let $B$ be a bialgebra over a field (i.e. associative, coassociative, unitary and counitary, maybe it has an antipode or maybe not). If $b\in B$ acts by zero on every finite dimensional representation of $B$, then necesarily $b=0$?
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No, it may not even have any non-trivial finite dimensional representations. For example, take the group algebra $kG$, where $G$ is a simple group with cardinality greater than that of the infinite field $k$.
answered Apr 24, 2017 at 14:50
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$\begingroup$ but assuming something in the bialgebra? for instance if the bialgebra is finitely generated as algebra? $\endgroup$ Commented Apr 24, 2017 at 16:31
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$\begingroup$ Since finitely generated linear groups are residually finite by a theorem of Malcev, if you take an infinite finitely generated simple group (like Thompson's group V), then all its finite dimensional representations collapse the group. $\endgroup$ Commented Apr 24, 2017 at 17:46
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$\begingroup$ For a finitely generated group or monoid, the finite dimensional representations separate points of the monoid algebra iff the monoid is residually finite. $\endgroup$ Commented Apr 24, 2017 at 17:46
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1$\begingroup$ thanks for this answer, now I have a better view. Also Peter Schauenburg pointed out to me that the relation $ab\sim 1$ (in the free algebra generated by a and b) do not imply $ba\sim 1$, so the element ba-1 is not zero in $k\{a,b\}/(ba\sim 1)$, but maps as zero to every f.d. representation. $\endgroup$ Commented Apr 25, 2017 at 12:53