It turns out that the vectors ${\bf p}_i$ for this case only depend on the type $\t$ and the orientation $a$. The resulting nine vectors are given in Table 1.

  ${\bf p}_{(\t ,a)}$   $a=1$               $a=2$               $a=3$
  --------------------- ------------------- ------------------- -------------------
  $\t = \a$             $(1,0,0,0,0,1,0)$   $(0,1,0,0,1,0,0)$   $(0,0,1,1,0,0,0)$
  $\t = \b$             $(1,0,0,0,0,0,1)$   $(0,0,1,0,1,0,0)$   $(0,1,0,1,0,0,0)$
  $\t = \g$             $(1,0,0,1,0,0,0)$   $(0,0,1,0,0,1,0)$   $(0,1,0,0,0,0,1)$

  : Values of vectors ${\bf p}_{(\t ,a)}$ which define the moduli Kähler potential for the $G_2$ manifold based on $T^7/\ZM_2^3$.

It will be useful for the subsequent discussion to have an interpretation for the moduli and the various parts of the Kähler potential  in terms of the geometry of the underlying $G_2$ manifold. To this end, one notes that the moduli Kähler potential for M-theory on a $G_2$ manifold is generally related to the volume $\cV$ of this manifold by [@Beasley:2002db] $$K = -3\ln\left(\frac{\cV}{2\p^2}\right)\; . \label{KV}$$ The first, $T^A$–dependent, part of the Kähler potential  then corresponds to the volume $\cV_0$ of the orbifold, while the second, $U^i$–dependent, part measures the reduction of this volume due to the blow-ups. For the orbifold volume $\cV_0$ one easily finds from Eqs.  and that $$\cV_0 = \frac{1}{8}\exp\left(\frac{1}{3}\sum_A\beta_A\right)\; ,
 \label{V0}$$ where we have used the value $c=6\ln 