C}(X\otimes Y,X^{\prime }\otimes Y^{\prime }).$$

This is the enriched version of the familiar operation

$$(f,g)⟼f\otimes g.$$

It is this morphism that will eventually become the multiplication underlying convolution.

The enriched hom functor

From now on, assume additionally that $\mathcal{V}$$\mathcal V$ is symmetric monoidal closed.

Thus for $U,V\in \mathcal{V}$$U,V\in\mathcal V$ there is an internal hom object

$$[U,V]\in \mathcal{V}$$

characterized by the adjunction

$${\mathrm{Hom}}_{\mathcal{V}}(T\otimes U,V)\cong {\mathrm{Hom}}_{\mathcal{V}}(T,[U,V]).$$

In particular, $\mathcal{V}$$\mathcal V$ is naturally enriched over itself, with

$$\underset{―}{\mathcal{V}}(U,V)=[U,V].$$

We can therefore ask whether the assignment

$$(X,A)⟼\mathcal{C}(X,A)$$

defines a $\mathcal{V}$$\mathcal V$-functor

$$H=\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}.$$

It does.

Let

$$P=(X,A),Q=(Y,B).$$

The hom-object between them in ${\mathcal{C}}^{op}⊠\mathcal{C}$$\mathcal C^{op}\boxtimes\mathcal C$ is

$$({\mathcal{C}}^{op}⊠\mathcal{C})(P,Q)=\mathcal{C}(Y,X)\otimes \mathcal{C}(A,B).$$

To define a $\mathcal{V}$$\mathcal V$-functor we therefore need a morphism

$$\mathcal{C}(Y,X)\otimes \mathcal{C}(A,B)⟶[\mathcal{C}(X,A),\mathcal{C}(Y,B)].$$

By closedness, it is equivalent to construct its transpose

$$\mathcal{C}(Y,X)\otimes \mathcal{C}(A,B)\otimes \mathcal{C}(X,A)⟶\mathcal{C}(Y,B).$$

Using symmetry, rearrange the factors as

$$\mathcal{C}(A,B)\otimes \mathcal{C}(X,A)\otimes \mathcal{C}(Y,X).$$

Now compose twice:

$$\mathcal{C}(A,B)\otimes \mathcal{C}(X,A)⟶\mathcal{C}(X,B),$$

followed by

$$\mathcal{C}(X,B)\otimes \mathcal{C}(Y,X)⟶\mathcal{C}(Y,B).$$

Thus we obtain

$$\mathcal{C}(Y,X)\otimes \mathcal{C}(A,B)⟶[\mathcal{C}(X,A),\mathcal{C}(Y,B)].$$

In ordinary notation this is simply

$$(p,q)⟼(h\mapsto q\circ h\circ p).$$

This notation should only be regarded as a mnemonic: the actual construction takes place entirely inside $\mathcal{V}$$\mathcal V$.

The enriched identity axiom follows because

$$1_A\circ h\circ 1_X=h,$$

which is exactly the enriched unit law.

Likewise, compatibility with composition follows from enriched associativity. If, symbolically,

$$p:Y\to X,r:Z\to Y,$$

and

$$q:A\to B,s:B\to D,$$

then both ways of composing act on $h:X\to A$$h:X\to A$ as

$$s\circ q\circ h\circ p\circ r.$$

The equality of these two composites is precisely the enriched associativity axiom.

Hence

$$\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}$$

is indeed a $\mathcal{V}$$\mathcal V$-functor.

The enriched hom functor is lax monoidal

We now come to the central observation.

The category

$${\mathcal{C}}^{op}⊠\mathcal{C}$$

is monoidal componentwise:

$$(X,A)\otimes (Y,B)=(X\otimes Y,A\otimes B),$$

with tensor unit

$$({\mathbb{1}}_{\mathcal{C}},{\mathbb{1}}_{\mathcal{C}}).$$

We claim that

$$H=\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}$$

is lax monoidal.

For two objects $(X,A)$$(X,A)$ and $(Y,B)$$(Y,B)$, define the lax monoidal structure map by

$${\mu }_{(X,A),(Y,B)}:\mathcal{C}(X,A)\otimes \mathcal{C}(Y,B)⟶\mathcal{C}(X\otimes Y,A\otimes B),$$

where $\mu$$\mu$ is exactly the hom-object map induced by the enriched tensor functor

$$\otimes :\mathcal{C}⊠\mathcal{C}\to \mathcal{C}.$$

Thus, in ordinary notation,

$$\mu (f,g)=f\otimes g.$$

The unit morphism is the enriched identity of the tensor unit:

$${\mu }_0:\mathbb{I}⟶\mathcal{C}({\mathbb{1}}_{\mathcal{C}},{\mathbb{1}}_{\mathcal{C}}).$$

We must check that these maps are $\mathcal{V}$$\mathcal V$-natural and satisfy the lax monoidal coherence conditions.

