Fall 2017
PHY680
Homework 12 Solutions Problem 1 (a) [[Mmn , Kk ], Pl ] + [[Kk , Pl ], Mmn ] + [[Pl , Mmn ], Kk ] = ηnk [Km , Pl ] − ηmk [Kn , Pl ] − [(2Mkl + 2ηkl D), Mmn ] + ηlm [Pn , Kk ] − ηln [Pm , Kk ]
(1.1)
The term with [D, Mmn ] vanishes. The four terms with [K, P ] commutators give four terms with 2D which cancel pairwise, and four terms with 2Mmn which cancel the four terms with 2Mmn which are produced by [−2Mkl , Mmn ]. So we trust the constants in (1.1). (b) The generators Hi = {D, Mmn , Km } form a subalgebra: [Hi , Hj ] = fij k Hk . The decomposition into coset generators Pm , and subgroup generators (D, Mmn , Km ) is not reductive because [Pm , Kn ] lies in Hi . It is symmetric since [Pm , Pn ] lies in Hi (in fact [Pm , Pn ] vanishes). (c) and (d) ea·P ex·P = e(x+a)·P so δ(a)xµ = aµ . This is generated by Pm = ∂x∂ µ . We already explained δ(λD )xµ = −λD xµ so D = −xν ∂/∂xν (because [λD D, xµ ] = λD [−xν ∂/∂xν , xµ ] = −λD xν ). For Mmn we follow the same steps as for D 1
e2λ
mn M
mn
1 mn 1 mn 1 mn ex·P = e 2 λ Mmn ex·P e− 2 λ Mmn e+ 2 λ Mmn
(1.2)
Since [Mmn , Pk ] ∼ P we find for the first factor i h 1 mn 1 m0 n0 exp e 2 λ Mmn (x · P )e− 2 λ Mm0 n0 = exp x · P + 21 λmn [Mmn , Pk ]xk + O(λ2 ) = exp x · P + λm k xk Pm , so δ(λmn )xµ = λµ ν xν (1.3) and this is obtained if we take Mmn = −xm ∂n + xn ∂m (where xm = ηmn xn in Minkowski space) because then [ 12 λmn Mmn , xµ ] = −λmµ xm = λµ ν xν 1
(1.4)
Fall 2017
PHY680 Homework 12 Solutions
For ek·K ex·P the nonreductivity implies that if we follow the same steps by writing ek·K ex·P = exp ek·K (x · P )e−k·K ek·K
(1.5)
then a complications arises: [k · K, x · P ] does not lie in the space of coset generators. In fact we obtain, using [Km , Pn ] = −2Mmn − 2ηmn D exp ek·K (x · P )e−k·K = exp x · P + [k · K, x · P ] + O(k 2 ) = exp x · P − 2k m xn Mmn − 2k · xD + O(k 2 )
(1.6)
We must write this as follows mn M mn +bD
ex·P +δx·P ea
(1.7)
where amn and b are proportional to k m , and everything is to order k (but not k 2 ). To find δxµ (and in the process also amn and b), we use that to linear order in B, 1
1
eA eB = eA+B+ 2 [A,B]+ 3! [A,[A,B]]+···
(1.8)
Then we get for (1.7) to order k 1 mn mn exp x · P + δx · P + a Mmn + bD + [x · P, a Mmn + bD] + · · · 2
(1.9)
We do not need to evaluate the double commutator because [P, M ] and [P, D] are proportional to P , and then the double commutator is of the form 3!1 [x · P, P ] which vanishes. So we get exp x · P + δx · P + amn Mmn + bD + ak n xk Pn + 12 bx · P
(1.10)
Equating this to (1.6) yields 3 relations x · P = x · P + δx · P + ak n xk Pn + 12 bx · P −2k m xn Mmn = amn Mmn −2k · xD = bD
2
(1.11)
Fall 2017
PHY680 Homework 12 Solutions
Hence b = −2k · x and amn = −k m xn + k n xm , and then we get 1 δ(k)xµ = −aν µ xν − bxµ 2 µ ν = kν x x − k µ x2 + k · xxµ = 2k · xxµ − k µ x2
(1.12)
This is a result we obtained earlier in class when we discussed (super)conformal symmetries. Hence k · K = (2k · xxµ − k µ x2 )∂µ
(1.13)
We have almost found a representation of the conformal generators in terms of differential operators acting on xµ . One might guess that it reads Pµ =
∂ , ∂xµ
D = −xν ∂ν ,
Mmn = −xµ ∂ν + xν ∂µ ,
Kµ = (2xµ xν − δµ ν x2 )∂ν
(1.14)
But already [Pµ , D] = −Pµ is off by a sign. Actually: minus these operators form a representation Pµ = −∂µ , D = xν ∂ν ,
Mµν = xµ ∂ν − xν ∂µ ,
Kµ = −2xµ x · ∂ + x2 ∂µ
(1.15)
Now, of course, [Pµ , D] comes out right, namely [Pµ , D] = Pµ , but also all other commutators are correct, for example [Pµ , Kν ] = ∂µ (2xν x · ∂ − x2 ∂ν ) = 2ηµν x · ∂ + 2xν ∂µ − 2xµ ∂ν = −2Mµν + 2ηµν D
(1.16)
However, it is (1.14) and not (1.15) which is used in the theory of induced representations (where these operators are the ordinary Killing vectors, denoted by f µ = M fM µ in class.) Comment 1: The deeper reason for this minus sign follows from the theory of induced representations. (Some) physicists summarize this theory by saying “coordinates transform opposite to fields”. Comment 2: One can use the same approach to superconformal algebras. One starts from 3
Fall 2017
PHY680 Homework 12 Solutions α
the coset elements ex·P +θ Qα with D, Mmn , Km , Sα , and A relegated to the subgroup, but to find δθα for S and K transformations is quite tedious. Comment 3: In field theory, the orbital terms in the transformation laws of fields are given by the differential operators above. The spin terms can also be obtained in a similar way. One begins by writing a (set of) fields ϕ (with indices suppressed) as ϕ(x) = eP ·x ϕ(0)e−P ·x
(1.17)
(For susy, one needs a multiplet of fields.) Then one chooses a matrix representation M ma of the generators which acts on the fields at the origin δ(λ)ϕ(x) = λM ϕ(x) = λM eP ·x ϕ(0)e−P ·x = [λM, eP ·x ]ϕ(0)e−P ·x + eP ·x (λM ma ϕ(0))e−P ·x + eP ·x ϕ(0)[λM, e−P ·x ]
(1.18)
This implies the second term yields the spin terms δϕ(x) = eP ·x λM ma ϕ(0)e−P ·x = λM ma ϕ(x)
(1.19)
Each commutator [x · P, M ] ∼ P yields the orbital terms. See P. van Nieuwenhuizen, Phys. Reports 68B. (e) Pm = ajm5 + bjm6 Km = cjm5 + djm6
D = j56 Mmn = jmn
(1.20)
From [Pm , D] = Pm we learn that a = b [ajm5 + bjm6 , j56 ] = ajm5 + bjm6
(1.21)
ajm6 + bjm6 = ajm5 + bjm6 Similarly [Km , D] yields c = −d. As expected [Mmn , Pk ] and [Mmn , Kk ] give no information.
