## Covariant Differentiation A-La-Rund

Covariant Differentiation A-La-Rund

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 En, 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 2-sphere. (There are, however, parallel fields on the 3-sphere...)

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 Xj was such a field, we should check for its parallelism by taking the derivatives dXj/dt (or just dXj). The problem is, since

 j = ixh Xh ,

one has, by the product rule,

 dj = 2jxhxk Xhdxk + ixh dXh , ........... (I)

showing that, unless the second derivatives vanish, dX does not transform as a vector field. What this means in practical terms is that we cannot check for parallelism at present-even in E3 if the coordinates are not linear. We shall now, in the words of Hanno Rund, "endeavor to eliminate the second derivatives which appear on the right-hand side of (I) by means of the gij in the following manner." First recall that the gij transform according to

 ijij = ijXhYk ixh jxk

(Note that this is the reverse transformation. . .) Now, since we are interested in second derivatives, let us take /xl of both sides: (here we are following p. 78 in Rund's book)

ghk

xl
=
ij

r
r

xl
i

xh
j

xk
+  ij 2ixhxl jxk
+  ij ixh 2jxkxl
.....     (a)

Look now at what happens to the indices: the indices h, k, and l all wind up on the bottom. If we permute them (they're just letters, after all) cyclically in the above formula (that is, hkl, we get two more formulas.

gkl

xh
=
ij

r
r

xh
i

xk
j

xl
+  ij 2ixkxh jxl
+  ij ixk 2jxlxh
.....     (b)

glh

xk
=
ij

r
r

xk
i

xl
j

xh
+  ij 2ixlxk jxh
+  ij ixl 2jxhxk
.....     (c)

Notice that the terms with corresponding color boxes agree, since we can switch the i and j's around any way we want, because ij is symmetric and i and j are dummy indices. Thus, we can cancel as many terms as possible by adding (b) and (c) and subtracting (a) (this will eliminate all boxed terms except for those in the yellow boxes).

 gklxh + glhxk - ghkxl
=
ij

r
r

xh
i

xk
j

xl
+
ij

r
r

xk
i

xl
j

xh
-
ij

r
r

xl
i

xh
j

xk
+  2ij ixl 2jxhxk
(ijr) (irj)

so that, dividing by 2 and doing the above-indicated permutations of dummy indices,

 12 gklxh + glhxk - ghkxl = 12 ijr + jri - rij rxh ixk jxl + ij 2ixkxh jxl .

Noticing the indices in the parenthetical terms, we make the following definition.

Definition 7.1 The quantities
 [hk, l] = 12 gklxh + glhxk - ghkxl

(note that indices are taken from term with the minus sign) are called Christoffel symbols of the first kind.

By the formula above the definition, these gizmos transform as follows.

Transformation Law for Christoffel Symbols of the First Kind
 [hk, l] = [ri, j] rxh ixk jxl + ij 2ixhxk jxl

(Hint to remembering indices: notice that the sequences of indices in the partial derivative terms are (h, k, l) and (r, i, j))

Note This is like the transformation law for a type (0, 3) tensor except for the extra term added on the right.

Now, the whole purpose of the game it to try and get those offending 2nd derivatives by themselves, which means we must first eliminate the terms ij.

Question How do we do this?
Answer By using the inverse gij of gij (discussed in the exercise set) as follows:

First, multiply the boxed formula above by glp. (Notice that the agreement of one of the indices l means that we are summing as well!) This gives

 glp [hk,l] = [ri, j] rxh ixk glp jxl + ij 2ixkxh glp jxl ........... (II)

To get these to do any good, we need to express the g's on the right-hand side in terms of the 's. That is why we have parenthesized the formulas. Indeed, since

 glp = st xls xpt ,

 multiplying both sides by jxl gives

 glp jxl = st xls xpt jxl = st js xpt = jt xpt

Substituting this in (II) and then using the identity ij jt = ti gives

glp [hk,l] =  [ri, j] rxh ixk jt xpt + ij 2ixkxh jt xpt ........... (II)
=  jt [ri, j] xpt rxh ixk + xpt 2txkxh

Notice the correspondence of the indices gip [hk,l] to its barred counterpart. If we now let

 ph k = glp [hk,l] ,

the above equation reduces to

 ph k
=  tr i
xp

t
r

xh
i

xk
+
xp

t
2t

xhxk
......     (III)

Definition 7.2 Christoffel Symbols of the Second Kind

 ph k = glp [hk,l] = 12 gklxh + glhxk - ghkxl

By the formula above the definition, these gizmos transform as follows.

