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:


1. All paths of length greater than 3.
2. All paths of length 3, except $yh\_1x$ and $xh\_2y$, and $yh\_1x+xh\_2y$.
3. 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,x*y,y*x,h\_1*h\_2,h\_2*h\_1,x*h\_1*x,x*h\_1*y,y*h\_1*y,x*h\_2*x,y*h\_2*x,y*h\_2*y,y*h\_1*x+x*h\_2*y,h\_1*x*h\_1,h\_1*x*h\_2,h\_2*x*h\_1,h\_2*x*h\_2,h\_1*y*h\_1,h\_1*y*h\_2,h\_2*y*h\_1,h\_2*y*h\_2];  
> 
> gb := GBNPGroebnerBasis(relations,R);  
> 
> I:=Ideal(R,gb);  
> 
> GroebnerBasis(I,gb);  
> 
> A:=R/I;  
> 
> M := AlgebraAsModuleOverEnvelopingAlgebra(A);  
> 
> HH2 := ExtOverAlgebra(NthSyzygy(M, 1), M);
> 
> 
> 



Checking if Hochschild cohomology $\mathit{HH}^2(A)=0$
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.)


