Is the dual of a module naturally a comodule?

Question

This question is basically an extension of the following fact: given a finite dimensional, associative, unital $k$-algebra $A$, then the vector dual $A^*$ is a coassociative, counital coalgebra with coproduct $\Delta : A^* \rightarrow A^*\otimes A^*$ given by $\Delta(f)(a\otimes b) = f(ab)$. This fact depends on $(A\otimes A)^*$ being isomorphic to $A^*\otimes A^*$ as vector spaces, as seen by evaluating the coproduct $\Delta(f)$ on $A\otimes A$.

I am starting with a finite dimensional right $A$-module $Q$, over finite dimensional algebra $A$, and constructing a right $A^*$-comodule structure on $Q^*$ as $\rho : Q^* \rightarrow Q^*\otimes A^*$ defined by $\rho(f)(x\otimes a) = f(x\cdot a)$. Again we have an isomorphism between $(Q\otimes A)^*$ and $Q^*\otimes A^*$ and freely evaluate $\rho(f)$ on $Q\otimes A$. As far as I see this is a coassociative comodule when Q is an associative module. Have I made an error in reasoning? Why does no source I know of on Hopf algebras mention such a construction?

Thanks for any help.

Answer 1

Let me add another proof of the fact claimed in OP i.e. of the fact that:

The pair $(Q^*, \rho)$ is a comodule over the coalgebra $(A^*, \Delta, \epsilon)$

I am actually going to show that the above comodule comes from synthesizing (composing) two well known constructions (functors):

Construction I: $\ \ \mathbf{Q_A\longrightarrow {}_AQ^*}$
i.e. if $Q$ is a right $A$ module then its dual space $Q^*$ becomes a left $A$ module through: $$ (a\cdot g)(x)=g(x\cdot a) $$ Construction II: $\ \ \mathbf{{}_AQ^*\longrightarrow (Q^*)^{A^*}}$
i.e. if $Q^*$ is a left $A$ module then $Q^*$ is a right $A^*$ comodule through: $$ \begin{array}{c} \rho:Q^*\rightarrow Q^*\otimes A^* \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g \ \ \mapsto\rho(g)=\Sigma_{i=1}^{k(g)}g_i\otimes f_i \end{array} $$ where $a\cdot g=\Sigma_{i=1}^{k(g)}f_i(a)g_i$, with: $a\in A $, $\ x\in Q$, $\ g_i,g\in Q^* $, $\ f_i \in A^*$.
(In the above, $g_i$, $\ i=1,...,k(g)$, are a $k$-basis of the submodule $A\cdot g$ generated by $g\in Q^*$. $k(g)$ is the $k$-dimension of $A\cdot g$ -which is necessarily finite- and it can be shown that the above definition of $\rho$ is independent of the choice of the $g_i$ (i.e. independent of the choice of basis of $A\cdot g$). $k$ is the field.)

Now, synthesizing the above constructions $I$, $II$, you can start with a right $A$ module $Q_A$ and end up with a right $A^*$ comodule $(Q^*)^{A^*}$: $$ \mathbf{Q_A\stackrel{I}{\longrightarrow} {}_AQ^*\stackrel{II}{\longrightarrow} (Q^*)^{A^*}} $$ I will now show that this synthesis, results in exactly the same comodule structure described in OP:
Supressing the $\Psi$ isomorphism -for simplicity in the notation- we get: $$ \rho(g)\big(x\otimes a\big)\stackrel{II}{=}\sum_{i=1}^{k(g)}g_i(x)f_i(a) \stackrel{II}{=} \big(a\cdot g\big)(x) \stackrel{I}{=} g(x\cdot a) $$ for all $x\in Q$ and $a\in A$, which concludes the proof.

P.S.: For details on the construction $I$ you can look at Drozd-Kirichenko's book on Finite dimensional algebras (ch. 9, p. 159). Construction $II$ is valid for any left $A$ module $M$ (i have just set $M=Q^*$ above). Such constructions can be found in Dascalescu's et al book on Hopf algebras. However, there, it is written for the general case of infinite dimensional Hopf algebras and their restricted duals. (See also these notes, sect. 4.3, p.14-15). If you are interested in a simplified form of the proof for the finite dimensional case, I can provide it here in detail.
Finally, it is interesting to note that both of these constructions are functorial, and they provide isomorphisms between the corresponding categories of (f.d.) right modules, left modules and right comodules. Such functors are frequently called dualities in the literature.

Answer 2

A right comodule $N$ over a coalgebra $(C, \Delta, \epsilon)$ is a pair $(N, \rho)$ where $N$ is a linear space and $\rho : N \rightarrow N\otimes C$ is a linear map such that

• The coassociativity condition is satisfied, $$(\rho \otimes id)\circ \rho = (id \otimes \Delta)\circ \rho$$
• The counit condition is satisfied, $$\iota_{N} \circ(id \otimes \epsilon)\circ \rho = id,$$ where $\iota_{N} : N\otimes k \rightarrow N: n\otimes \lambda \mapsto \lambda n$ is the natural linear isomorphism.

