In Ginzburg's notes Lectures on Noncommutative Geometry, he claim that the path algebra of a quiver is formally smooth (See Section 19.2).

I have two questions. First, how to show this claim and which criterion of formal smoothness we are going to check? Secondly, does this claim hold for the path algebra of a quiver with relations as well?

Any help or hints would be very appreciated.


2 Answers 2


Assuming standard results on lifting idempotents, it's not hard to check that a path algebra $kQ$ satisfies the lifting property that Ginzburg uses to define formal smoothness in Definition 19.1.1.

If $B$ is an algebra with a nilpotent ideal $I$, we need to check that every homomorphism $\varphi:kQ\to B/I$ lifts to a homomorphism $\tilde{\varphi}:kQ\to B$.

Let $\{e_1,\dots,e_n\}$ be the idempotents in $kQ$ corresponding to the vertices of $Q$. Then $\{\varphi(e_1),\dots,\varphi(e_n)\}$ is a set of orthogonal idempotents in $B/I$ whose sum is $1$.

These lift to a set $\{f_1,\dots,f_n\}$ of orthogonal idempotents in $B$ whose sum is $1$.

For each arrow $a_s$ of $Q$ from vertex $i$ to vertex $j$, pick an arbitrary lift $\widetilde{\varphi(a_s)}$ of $\varphi(a_s)$, and let $b_s=f_i\widetilde{\varphi(a_s)}f_j$, so $b_s$ is also a lift of $\varphi(a_s)$.

Then there is a unique map $\tilde{\varphi}:kQ\to B$ with $\tilde{\varphi}(e_i)=f_i$ and $\tilde{\varphi}(a_s)=b_s$ for all $i$ and $s$, and this is a lift of $\varphi$.

This doesn't work for quivers with relations. For example, if $Q$ is an acyclic quiver and $A=kQ/I$ with $I$ a nonzero admissible ideal, then the identity map $A\to kQ/I$ does not lift to a homomorphism $A\to kQ$.


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.

  • 2
    $\begingroup$ The resolution of length one that gives your claim on HH works in the general case of quiver algebras with no relations too. $\endgroup$
    – Pedro
    Jun 17, 2018 at 15:24

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