6.10 The Einstein and Ernst Equations 297Proof. Proof of (a).Denote the determinant on the left byWm.
wi+j− 2 +wi+j=2 n
∑k=1ykxi+j− 2
k,
where
yk=(−1)k+1
εkMk(c). (6.10.33)Hence, applying the lemma in Section 4.1.7 withN→ 2 nandn→m,
Wm=∣
∣
∣
∣
∣
2 n
∑k=1ykxi+j− 2
k∣ ∣ ∣ ∣ ∣ =
2 n
∑k 1 ,k 2 ,...,km=1Ym(
m
∏r=2xr− 1
kr)
∣
∣
xj− 1
ki∣
∣
m,
where
Ym=m
∏r=1yk
r. (6.10.34)
Hence, applying Identity 4 in Appendix A.3,
Wm=1
m!2 n
∑k 1 ,k 2 ,...,km=1Ymk 1 ,k 2 ,...km
∑j 1 ,j 2 ,...,jm(
m
∏r=2xr− 1
jr)
V(xj 1 ,xj 2 ,...,xjm).(6.10.35)
Applying Theorem (b) in Section 4.1.9 on Vandermondian identities,Wm=1
m!2 n
∑k 1 ,k 2 ,...,km=1Ym{
V(xk
1
,xk
2
,...,xk
m)
} 2
. (6.10.36)
Due to the presence of the squared Vandermondian factor, the conditions of
Identity 3 in Appendix A.3 withN→ 2 nare satisfied. Also, eliminating the
x’s using (6.10.26) and (6.10.28) and referring to Exercise 3 in Section 4.1.2,
{
(V(xk
1
,xk
2
,...,xk
m)
} 2
=ρ−m(m−1){
V(ck
1
,ck
2
,...,ck
m)
} 2
. (6.10.37)
Hence,
Wm=ρ−m(m−1)∑
1 ≤k 1 <k 2 <...<km≤ 2 nYm{
V(ck 1 ,ck 2 ,...,ckm)} 2
. (6.10.38)
From (6.10.33) and (6.10.34),
Ym=(−1)K
Emm
∏r=1Mkr(c),