3.3. ORTHOGONAL DECOMPOSITION FOR TENSOR FIELDS 141
Theorem 3.20.Let u∈L^2 (T 20 M)be symmetric, i.e. uij=uji, and the first Betti number
β 1 (M) = 0 forM. Then the following assertions hold true:
1) u has a unique orthogonal decomposition if and only if there is a scalar functionφ∈
H^2 (M)such that u can be expressed as(3.3.28)
uij=vij+DiDjφ,
vij=vij, Divij= 0.2) If vijin (3.3.26) is symmetric: vij=vji, then u can be expressed by (3.3.28).3) u can be orthogonally decomposed in the form (3.3.28) if and only if the following
differential equations have a solutionφ∈H^2 (M), andφis the scalar field in (3.3.28):(3.3.29)
∂
∂xi∆φ+Rki∂ φ
∂xk=−Djuji for 1 ≤i≤n,where Rki=gk jRijand Rijare the Ricci curvature tensors, and∆is the Laplace oper-
ator for scalar fields as defined by (3.2.28).Proof.We only need to prove Assertions (2) and (3).
We first prove Assertion (2). Sincevijin (3.3.26) is symmetric, then we have
(3.3.30) Diφj=Djφi.
Note that
Diφj=∂ φj
∂xi−Γkijφk,andΓkij=Γkji. We infer then from (3.3.30) that
(3.3.31)
∂ φj
∂xi=
∂ φi
∂xj.
By assumption, the 1-dimensional homology ofMis zero,
H 1 (M) = 0 ,and by the de Rham theorem (Ma, 2010 ), it follows that all closed 1-forms are complete
differentials, i.e. for any
ω=ψidxi, dω= 0 ,
there is a scalar functionψsuch that
ψi=∂ ψ/∂xi for 1≤i≤n.In view of (3.3.31), it implies that the 1-form
ω=φkdxk