152 CHAPTER 3. MATHEMATICAL FOUNDATIONS
We are now in position to consider the variation with divA-free constraints. We know that
an extremum pointgijof a metric functional is a solution of the equation
(3.4.37) δF(gij) = 0 ,
in the sense that for anyXkl=Xlk∈L^2 (T 20 M),
〈δF(gij),X〉=d
dλ∣
∣
∣
λ= 0F(gij+λXij)|λ= 0 =∫M(δ∗F(gij))klXkl√
(3.4.38) −gdx= 0.
Note that the solutiongijof (3.4.37) in the usual sense should satisfy
(3.4.39) 〈δF(gij),X〉= 0 ∀X∈L^2 (T 20 M).
Notice that (3.4.38) has a symmetric constraint on the variational elementsXij:Xij=Xji.
Therefore, comparing (3.4.38) with (3.4.39), we may wonder if a solutiongijsatisfying
(3.4.38) is also a solution of (3.4.39). Fortunately, note thatL^2 (T 20 M)can be decomposed
into a direct sum of symmetric and anti-symmetric spaces as follows
L^2 (T 20 M) =Ls^2 (T 20 M)⊕L^2 c(T 20 M),
L^2 s(T 20 M) ={u∈L^2 (T 20 M)|uij=uji},
L^2 c(T 20 M) ={u∈L^2 (T 20 M)|uij=−uji},andL^2 s(T 20 M)andL^2 c(T 20 M)are orthogonal:
∫
Mgikgjluijvkl√
−gdx=−∫Mgikgjluijvlk√
−gdx= 0 ∀u∈L^2 s(T 20 M), v∈L^2 c(T 20 M).Thus, due to the symmetry ofδF(gij), the solutiongijof (3.4.37) satisfying (3.4.38) must
also satisfy (3.4.39). Hence the solutions of (3.4.37) in the sense of (3.4.38) are the solutions
in the usual sense.
However, if we consider the variations ofFunder the divA-free constraint, then the ex-
tremum points ofFare not solutions of (3.4.37) in the usual sense. Motived by physical
considerations, we now introduce variations with divA-free constraints.
Definition 3.25.Let F(u)be a functional of tensor fields u. We say that u 0 is an extremum
point of F(u)under thedevA-free constraint, if
(3.4.40) 〈δF(u 0 ),X〉=
d
dλ∣
∣
∣
λ= 0F(u 0 +λX) = 0 ∀divAX= 0 ,wheredivAis as defined in (3.2.38).
In particular, if F is a functional of Riemannian metrics, and the solution u 0 =gijis a
Riemannian metric, then the differential operator DAindivAX in (3.4.40) is given by
(3.4.41) DA=D+A, D=∂+Γ,
and the connectionΓis taken at the extremum point u 0 =gij.