### hw3 - Caltech Particle Theory

```Ph 250, Spring 2015: HW 3
1. BBS 5.9
(i) We recall that the field strength is a Lie-algebra-valued two-form, given by
F = dA + A ∧ A,
and changing under a gauge transformation by
δA = dλ + [A, λ],
δF = [F, λ].
Let’s expand the gauge potential in terms of a set of generators T a for the Lie
algebra, chosen to be orthonormal with respect to the Killing form:
tr(T a T b ) = δ ab .
Now, we have that
d tr F ∧ F = 2 tr dF ∧ F = 2 tr [(dA ∧ A − A ∧ dA) ∧ F ] .
We now observe that dF = F ∧ A − A ∧ F , since the term cubic in A cancels out.
As such, we can write
d tr F ∧ F = 2 tr (F ∧ A ∧ F − A ∧ F ∧ F ) .
Expanding this in generators,
d tr F ∧ F = 2(F a ∧ Ab ∧ F c − Aa ∧ F b ∧ F c ) tr T a T b T c
= 2(Ab ∧ F c ∧ F a − Aa ∧ F b ∧ F c ) tr T a T b T c
= 0,
where the last equality follows from the cyclic property of the trace. Gauge invariance follows from a similar calculation:
δ tr F ∧ F = tr([F, λ] ∧ F ) + tr(F ∧ [F, λ])
= λa F b ∧ F c tr [T b , T a ]T c + T b [T c , T a ]
= λa F b ∧ F c tr T b T a T c − T a T b T c + T b T c T a − T b T a T c
= 0.
1
(ii) It is easy to check that
2
dω3 = tr dA ∧ dA + (dA ∧ A ∧ A − A ∧ dA ∧ A + A ∧ A ∧ dA) ,
3
2
dAa ∧ Ab ∧ Ac − Aa ∧ dAb ∧ Ac + Aa ∧ Ab ∧ dAc tr(T a T b T c ),
= tr(dA ∧ dA) +
3
= tr (dA ∧ dA + 2 dA ∧ A ∧ A) ,
where in writing the last line we have used the cyclicity of the trace, and canceled the
negative sign of the middle term by exchanging Aa and Ac . It is trivial to see that
tr(F ∧ F ) is equal to the last line above. One need only check that tr A4 vanishes,
which follows by writing the expression as a sum over Lie-algebra generators, moving
Aa to the back (acquiring a negative sign since it passes through three odd forms),
and then using the cyclic property of the trace.
(iii) It is clear that tr F 4 will have terms of total degree eight (where both d and A carry
degree one). ω7 should therefore have the general form
ω7 = tr A ∧ dA ∧ dA ∧ dA + c2 A3 ∧ dA ∧ dA + c02 A2 ∧ dA ∧ A ∧ dA + c1 A5 ∧ dA + c0 A7 .
The c2 and c02 terms are different because we only have the freedom to cyclically
rearrange the various forms. The derivative of this expression should match
tr F 4 = tr dA4 + 4A2 ∧ dA3 + 4A4 ∧ dA2 + 2A2 ∧ dA ∧ A2 ∧ dA + 4A6 ∧ dA .
Here we have dropped the A8 term because the trace of any even exterior power of
the vector potential must vanish by the argument given for A4 above. It is easy to
see that
d tr(A ∧ A ∧ A ∧ dA ∧ dA) = 2 tr(A ∧ A ∧ (dA)3 ) − tr(A ∧ dA ∧ A ∧ (dA)2 ).
Similarly,
d tr(A ∧ A ∧ dA ∧ A ∧ dA) = 2 tr(A ∧ dA ∧ A ∧ (dA)2 ) + tr(A ∧ A ∧ (dA)3 ).
Therefore, to match these terms, we should have 2c2 + c02 = 4 and 2c02 − c2 = 0. The
solutions are c02 = 4/5, c2 = 8/5. Next, we observe that
d tr(A5 ∧ dA) = tr(2A4 ∧ (dA)2 + A2 ∧ dA ∧ A2 ∧ dA).
It follows that c1 = 2. Lastly,
d tr A7 = 7 tr(A6 ∧ dA),
and so we set c0 = 4/7.
2. BBS 8.2
The equations of motion for 11-dimensional supergravity, as derived from the action given
in the text, are
1
1
1
2
Rµν − Gµν R − |F4 | − Fµ Fν = 0,
2
2
12
µ1 ν1 µ2 ν2 µ3 ν3
where Fµ Fν = G
G
G
Fµµ1 µ2 µ3 Fνν1 ν2 ν3 , and
1
d(?F4 ) + F4 ∧ F4 = 0.
2
2
We now use the given ansatz and check that it satisfies these equations. Since the metric is
block-diagonal, the Ricci scalar is
R = Rµν g µν + Rij g ij = −4M42 + 7M72 .
Here and in the sequel, Greek indices will refer to the four-dimensional factor of the spacetime
and Latin indices to the seven-dimensional factor.
The four-form field strength is given by
where (vol4 )µνρλ =
√
F4 = M · vol4 ,
−gεµνρλ is the Riemannian volume form. From this, we calculate
M2
(−g)g µ1 ν1 · · · g µ4 ν4 µ1 µ2 µ3 µ4 ν1 ν2 ν3 ν4 = −M 2 .
4!
Looking at the ij-components of the equations of motion for the metric, we see that they
1
1
2
Rij − Gij R − |F4 | = 0.
2
2
This reduces to
M2
1
2
2
2
7M7 − 4M4 −
⇒ M 2 + 10M72 = 8M42 .
0 = gij M7 −
2
2
When we consider the µν-components of the equation of motion, the Fµ Fν term will not
vanish. It will be given by
M2
1 µ2 ν2 µ3 ν3 µ4 ν4
M2
(−3!)gµν = +
gµν .
g
− g
g
Fµµ2 µ3 µ4 Fνν2 ν3 ν4 = −
12
12
2
The equation of motion then gives
M2
1 2
1
2
2
2
0 = gµν −M4 −
7M7 − 4M4 −
+ M
⇒ M 2 + M42 = 14M72 .
2
2
2
Solving these two equations simultaneously gives the required values of the parameters:
1
1
M42 = M 2 , M72 = M 2 .
3
6
The equation of motion for the four-form field is satisfied trivially.
|F4 |2 =
3
```