THE TENSOR CHARACTERIZATION LEMMA
Lemma. For a smooth manifold M let X(M ) denote the set of smooth vector fields
on M and C ∞ (M ) the set of smooth functions M → R. A function
A : X(M ) × . . . × X(M ) → C ∞ (M )
|
{z
}
or X(M )
s
determines a (0, s)- or (1, s)-tensor field on M , respectively, if and only if A is multilinear over R and for any f1 , . . . , fs ∈ C ∞ (M ) and X1 , . . . , Xs ∈ X(M ),
A(f1 X1 , . . . , fs Xs )(p) = f1 (p) · · · fs (p) A(X1 , . . . , Xs )(p)
(1)
for all p ∈ M .
Proof. Any (0, s)- or (1, s)-tensor field A on M defines a map on vector fields by:
A(X1 , . . . , Xs )(p) = Ap (X1 |p , . . . , Xs |p ) for all p ∈ M
It is straighforward to check in local coordinates that the function or vector field
so-defined is smooth.
Now suppose that A is a function on X(M ) × . . . × X(M ) which is multilinear over
R and has property (1) above. I’ll prove in the case that A maps to X(M ) that it
determines a unique (1, s)-tensor field. The following fact is key:
Fact. For some p ∈ M , if Xi ≡ Yi for each i on a neighborhood U of p in M then
A(X1 , . . . , Xs )(p) = A(Y1 , . . . , Ys )(p).
Proof of Fact. Let f ∈ C ∞ (M ) satisfy supp(f ) ⊂ U and f (q) = 1 for all q in an open
set U0 ⊂ U with p ∈ U0 . (Such a function f is called a bump function; you can prove
it exists using partitions of unity.) The sequence of equations below proves the fact:
A(f X1 , . . . , f Xs )(p) = f (p)s A(X1 , . . . , Xs )(p) = A(X1 , . . . , Xs )(p)
A(f Y1 , . . . , f Ys )(p) = f (p)s A(Y1 , . . . , Ys )(p) = A(Y1 , . . . , Ys )(p)
A(f X1 , . . . , f Xs ) ≡ A(f Y1 , . . . , f Ys )
The last equation holds because f Xi is identical to f Yi on all of M for each i — they
agree on U since Xi and Yi agree there, and outside U each is identically 0.
Now fix a chart φ = (x1 , . . . , xn ) : U → Rn for an open set U ⊂ M , and for p ∈ U
let f be a bump function with the properties listed in the proof of the Fact. For
q ∈ U0 and i, j1 , . . . , js ∈ {1, . . . , n} define Aij1 ,...,js (q) by
n
X
∂
∂
∂ i
A f j1 , . . . , f js (q) =
Aj1 ,...,js (q) i ∂x
∂x
∂x q
i=1
The reason we are multiplying the ∂x∂ji ’s by f here is to produce smooth vector
fields on all of M (which are 0 outside U ). The functions Aij1 ,...,js so-defined on U0
are smooth since A sends s-tuples of smooth vector fields to smooth vector fields.
Applying this procedure at each p ∈ U yields well-defined (by the Fact) functions
1
2
THE TENSOR CHARACTERIZATION LEMMA
Aij1 ,...,js on all of U . Moreover, for any smooth vector fields X1 , . . . , Xs on M with Xi
P
j ∂
given in local coordinates by Xi = nj=1 ξ(i)
, for any p ∈ U we have:
∂xj
A(X1 , . . . , Xs )(p) = A(f X1 , . . . , f Xs )(p)
X j
∂
∂
js
1
=
ξ(1) (p) · · · ξ(s) (p)A f j1 , . . . , f js (p)
∂x
∂x
j ,...,j
s
1
The former equation follows from the Fact, the latter from multilinearity of A. It
now follows from the definition of the Aij1 ,...,js that in local coordinates on U ,
X
∂
A=
Aij1 ,...,js i ⊗ dxj1 ⊗ . . . ⊗ dxjs
∂x
i,j ,...,j
1
s
More precisely: the formula on the right defines a tensor AU on U which is unique
with the property that for any smooth vector fields X1 , . . . , Xs on M ,
AU (X1 |U , . . . , Xs |U ) = A(X1 , . . . , Xx )|U
Here uniqueness stems from the fact that the coefficient functions Aij1 ,...,js are uniquely
determned. The fact that AU is uniquely determined by A ensures that the tensor
AV analogously determined by A on a different chart domain V agrees with AU on
the overlap U ∩ V , hence that the local tensors AU patch together to determine a
well-defined tensor on M .