2 [2me(0++), me(−++)] + [2me(0++), −e(0+)e(−+)]
+ h
2me(0++), −α
2he(0+)i
+ [−e(0+)e(0+), me(−++)]
+ [−e(0+)e(0+), −e(0+)e(−+)] + h
−e(0+)e(0+), −α
2he(0+)i
!
= −
√ 2
−6αm2e(0+++) − 2αe(0+)e(0+)e(0+) +5αme(0+)αe(0++) + αme(0++)e(0+)
= −
√ 2α
−6m2e(0+++) + 6me(0+)e(0++)
−h[e(0+),e(0++)] − 2e(0+)e(0+)e(0+)
=2
√
2α 3m2e(0+++) − 3me(0+)e(0++) + e(0+)e(0+)e(0+) .
Note that the quantum correction is zero again because the underlined bracket vanishes.
3.4 Leading terms of exceptional elements
After the above review of the basic background of quantization, this section introduces some new results on the leading terms of some exceptional elements. As it was mentioned before, an exceptional elementLlgin the Poisson algebra of invariantshcannot be generated from invariantsh,h0of lower degree because if it could be generated that way, the leading term of leadLlgcould be written as a Lie bracket of leading terms of invariantshandh0, which leads to a contradiction [PR86, p. 622]. While the argument itself does not exclude the possibility that leadLlgcould be written as a Lie bracket of anytwo Euler-Lyndon elements, it leaves open the question whether this is possible. If the answer were no, thengwould not be finitely generated because then there would be an infinite number of elements leadLlgthat could not be rewritten as Lie brackets.
Surprisingly, e(−000+) = leadL3g, was found in quantum corrections in the stratum ˆh5 of the quantum algebra of invariants. A calculation indicating this was first performed by Meusburgerin the course of herDiplomarbeitand later verified in theDiplomarbeit[Han09].
Since the quantum correction was calculated using the Pohlmeyer-Rehren Lie algebrag, this points towards a positive answer to the question given above.
Example 3.4.1. In order to understand the occurrence of leadL3gbetter, we calculate the contributions of the quantum lifting [[ˆL2,Lˆ1],[ˆL−2,Lˆ−1]]+ of the invariant {L2,L1}{L−2,L−1} to leadL3g. The term “contribution” to leadL3gis a shorthand for the projection onto the subspace of ˆhspanned by elements that have a factor of an Euler word with the same letter content as leadL3g, i.e. three zeroes and one plus and minus each. In example 3.3.11, we calculated
[ˆL2,Lˆ1] =2
√
2α 3m2e(0+++) − 3me(0+)e(0++) + e(0+)e(0+)e(0+) .
Analogously, we calculate [ˆL−2,Lˆ−1] =2
√
2α 3m2e(0−−−) − 3me(0−)e(0−−) + e(0−)e(0−)e(0−) .
The anticommutator of these terms can be rewritten using linearity as a sum of nine anti-commutators of the individual summands.
• The anticommutator 72α2m4[e(0+++),e(0−−−)]+ has Euler elements with two zeroes total, so its quantum correction cannot contribute to leadL3g.
• First note that the Euler words in the anticommutator 72α2m3[e(0+++),e(0−)e(0−−)]+ contain only three zeroes in total. So only the [·,·]± part of the decomposition [·,·]=[·,·]±+[·,·]0of the bracket obtained from reordering the noncommutative fac-tors can contribute to leadL3gand will henceforth be considered because [·, ·]0reduces the number of zeroes. Now, using Leibniz’s rule on the bracket, we obtain
36α2m3[e(0+++),e(0−)e(0−−)]±
=36α2m3([e(0+++),e(0−)]±e(0−−) + e(0−)[e(0+++),e(0−−)]±) .
Observe that the bracket in the right hand summand contains only two zeroes and therefore cannot contribute to leadL3g. The remaining term is
36α2m3[e(0+++),e(0−)]±e(0−−)
= −36α3m3he(00++)e(0−−)
= −36α3m3h(e(0−−)e(00++)+[e(0−−),e(00++)]) .
The first summand is in ascending order again, and the quantum correction stems from the second summand. Because only three zeroes occur, its contribution to leadL3gis contained in
−36α3m3h[e(0−−),e(00++)]±=36α3m3h2(e(000−+)+e(000+−)+e(00+0−)).
• Analogously, the contribution of the quantum correction of the anticommutator 72α2m3[e(0+)e(0++),e(0−−−)]+to leadL3gis
36α3m3h2(e(000−+)+e(000+−)+e(00−0+)).
• Applying Leibniz’s rule to brackets arising in the reordering of the remaining six anti-commutators produces Euler elements that cannot contribute to leadL3g∈g3, because they are of degreel≤2.
We have shown that the sum of the contributions to leadL3gfrom all anticommutators is 36α3m3h2(2e(000−+)+2e(000+−)+e(00+0−)+e(00−0+))=36α3m3h2e(−000+), which actually is a multiple of leadL3g.
