Rigid Monoidal Category via Walking Adjunction.

Rigid Monoidal Categories: Duality as Adjunction1. A Monoidal Category as a One-Object Bicategory2. The Walking Adjunction3. Duality as an Adjunction in $B\mathcal C$4. Left Duals and Right Duals5. Left Rigid, Right Rigid, and Rigid6. The Walking Adjoint Triple7. The Pseudofunctor $F:B\mathcal C\to\mathbf{Cat}$8. The Walking Adjunction Produces Adjoint Functors9. The Tensor-Hom Adjunction10. Left Closed and Right Closed Monoidal Categories11. Why Rigidity Implies Closedness12. Finite-Dimensional Vector Spaces13. Left and Right Duals as an Adjoint Chain

 

Rigid Monoidal Categories: Duality as Adjunction

A useful way to understand rigid monoidal categories is to regard duality as a special case of adjunction.

The central idea is

$$\text{duality in a monoidal category}=\text{adjunction in its one-object bicategory}.$$

The passage from dual objects to ordinary adjoint functors is then explained by the canonical action

$$F:B\mathcal{C}\to \mathbf{Cat},X⟼X\otimes -.$$

Thus there are two levels:

$$X⊣X^{\vee }\text{in }B\mathcal{C},$$

and

$$X\otimes -⊣X^{\vee }\otimes -\text{in }\mathbf{Cat}.$$

The second adjunction is obtained from the first by applying the pseudofunctor $F$$F$.

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1. A Monoidal Category as a One-Object Bicategory

Let

$$(\mathcal{C},\otimes ,\mathbf{1})$$

be a monoidal category.

Its delooping

$$B\mathcal{C}$$

is the bicategory with a single $0$$0$-cell

$$\ast .$$

The correspondence is

$$\begin{array}{cc}\mathcal{C}& B\mathcal{C}\\ \text{object }X& \text{1-cell }X:\ast \to \ast \\ \text{morphism }f:X\to Y& \text{2-cell }f:X\Rightarrow Y\\ X\otimes Y& \text{composition of 1-cells}\\ \mathbf{1}& 1_{\ast }\end{array}$$

Thus an object $X\in \mathcal{C}$$X\in\mathcal C$ becomes a $1$$1$-morphism

$$X:\ast \to \ast$$

in $B\mathcal{C}$$B\mathcal C$.

Therefore, when we speak of an adjoint of an object $X$$X$ in a monoidal category, what we really mean is an adjoint of the corresponding $1$$1$-morphism in $B\mathcal{C}$$B\mathcal C$.

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2. The Walking Adjunction

The walking adjunction $\mathbf{Adj}$$\mathbf{Adj}$ is the $2$$2$-category freely generated by an adjunction

$$L⊣R.$$

It has two objects

$$0,1,$$

generating $1$$1$-morphisms

$$L:0\to 1,R:1\to 0,$$

and generating $2$$2$-morphisms

$$\eta :1_0\Rightarrow RL,$$

$$\epsilon :LR\Rightarrow 1_1,$$

subject to the triangle identities

$$(\epsilon L)\circ (L\eta )=1_L,$$

$$(R\epsilon )\circ (\eta R)=1_R.$$

Since $\mathbf{Adj}$$\mathbf{Adj}$ is freely generated under composition, it also contains composites such as

$$RL,LR,LRL,RLR,\dots$$

together with the $2$$2$-morphisms generated from $\eta$$\eta$ and $\epsilon$$\varepsilon$.

Its universal property is

$$2\text{-functors }\mathbf{Adj}\to \mathcal{B}⟷\text{adjunctions in }\mathcal{B}.$$

Thus a map out of the walking adjunction is precisely a realization of an adjunction.

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3. Duality as an Adjunction in $B\mathcal{C}$$B\mathcal C$

Suppose $X^{\vee }$$X^\vee$ is a right dual of $X$$X$.

Then there are morphisms

$${\mathrm{coev}}_R:\mathbf{1}\to X^{\vee }\otimes X,$$

and

$${\mathrm{ev}}_R:X\otimes X^{\vee }\to \mathbf{1},$$

satisfying the snake identities.