Naturality of the laxator

The essential identity is the enriched form of the interchange law.

Suppose, symbolically, that we have

$$x:X^{\prime }\to X,a:A\to A^{\prime },$$

and

$$y:Y^{\prime }\to Y,b:B\to B^{\prime }.$$

If

$$f:X\to A,g:Y\to B,$$

then applying the enriched hom functor first gives

$$a\circ f\circ x$$

and

$$b\circ g\circ y.$$

Tensoring them gives

$$(a\circ f\circ x)\otimes (b\circ g\circ y).$$

On the other hand, tensor first:

$$f\otimes g:X\otimes Y\to A\otimes B,$$

and then act by

$$x\otimes y$$

and

$$a\otimes b.$$

This gives

$$(a\otimes b)\circ (f\otimes g)\circ (x\otimes y).$$

These are equal:

$$(a\circ f\circ x)\otimes (b\circ g\circ y)=(a\otimes b)\circ (f\otimes g)\circ (x\otimes y).$$

In the enriched setting this equality is not an additional assumption. It is exactly the statement that

$$\otimes :\mathcal{C}⊠\mathcal{C}⟶\mathcal{C}$$

is a $\mathcal{V}$$\mathcal V$-functor.

Equivalently, tensoring morphisms is compatible with enriched composition.

Therefore the morphisms

$${\mu }_{(X,A),(Y,B)}$$

form a $\mathcal{V}$$\mathcal V$-natural transformation.

Associativity coherence

Now take three pairs

$$(X,A),(Y,B),(Z,D).$$

Starting from

$$\mathcal{C}(X,A)\otimes \mathcal{C}(Y,B)\otimes \mathcal{C}(Z,D),$$

we can apply the laxator in two ways.

The first produces, up to associators,

$$(f\otimes g)\otimes h,$$

while the second produces

$$f\otimes (g\otimes h).$$

The associator of the monoidal $\mathcal{V}$$\mathcal V$-category $\mathcal{C}$$\mathcal C$ is a $\mathcal{V}$$\mathcal V$-natural isomorphism

$${\alpha }_{X,Y,Z}:(X\otimes Y)\otimes Z⟶X\otimes (Y\otimes Z).$$

Its naturality says precisely that

$${\alpha }_{A,B,D}\circ ((f\otimes g)\otimes h)=(f\otimes (g\otimes h))\circ {\alpha }_{X,Y,Z}.$$

This is exactly the lax monoidal associativity coherence diagram for $H$$H$.

Thus the associativity of the lax monoidal structure on the hom functor comes from the $\mathcal{V}$$\mathcal V$-naturality of the associator of $\mathcal{C}$$\mathcal C$.

Unit coherence

The same phenomenon occurs for the unit.

Let

$$f:X\to A.$$

Tensoring with the identity of the monoidal unit gives

$$1_{\mathbb{1}}\otimes f:\mathbb{1}\otimes X⟶\mathbb{1}\otimes A.$$

Naturality of the left unitor gives

$${\lambda }_A\circ (1_{\mathbb{1}}\otimes f)=f\circ {\lambda }_X.$$

Likewise, naturality of the right unitor gives

$${\rho }_A\circ (f\otimes 1_{\mathbb{1}})=f\circ {\rho }_X.$$

These are precisely the two lax unit coherence conditions.

Therefore we have proved:

Proposition.
Let $\mathcal{V}$$\mathcal V$ be symmetric monoidal closed and let $\mathcal{C}$$\mathcal C$ be a monoidal $\mathcal{V}$$\mathcal V$-category. Then the enriched hom functor

$$\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}$$

is a lax monoidal $\mathcal{V}$$\mathcal V$-functor.

The structural morphism is simply

$$\mathcal{C}(X,A)\otimes \mathcal{C}(Y,B)⟶\mathcal{C}(X\otimes Y,A\otimes B),$$

the enriched version of tensoring two morphisms.

