Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $<\rho(v^*, w^*), v\otimes w>=<v^*, v><w^*,w>$.

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 $<M^*(c^*), a\otimes b>=<c^*, M(a\otimes b)>=c^*(ab)$. 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!

share|improve this question
1  
One little comment: when doing those angled brackets , \langle and \rangle look nicer than < and >. –  Alex Petzke Jun 18 '12 at 12:59
    
This is the kind of question that makes me wish we had xymatrix on M.SE. –  darij grinberg Mar 10 '13 at 21:46

1 Answer 1

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 $\rho (d_1^* \otimes d_2^*)=d^*$.

share|improve this answer
    
I am trying to check this $\rho (d_1^* \otimes d_2^*)=d^*$: $$ <\rho (d_1^* \otimes d_2^*), a\otimes b>=<d_1^*, a><d_2^*,b>=d^*(a\otimes 1)\cdot d^*(1\otimes b)=c^*(a)\cdot c^*(b) \\ <d^*, a\otimes b>=c^*(ab)$$ and so $c^*(ab)$ must be equal to $c^*(a)\cdot c^*(b)$ what, I beleive, is not always right. Or I misunderstood something? –  grozhd Apr 9 '12 at 10:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.