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8. Covariant Differentiation
Intuitively, by a parallel vector field, we mean a vector field with the property that the vectors at different points are parallel. Is there a notion of a parallel field on a manifold? For instance, in E_{n}, there is an obvious notion: just take a fixed vector v and translate it around. On the torus, there are good candidates for parallel fields (see the figure) but not on the 2sphere. (There are, however, parallel fields on the 3sphere...)
Let us restrict attention to parallel fields of constant length. Usually, we can recognize such a field by taking the derivatives of its coordinates, or by following a path, and taking the derivative of the vector field with respect to t: we should come up with zero. The problem is, we won't always come up with zero if the coordinates are not rectilinear, since the vector field may change direction as we move along the curved coordinate axes.
Technically, this says that, if X^{j} was such a field, we should check for its parallelism by taking the derivatives dX^{j}/dt along some path x^{i} = x^{i}(t). However, there are two catches to this approach: one geometric and one algebraic.
Geometric Look, for example, at the filed on either torus in the above figure. Since it is circulating and hence nonconstant, dX/dt 0, which is not what we want. However, the projection of dX/dt parallel to the manifold does vanish  we will make this precise below.
Algebraic Since
^{j} =  x^{h} 
X^{h}, 
dt 
=  x^{k}x^{h} 
X^{h}  dt 
+  x^{h} 
dt 
... (I) 
showing that, unless the second derivatives vanish, dX/dt does not transform as a vector field. What this means in practical terms is that we cannot check for parallelism at present  even in E_{3} if the coordinates are not linear.
The projection of dX/dt along M will be called the covariant derivative of X (with respect to t), and written DX/dt. To compute it, we need to do a little work. First, some linear algebra.
Lemma 8.1 (Projection onto the Tangent Space)
Let M be a Riemannian nmanifold with metric g, and let V be a vector in E_{s},. The projection V of V onto T_{m} has (local) coordinates given by (V)^{i} = g^{ik}(V^{.}/x^{k}),where [g^{ij}] is the matrix inverse of [g_{ij}], and g_{ij} = (/x^{i}).(/x^{j}) as usual. 
Proof We can represent V as a sum,
V = V + W,where W is the component of V normal to T_{m}. Now write /x^{k} as e_{k}, and write
V = a^{1}e_{1} + ... + a^{n}e_{n},where the a^{i} are the desired local coordinates. Then
V  =  V + W 
=  a^{1}e_{1} + ... + a^{n}e_{n} + W 
V^{.}e_{1}  =  a^{1}e_{1}^{.}e_{1} + ... + a^{n}e_{n}^{.}e_{1} + 0 
V^{.}e_{2}  =  a^{1}e_{1}^{.}e_{2} + ... + a^{n}e_{n}^{.}e_{2} 
...  
V^{.}e_{n}  =  a^{1}e_{1}^{.}e_{n} + ... + a^{n}e_{n}^{.}e_{n} 
= 
[V^{.}e_{i}] = [a^{i}]g_{**}whence
[a^{i}] = [V^{.}e_{i}]g^{**}.Finally, since g^{**} is symmetric, we can transpose everything in sight to get
[a^{i}] = g^{**}[V^{.}e_{i}],as required.
For reasons that will become clear later, let us now look at some partial derivatives of the fundamental matrix [g_{**}] in terms of ambeint coordinates.
x^{p}  [g_{qr}]  = 


= 

g_{qr,p} = y_{s,pq }y_{s,r} + y_{s,rp }y_{s,q}Look now at what happens to the indices q, r, and p if we permute them (they're just letters, after all) cyclically in the above formula (that is, pqr), we get two more formulas.
g_{qr,p}  = 

+ 

(Original formula)  
g_{rp,q}  = 

+ 


g_{pq,r}  = 

+ 

y_{s,pq }y_{s,r}  =  1 2 
[ g_{qr,p} + g_{rp,q}  g_{pq,r} ]. 
Definition 8.2 Christoffel Symbols
We make the following definitions.

Neither of these gizmos are tensors, but instead transform as follows (Which you will prove in the exercises!)
Transformation Law for Christoffel Symbols of the First Kind
Transformation Law for Christoffel Symbols of the Second Kind
(Look at how the patterns of indices match those in the Christoffel symbols...) 
We can now obtain a formula for the covariant derivative.
Proposition 8.2 (Formula for Coavariant Derivative)

Proof By definition,
dt  =  dt 
dt 
=  g^{ir }  dt 
^{.}  x^{r} 
. 
dt  X^{p}  x^{p} 
=  dt  x^{p} 
+  X^{p}  x^{p}x^{q} 
dt  . 


