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B Auxiliary Tools for p-Brane Actions																				415
and nice 4 × 4 matrices. Explicitly we get:													0		0 1 0
γ01	0 −1					0		0				γ02	0		0 1 0
	=	−1		0		0		0		;			=	0		0	0	1
	=	0		0 1				0		;			=	0	1 0 0
		0		0 0				1						1	0 0 0

γ03		0	0		0	−i						γ12		0			0	1	0	(A.5.3)
		0	0		−i		0							0			0	0	1
	= i		0		0		0	;					=	0 −1 0 0
		0 i			0		0								1 0 0 0
														−
γ13	0			0		0 −i						γ23	−i				0	0 0
	=	0		0		−i			0		;		=	0			−i	0	0
	=	−i		0		0			0		;		=	0			0 0 i
		0		−	i	0			0					0			0 i		0
				−

Let us mention some relevant formulae that are easily veriﬁed in the above basis:

and if we ﬁx the convention:	γ0γ1γ2γ3 = iγ5	(A.5.4)

we obtain:	ε0123 = +1	(A.5.5)

1	εabcdγa γbγcγd = −iγ5	(A.5.6)

24

Appendix B: Auxiliary Tools for p-Brane Actions

In this appendix we collect some auxiliary calculations and algebraic tools relevant to the discussion of p-brane world volume actions presented in Chap. 7.

B.1 Notations and Conventions

General adopted notations for ﬁrst order world volume actions are the following ones:

d = dimension of the world-volume Wd

D = dimension of the bulk space-time MD

V a = vielbein 1-form of bulk space-time

416	10 Conclusion of Volume 2
Πia = D × d matrix. 0-form auxiliary ﬁeld	(B.1.1)

hij = d × dsymmetric matrix. 0-form auxiliary ﬁeld e = vielbein 1-form of the world-volume

ηab = diag{+, −, . . . , −} = ﬂat metric on the bulk

D−1 times

ηij = diag{+, −, . . . , −} = ﬂat metric on the world-volume

		−
d			1 times

B.2 The κ -Supersymmetry Projector for D3-Branes

In this appendix we present the derivation of the κ-supersymmetry projector utilized in Sect. 7.5 to establish the κ-susy invariance of the D3-brane action. In particular we refer to (7.5.23).

Let us begin with property (a) and consider the ansatz in (7.5.26). By direct calculation we ﬁnd:

ω[24] = α42(4!)2

(B.2.1)

(α2)2ω[0]ω[4]

ω[2] =

+ 8(α2)

Tr F

α0α4

so that we get:

(4 α4)2

ω[4]

(α

Γ 2

ω[0]

8(α2)2 Tr F

(B.2.2)

N 2

4 +

so we obtain Γ 2 = 1 if the normalization factor N is chosen as in (7.5.27) and if the coefﬁcients are chosen as in (7.5.28). This conclusion is easily reached using the identity (7.5.30) of the main text.

Let us now turn to property (b), namely to the condition

	Γ Ak = Ak								(B.2.3)
To implement it we need to calculate some γ matrix products:
	1		γ˜k
ω[4]γk =
ω[4]γk =		6
ω[4]γ˜k = 6γk									(B.2.4)
	= −			1			1		(B.2.4)
				1			1
				1 7			≡ −1
ω[4]Πk					F ij γij k				k
ω[4]Πk				2	F ij γij k		2		k
ω[4]Πk = −					ij	γij k	≡ −
					F7				k
				2	F7			2	k

B Auxiliary Tools for p-Brane Actions					417
	i
ω[2]γk =			k + iΠk
ω[2]γk =	2		k + iΠk
ω[2]γ˜k = −3iΔk − 6iΠk
		i			(B.2.5)
ω[2]Πk = −		i		F 2 kl γ˜ l − iω[0]γk

		6
ω[2]Πk = i F				7	i
ω[2]Πk = i F				2 kl γ l +	6 ω[0]γ˜k
	7
Now we impose (B.2.3) and we obtain the following equations.
• The contributions from k and	k are:

i	+ 3f2	σ1 = 0
mhmk 2 f4

(B.2.6)

i i

k− 2 f3 + 2 f1 σ2 = 0

•For the contributions with γk we have two equations, one proportional to σ3 and one proportional to 1, namely:

γk ω[0](f1 − f3) σ3 = 0

and

1 6f214×4 + if4F72 h 12×2 = if114×4 12×2

For:
			6f2 = if1
			if1 = −if4
and using the property (7.5.31) we obtain:
				N −1 1 − F				2 h = 1
N	1	1	− F	2	N	1	7		2	= 1
N	−	1	− F		N	−	1 − F			= 1
			7					7
	1 − F 2			− F		2 + F 2F			2	= N 2
			7		7		7	7

1 1

28 ij Fkl εij kl 1 = N 2

•For the contributions with γ˜k we get the following equations:

γ˜mhmk ω[0] f2 +

12×2

= 0

−1γm 6 f1δmk

6 f3 F 2 mk

σ3

f2γmhmk

σ3

−

= ˜

(B.2.7)

(B.2.8)

(B.2.9)

(B.2.10)

(B.2.11)

418				10 Conclusion of Volume 2
Then if:
f2 = −		i
			f4
		6	f4
f1	= f3			(B.2.12)
f1	= 6f2

we obtain that the second of equations (B.2.11) as a matrix equation becomes:

N	−1 1 − F 2 = h	(B.2.13)
	7

and just coincides with the solution (7.4.51) for the auxiliary ﬁeld h in terms of the physical ones.

• Now we consider Π and Π .

The equation proportional to σ1 is:
			αω[0]Π mhmk + βΠ mhmk = N γ Π k																	(B.2.14)
		αω[0]F7lmhmk γl + βF7lmhmk γl = γ N F7lk γl																		(B.2.14)
		αω[0]F7lmhmk γl + βF7lmhmk γl = γ N F7lk γl
in matrix form we have:					αω[0](F7h) + β(F7h) = γ N F7
					αω[0](F7h) + β(F7h) = γ N F7
αω[0]F	1 − F 2 N −1 + βF 2 1 − F 2 N −1											= γ N F
7	7				7			7					7
	αω[0]F	1 − F		2 + βF			2 1 − F 2					= γ N 2F								(B.2.15)
	7		7	2		7	2	1	7	2			γ 1	7	1		Tr F	2		ω20 F
	αω[0]F 1		F	2		βF	2	1	F	2			γ 1			2	Tr F	2		ω20 F
	7		− 7		+	7		−	7			=		−	γ		7		+	[ ] 7
	αω[0]F − αω[0](F F )F + βF − βF									2F =			γ F −		γ		Tr F 2 F + γ ω[20]F

															2
if:	7		77 7				7		7		7		7				7		7	7
if:
							β = γ													(B.2.16)

and using (7.5.31), than (B.2.15) become:

αω[0]F + αω[20]F − βF			2F = −	γ	Tr F	2 F + γ ω[20]F (B.2.17)

				2
7	7	7 7			7	7	7
if:
			α = γ				(B.2.18)

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