Showing $\lambda: A \otimes C^* \rightarrow \text{Hom}(C,A)$ is a morphism of algebras

Question

Show that $\lambda: A \otimes C^* \rightarrow \text{Hom}(C,A)$ is a morphism of algebras.

Let either $C^*$ or $A$ be finite dimensional, and let $\lambda$ be the isomorphism $\lambda: A \otimes C^* \rightarrow \text{Hom}(C,A)$ defined by $$\lambda(a,\gamma)(x)=a\gamma(x)$$ for $a \in A, \gamma \in C^*$ and $x \in C$.

Let $(C,\Delta,\epsilon)$ be a colalgebra and $(A, \mu, \nu)$ be an algebra where $\Delta, \mu$ are the coproduct and product whilst $\epsilon, \nu$ are the counit and unit.

Define the convolution $@$ for $f,g \in \text{Hom}(C,A)$ by $$(f @ g)(x) = \mu (f \otimes g) \Delta(x)$$

Then we have the following string of equalities for $a,b \in A$ and $\alpha,\beta \in C^*$

$$\lambda(a \otimes \alpha) @ \lambda(b \otimes \beta)(x)=\Sigma_{(x)}\alpha(x')\beta(x'')ab= (\alpha \beta)(x)ab=(\lambda(ab \otimes \alpha \beta))(x)$$

Can somebody explain to me the equailty $\Sigma_{(x)}\alpha(x')\beta(x'')ab= (\alpha \beta)(x)ab$ please?

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Answer 1

The dual space $C^*$ is an algebra because $C$ is a coalgebra. The multiplication of $C^*$ comes from “dualizing” the comultiplication of $C$.

Let us be more precise. The comultiplication $$ \Delta \colon C \to C \otimes C $$ induces a linear map $$ \Delta^* \colon (C \otimes C)^* \to C^* \,, \quad \Delta^*(\varphi)(x) = \varphi( \Delta(x) ) \,. $$ We also have a linear map $$ \Phi \colon C^* \otimes C^* \to (C \otimes C)^* $$ given by $$ \Phi(\alpha \otimes \beta)(x \otimes y) = \alpha(x) \beta(y) $$ for all $\alpha, \beta \in C^*$ and $x, y \in C$. The composite $$ \Delta^* \circ \Phi \colon C^* \otimes C^* \to C^* $$ is the induced multiplication on $C^*$, which makes $C^*$ into an algebra. This composite is given by \begin{align*} (\Delta^* \circ \Phi)(\alpha \otimes \beta)(x) &= \Delta^*( \Phi( \alpha \otimes \beta) )(x) \\ &= \Phi( \alpha \otimes \beta )( \Delta(x) ) \\ &= \Phi( \alpha \otimes \beta )\left( \sum_{(x)} x' \otimes X'' \right) \\ &= \sum_{(x)} \Phi( \alpha \otimes \beta )( x' \otimes x'' ) \\ &= \sum_{(x)} \alpha(x') \beta(x'') \end{align*} for all $\alpha, \beta \in C^*$ and $x \in C$.

The above calculation explains that the multiplication on the algebra $C^*$ is given by $$ (\alpha \beta)(x) = \sum_{(x)} \alpha(x') \beta(x'') $$ for all $\alpha, \beta \in C^*$ and $x \in C$. The identity in question, $$ (\alpha \beta)(x) ab = \sum_{(x)} \alpha(x') \beta(x'') ab \,, $$ follows directly from this.