Comonoids and monoids

Now let $C$$C$ be a comonoid object in $\mathcal{C}$$\mathcal C$:

$$\mathrm{\Delta }:C\to C\otimes C,\epsilon :C\to {\mathbb{1}}_{\mathcal{C}},$$

and let $A$$A$ be a monoid object:

$$m:A\otimes A\to A,u:{\mathbb{1}}_{\mathcal{C}}\to A.$$

Passing to the opposite category reverses the arrows. Thus the comultiplication

$$C\to C\otimes C$$

becomes a multiplication

$$C\otimes C\to C$$

inside ${\mathcal{C}}^{op}$$\mathcal C^{op}$, while the counit becomes a unit.

Hence $C$$C$ is a monoid object in ${\mathcal{C}}^{op}$$\mathcal C^{op}$.

Since $A$$A$ is a monoid object in $\mathcal{C}$$\mathcal C$, the pair

$$(C,A)$$

is a monoid object in

$${\mathcal{C}}^{op}⊠\mathcal{C}.$$

Its multiplication is the pair

$$(\mathrm{\Delta },m):(C\otimes C,A\otimes A)⟶(C,A),$$

where the first component is understood in ${\mathcal{C}}^{op}$$\mathcal C^{op}$.

Its unit is

$$(\epsilon ,u):(\mathbb{1},\mathbb{1})⟶(C,A).$$

Now use the general principle that lax monoidal functors preserve monoid objects.

Applying

$$\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}$$

to the monoid $(C,A)$$(C,A)$ gives a monoid object

$$\mathcal{C}(C,A)$$

in $\mathcal{V}$$\mathcal V$.

Thus:

Enriched Convolution Theorem.
Let $\mathcal{V}$$\mathcal V$ be symmetric monoidal closed and let $\mathcal{C}$$\mathcal C$ be a monoidal $\mathcal{V}$$\mathcal V$-category. If $C$$C$ is a comonoid object of $\mathcal{C}$$\mathcal C$ and $A$$A$ is a monoid object of $\mathcal{C}$$\mathcal C$, then

$$\mathcal{C}(C,A)$$

is canonically a monoid object in $\mathcal{V}$$\mathcal V$.

Its multiplication is

$$\mathcal{C}(C,A)\otimes \mathcal{C}(C,A)\stackrel{\mu }{\to }\mathcal{C}(C\otimes C,A\otimes A)\stackrel{\mathcal{C}(\mathrm{\Delta },m)}{\to }\mathcal{C}(C,A).$$

The first morphism tensors two morphisms, while the second precomposes with $\mathrm{\Delta }$$\Delta$ and postcomposes with $m$$m$.

Symbolically,

$$f\ast g=m\circ (f\otimes g)\circ \mathrm{\Delta }.$$

The unit is the generalized element

$$\mathbb{I}⟶\mathcal{C}(C,A)$$

corresponding to

$$C\stackrel{\epsilon }{\to }{\mathbb{1}}_{\mathcal{C}}\stackrel{u}{\to }A.$$

No separate proof of associativity of convolution is needed.

Associativity follows formally because $(C,A)$$(C,A)$ is already a monoid object and the enriched hom functor is lax monoidal.

The self-enriched case

The familiar internal-hom statement is now just a special case.

Take

$$\mathcal{C}=\mathcal{V}.$$

Since $\mathcal{V}$$\mathcal V$ is closed, it is enriched over itself with

$$\mathcal{V}(C,A)=[C,A].$$

The enriched convolution theorem therefore becomes:

If $C$$C$ is a comonoid and $A$$A$ is a monoid in a symmetric monoidal closed category $\mathcal{V}$$\mathcal V$, then

$$[C,A]$$

is canonically a monoid object in $\mathcal{V}$$\mathcal V$.

Its multiplication is

$$[C,A]\otimes [C,A]⟶[C\otimes C,A\otimes A]\stackrel{[\mathrm{\Delta },m]}{\to }[C,A].$$

Thus the lax monoidality of the internal hom

$$[-,-]:{\mathcal{V}}^{op}⊠\mathcal{V}⟶\mathcal{V}$$

is itself merely the self-enriched instance of the more general lax monoidality of

$$\mathcal{C}(-,-).$$

Example: convolution algebras

Take

$$\mathcal{V}=\mathcal{C}={\mathbf{Vect}}_k.$$

The enrichment is

$$\mathcal{C}(C,A)={\mathrm{Hom}}_k(C,A).$$

A comonoid object is a coalgebra and a monoid object is an algebra.

Hence

$${\mathrm{Hom}}_k(C,A)$$

is naturally an algebra.

If

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

then its multiplication is

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

This is the usual convolution algebra.

