Page 351 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Let Y T U, the covariant derivative of Z along Y is:

u
i
 Y Z  y Z .
;i
We write the components of  Y Z as:
k
j
z j ;i  z   j ik z , ...(10)
,i
i
j
.
so that  Y Z  y z  Furthermore, note that
;i
j
k
     k . ...(11)
i
ij
j
Our first task is to show that covariant differentiation is reparametrization invariant. However,
since the metric g was used in the definition of the covariant derivative, it stands to reason that
it would be invariant only under those reparametrization which preserve the metric, i.e., under
isometries.


Proposition 3. Let  : U,g     (U,g) be an isometry. Let Y T U, u   and let Z be a vector field on
U. Then

  .
 
d    Y Z    d Y    d Z  ...(12)


Proof. This proof, although tedious, is quite straightforward, and is relegated to the exercises.

Note that on the left hand-side of (12), the covariant derivative  is that obtained from the
metric g. 
Our next observation, which follows almost immediately, gives an interpretation of the covariant
derivative when the metric g is induced by a parametric surface X.
Proposition 4. Let the Riemannian metric g be induced by the parametric surface X. Then the

image under dX of the covariant derivative dX  i  Z is the projection of  i Z onto the tangent
space.

 
j
Proof. Note that dX   X . Thus, if Z  z  j then we find:
i
i
j
j
k
dX  i Z   z X  z X   ik j z X   i  z X j  k z N, ij j
j
;i
j
j
j
;i
which proves the proposition.
We now show that covariant differentiation is in addition well-adapted to the metric g.
Proposition 5. Let (U, g) be a Riemannian surface, and let Y and Z be vector fields on U. Then, we
have
 ig (Y,Z)   g  i Y,Z  g  i Y,Z . ...(13)
Proof. We first note that, as in the proof of Theorem 2.29, the definition of the Christoffel
symbols (3) :
g ij,l   k il g   k jl g . ...(14)
kj
ki

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