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Removed part that didn't make too much sense.
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pbelmans
pbelmans
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Let me answer this for acyclic quivers, if there are cycles it should be the same, but to be on the safe side let me not claim it in that generality.

The path algebra is hereditary (i.e. of global dimension 1), from which it follows that the Hochschild cohomology vanishes above the global dimension. Hence $\operatorname{HH}^2(A,M)=0$.

If there are relations, the global dimension can jump up. It is still possible that the Hochschild dimension is 1, as Hochschild cohomology is a derived invariant, and global dimension is not: one can for instance take a quiver without relations and consider its path algebra, and find a derived equivalent path algebra with relations. But only in this very special case will $\mathrm{HH}^2(A,M)=0$ups.

Let me answer this for acyclic quivers, if there are cycles it should be the same, but to be on the safe side let me not claim it in that generality.

The path algebra is hereditary (i.e. of global dimension 1), from which it follows that the Hochschild cohomology vanishes above the global dimension. Hence $\operatorname{HH}^2(A,M)=0$.

If there are relations, the global dimension can jump up. It is still possible that the Hochschild dimension is 1, as Hochschild cohomology is a derived invariant, and global dimension is not: one can for instance take a quiver without relations and consider its path algebra, and find a derived equivalent path algebra with relations. But only in this very special case will $\mathrm{HH}^2(A,M)=0$.

Let me answer this for acyclic quivers, if there are cycles it should be the same, but to be on the safe side let me not claim it in that generality.

The path algebra is hereditary (i.e. of global dimension 1), from which it follows that the Hochschild cohomology vanishes above the global dimension. Hence $\operatorname{HH}^2(A,M)=0$.

If there are relations, the global dimension jump ups.

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pbelmans
pbelmans
  • 1.5k
  • 1
  • 16
  • 26

Let me answer this for acyclic quivers, if there are cycles it should be the same, but to be on the safe side let me not claim it in that generality.

The path algebra is hereditary (i.e. of global dimension 1), from which it follows that the Hochschild cohomology vanishes above the global dimension. Hence $\operatorname{HH}^2(A,M)=0$.

If there are relations, the global dimension can jump up. It is still possible that the Hochschild dimension is 1, as Hochschild cohomology is a derived invariant, and global dimension is not: one can for instance take a quiver without relations and consider its path algebra, and find a derived equivalent path algebra with relations. But only in this very special case will $\mathrm{HH}^2(A,M)=0$.