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It is known that a finite dimensional Hopf algebra $H$ over a field $k$ is a Frobenius algebra. Thus there is an isomorphism $H \cong H^\ast$ of left $H$-modules.

Question: Is it possible to write down such an isomorphism entirely in terms of the Hopf algebra, i.e. by product, coproduct, unit, counit and involution ?

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Do you consider cointegrals to be part of The Hopf Algebra™? They are uniquely determined (up to scalars) from equations which can be written down in terms of the Hopf algebra structure maps, just as the antipode is... If you do, then yes. –  Mariano Suárez-Alvarez Jan 23 '12 at 19:30
    
How would the isomorphism look then ? –  user18951 Jan 23 '12 at 19:36
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I don't think this can be done using left integrals of $H$ itself. But if $\lambda$ is a left integral of $H^{\ast}$, then the linear map $H\to H^{\ast},\ a\mapsto a\lambda$ is an $H$-left linear isomorphism. –  darij grinberg Jan 23 '12 at 19:51
    
But the existence of such an integral is guaranteed by the Hopf Algebra axioms and the finite dimensionality, isn't it ? So isn't it possible to express this integral (and hence the isomorphism) in terms of the Hopf algebra's structure maps ? –  user18951 Jan 23 '12 at 20:19
    
It is a theorem that a finite dimensional Hopf algebra has (up to scalars) a unique integral in $H^*$; the space it spans can be described as the coinvariant subspace in $H^*$ for some Hopf module structure (this is explained in standard placed, like Sweedler's book or Susan Montgomery's)... It is not obvious what you mean by express, so it is hard to answer your question—I do doubt there is a universal expression for the isomorphism in the category generated by the structure maps of the Hopf algebra, though. –  Mariano Suárez-Alvarez Jan 23 '12 at 21:11

1 Answer 1

up vote 1 down vote accepted

This isn't an answer but a lengthy comment.

The proofs for the $H$-modul ismorphism $H \cong H^\ast$ I know of use (some variant of) the $H$-module isomorphism $I_L(H^\ast) \otimes_k H \cong H^\ast$ where $$I_L(H^\ast) = \lbrace g \in H^\ast \mid \forall f\in H^\ast: f\ast g = f(1) \cdot g \rbrace$$ is the space of left integrals of $H^\ast$ ($f \ast g$ is the convolution). A dimension argument shows that $I_L(H^\ast)$ is one-dimensional. If $\lambda$ is a non-zero left integral then $I_L(H^\ast) = k\cdot \lambda \cong k$ yields $H \cong H^\ast$ as left $H$-modules. This is just the isomorphism described by darij.

The problem is that the proof doesn't yield a formula for such a $\lambda$. Choose a $k$-basis $\lbrace e_1=1,e_2,...,e_n\rbrace$ of $H$ and suppose $\Delta(e_k) = \sum_{i,j}d_{ij}^{(k)}\cdot e_i \otimes e_j$. Set $D^{(k)} = (d_{ij}^{(k)}-\delta_{i,1}\cdot \delta_{kj})_{i,j=1,\dots,n}$. Expressing $\lambda$ as linear combination of the dual basis then leads to the system of linear equations $D^{(k)} x = 0$ $(k=1,...,n)$. As pointed out by Mariano it has a unique solution (up to scalars). Again, this is no closed formula, but it can be used in practise to compute integrals.

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