Assume all algebras are finite dimensional quiver algebras over a field (no restriction of generaltiy if the field is algebraically closed).



> 
> Let A be a local Frobenius algebra.
> Is A isomorphic to its opposite algebra?
> 
> 
> 


For non-Frobenius algebras this is false, see [Do you know which is the minimal local ring that is not isomorphic to its opposite?](https://mathoverflow.net/questions/376454/do-you-know-which-is-the-minimal-local-ring-that-is-not-isomorphic-to-its-opposi/376461?noredirect=1#comment1069416_376461) (where the current question remained open, see the answer and comment).


Is there an easy example of a (not necessarily local) Frobenius algebra that is not isomorphic to its opposite algebra?



Local Frobenius algebras and their opposite algebras
No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of *arXiv:0709.3029*), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.


Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX+r$ and $f(Y) = bY + s$, where $a,b\in k\setminus\{0\}$ and $r,s\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now
$$f(XY) = f(Y)f(X) = (bY + s)(aX +r) = abq^{-1}XY + t,\quad t\in\mathop{\rm rad}\nolimits^3(A)$$
but also
$$f(XY) = f(qYX) = qf(X)f(Y) = q(aX + r)(bY + s) = qabXY + u,\quad u\in\mathop{\rm rad}\nolimits^3(A)$$


Since $XY\notin\mathop{\rm rad}^3(A^{\rm op})$, this means that
$abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).


So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.