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Fall 2017
PHY680 Homework 12 Solutions
They are covariant for any values of a and b. We need to satisfy [Pm , Pn ] = 0,
[Pm , Kn ] = −2Mmn + 2ηmn D
[Km , Kn ] = 0;
(1.22)
First [Pm , Pn ]. We get [Pm , Pn ] = [ajm5 + bjm6 , ajn5 + bjn6 ] = (−a2 + b2 )Mmn
(1.23)
So a = ±b. Similarly from [Km , Kn ] = 0 we get c = ±d. Finally [Pm , Kn ] = [ajm5 + bjm6 , cjn5 + djn6 ] = −acjmn − adηmn j56 − bcηmn j65 + bdjmn = −2jmn + 2ηmn j56
(1.24)
So −ac + bd = −2 and
− ad + bc = 2
(1.25)
Since a = +b we get a(c − d) = 2, so c = −d, and then ac = 1, so 1 Km = (jm5 − jm6 ) a
Pm = a(jm5 + jm6 );
(1.26)
The algebra is unaffected if Pm and Kn scale oppositely, so we may set a = 1. Then Pm = jm5 + jm6
;
Km = jm5 − jm6
(1.27)
(f ) Since {Q, Q} ∼ P and [D, P ] = −P , we get [D, Q] = − 12 Q. Since {S, S} ∼ K and [D, K] = +K we get [D, S] = + 12 S. Also [Mmn , Qα ] = 21 λmn (γmn )α β Qβ and idem for S α . For D we get now eλD ex
µP
µ +θ
αQ
α
= exp eλD D (x · P + θ · Q)e−λD D eλD D = exp x · P − λD x · P + θ · Q − 21 λD θ · Q eλD D
(1.28)
So δxµ = −λD xµ as we already found earlier, and δθα = − 12 λD θα . So, 1 ∂ D = −xν ∂ν − θα α 2 ∂θ 5
;
A = γ5 θα
∂ ∂θα
(1.29)
Fall 2017
PHY680 Homework 12 Solutions
(g) The plane (Y 6 )2 − (Y 5 )2 −
3 X
(Y k )2 + (Y 0 )2 = 0
(1.30)
k=1
looks like our anti-de Sitter space with R = 0. Introducing light cone coordinates for the extra space and time coordinate, Y ± = Y 6 ± Y 5 , we can solve for Y − Y− =
Y a ηab Y b Y+
(1.31)
Then we can introduce coordinates xµ , Y + (µ = 0, 1, 2, 3 as usual) which parametrize this space as follows xµ =
Yµ Y+
(µ = 0, 1, 2, 3)
(1.32)
Then (xµ )2 µ (Y , Y , Y ) = Y 1, + 2 , x (Y ) = Y + 1, x2 , xµ with x2 = xµ ηµν xν
(1.33)
wM = (1, x2 , xµ )
(1.34)
+
−
µ
+
The vector
has zero length. To show this write η66 (Y 6 )2 + η55 (Y 5 )2 = −(Y 6 )2 + (Y 5 )2 = −Y + Y − = η+− Y + Y − + η−+ Y − Y +
(1.35)
So η+− = η−+ = −1/2. Then indeed wM ηM N wN = 2w+ η+− w− + wµ ηµν wν = −x2 + x2 = 0
(1.36)
The relation wM wM = 0 implies dwM wM = 0. Then the calculation of (ds)2 becomes very
6
Fall 2017
PHY680 Homework 12 Solutions
simple (ds)2 = dY M dYM = d(Y + wM )ηM N d(Y + wN ) = (Y + )2 dwM ηM N dwN = (Y + )2 dwµ ηµν dwν = (Y + )2 (dxµ )2 ≡ e(dxµ )2
(1.37)
where e = (Y + )2 is an arbitrary function. M
M
M
µ
(h) From e TM ex·P = e(x + fM )Pµ h one finds the Killing vectors fM µ , two of which are given in (1.4) and (1.12). They leave the coset vielbein field eµ m (x) “physically” (i.e., up to an H-gauge transformation) invariant: LfM eµ m = 0. (i) The Killing spinors are solutions of the Killing spinor equation Dα η ± i 2c γα η = 0. One could work this further out.
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