Inside $B\mathcal{C}$$B\mathcal C$, these are precisely the unit and counit of an adjunction

$$X⊣X^{\vee }.$$

Equivalently, there is a realization of the walking adjunction

$$D:\mathbf{Adj}\to B\mathcal{C}$$

such that

$$D(0)=D(1)=\ast ,$$

$$D(L)=X,D(R)=X^{\vee },$$

and

$$D(\eta )={\mathrm{coev}}_R,D(\epsilon )={\mathrm{ev}}_R.$$

The triangle identities in $\mathbf{Adj}$$\mathbf{Adj}$ become the snake identities in $\mathcal{C}$$\mathcal C$.

Thus

$$\text{right dual of }X=\text{right adjoint of }X\text{ in }B\mathcal{C}.$$

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4. Left Duals and Right Duals

In a general monoidal category, left and right duals must be distinguished.

A left dual of $X$$X$ is denoted

$${}^{\vee }X$$

and comes with

$${\mathrm{ev}}_L:{}^{\vee }X\otimes X\to \mathbf{1},$$

and

$${\mathrm{coev}}_L:\mathbf{1}\to X\otimes {}^{\vee }X.$$

In $B\mathcal{C}$$B\mathcal C$, this is the adjunction

$${}^{\vee }X⊣X.$$

A right dual $X^{\vee }$$X^\vee$ comes with

$${\mathrm{ev}}_R:X\otimes X^{\vee }\to \mathbf{1},$$

and

$${\mathrm{coev}}_R:\mathbf{1}\to X^{\vee }\otimes X,$$

corresponding to

$$X⊣X^{\vee }.$$

Hence a rigid object sits in an adjoint triple

$${}^{\vee }X⊣X⊣X^{\vee }.$$

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5. Left Rigid, Right Rigid, and Rigid

A monoidal category $\mathcal{C}$$\mathcal C$ is left rigid if every object has a left dual:

$$\mathcal{C}\text{ is left rigid}⟺\mathrm{\forall }X,{}^{\vee }X⊣X.$$

Equivalently, every $1$$1$-cell in $B\mathcal{C}$$B\mathcal C$ has a left adjoint.

Similarly,

$$\mathcal{C}\text{ is right rigid}⟺\mathrm{\forall }X,X⊣X^{\vee }.$$

Equivalently, every $1$$1$-cell in $B\mathcal{C}$$B\mathcal C$ has a right adjoint.

A rigid monoidal category has both:

$$\mathcal{C}\text{ is rigid}⟺\mathrm{\forall }X,{}^{\vee }X⊣X⊣X^{\vee }.$$

Thus rigidity can be expressed entirely in bicategorical language:

$$\text{every 1-cell in }B\mathcal{C}\text{ has both a left and a right adjoint}.$$

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6. The Walking Adjoint Triple

For a single right dual, the walking adjunction

$$L⊣R$$

is sufficient.

To describe both the left and right dual of a fixed object $X$$X$, it is more natural to consider the $2$$2$-category freely generated by an adjoint triple

$$L⊣M⊣R.$$

A realization

$${\mathcal{W}}_{\mathrm{tri}}\to B\mathcal{C}$$

may send

$$L\mapsto {}^{\vee }X,$$

$$M\mapsto X,$$

$$R\mapsto X^{\vee }.$$

Thus the universal adjoint triple

$$L⊣M⊣R$$

becomes

$${}^{\vee }X⊣X⊣X^{\vee }.$$

For a fixed object, this walking adjoint triple captures both sides of rigidity.

Rigidity of the whole category means that every $X\in \mathcal{C}$$X\in\mathcal C$ can occur as the middle term of such an adjoint triple.