= 

dt 
= 


= 


= 


= 

In the exercises, you will check directly that the covariant derivative transforms correctly.
This allows us to say whether a field is parallel and of constant length by seeing whether this quantity vanishes. This claim is motivated by the following.
Proposition 8.3 (Parallel Fields of Constant Length)
X^{i} is a parallel field of constant length in E_{n} iff DX^{i}/dt = 0 for all paths in E_{n}. 
Proof Designate the usual coordinate system by x^{i}. Then X^{i} is parallel and of constant length iff its coordinates with respect to the chart x are constant; that is, iff
dt 
=  0. 
But, since for this coordinate system, g_{ij} = _{ij}, the Christoffel symbols clearly vanish, and so
dt 
=  dt 
=  0. 
But, if the contravariant vector DX^{i}/dt vanishes under one coordinate system (whose domain happens to be the whole manifold) it must vanish under all of them. (Notice that we can't say that about things that are not vectors, such as dX^{i}/dt.)
Partial Derivatives
Let us make the following definition.
Definitions 8.4 The covariant partial derivative of the contravariant field X^{p} is the type (1, 1) tensor given by
Covariant Partial Derivative of X^{p}
Covariant Partial Derivative of Y_{p}

Question How do we know that these things are second order tensors as claimed?
Answer Some of these will be in the exercises. Click here for a proof that X^{p}_{k} is a type (1, 1) tensor.
Notes
1. All these forms of derivatives satisfy the expected rules for sums and also products. (See the exercises.)
2. If C is a path on M, then we obtain the following analogue of the chain rule:
dt 
=  X^{p}_{k}  dt 
(Again, see the exercises).
1.(a) Show that _{}_{j}^{i}_{k}_{} = _{}_{k}^{i}_{j}_{}.
(b) If _{j}^{i}_{k} are functions that transform in the same way as Christoffel symbols of the second kind (called a connection) show that _{j}^{i}_{k}  _{k}^{i}_{j} is always a type (1, 2) tensor (called the associated torsion tensor).
(c) If a_{ij} and g_{ij} are any two symmetric nondegenerate type (0, 2) tensor fields with associated Christoffel symbols _{}_{j}^{i}_{k}_{}_{a} and _{}_{ j}^{i}_{k}_{}_{g} respectively. Show that _{}_{j}^{i}_{k}_{}_{a}  _{}_{j}^{i}_{k}_{}_{g} is a type (1, 2) tensor.
2. Covariant Differential of a Covariant Vector Field Use the results and analysis of the section (and look at, eg. Rund) to show that, if Y_{i} is a covariant vector, then DY_{p} = dY_{p}  _{}_{p}^{i}_{q}_{} Y_{i} dx^{q}. are the components of a covariant vector field.
3. (See Rund, pp. 7273) Covariant Differential of a Tensor Field We can again use the same analysis to obtain, for a type (1, 1) tensor, DT^{h}_{p} = dT^{h}_{p} + _{}_{p}^{h}_{q}_{}T^{r}_{p}dx^{q}  _{}_{p}^{i}_{q}_{} T^{h}_{i} dx^{q}.
4. Obtain the transformation equations for Chritstoffel symbols of the first and second kind. (You might wish to consult an earlier printing of these notes or Rund's book...)
5. Show directly that the coordinates of DX^{p}/dt transform as a contravariant vector.
6. Show that, if X^{i} is any vector field on E_{n}, then its ordinary partial derivatives agree with X^{p}_{k}.
7. Show that, if X^{i} and Y^{j} are any two (contravariant) vector fields on M, then
8. Show that, if C is a path on M, then
dt 
=  X^{i}_{k}  dt 
. 
9. Show that, if X and Y are vector fields, then
dt 
X, Y  =  dt 
,  Y  +  X  dt 
, 
10. (a) What is _{i} if is a scalar field?
(b) Give a definition of the "contravariant" derivative, X^{ab} of X^{a} with respect to x^{b}, and show that X^{ab} = 0 if and only if X^{a}_{b} = 0.
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