A question suggesting itself in the context of the discovery of leadL3gin a quantum correction is if leading terms of other exceptional elements can also be found in quantum corrections of elements of higher degree. Being the only known obstacle, this would be a necessary condition for finite generation of the quantum algebra of invariants ˆh.
While a general statement for all Llgcould not be reached, we do have some tentative results that lead to a general conjecture.
Conjecture 3.4.2. The quantum corrections of the following terms contain the leading term of the exceptional elementsLlgfor alll=2n+3, n∈N0:
{. . .{{L−2,L−1},L0}, . . . ,L0}
| {z }
n
· {. . .{{L2,L1},L0}, . . . ,L0}
| {z }
n
∈h2n+5 . (3.44)
This conjecture is supported by two observations:
1. The conjecture has been verified inMathematicaup tol=9. For higherl, the compu-tation was not possible due to lack of RAM.
2. In the rest frame, the Li are sums of terms of the form me(abc) ande(ab)e(cd), where a,b,c,d ∈ X and by linearity, the product above can be expanded into summands that inherit only one of those terms from each Li. The sum can then be developed into a linear combination of linearly independent vectors including leadLlg. Finding the coefficient of leadLlgis a straightforward but laborious calculation (in essence a generalized version of example 3.4.1) described in appendix C. A closed term for the coefficient could be found, but being composed of a large number of summations, binomial coefficients, powers of−1 and Kronecker delta functions all with arguments depending on each other, it was so unwieldy that it resisted attempts to be proven to be nonzero in the case of arbitraryl.
We now leave the question of finding leading terms of exceptional elements in quantum corrections in ˆhand conclude the chapter with a positive answer to the question stated at the beginning of this section if is possible to generate the leading term of each of the exceptional elementsLlging.
Proposition 3.4.3. The leading terms of the exceptional elements Llgwith l =2n+1and n ∈ N+ can be generated the following way:
leadLlg= e(−0l+2+) = −1
2l[e(0−0+),e(−0l+)] + z, (3.45) where z is a linear combination of Euler-Lyndon elements with l zeroes.
Proof. Because of the construction of the Lie algebras [·, ·]±and [·,·]0in proposition 1.8.1, [e(0−0+),e(−0l+)] = [e(0−0+), e(−0l+)]±+[e(0−0+),e(−0l+)]0.
We begin by noting that the latter summand, which we call 2lz, is a linear combination of Euler-Lyndon elements withlzeroes because the Lie bracket [·,·]0diminishes the number of zeroes by two.
Considering the fact thatlis odd, we calculate [e(0−0+),e(−0l+)]±
= −2·
(−1)4+4+l+2e
+00 0l−
+(−1)4+4+1e(0−0l+1+)
=2·
e
0+0l+1− +2e
+0l+2−
+(−1)l+2+2−2e
0 +0l+1−
=2·(e
0+0l+1− +2e
+0l+2−
−e
0+0l+1−
−(l+2)e
+0l+2− )
= −2l·e
+0l+2−
= −2l·e
−0l+2+ .
It should be noted that this result seems to point in the direction of finite generation of g, but it does not prove it because neither are the elements considered the only exceptional elements inhnor is it clear if another infinite series of elements ofgexists the entries of which cannot be written as a Lie bracket of elements ofgof lower degree.
Computational Applications of
Hall-Bases in Non-Free Graded Lie Algebras
4.1 Motivation
This chapter is motivated by two related problems that arise when studying the Pohlmeyer-Rehren Lie algebrag: the search forgeneratorsand a systematic exploration of therelations between the multiple Lie brackets of those generators. We know the structure ofgand its strataglas vector spaces and can calculate the Lie bracket on a basis ofgby using formula (1.54). We don’t however have a way to calculate the relations between multiple Lie brackets of some set of possible generators directly, i.e. without reference to this basis. Regarding generators ofg, we have the following conjectures1:
Conjecture 4.1.1. gis finitely generated.
More specifically, ford≥3 (in contrast to proposition 1.5.3)
Conjecture 4.1.2. The subalgebrag≥1of the Pohlmeyer-Rehren Lie algebragis generated by a basis of its first stratum,g1.
Since the entirety ofg1is the direct sum of its irreducibleg0-modules that in turn are the orbits of their respective highest weight vectors under the action ofg0 (a direct application of theorem 2.3.3:1), the following conjecture would imply 4.1.2 although the number of generators is lower:
Conjecture 4.1.3. g≥0is generated by generators ofg0and the highest weight vectors ing1. Computer algebra can be a useful tool insofar as it can be used to falsify specific candidates for sets of generators. Since the number of elements of g that can be checked by direct computation for linear dependence from multiple brackets is necessarily finite (and this set of elements trivially generates itself), we cannot hope to falsify conjecture 4.1.1 this way. It could, however, be possible to find counterexamples to conjectures 4.1.2 and 4.1.3 this way.
1Note that the restriction tog≥1org≥0is mostly cosmetic as the stratumg−1is central.