String Theory and M-Theory

154 Strings with space-time supersymmetry
which proves the desired result.
More generally, the same reasoning gives
where we define
¯Θ1Γµ1···µnΘ2 = (−1) n(n+1)/2 ¯ Θ2Γµ1···µnΘ1,
Γµ1µ2···µn = Γ [µ1 Γµ2 · · · Γ µn]
and square brackets denote antisymmetrization of the enclosed indices. ✷
EXERCISE 5.2
Check explicitly that the commutator of two supersymmetry transformations
gives the result claimed in Eq. (5.5).
SOLUTION
Under the supersymmetry transformations in Eqs (5.3) and (5.4) the fermionic
coordinate transformation is δΘA = εA . Therefore, δ1δ2ΘA = δ1εA 2 = 0,
which implies that [δ1, δ2] ΘA = 0. Similarly,
δ1δ2X µ A
= δ1 ¯ε 2 Γ µ Θ A = ¯ε A 2 Γ µ ε A 1 .
As a result,
[δ1, δ2] X µ = ¯ε A 2 Γ µ ε A 1 − ¯ε A 1 Γ µ ε A 2 = −2¯ε A 1 Γ µ ε A 2 ,
where we have used the result of the previous exercise. ✷
EXERCISE 5.3
Show that Π µ
0 , as defined in Eq. (5.6), is invariant under the supersymmetry
transformations in Eqs (5.3) and (5.4).
SOLUTION
From the definition of Π µ
0
it follows that
δ( ˙ X µ − ¯ Θ A Γ µ ˙ Θ A ) = d
dτ (¯εA Γ µ Θ A ) − ¯ε A Γ µ ˙ Θ A − ¯ Θ A Γ µ ˙ε A
= ¯ε A Γ µ ˙ Θ A − ¯ε A Γ µ ˙ Θ A = 0.
EXERCISE 5.4
Derive the equations of motion for X µ and Θ A obtained from the action S1.
✷