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Let $A$ be the $K$-algebra defined as the quotient of the non-commutative polynomial ring in variables a,b,c,d,e,f,g,h,z modulo the relations

[ (1)*a^2, (1)*a*b+(1)*b*a, (1)*b^2, (1)*a*c+(1)*c*a, (1)*b*c+(1)*c*b, (1)*c^2, (1)*a*d+(1)*d*a, (1)*c*d+(1)*d*c, (1)*d^2, (1)*a*e+(1)*e*a, (1)*b*e+(1)*e*b, (1)*c*e+(1)*e*c, (1)*d*e+(1)*e*d, (1)*e^2, (1)*a*f+(1)*f*a, (1)*b*d+(1)*b*f+(1)*d*b+(1)*f*b, (1)*c*f+(1)*f*c, (1)*d*f+(1)*f*d, (1)*e*f+(1)*f*e, (1)*f^2, (1)*a*g+(1)*g*a, (1)*b*g+(1)*g*b, (1)*c*g+(1)*g*c, (1)*d*g+(1)*g*d, (1)*e*g+(1)*g*e, (1)*f*g+(1)*g*f, (1)*g^2, (1)*h*g, (1)*a*h+(1)*h*a, (1)*b*h+(1)*h*b, (1)*c*h+(1)*h*c, (1)*d*h+(1)*h*d, (1)*e*h+(1)*h*e, (1)*f*h+(1)*h*f, (1)*g*h, (1)*h^2, (1)*a*z+(1)*b*d+(1)*d*b+(1)*f*b+(-2)*g*a+(1)*h*a+(1)*z*a, (-1)*c*b+(1)*d*z+(1)*g*d+(-2)*h*d+(1)*z*d, (1)*b*z+(1)*d*a+(1)*e*z+( -1)*f*c+(-1)*g*b+(1)*g*e+(-1)*h*b+(1)*h*e+(1)*z*b+(1)*z*e, (-1)*e*a+(1)*f*z+(-1)*g*f+(2)*h*f+(1)*z*f, (1)*c*a+(1)*e*b+(1)*g*z+( 1)*z*g, (1)*e*b+(1)*f*d+(1)*h*z+(1)*z*h, (1)*z^2 ]

Question: Is there a computer algebra system (or modern AI methods) that can compute whether $A$ is finite dimensional and give the vector space dimension+Loewy length of $A$ and its center in case it is finite dimensional?

I tried using GAP/QPA, but it did not finish the computation in a reasonable time. Also Magma could not do it.

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  • $\begingroup$ Perhaps you tried this already, but I would try this inductively. Up to the algebra generated by {a,b,c} with only the relations involving those variables, the answer seems obvious. Then things begin to change, but does the computation finish for {a,b,c,d}? For {a,b,c,d,e}? Then you would at least get a sense of how far out your problem is from being doable. $\endgroup$ Commented Jun 11 at 11:24
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    $\begingroup$ If you quotient out all your variables except $b$, $d$ and $f$, you already get an infinite dimensional algebra. $\endgroup$ Commented Jun 11 at 11:33

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Additionally to the very apt comment of Dave Benson, and addressing your question on software, I computed the beginning of the Hilbert series using the "bergman" program (https://servus.math.su.se/bergman/), and it is $$ 1+9z+38z^2+91z^3+128z^4+110z^5+81z^6+88z^7+116z^8+156z^9+212z^{10} +292z^{11}+404z^{12}+564z^{13}+788z^{14}+1108z^{15}+O(z^{16}), $$ which suggests that the algebra is very likely infinite-dimensional.

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    $\begingroup$ Thanks. I have some reasons to believe that the algebra I intended to look at is finite.-dimensional, but probably I made some mistakes in the relations. But the main question is answered and I think this bergman program will probably also be helpful after I found the mistake in the relations. $\endgroup$ Commented Jun 11 at 11:57
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    $\begingroup$ @Mare alright ! Let me know if you need to run bergman on some other relations (I have been using it since 2006 so I have a reasonable grasp of it, but I know that some people have been having trouble installing or running it). $\endgroup$ Commented Jun 11 at 12:17
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    $\begingroup$ @Mare If you have a genuinely finite dimensional algebra, Magma should be able to tell you so very quickly unless either the dimension is very large or there is very distant collapse. $\endgroup$ Commented Jun 11 at 16:56
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Factor out the variables $a$ and $b$ in addition to the given relations. Then the ideal generated by these elements have a finite Groebner basis with the following tips:

gap> List(gb!.relations,Tip); [ (1)*a, (1)*b, (1)*c^2, (1)*d*c, (1)*d^2, (1)*e*c, (1)*e*d, (1)*e^2, (1)*f*c, (1)*f*d, (1)*f*e, (1)*f^2, (1)*g*c, (1)*g*d, (1)*g*e, (1)*g*f, (1)*g^2, (1)*g*h, (1)*h*c, (1)*h*d, (1)*h*e, (1)*h*f, (1)*h*g, (1)*h^2, (1)*z*d, (1)*z*e, (1)*z*f, (1)*z*g, (1)*z*h, (1)*z^2 ]

Here we see that, for instance, that $z*c$ or $c*z$ don't occur as a tip of any Groebner basis element. The elements $(z*c)^n$ will not be reduce by the Groebner basis for any $n\geq 1$. They will be non-zero and all different in the quotient and therefore also in the original algebra $A$.

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  • $\begingroup$ So this proves that the algebra is indeed infinite dimensional. Nice! $\endgroup$ Commented Jun 15 at 11:50

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