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One defines the finite dual of a Hopf algebra $A$ as $$ H^o := \{f \in H^* ~|~ f(I) = 0, \text{ for some ideal $I$ of $H$ with } \dim_C(H/I) < \infty \}. $$ As is well-known, $H^o$ has a well-defined Hopf algebra structure obtained by dualizing the Hopf structure of $H$.

On the other hand, for any finite-dimensional $H$-module $V$, and element $v \in V$, and a functional $f \in V^*$, we can define a functional $c_{f,v} \in H^*$ according to $$ c_{f,v}(h) := f(hv). $$ One usually calls any such functional a matrix coefficient of $H$. It is not difficult to see that the set of matrix coefficients of $H$ forms a Hopf subalgebra of $H^o$, which we will denote by Mat$(H)$.

What I would like to know is when do we have the equality $$ H^o = \text{Mat}(H)? $$

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The equality always holds, for any algebra (not just Hopf algebra) $A$. See Kapitel IV, Bemerkung 3.4 5) in sites.google.com/site/darijgrinberg/hopfalgebren/… if you can read German (which I assume you can, given your location); it is currently on page 480. Sorry for the bad formatting... –  darij grinberg Apr 6 '13 at 16:13
    
Danke! Put your remark as an answer and I can accept it. –  Janos Erdmann Apr 6 '13 at 16:38
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up vote 2 down vote accepted

Typo: $A$ should be $H$ (or vice versa).

The equality always holds, for any algebra (not just Hopf algebra) $H$. See Kapitel IV, Bemerkung 3.4 5) in Schneider's Hopf algebra class if you can read German (which I assume you can, given your location); it is currently on page 479 (as of version 0.98). Sorry for the bad formatting...

Yes, I just updated the notes because I looked at the proof again and couldn't understand it. If you find similar issues elsewhere in the notes, please let me know!

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