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There is a function μ: ℕ → ℕ, such that any Lie algebra, all of whose subalgebras are n-step subideals, is nilpotent of class ≤ μ(n).[1] It is defined by:

  • μ₁(c,d) = cd + (c-1)(d-1)
  • μ₂(r,s) = s + s² + … + sʳ (equal to s(sʳ-1)/(s-1) for s>1, and to r for s=1)
  • μ₃(n) = μ₂(n²,n) (equal to n(n-1)/(n-1) for n>1, and to 1 for n=1)
  • μ₄(n,1) = 1
  • μ₄(n,2) = μ₃(n)
  • μ₄(n,d) = ab + (a-1)(b-1), where a = μ₃(n) and b = μ₄(n,d-1)
  • μ₅(n) = n - 1 + μ₂(n,n) (equal to n - 1 + n(nⁿ-1)/(n-1) for n>1, and to 1 for n=1)
  • μ₆(n) = n μ₅(n) (equal to n² - n + n²(nⁿ-1)/(n-1) for n>1, and to 1 for n=1)
  • μ(1) = 1
  • μ(n) = μ₄(n, μ₆(1 + μ₄(n, (n-1) μ(n-1))))

By induction on d, it can be shown that μ₄(n,d) = ((2μ₃(n)-1)ᵈ⁻¹+1)/2 (equal to ((2n(n-1)/(n-1)-1)ᵈ⁻¹+1)/2 for n>1, and to 1 for n=1).

Analysis

With μ₃(2) = μ₂(4, 2) = 2(2⁴-1)/(2-1) = 30, we have μ(2) = μ₄(2, μ₆(1 + μ₄(2, μ(1)))) = μ(2) = μ₄(2, μ₆(1 + μ₄(2, 1))) = μ₄(2, μ₆(2)) = μ₄(2, 14) = ((2μ₃(2)-1)¹³+1)/2 = (59¹³+1)/2 = 52,486,323,838,066,215,147,990.

For large n, log μ₃(n) is from the order n2+ε. This implies that log μ₄(n,d) is from the order n2+εd. Furthermore, log μ₆(n) is from the order n1+ε. Therefore, μ(n) can be approximated as exp³(n³μ(n-1)) for n>2. So μ(n) ≈ exp³ⁿ⁻⁶(10²⁴) for n>2.

Sources

  1. Stewart, Ian. Subideals of Lie algebras. Retrieved 2016-10-16.
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