Algebra to Coalgebra Construction

Given two arbitrary $k$-vector spaces $X$ and $Y$ define the linear mapping $$\psi_{X,Y} : X^* \otimes Y^* \rightarrow (X\otimes Y)^*$$ by $\psi_{X,Y}(g\otimes h)(x\otimes y) = g(x)h(y)$. This is well-known to be an isomorphism whenever $X$ and $Y$ are finite dimensional. Being a monomorphism trivially, and an epimorphism by counting dimensions.

As mentioned in the original post the whole intuition for my question came from the following well known fact: given an algebra $(A, \mu, 1)$ the finite dimensional vector space $A^*$ has a coalgebra structure $\Delta : A^* \rightarrow A^* \otimes A^*$ defined in Sweedler notation by $\Delta(f) = \sum f_{(1)} \otimes f_{(2)}$ such that $\sum f_{(1)}(x)f_{(2)}(y) = f(xy)$. This is well-defined because $\Delta(f) = \psi_{A,A}^{-1}(f\circ \mu)$. Justification that $(A^*,\Delta, \epsilon)$ is a coalgebra, with counit $\epsilon(f) = f(1)$, can be found in Susan Montgomery's book for example.

Module to Comodule Construction

I have never seen the previous fact generalised to a module to comodule construction. We assume that $(M, \mathfrak{r})$ is a finite dimensional right $A$-module where the module action is explicitly $\mathfrak{r} : M\otimes A \rightarrow M$, and is associative $\mathfrak{r}\circ (\mathfrak{r} \otimes id) = \mathfrak{r} \circ (id \otimes \mu)$, and unital $\mathfrak{r}(m\otimes 1) = m$, $\forall m \in M$. From now on write $\mathfrak{r}(m\otimes a)$ as $m\cdot a$ interchangeably. The candidate for comultiplication is $\rho : M^* \rightarrow M^* \otimes A^*$ written in Sweedler notation as $\rho(f) = \sum f^{(0)}\otimes f^{(1)} \in M^* \otimes A^*$ such that $\sum f^{(0)}(m)f^{(1)}(a) = f(m\cdot a)$, $\forall m \in M, \forall a \in A$. This is well defined by $\rho(f) = \psi_{M,A}^{-1}(f \circ \mathfrak{r})$ because $f \circ \mathfrak{r} \in (M\otimes A)^*$. I shall provide a proof of the following:

The pair $(M^*, \rho)$ is a comodule over the coalgebra $(A^*, \Delta, \epsilon)$.

Showing Coassociativity

We must show that $(\rho \otimes id)\circ \rho = (id \otimes \Delta) \circ \rho$. Notice that the codomain of both left and right hand side is $M^* \otimes A^* \otimes A^*$, isomorphic to $(M\otimes A \otimes A)^*$ by the mapping $\Psi = \psi_{M\otimes A, A} \circ (\psi_{M,A} \otimes id)$, then $\Psi(f\otimes g \otimes h)(m\otimes a \otimes b) = f(m)g(a)h(b)$. In Sweedler notation we have for $f \in M^*$ \begin{align} [(\rho \otimes id)\circ \rho] (f) = \sum \rho(f^{(0)}) \otimes f^{(1)} = \sum (f^{(0)})^{(0)} \otimes (f^{(0)})^{(1)} \otimes f^{(1)} \end{align} for the left hand side, and for the right hand side \begin{align} [(id \otimes \Delta) \circ \rho](f) = \sum f^{(0)} \otimes \Delta(f^{(1)}) = \sum f^{(0)} \otimes (f^{(1)})_{(0)} \otimes (f^{(1)})_{(1)} \end{align}

Now $\Psi$ applied to the first equation gives us \begin{align*} \Psi([(\rho \otimes id)\circ \rho] (f))(m\otimes a \otimes b) &= \sum (f^{(0)})^{(0)}(m)(f^{(0)})^{(1)}(a) f^{(1)}(b) \\ &= \sum f^{(0)}(m\cdot a)f^{(1)}(b) \\ &= \sum f((m\cdot a) \cdot b). \end{align*}

Moreover $\Psi$ applied to the second one gives \begin{align*} \Psi([(id \otimes \Delta) \circ \rho](f))(m\otimes a \otimes b) &= \sum f^{(0)}(m) (f^{(1)})_{(0)}(a) (f^{(1)})_{(1)}(b) \\ &= \sum f^{(0)}(m) f^{(1)}(ab) \\ &= \sum f(m\cdot ab) \end{align*}

Comparing the last two lines of the previous equations and applying $\Psi^{-1}$ we see that, because $M$ is an associative module, both $(\rho \otimes id)\circ \rho$ and $(id \otimes \Delta) \circ \rho$ are equal. We have shown that $\rho$ is coassociative.

Counit Condition

Here we must prove that $\iota_{M}\circ (id\otimes \epsilon)\circ \rho = id$. Since $\rho(f) = \sum f^{(0)} \otimes f^{(1)}$, applying $id \otimes \epsilon$ gives us $\sum f^{(0)}\otimes f^{(1)}(1)$, which when applying $\iota_M$ gives us $\sum f^{(0)}f^{(1)}(1)$. Now $\sum f^{(0)}(m)f^{(1)}(1) = f(m\cdot 1) = f(m)$, $\forall m\in M$ so that $\rho$ satisfies the counit condition.