Images of Golod-Shafarevich algebras with small growth. (Q2922430)

The authors consider so called Golod-Shafarevich algebras named after famous papers: [\textit{E. S. Golod}, Transl., Ser. 2, Am. Math. Soc. 48, 103-106 (1965); translation from Izv. Akad. Nauk SSSR, Ser. Mat. 28, 273-276 (1964; Zbl 0215.39202)], [\textit{E. S. Golod} and \textit{I. R. Shafarevich}, Izv. Akad. Nauk SSSR, Ser. Mat. 28, 261-272 (1964; Zbl 0136.02602)].NEWLINENEWLINE The authors prove the following results.NEWLINENEWLINE Theorem A (Golod-Shafarevich algebras). Let \(K\) be an algebraically closed field and let \(A=K\langle x,y\rangle\) be the free associative non-commutative algebra generated (in degree 1) by elements \(x,y\). Let \(d\geq 50\) be given; let a sequence \((D(k))_{k\in\mathbb N}\) of subspaces of \(A\) be given, with \(D(k)\) homogeneous of degree \(k\), such that (A.1) \(D(k)=0\) if \(k/(\log k)^2<200d^2\); (A.2) \(D(k)=0\) if \(2^n-2^{n-3}<k\leq 2^n+2^{n-1}\) for some \(n\in\mathbb N\); (A.3) \(D(k)=0\) if there exist \(j,n\in\mathbb N\) such that \(D(j)\neq 0\) and \(j<2^n<k<\max\{j^{1000},12dnj\}\); (A.4) \(\dim D(k)\leq k^d\) for all \(k\in\mathbb N\). Then the factor algebra \(A/\langle D(k):k\in\mathbb N\rangle\) can be homomorphically mapped onto an infinite-dimensional algebra with Gelfand-Kirillov dimension at most \(45d\).NEWLINENEWLINE Theorem B (Finitely presented Golod-Shafarevich algebras). With the same assumptions and notation as in Theorem A, it is assumed that \(D(k)=0\) for almost all \(k\). Then \(A/\langle D(k):k\in\mathbb N\rangle\) can be mapped onto an infinite-dimensional algebra with at most quadratic growth.NEWLINENEWLINE The authors also construct algebras with prescribed growth functions as follows.NEWLINENEWLINE Theorem C. Let \(f\colon\mathbb N\to\mathbb N\) be a submultiplicative and increasing function, that is, \(f(m+n)\leq f(m)f(n)\) for all \(m,n\) and \(f(n+1)\geq f(n)\). Then there exists a finitely generated algebra \(B\) whose growth function \(\nu(n)\) satisfies \(f(2^n)\leq\dim B(2^n)\leq 2^{2n+3}f(2^{n+1})\). Furthermore, \(B\) may be chosen to be a monomial algebra.NEWLINENEWLINE The hypotheses are satisfied by ``any sufficiently regular functions'' that grow at least as fast as \(n^{\log n}\).