The unit is

$$c⟼\epsilon (c)1_A.$$

The familiar formula is therefore simply what the enriched convolution theorem looks like in ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$.

Example: the dual algebra of a coalgebra

Take

$$A=k.$$

Then

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

The general convolution theorem gives $C^{\ast }$$C^*$ its usual algebra structure:

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

Thus the statement

$$\text{the dual of a coalgebra is an algebra}$$

is a very special case of the enriched convolution theorem.

The hierarchy is

$$C^{\ast }=[C,k]\subset [C,A]\subset \mathcal{C}(C,A)$$

not as literal inclusions in general, but as increasingly general instances of the same categorical mechanism.

The dual algebra is therefore best regarded not as an isolated construction, but as convolution with the tensor unit.

Example: convolution of equivariant maps

The genuinely enriched statement becomes clearer when $\mathcal{C}$$\mathcal C$ is not equal to $\mathcal{V}$$\mathcal V$.

Let

$$\mathcal{V}={\mathbf{Vect}}_k$$

and let

$$\mathcal{C}={\mathrm{Rep}}_k(G)$$

for a group $G$$G$.

This is a monoidal ${\mathbf{Vect}}_k$$\mathbf{Vect}_k$-category. Its hom-object

$$\mathcal{C}(X,Y)$$

is the vector space

$${\mathrm{Hom}}_G(X,Y)$$

of $G$$G$-equivariant linear maps.

Suppose $C$$C$ is a coalgebra object in ${\mathrm{Rep}}_k(G)$$\operatorname{Rep}_k(G)$ and $A$$A$ is an algebra object.

Then the enriched convolution theorem says that

$${\mathrm{Hom}}_G(C,A)$$

is naturally an algebra.

For equivariant maps $f,g:C\to A$$f,g:C\to A$,

$$f\ast g=m\circ (f\otimes g)\circ \mathrm{\Delta }.$$

The important point is that no additional argument is required to show that $f\ast g$$f*g$ is again $G$$G$-equivariant.

That closure is already built into the enriched monoidal structure of

$${\mathrm{Rep}}_k(G).$$

This example shows why the enriched formulation is stronger than merely studying

$${\mathrm{Hom}}_k(C,A).$$

The hom-object already remembers the particular class of morphisms appropriate to the category.

Example: dg convolution

Take

$$\mathcal{V}=\mathcal{C}={\mathbf{Ch}}_k.$$

A monoid object is a dg-algebra and a comonoid object is a dg-coalgebra.

The internal hom

$$[C,A]=\underset{―}{\mathrm{Hom}}(C,A)$$

is a chain complex, and the theorem makes it into a monoid object of ${\mathbf{Ch}}_k$$\mathbf{Ch}_k$, hence a dg-algebra.

For homogeneous $f,g$$f,g$, the lax monoidal structure contains the symmetry of chain complexes. Thus

$$(f\otimes g)(x\otimes y)=(-1{)}^{|g||x|}f(x)\otimes g(y).$$

Consequently, if

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

then the convolution product is

$$(f\ast g)(c)=\sum (-1{)}^{|g||c_{(1)}|}f(c_{(1)})g(c_{(2)}).$$

The Koszul sign is not added as an extra convention.

It is forced automatically by the symmetry of the enriching category.

This is one of the main advantages of the categorical formulation: changing the ambient symmetric monoidal category automatically changes convolution in the correct way.

The general picture

The whole construction can now be compressed into one chain of ideas.

First,

$$C\text{ comonoid in }\mathcal{C}$$

means

$$C\text{ monoid in }{\mathcal{C}}^{op}.$$

Therefore

$$(C,A)$$

is a monoid object in

$${\mathcal{C}}^{op}⊠\mathcal{C}.$$

Second, the enriched hom functor

$$\mathcal{C}(-,-):{\mathcal{C}}^{op}⊠\mathcal{C}⟶\mathcal{V}$$

is lax monoidal.

Finally, lax monoidal functors preserve monoid objects.

Hence

$$\mathcal{C}(C,A)$$

is a monoid object in $\mathcal{V}$$\mathcal V$.

In the self-enriched case this becomes

$$[C,A],$$

and in vector spaces it becomes

$${\mathrm{Hom}}_k(C,A).$$

So the familiar formula

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

is only the linear shadow of a much more general statement:

$$\text{comonoid source}+\text{monoid target}+\text{lax monoidal enriched hom}$$

$$⟹$$

$$\text{monoid-valued hom-object}.$$

Convolution is not fundamentally a formula on functions.

It is a consequence of the monoidal structure of morphisms themselves.