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During a classification problem I came across a set of algebras given as the path algebra of a quiver with relations. As an example the local ones: $k\langle x,y\rangle/x^2,y^2, xy-qyx$, where $q\in k$, sometimes called truncated quantum plane.

My question is whether there are necessary/sufficient criteria for (these) algebras to be a block of a Hopf algebra (up to Morita equivalence). For example, $q$ obviously has to be a root of unity, because otherwise the Nakayama automorphism will not be of finite order. A sufficient criteria is $q=1$, but I don't know of other instances. (Maybe it also depends on the characteristic of the underlying field).

Weaker results would also be of interest to me, for example [Farnsteiner: Polyhedral groups, McKay quivers, and the finite algebraic groups with tame principal blocks, Proposition 7.4.3] shows that they will not be the principal block of a cocommutative Hopf algebra if $char k \geq 3$.

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If $B$ is a block, then shouldn't $Ext_B^*(k,k)$ be graded commutative? In the particular example you gave, it looks like that would force $q$ to be $-1$, not 1. But I may be completely misunderstanding... –  John Palmieri Sep 2 '11 at 15:16
    
I haven't checked your argument about graded commutativity yet. Let me give the example, where to find, that $k[x,y]/x^2,y^2$ is the block of a Hopf algebra. In [Xiao: Finite dimensional representations of $U_t(sl(2))$ at roots of unity] the example $k[a,b]/a^2-b^2,ab$ is given, which is isomorphic to the above stated algebra via the isomorphism $a+ib\mapsto x, a-ib\mapsto y$. –  Julian Kuelshammer Sep 2 '11 at 19:51
    
Could you provide references for your arguments? Why is the Ext-algebra graded commutative if $B$ is not the principal block? Is it calculated somewhere in this particular example? –  Julian Kuelshammer Sep 5 '11 at 9:26
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