Similarly, integration of Equations 4-9, as carried out in Appendix C, leads to the
following solution for the components of the residual gauge vector:
Cl = Clo
11 1\=) 1~ o ~)"
to = 6 O + -+-5(o 40 21
2p p 2
1 (4 17)
(m = _(mo o10
where 920 iS related to the background curvature via 92 20 3~p. In order to use this
residual gauge freedom to impose the full IRG, we return to the gauge transformation for
hmm (Equation 4-10) which becomes, after some manipulation (using Equations C-6C-9
and Equation C-13),
pI ~;6" p ~; 1a, + +(~ )-aa-p' -- '>" n] 4
kmm 8 (moP o+ kopo__g opoa .(-8
p p 2
In this form it is clear that we can impose the trace condition (Equation 4-6) of the full
IRG if we choose our gauge vector so that
-I 1-I --
B (mo + o a ,o( 8- I ploo a on = bo. (4-19)
We have now shown by construction that the condition : = 0 is both necessary and
sufficient for imposing the full IRG in a type II background. We turn next to discussing
the complete extent of the residual gauge freedom in more detail.
4.3 Remaining Gauge Freedom
Although Equations 4-19 involve three real degrees of freedom (ao is complex),
it turns out that only two real degrees of gauge freedom are required to fully remove
any solution of Equation 4-13 for the trace hmm. To see this we introduce the following
identity:
p~~~ p p+p (p + p)Ro, (4-20)
p p p p