12.07.2015 Views

Contents

Contents

Contents

SHOW MORE
SHOW LESS
  • No tags were found...

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

34.2 Prolongation structure for the Ernst equation 531As in (10.51) we introduce an as yet unspecified number of new variables,the pseudopotentials y, bydy = F (M i, M ∗i ,y)dζ + G(M i, M ∗i ,y)dζ, (i =2, 3, 4). (34.41)(ζ and ζ do not appear in (34.40) and we therefore neglect the dependenceon them in F and G. We also suppress the index on y, F and G,the last two being vector fields with respect to y.) The following calculationsare somewhat long albeit straightforward and we shall thus onlydescribe them. The integrability condition ddy = 0 yields an equationwhich contains expressions proportional to dM i ∧dζ, dM ∗i ∧dζ, dM ∗i ∧dζand dM i ∧dζ. The former two expressions can be replaced by (34.40)to yield terms proportional to dζ∧ dζ, while the latter two arising from∂ M ∗iF and ∂ M iG cannot be replaced. As these terms have to vanish, weconclude that F is independent of M ∗i while G does not depend on M i .Moreover, the two terms ∂ y F dy∧ dζ +∂ y G dy∧ dζ yield via (34.41) a term(F∂ y G−G∂ y F )dζ∧dζ =[F, G]dζ ∧ dζ. The resulting equation is proportionalto dζ ∧ dζ and is linear in M i and M ∗i . By repeated differentiationit can be shown that all second derivatives of F and G with respect toM i resp. M ∗i commute; it suffices to assume F and G to be linear (fora more general treatment see Finley and McIver (1995)). With (for lateruse we replaced the letter ‘X’ of (10.56) by an ‘A’)F = M 2 A 1 + M 3 A 2 + M 4 A 3 ,G = M ∗2 A 4 + M ∗3 A 5 + M ∗4 A 6 ,(34.42)where the A k s depend on the ys only, we get after sorting with respect toM i and M ∗i the commutator relations[A 1 ,A 4 ]=A 1 − A 4 , [A 1 ,A 6 ]= 1 2 (A 4 − A 1 ), [A 3 ,A 4 ]= 1 2 (A 4 − A 1 ),[A 1 ,A 5 ]=A 5 − A 1 , [A 3 ,A 6 ]=A 6 − A 3 , [A 2 ,A 4 ]=A 4 − A 2 , (34.43)[A 2 ,A 5 ]=A 2 − A 5 , [A 2 ,A 6 ]= 1 2 (A 5 − A 2 ), [A 3 ,A 5 ]= 1 2 (A 5 − A 2 ).The problem is now to determine that free algebra generated by the abovecommutators which still ‘remembers’ the original equations (34.40), cp.the remarks below equation (10.57).In terms of the semidirect product A 1 1⊗V of the loop algebra of SU(1,1)and the Virasoro algebra, i.e.[X i ,Y k ]=Y i+k , [X i ,Z k ]=−Z i+k , [Y i ,Z k ]=2X i+k,[X i, X k ]=0, and Y,Z, [V i, X k ]=kX i+k and Y,Z,[V i ,V k ]=(k − 1)V i+k ,(34.44)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!