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7. The Pseudofunctor $F:B\mathcal{C}\to \mathbf{Cat}$$F:B\mathcal C\to\mathbf{Cat}$

The monoidal structure is originally a bifunctor

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

Fixing the first variable gives, for every $X\in \mathcal{C}$$X\in\mathcal C$, an endofunctor

$$X\otimes -:\mathcal{C}\to \mathcal{C}.$$

These endofunctors assemble into a pseudofunctor

$$F:B\mathcal{C}\to \mathbf{Cat}.$$

It is defined by

$$F(\ast )=\mathcal{C},$$

and for a $1$$1$-cell $X:\ast \to \ast$$X:*\to*$,

$$F(X)=X\otimes -.$$

For a $2$$2$-cell

$$f:X\Rightarrow Y,$$

that is, a morphism $f:X\to Y$$f:X\to Y$ in $\mathcal{C}$$\mathcal C$, we obtain the natural transformation

$$F(f)=f\otimes 1:X\otimes -\Rightarrow Y\otimes -.$$

Compatibility with composition is supplied by the associator:

$$F(X)\circ F(Y)=X\otimes (Y\otimes -)\cong (X\otimes Y)\otimes -=F(X\otimes Y).$$

Thus

$$X⟼X\otimes -$$

is the action on $1$$1$-cells of the pseudofunctor

$$F:B\mathcal{C}\to \mathbf{Cat}.$$

It is induced by the tensor bifunctor

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

This is the left regular action of $\mathcal{C}$$\mathcal C$ on itself.

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8. The Walking Adjunction Produces Adjoint Functors

Suppose $X^{\vee }$$X^\vee$ is a right dual of $X$$X$.

The dual pair gives a realization of the walking adjunction

$$D:\mathbf{Adj}\to B\mathcal{C}$$

with

$$L\mapsto X,R\mapsto X^{\vee }.$$

Now compose this with

$$F:B\mathcal{C}\to \mathbf{Cat}.$$

We obtain

$$\mathbf{Adj}\stackrel{D}{\to }B\mathcal{C}\stackrel{F}{\to }\mathbf{Cat}.$$

Since $\mathbf{Adj}$$\mathbf{Adj}$ is the walking adjunction, the composite

$$F\circ D:\mathbf{Adj}\to \mathbf{Cat}$$

is precisely the data of an ordinary adjunction of functors.

The generating $1$$1$-morphisms are sent to

$$F(D(L))=F(X)=X\otimes -,$$

and

$$F(D(R))=F(X^{\vee })=X^{\vee }\otimes -.$$

Hence

$$X\otimes -⊣X^{\vee }\otimes -.$$

Schematically,

$$\mathbf{Adj}⟶B\mathcal{C}⟶\mathbf{Cat}$$

gives

$$L⊣R$$

$$\Downarrow$$

$$X⊣X^{\vee }$$

$$\Downarrow$$

$$X\otimes -⊣X^{\vee }\otimes -.$$

Thus the adjunction of tensor functors is not an independent fact. It is the image of the bicategorical adjunction

$$X⊣X^{\vee }$$

under the pseudofunctor

$$F:B\mathcal{C}\to \mathbf{Cat}.$$

The walking-adjunction principle is exactly what turns the dual pair into an ordinary pair of adjoint functors.

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9. The Tensor-Hom Adjunction

The adjunction

$$X\otimes -⊣X^{\vee }\otimes -$$

means that there is a natural isomorphism

$${\mathrm{Hom}}_{\mathcal{C}}(X\otimes A,B)\cong {\mathrm{Hom}}_{\mathcal{C}}(A,X^{\vee }\otimes B).$$

Given

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

the corresponding morphism is

$$A\cong \mathbf{1}\otimes A\stackrel{{\mathrm{coev}}_R\otimes 1}{\to }{X}^{\vee }\otimes X\otimes A\stackrel{1\otimes f}{\to }{X}^{\vee }\otimes B.$$

Conversely, given

$$g:A\to X^{\vee }\otimes B,$$

we obtain

$$X\otimes A\stackrel{1\otimes g}{\to }X\otimes X^{\vee }\otimes B\stackrel{{\mathrm{ev}}_R\otimes 1}{\to }B.$$

These constructions are mutually inverse because the evaluation and coevaluation satisfy the snake identities.

