Duals of a finite dimensional eveloping coalgebra

Question

Let $C^e$ be the enveloping $k$-coalgebra of a $k$-coalgebra $C$ and denote by ${C^e}^{\star}:=\mathrm{Hom}\,_{k}(C^e,k)$.

Then is ${C^e}^{\star} \cong {C^{\star}}^e$?

Answer 1

(Note, all homs and tensors will be over $k$)

Let $C$ be a coalgebra. The enveloping coalgebra of $C$ is $C^e:=C\otimes C^{op}$. Similarly, we can define the enveloping algebra of an algebra by $A^e:=A\otimes A^{op}$. We have that op commutes with dualization, in the sense that if $C$ is a coalgebra, then $(C^*)^{op}\cong (C^{op})^*$ as algebras.

Given vector spaces $V, W$, we have a natural map $V^*\otimes W^* \to (V\otimes W)^*$ given sending $\phi \otimes \psi$ to the map $v\otimes w \mapsto (\phi(v)\otimes \psi(w))$. If $V$ and $W$ are finite dimensional, this map is an isomorphism. One proof is to take a basis for $V,W$, and then examine the corresponding dual bases.

However, if $V$ and $W$ are both infinite, the map is no longer an isomorphism. We can factor the natural map as $$\hom(V,k)\otimes \hom(W,k)\to \hom(V,\hom(W,k))\cong \hom(V\otimes W,k).$$ Since the last maps is an isomorphism, the natural map is an isomorphism if and only if the first map is. We assert that it is not.

Let $Z=\hom(W,k).$ An element of $\hom(V,k)\otimes Z$ is a finite sums of the form $\sum \phi_i \otimes z_i$, which under our map is sent to $v\to \sum \phi(i)v z_i$. The image of such a map is finite dimensional. Since $V$ and $Z$ are both infinite dimensional, there are maps in $\hom(V,Z)$ with infinite image, and therefore our map cannot be surjective.

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It is worth pointing out that the failure of tensor products to commute with dualization means that while the dual of a coalgebra is an algebra, the dual of an infinite dimensional algebra is NOT in general a coalgebra.

Suppose we have a map $\Delta:C\to C\otimes C$. Dualizaing and precomposing with the natural map, we have $C^* \otimes C^* \to (C\otimes C)^* \stackrel{\Delta^*}{\to} C^*$.

However if we have a map $\mu:A\otimes A\to A$, we cannot in general factor $\mu^*:A^*\to (A\otimes A)^*$ through $A^*\otimes A^*$.

One solution to this is to view dualization not as a controvatiant functor from vector spaces to itself, but rather a pair of controvariant functors between vector spaces and topological vector space with the pro-finite topology.