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Let $(A, M, u)$ be a finite dimensional algebra where $M: A\otimes A \rightarrow A$ denotes multiplication and $u: k \rightarrow A$ denotes unit.

I want to prove that $(A^*, \Delta, \varepsilon) $ is a colagebra where $\Delta: A^*\rightarrow A^* \otimes A^*$ is a composition: $A^* \overset{M^*}{\rightarrow}(A\otimes A)^* \overset{\rho^{-1}}{\rightarrow}A^*\otimes A^*$

And $\rho: V^*\otimes W^* \rightarrow (V\otimes W)^*$ is given by $<\rho(v^*, w^*), v\otimes w>=$.

I have proven that $\rho$ is injective and since $A$ is finite dimensional $\rho$ is also bijective and we can take the inverse $\rho^{-1}$.

But I have problems understanding how does $\Delta$ work.

By definition we have $==c^*(ab)$. But I can't understand what is $\rho^{-1}(M^*(c^*))$, or in other words which element of $A^*\otimes A^*$ can act like $M^*(c^*)$ via $\rho$?

P.S. Please correct me if I have grammar mistakes. Thanks!

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    This is the kind of question that makes me wish we had xymatrix on M.SE.2013-03-10

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Given $M^*c^*=:d^* \in (A \otimes A)^*$, $\rho^{-1}(d^*)=d_1^* \otimes d_2^* \in A^* \otimes A^*$, where $d_1^*(a)=d^*(a \otimes 1)$ and $d_2^*(a)=d^*(1 \otimes a)$. Notice that $\rho (d_1^* \otimes d_2^*)=d^*$.

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    I am trying to check this $\rho (d_1^* \otimes d_2^*)=d^*$: $$ <\rho (d_1^* \otimes d_2^*), a\otimes b>==d^*(a\otimes 1)\cdot d^*(1\otimes b)=c^*(a)\cdot c^*(b) \\ =c^*(ab)$$ and so $c^*(ab)$ must be equal to $c^*(a)\cdot c^*(b)$ what, I beleive, is not always right. Or I misunderstood something?2012-04-09