String Theory and M-Theory

182 Strings with space-time supersymmetry
Finally, the local counterterms that complete the anomaly analysis have the
structure
i B (i) (i)
∧ X 8 , where the integral is over the ith boundary. The
field B (i) is obtained from the M-theory three-form field Aµνρ by setting
one index equal to 11 (the compact direction) and restricting to the ith
boundary.
EXERCISES
EXERCISE 5.9
Let us consider supergravity theories in six dimensions with N = 1 supersymmetry.
Let us further assume that the minimal supergravity multiplet is
coupled to a tensor multiplet as well as nH hypermultiplets and nV vector
multiplets. Show that a necessary condition for anomaly cancellation is
SOLUTION
nH − nV = 244.
The fields of the gravity and tensor multiplets combine to give a graviton
gµν, a two-form Bµν, a scalar, a left-handed gravitino and a right-handed
dilatino. The reason for combining these two multiplets is that one of them
gives the self-dual part of H = dB and the other gives the anti-self-dual part.
A vector multiplet contains a vector gauge field and a left-handed gaugino.
A hypermultiplet contains four scalars and a right-handed hyperino. Therefore,
the total purely gravitational anomaly is given by the eight-form part
of
I 3/2(R) + (nV − nH − 1)I 1/2(R).
Using the formulas in the text, the eight-form parts of I1/2(R) and I3/2(R) are
I (8) 1
1/2 (R) =
128 · 180 (4 trR4 + 5(trR 2 ) 2 ),
I (8)
3/2 (R) =
1
128 · 180 (980 trR4 − 215(trR 2 ) 2 ).
By the same reasoning as in the text, a necessary requirement for anomaly
cancellation is that the total anomaly factorizes into a product of two fourforms.
A necessary requirement for this to be possible is the cancellation of