g p! where ω is a p-form. The operator acts on forms, not on components. Example: Consider R 3 with metric +++, i.e. g µν =

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1 Chapter 17 Hodge duality We will ext defie the Hodge star operator. We will defieit i a chart rather tha abstractly. The Hodge star operator, deoted i a -dimesioal maifold is a map from p-forms to ( p)-forms give by ( ω) µ1 µ p ɛ µ1 µ p! g µ p+1ν1 g µνp ω ν1 ν p, (17.1) where ω is a p-form. The operator acts o forms, ot o compoets. Example: Cosider R 3 with metric +++, i.e. g µν = diag(1, 1, 1). The g = 1, g µν diag(1, 1, 1). Write the coordiate basis 1-forms as dx, dy, dz. Their compoets are clearly (dx) i = δ 1 i, (dy) i = δ 2 i, (dz) i = δ 3 i, (17.2) the δ s o the right had sides are Kroeecker deltas. So ( dx) ij = ɛ ijk g kl (dx) l = ɛ ijk g kl δ 1 l = ɛ ijk g k1 dx = 1 2! ( dx) ijdx i dx j = 1 2! ɛ ijkg k1 dx i dx j dx = 1 2! g k1 = 1 for k = 1, 0 otherwise ( dx 2 dx 3 dx 3 dx 2) = dx 2 dx 3 = dy dz. (17.3) Similarly, dy = dz dx, dz = dx dy. 65

2 66 Chapter 17. Hodge duality Example: Cosider p = 0 (scalar), i.e. a 0-form ω i dimesios. ( ω) µ1 µ = ɛ µ1 µ ω ( 1) µ1 µ = ɛ µ1 µ ( 1) = ɛ µ1 µ! dx µ = dv (17.4) Example: p =. The ( ω) = ɛ µ1 µ! g µ 1ν1 g µν ω ν1 ν. (17.5) For the volume form, dv = ɛ µ1 µ! dx µ (dv ) ν1 ν = ( dv ) =! ɛ µ 1 µ g µ 1ν1 g µν =!!(det g) 1 =!! g = sig(g) = ( 1)s,(17.6) where s is the umber of ( 1) i g µν. So we fid that ad ( 1) = dv = ( 1) s, (17.7) ( dv ) = ( 1) s ( 1) = ( 1) s dv, (17.8) i.e., ( ) 2 = ( 1) s o 0-forms ad -forms. I geeral, o a p-form i a -dimesioal maifold with sigature (s, s), it ca be show i the same way that ( ) 2 = ( 1) p( p)+s. (17.9) I particular, i four dimesioal Mikowski space, s = 1, = 4, so ( ) 2 = ( 1) p(4 p)+1. (17.10)

3 67 It is useful to work out the Hodge dual of basis p-forms. Suppose we have a basis p-form dx I 1 dx Ip, where the idices are arraged i icreasig order I p > > I 1. The its compoets are p!δ I 1 µ 1 δµ Ip p. So ( dx I 1 dx Ip) ν 1 ν p = p! p µ 1 µ p g µ 1µ 1 g µ pµ p p! δ I 1 δ Ip µ µ 1 p = p µ 1 µ p g µpip. (17.11) We will use this to calculate ω ω. For a p-form ω, we have ω = 1 p! ω µ 1 µ p dx µp = I dx I 1 dx Ip (17.12) where the sum over I meas a sum over all possible idex sets I = I 1 < < I p, but there is o sum over the idices {I 1,, I p } themselves, i a give idex set the I k are fixed. Usig the dual of basis p forms, ad Eq. (13.13), we get ω = I = I (dx I 1 dx Ip ) ( p)! ɛ ν 1 ν p µ 1 µ p g µpip dx ν p. (17.13) The sum over I is a sum over differet idex sets as before, ad the Greek idices are summed over as usual. Thus we calculate ( p)! p µ 1 µ p g µpip = dx ν p ( ω J1 J p dx J 1 dx Jp) ( p)! p µ 1 µ p g µpip ω J1 J p dx ν p dx J 1 dx Jp (17.14)

4 68 Chapter 17. Hodge duality We see that the set {ν 1,, ν p } caot have ay overlap with the set J = {J 1,, J p }, because of the wedge product. O the other had, {ν 1,, ν p } caot have ay overlap with {µ 1,, µ p } because ɛ is totally atisymmetric i its idices. So the set {µ 1,, µ p } must have the same elemets as the set J = {J 1,, J p }, but they may ot be i the same order. Now cosider the case where the basis is orthogoal, i.e. g µν is diagoal. The g µ ki k = g I ki k etc. ad we ca write ( p)! p I 1 I p g I 1I1 g IpIp ω J1 J p dx ν p dx J 1 dx Jp. (17.15) We see that i each term of the sum, the idices {I 1 I p } must be the same as {J 1 J p } because both sets are totally atisymmetrized with the idices {ν 1 ν p }. Sice both sets are ordered, it follows that we ca replace J by I, ( p)! p I 1 I p g I 1I1 g IpIp = I dx ν p dx Ip ( p)! p I 1 I p I dx ν p dx Ip. (17.16) I each term of this sum, the idices {ν 1 ν p } are completely determied, so we ca replace them by the correspodig ordered set K = K 1 < < K p, which is completely determied by the set I, so that I ɛ K1 K p I 1 I p dx K 1 dx K p dx Ip (17.17). The idices o this ɛ are a permutatio of {1,, }, so ɛ is ±1. But this sig is the same as that for the permutatio to brig the basis to the order dx 1 dx, so the overall sig to get both to

5 69 the stadard order is positive. Thus we get I ɛ 1 dx 1 dx = 1 p! ωµ 1 µ p ω µ1 µ p dx 1 dx = 1 p! ωµ 1 µ p ω µ1 µ p (vol) (17.18) If we are i a basis where the metric is ot diagoal, it is still symmetric. So we ca diagoalize it locally by goig to a appropriate basis, or set of coordiates, at each poit. I this basis, the compoets of ω may be ω µ 1 µ p, so we ca write ( ) 1 p! ωµ 1 µ pω µ 1 µ (vol ) (17.19) p But both factors are ivariat uder a chage of basis. So we ca ow chage back to our earlier basis, ad fid Eq. (17.18) eve whe the metric is ot diagoal. Note that the metric may ot be diagoalizable globally or eve i a exteded regio.