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I encountered the following quiver algebra $k Q/I$ ($k$ algebraically closed with $\textsf{char}(k) = 0$):

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where $I = (\alpha_1^n, \alpha_2^n, (\beta_1\beta_2)^n, (\beta_2\beta_1)^n)$ with $n \geq 1$.

Does anyone see this quiver algebra somewhere? I can show that $A: = kQ/I$ is a Frobenius algebra. What else do we know about this algebra? I want to know when a quotient algebra $A/J$ ($J$ is a two-sided ideal of $A$) is also Frobenius.

Thank you!

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    Out of curiosity, how did you find this algebra in the first place? – 2017-02-14
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    I tried to study to quiver of the algebra $A/(x^{2n}, (y^2-1)^n)$, where $A$ is the skew polynomial ring $k_{-1}[x, y] = k\langle x, y\rangle/(xy + yx)$ and $n \geq 1$. This is a Frobenius algebra.I think the quiver of that algebra is the one I draw. – 2017-02-19
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    The algebra in the question is infinite dimensional as $\alpha_1\beta_1\alpha_2\beta_2$ is not nilpotent. So I think that it cannot be the algebra described in the above comment. – 2018-11-26

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