148 CHAPTER 3. MATHEMATICAL FOUNDATIONS
It follows that
(3.4.19) g ̃ij=−gikgjlg ̃kl.
Thus, (3.4.18) is rewritten as
(3.4.20)
d
dλ∣
∣
∣
λ= 0det(gij+λ ̃gij) =−ggijg ̃ij.For the Einstein-Hilbert functional (3.4.12), we haved
dλ∣
∣
∣
λ= 0
F(gij+λg ̃ij)=
∫M[
Rij ̃gij√
−g−1
2
√
−gR
d
dλdet(gij+λ ̃gij)+gijd
dλRij(gij+tg ̃ij)√
−g]
dx∣
∣
∣
λ= 0
=(by( 3. 4. 20 ))=∫M(Rij−1
2
gijR) ̃gij√
−gdx+∫Mgijd
dλ∣
∣
∣
λ= 0Rij(gij+λ ̃gij)√
−gdx.In view of (3.4.5) and
〈δF,g ̃ij〉=∫MδF ̃gij√
−gdx,d
dλ∣
∣
∣
λ= 0Rij(gij+λ ̃gij) =∂Rij
∂gkl̃gkl,we arrive at
∫M[
δF−(Rij−1
2
gijR)]
̃gij√
−gdx=∫Mgij∂Rij
∂gkl
̃gkl√
−gdx.To verify (3.4.15), it suffices to prove that
(3.4.21)
∫MgijδRij√
−gdx= 0 ,whereδRijis the variational element
δRij=Rij(gkl+δgkl)−Rij(gkl),which are equivalent to the following directional derivative:
d
dλ∣
∣
∣
λ= 0Rij(gij+λ ̃gij).To get (3.4.21), we takeRijin the form
(3.4.22) Rij=
∂Γkki
∂xj−
∂Γkij
∂xk+ΓlikΓkl j−ΓlijΓklk.