Duality and Rigidity in Monoidal Categories - Examples

Theorem Finite-dimensional vector spaces are rigid proves that these maps make $V^{\ast }$ both a left dual and a right dual of $V$ in ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_k$.

So the familiar linear-algebra dual object is exactly a categorical dual in the monoidal sense.

By Finite-dimensional vector spaces are rigid, the categorical dual data are $V^{\ast }$, the evaluation pairing, and the coevaluation $1\mapsto {\sum }_i{v}_i\otimes v_i^{\ast }$.

On a basis vector $v_j$, the first zig-zag gives ${\sum }_i{v}_i^{\ast }(v_j)v_i={\sum }_i{\delta }_{ij}{v}_i=v_j$. On a dual basis vector $v_j^{\ast }$, the second zig-zag gives ${\sum }_i{v}_j^{\ast }(v_i)v_i^{\ast }={\sum }_i{\delta }_{ji}{v}_i^{\ast }=v_j^{\ast }$.

Hence the abstract zig-zag identities of The zig-zag identities reduce here to the familiar basis-and-dual-basis calculation.

The canonical map $J_V:V\to V^{\ast \ast }$ from The canonical evaluation map $J_V:V\to V^{\ast \ast }$ given by $J_V(v)(f)=f(v)$ is the standard pivotal comparison for finite-dimensional vector spaces, so $J_V\circ T$ is an input for the categorical trace of The categorical trace of a morphism into the double dual.

In a basis $(v_i)$ with dual basis $(v_i^{\ast })$, write $T(v_j)={\sum }_i{a}_{ij}{v}_i$. Then ${\mathrm{Tr}}_L(J_{V}T)={\sum }_i{J}_V(T(v_i))(v_i^{\ast })={\sum }_i{v}_i^{\ast }(T(v_i))={\sum }_i{a}_{ii}\mathrm{.}$

The right-hand side is exactly the ordinary trace of $T$ by The basis-independent trace of an endomorphism of a finite-dimensional vector space. So the categorical trace specializes to the matrix trace in finite-dimensional linear algebra.

Let $V$ have basis $(v_1,\dots ,v_n)$. With the usual duality from [L3], the canonical pivotal structure has component $J_V:V\to V^{\ast \ast }$ given by $J_V(v)(f)=f(v)$, and ${\dim }_J(V)$ is by definition the categorical trace of $J_V$ The dimension of an object relative to a pivotal structure.

Using the same basis-and-dual-basis computation as for categorical trace, ${\dim }_J(V)={\sum }_{i=1}^n{J}_V(v_i)(v_i^{\ast })={\sum }_{i=1}^n{v}_i^{\ast }(v_i)=n\cdot 1_k\in k\mathrm{.}$

The natural number $n$ is the ordinary dimension of $V$ by Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis. Hence the categorical dimension is its scalar image $n\cdot 1_k$ in $\mathrm{End}(k)\cong k$. In positive characteristic this scalar can differ from $n$ as a natural number; for example, a $p$-dimensional space has categorical dimension $0\in k$ when $\mathrm{char}k=p$.

The argument in Not every monoidal category is rigid already applies to the concrete infinite-dimensional space $k[x]$: any coevaluation $\mathrm{coev}(1)$ is a finite sum ${\sum }_i{v}_i\otimes f_i$.

The zig-zag identity would then force every polynomial to be a linear combination of the finitely many $v_i$, contradicting the infinite-dimensionality of $k[x]$.

Therefore $k[x]$ has no dual object in the categorical sense.

The currying theorem Currying gives the adjunction $-\times A⊣(-{)}^A$ in $\mathbf{S}\mathbf{e}\mathbf{t}$ says exactly that $(-)\times A⊣(-{)}^A$.

By A dual object in the endofunctor category is an adjoint functor, a right adjoint of an endofunctor is exactly its right dual in the composition monoidal category. Therefore $(-{)}^A$ is the right dual of $(-)\times A$.

So this ordinary adjunction is literally an instance of categorical duality in an endofunctor monoidal category.

By Finite-dimensional vector spaces are rigid, every finite-dimensional vector space $V$ has dual object $V^{\ast }$.

By The dual family of a finite basis is a basis of the dual space, with the same dimension, $\dim (V^{\ast })=\dim (V)$. Therefore Two finite-dimensional vector spaces over $F$ are linearly isomorphic if and only if they have the same dimension gives an isomorphism $V\cong V^{\ast }$.

Thus every object is self-dual up to isomorphism. The example is still only objectwise: the choice of isomorphism is generally noncanonical.