Finite-dimensional algebra-coalgebra duality

Let ${\theta }_{A,A}:A^{\ast }\otimes A^{\ast }\to (A\otimes A{)}^{\ast }$ be the isomorphism in [F3], applicable because $A$ is finite-dimensional and therefore has a finite basis by [F6]. Define ${\mathrm{\Delta }}_{A^{\ast }}:={\theta }_{A,A}^{-1}\circ m_A^{\ast }$ and ${\epsilon }_{A^{\ast }}(f):=f(1_A)$. For $f\in A^{\ast }$ and $a,b\in A$, evaluation through ${\theta }_{A,A}$ gives ${\theta }_{A,A}({\mathrm{\Delta }}_{A^{\ast }}f)(a\otimes b)=f(ab)$. The tensor-product pairing on three finite-dimensional factors, obtained by applying [F3] twice and reassociating, separates elements of $A^{\ast }\otimes A^{\ast }\otimes A^{\ast }$.

For arbitrary coalgebra $C$, define convolution by [F5]. It is bilinear because ${\mathrm{\Delta }}_C$ is linear and tensor evaluation is bilinear. For $f,g,h\in C^{\ast }$ and $c\in C$, the values $((f\ast g)\ast h)(c)$ and $(f\ast (g\ast h))(c)$ are evaluations of $f\otimes g\otimes h$ on $({\mathrm{\Delta }}_C\otimes \mathrm{id}){\mathrm{\Delta }}_C(c)$ and $(\mathrm{id}\otimes {\mathrm{\Delta }}_C){\mathrm{\Delta }}_C(c)$; coassociativity in [F2] makes them equal. The two counit equations give $({\epsilon }_C\ast f)(c)=f(c)=(f\ast {\epsilon }_C)(c)$. Hence $C^{\ast }$ is a unital associative algebra, without a finite-dimensionality hypothesis on $C$.

Let $\psi :C\to D$ be a coalgebra map. For $f,g\in D^{\ast }$ and $c\in C$, ${\psi }^{\ast }(f\ast g)(c)=(f\otimes g)({\mathrm{\Delta }}_D(\psi (c)))=(f\otimes g)((\psi \otimes \psi ){\mathrm{\Delta }}_C(c))=(({\psi }^{\ast }f)\ast ({\psi }^{\ast }g))(c)$. Also ${\psi }^{\ast }({\epsilon }_D)={\epsilon }_D\psi ={\epsilon }_C$. Therefore ${\psi }^{\ast }:D^{\ast }\to C^{\ast }$ is a unital algebra map.

For finite-dimensional coalgebra $C$, [F6] gives a finite basis of $C$; define ${\eta }_C:C\to C^{\ast \ast }$ by ${\eta }_C(c)(f)=f(c)$. A finite basis of $C$ and its dual show ${\eta }_C$ is a vector-space isomorphism. For $f,g\in C^{\ast }$, evaluating ${\mathrm{\Delta }}_{C^{\ast \ast }}{\eta }_C(c)$ against $f\otimes g$ gives ${\eta }_C(c)(f\ast g)=(f\ast g)(c)=(f\otimes g)({\mathrm{\Delta }}_C(c))$. The same evaluation of $({\eta }_C\otimes {\eta }_C){\mathrm{\Delta }}_C(c)$ has that value; finite tensor duality [F3] therefore gives equality of the coproducts. Also ${\epsilon }_{C^{\ast \ast }}({\eta }_C(c))={\eta }_C(c)({\epsilon }_C)={\epsilon }_C(c)$. Hence ${\eta }_C$ is a coalgebra isomorphism.

For $a,b,c\in A$, evaluation of $({\mathrm{\Delta }}_{A^{\ast }}\otimes \mathrm{id}){\mathrm{\Delta }}_{A^{\ast }}(f)$ on $a\otimes b\otimes c$ is $f((ab)c)$, while evaluation of $(\mathrm{id}\otimes {\mathrm{\Delta }}_{A^{\ast }}){\mathrm{\Delta }}_{A^{\ast }}(f)$ is $f(a(bc))$. These are equal by associativity in [F1]; separation from 1.1 gives coassociativity. Similarly, evaluation of $({\epsilon }_{A^{\ast }}\otimes \mathrm{id}){\mathrm{\Delta }}_{A^{\ast }}(f)$ at $a$ is $f(1_{A}a)=f(a)$ and evaluation of $(\mathrm{id}\otimes {\epsilon }_{A^{\ast }}){\mathrm{\Delta }}_{A^{\ast }}(f)$ at $a$ is $f(a{1}_A)=f(a)$. Thus both counit identities hold, so $A^{\ast }$ is a coalgebra.

