EQUATIONS OF ELASTIC HYPERSURFACES

2. OUTLINE OF DIFFERENTIAL GEOMETRY 11
Proof: Set
b j (x) = Xx j ,
U # = ∑ b j (x) ∂
∂x j
,
and consider V = U − U # . Then V is a derivation satisfying
V x j = 0 for all j = 1, . . . , n . (2.25)
Let us show that V f = 0 for all f ∈ C ∞ (Ω). Whatever V is a derivation and 1 · 1 = 1, we
get V 1 = 2V 1 and therefore V 1 = 0. Then the vector field V annihilates all constants
V c = cV 1 = 0 for all c = const (2.26)
and thus, all polynomials of degree ≤ 1 (cf. (2.25) and (2.26)).
Now we show that V f(p) = 0 for all p ∈ Ω. Without loss of generality, we can suppose
p = 0 coincides with the origin. Then
n∑
∫ 1
f(x) = f(0) + b j (x)x j with b j (x) = f(tx) dt .
j=1
As a consequence, V f(x) vanishes at 0
V f(0) = V f(0) +
∣
n∑ [ ∣∣∣∣
V b j (x)x j + b j (x)V x j = 0
x=0
j=1
(cf. (2.25) and (2.26)) and the assertion is proved.
Corollary 2.6 For a point p in R m the linear mapping Φ : R m → T p R m defined by Φ :
v ↦→ ∂ v is a vector space isomorphism. The system { } m
∂ , ∂ ek k=1 e k
= ∂/∂ xk , is a frame in
T p R m .
0
Proof: From Lemma 2.5 follows that the mapping Φ is surjective: to each derivation ∂ U
there corresponds a vector field U(x), which is a vector U(x 0 ) ∈ R n for a fixed x 0 ∈ Ω.
Let v, w ∈ R m such that v ≠ w. Choose an element u ∈ R m such that 〈u, v〉 ̸= 〈u, w〉
and define f : R m → R by f(x) = 〈u, x〉. Then
∂ v f = lim
t→0
〈u, x + tv〉 − 〈u, x〉
t
= 〈u, v〉 ̸= 〈u, w〉 = ∂ w f
so ∂ v ≠ ∂ w . This proves that the mapping Φ is injective.
The last claim about the frame follows from the representation (2.24).
The next theorem is a fundamental fact about vector fields: the field can be ”rectified”
in a neighborhood of each point where it does not vanishes.
Theorem 2.7 If U ∈ V (Ω) is a vector field and U(p) ≠ 0 for some p ∈ Ω, there exists a
local coordinate system { u 1 , . . . , u n
}
centered at p = 0 (so uj (p) = 0, j = 1, . . . , n) with
respect to which
∂ U = D n =
∂
∂u n
. (2.27)