Thus the snake identities are precisely the triangle identities of the corresponding adjunction.

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10. Left Closed and Right Closed Monoidal Categories

There are one-sided notions of closedness.

Using one common convention, $\mathcal{C}$$\mathcal C$ is called left closed if, for every $X$$X$,

$$X\otimes -$$

has a right adjoint:

$$X\otimes -⊣[X,-{]}_L.$$

Thus

$$\mathrm{Hom}(X\otimes A,B)\cong \mathrm{Hom}(A,[X,B{]}_L).$$

Similarly, $\mathcal{C}$$\mathcal C$ is called right closed if

$$-\otimes X$$

has a right adjoint:

$$-\otimes X⊣[X,-{]}_R,$$

so that

$$\mathrm{Hom}(A\otimes X,B)\cong \mathrm{Hom}(A,[X,B{]}_R).$$

If both exist, the monoidal category is biclosed.

The terminology left closed'' andright closed'' varies between authors, so it is often safer to specify explicitly which tensor functor is being considered.

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11. Why Rigidity Implies Closedness

Suppose $X$$X$ has a right dual

$$X⊣X^{\vee }$$

in $B\mathcal{C}$$B\mathcal C$.

Applying

$$F:B\mathcal{C}\to \mathbf{Cat}$$

gives

$$X\otimes -⊣X^{\vee }\otimes -.$$

Therefore the required internal Hom is explicitly

$$[X,B{]}_L\cong X^{\vee }\otimes B.$$

Hence

$$\text{right rigid}⟹\text{left closed}.$$

Similarly, if

$${}^{\vee }X⊣X,$$

then using the right regular action gives

$$-\otimes X⊣-\otimes {}^{\vee }X,$$

and hence

$$[X,B{]}_R\cong B\otimes {}^{\vee }X.$$

Therefore

$$\text{left rigid}⟹\text{right closed}.$$

Combining the two,

$$\text{rigid}⟹\text{biclosed}.$$

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12. Finite-Dimensional Vector Spaces

The basic example is

$${\mathrm{Vect}}_k^{fd}.$$

For a finite-dimensional vector space $X$$X$,

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

Since $X$$X$ is dualizable,

$$X\otimes -⊣X^{\vee }\otimes -.$$

Therefore

$$[X,Y]\cong X^{\vee }\otimes Y.$$

In ordinary linear algebra notation,

$${\mathrm{Hom}}_k(X,Y)\cong X^{\vee }\otimes Y.$$

The natural map

$$X^{\vee }\otimes Y\to {\mathrm{Hom}}_k(X,Y)$$

is

$$\phi \otimes y⟼(x\mapsto \phi (x)y).$$

If

$$e_1,\dots ,e_n$$

is a basis of $X$$X$, with dual basis

$$e^1,\dots ,e^n,$$

then a linear map

$$f:X\to Y$$

corresponds to

$$\sum _i{e}^i\otimes f(e_i).$$

In particular,

$$\mathrm{End}(X)\cong X^{\vee }\otimes X,$$

and

$$X^{\vee }\cong [X,\mathbf{1}].$$

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13. Left and Right Duals as an Adjoint Chain

In a rigid category we have

$${}^{\vee }X⊣X⊣X^{\vee }.$$

Since adjoints are unique up to canonical isomorphism,

$${}^{\vee }(X^{\vee })\cong X$$

and

$$({}^{\vee }X{)}^{\vee }\cong X.$$

This corresponds to taking an adjoint and then taking the adjoint of the opposite kind:

$$X\stackrel{\text{right dual}}{\to }{X}^{\vee }\stackrel{\text{left dual}}{\to }X.$$

By contrast,

$$X⟼X^{\vee }⟼X^{\vee \vee }$$

takes the right dual twice.

There is no general reason for this same-sided double dual to return $X$$X$.

Hence

$${}^{\vee }(X^{\vee })\cong X$$

and

$$X^{\vee \vee }\cong X$$

express different phenomena.