Let $\varphi :A\to B$ be a unital algebra map between finite-dimensional algebras, so [F6] supplies finite bases for both. Its transpose ${\varphi }^{\ast }:B^{\ast }\to A^{\ast }$ preserves counits since ${\epsilon }_{A^{\ast }}({\varphi }^{\ast }f)=f(\varphi (1_A))=f(1_B)={\epsilon }_{B^{\ast }}(f)$. For $a,a^{\mathrm{\prime }}\in A$, evaluation of ${\mathrm{\Delta }}_{A^{\ast }}({\varphi }^{\ast }f)$ at $a\otimes a^{\mathrm{\prime }}$ is $f(\varphi (a{a}^{\mathrm{\prime }}))=f(\varphi (a)\varphi (a^{\mathrm{\prime }}))$. Writing ${\mathrm{\Delta }}_{B^{\ast }}(f)={\sum }_r{f}_r\otimes g_r$, evaluation of $({\varphi }^{\ast }\otimes {\varphi }^{\ast }){\mathrm{\Delta }}_{B^{\ast }}(f)$ at $a\otimes a^{\mathrm{\prime }}$ is ${\sum }_r{f}_r(\varphi (a))g_r(\varphi (a^{\mathrm{\prime }}))={\mathrm{\Delta }}_{B^{\ast }}(f)(\varphi (a)\otimes \varphi (a^{\mathrm{\prime }}))=f(\varphi (a)\varphi (a^{\mathrm{\prime }}))$ by the defining transpose formula in 1.1. Since $A$ is finite-dimensional, [F3] separates the tensors, so ${\varphi }^{\ast }$ is a coalgebra map.

For finite-dimensional $A$, [F6] supplies a finite basis; define ${\eta }_A:A\to A^{\ast \ast }$ by ${\eta }_A(a)(f)=f(a)$. If $a_1,\dots ,a_n$ is a basis and $a_1^{\ast },\dots ,a_n^{\ast }$ its dual basis, then ${\eta }_A(a_i)$ is the dual basis to $a_i^{\ast }$; hence ${\eta }_A$ is a vector-space isomorphism. For $f\in A^{\ast }$, convolution in $A^{\ast \ast }$ and the defining equation for ${\mathrm{\Delta }}_{A^{\ast }}$ give $({\eta }_A(a)\ast {\eta }_A(b))(f)=\sum {\eta }_A(a)(f_1){\eta }_A(b)(f_2)={\mathrm{\Delta }}_{A^{\ast }}(f)(a\otimes b)=f(ab)={\eta }_A(ab)(f)$. The unit of $A^{\ast \ast }$ is the counit of $A^{\ast }$, and its value on $f$ is $f(1_A)={\eta }_A(1_A)(f)$. Thus ${\eta }_A$ is an algebra isomorphism.

For composable maps $u:V\to W$ and $v:W\to X$, precomposition gives $(v\circ u{)}^{\ast }=u^{\ast }\circ v^{\ast }$, so the transposes in steps 2.3 and 1.3 define contravariant functors. The evaluation maps are natural: for a linear map $u:V\to W$, $u^{\ast \ast }({\eta }_V(v))(f)={\eta }_V(v)(f\circ u)=f(u(v))={\eta }_W(u(v))(f)$. Steps 2.4 and 2.1 show these natural maps are structure-preserving isomorphisms on the finite subcategories, so the two dual functors are inverse equivalences there. The hypothesis concerns the whole algebra or coalgebra object; a finite-dimensional comodule over an infinite-dimensional coalgebra does not make that coalgebra finite-dimensional.

Steps 1.1 and 2.2 make $A^{\ast }$ a coalgebra when $A$ is finite; step 1.2 makes $C^{\ast }$ an algebra for every coalgebra; steps 2.3 and 1.3 prove reversal on algebra and coalgebra maps; steps 2.4, 2.1, and 3.1 prove the finite bidual equivalence and its scope.