Checking if Hochschild cohomology $\mathit{HH}^2(A)=0$ I am trying to compute the Hochschild cohomology of a particular bound quiver path algebra. The quiver $Q$ consists of one vertex and four loops $x,y, h_1,h_2$, and the relations $I$ are generated by:

*

*All paths of length greater than 3.

*All paths of length 3, except $yh_1x$ and $xh_2y$, and $yh_1x+xh_2y$.

*All paths of length 2, except $yh_1, h_1x, xh_2,h_2y$.

Basically, in this algebra I have $yh_1x=-xh_2y$, and only other nonzero paths are the subpaths of these 2.
I am interested in $\mathit{HH}^2(kQ/I)$. More specifically, I am interested in whether $\mathit{HH}^2(kQ/I)=0$ for some infinite field $k$. I couldn't find or come up with a direct way of computing it, and my attempt using GAP's QPA package ran into memory problems.  So I was wandering what are the tractable ways to compute this cohomology or prove that is zero or non-zero, either on paper or using computer algebra.
GAP code:

LoadPackage("qpa");
Q := Quiver(1, [[1,1,"x"],[1,1,"y"],[1,1,"h_1"],[1,1,"h_2"]]);
R := PathAlgebra(Rationals,Q);
gens:= GeneratorsOfAlgebra(R);
x:=gens[2];
y:=gens[3];
h_1:=gens[4];
h_2:=gens[5];
relations :=[x^2,y^2,h_1^2,h_2^2,xy,yx,h_1h_2,h_2h_1,xh_1x,xh_1y,yh_1y,xh_2x,yh_2x,yh_2y,yh_1x+xh_2y,h_1xh_1,h_1xh_2,h_2xh_1,h_2xh_2,h_1yh_1,h_1yh_2,h_2yh_1,h_2yh_2];
gb := GBNPGroebnerBasis(relations,R);
I:=Ideal(R,gb);
GroebnerBasis(I,gb);
A:=R/I;
M := AlgebraAsModuleOverEnvelopingAlgebra(A);
HH2 := ExtOverAlgebra(NthSyzygy(M, 1), M);

 A: I believe that there is a 2-dimensional cocycle $g$ such that:
$$
g(h_1 x \otimes h_2) = g(h_1 \otimes x h_2) = y h_1 x
$$
and $g(a \otimes b) = 0$ for all other paths $a$ and $b$.
To check that it's a cocycle, we have to verify that for all paths $a$, $b$, $c$, we have
$$
a g(b \otimes c) - g(a b \otimes c) + g(a \otimes bc) - g(a \otimes b)c = 0.
$$
If $a=1$ or $c=1$ this is true; if $a \neq 1$ and $c \neq 1$ then the value of $g$ is length-3 and so it's killed by any product, and so we just need to verify
$$
g(ab \otimes c) = g(a \otimes bc).
$$
This is automatic if $b=1$, and both sides are zero if $a$, $b$, or $c$ are length greater than 1. This identity just needs to be checked when $a$, $b$, and $c$ are paths of length $1$, where it is straightforward.
Finally, to verify that it's nonzero in Hochschild cohomology we need to verify that there is no function $f$ such that
$$
g(a \otimes b) = a f(b) - f(ab) + f(a)b.
$$
If we apply this to $h_1 x \otimes h_2$ we find we need
$$
h_1 x f(h_2) + f(h_1 x) h_2 = y h_1 x = -x h_2 y \notin I
$$
but multiplication on the left by $h_1$ or on the right by $h_2$ sends all length 2 paths in $kQ$ into the ideal $I$.
(The second Hochschild cohomology group often "detects primitive relations". This cocycle is detecting that the length-3 path $h_1 x h_2$ is zero and that this is not a consequence on the length-2 relations.)