Transformation Law for Christoffel Symbols of the Second Kind

 ph k
=  tr i
xp

t
r

xh
i

xk
+
xp

t
2t

xhxk

(Colors have been used to show how indices pop up..)

Note This is precisely the transformation rule for a type (1, 2) tensor (written back-to-front) with that annoying second derivative term added.

We now solve for that second derivative term. To do that, all we need now do is get rid of that little xp/xi in front, which is now a (relatively!) easy task: First rewrite the boxed above transformation law as

 ph k
=
xp

t
 tr i
r

xh
i

xk
+
2t

xhxk
.

Then multiply both sides by / xp (note that we must cancel the non-dummy index) and use

 xpt axp = at

to get

a

xp
p
h k
=  ar i
r

xh
i

xk
+
2a

xkxh

giving the following expression for the second derivatives:

2a

xhxk
=
a

xp
p
h k
-
 ar i
r

xh
i

xk

Now go back to formula (I):

 dj = 2jxhxk Xhdxk + ixh dXh ,

and substitute this value of the second derivative (with a replaced by l) to get

dj = 
j

xp
p
h k
-  jr i
r

xh
i

xk
Xhdxk+
i

xh
dXh
= 
j

xp
p
h k
Xhdxk -  jr i
r

xh
i

xk
Xhdxk+
i

xh
dXh
=
j

xp
p
h k
Xhdxk -  jr i
r di +
i

xh
dXh
(using the transformation rule for the type (2,0) tensor Xhdxk )

Taking all the barred things over to the left gives

dj +  jr i
r di =  jxp ph k Xhdxk + ixh dXh
=  jxp ph k Xhdxk + dXp (changing the dummy variable h to p )

But look at that! If we write

 DXi = dXi + ip q Xp dxq ,

the above formula says that

 Dj = jxi Dxi ,

showing that this quantity is a contravariant vector! This leads to the following definition.

Definition 7.3 If Xi is a vector field, then its covariant differential is the contravariant differential form defined by

Covariant Differential of Xi

 DXi = dXi + ip q Xp dxq

If now we have a path xi = xi(t) in M, and X is a contravariant vector field on M, we can now define the rate of change of X along the path by

 DXidt = dXidt ip q Xp dxqdt .

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 7.4 (Parallel Fields of Constant Length) Xi is a parallel field of constant length in En iff DXi/dt = 0 for all paths in En.

Proof Designate the usual coordinate system by xi. Then Xi is parallel and of constant length iff its coordinates with respect to the chart x are constant; that is, iff

 dXidt = 0

But, since for this coordinate system, gij = ij, the Christoffel symbols clearly vanish, and so

 DXidt = dXidt = 0

But, if the contravariant vector DXi/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 dXi/dt.)

Partial Derivatives

Let us make the following definition.

Definitions 7.5 The covariant partial derivative of the contravariant field Xp is the type (1, 1) tensor given by

Covariant Partial Derivative of Xp

 Xp|k = Xpxk + ph k Xh
(Some texts use kXp.) Similarly, the covariant partial derivative of the covariant field Yp is the type (0, 2) tensor given by

Covariant Partial Derivative of Yp

 Yp|k = Ypxk + hp k Yh

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 Xp|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:

 DXi dt = Xp|k dxk dt

(Again, see the exercises).

Exercise Set 7

1.(a) Show that jik = kij.

(b) If jik are functions that transform in the same way as Christoffel symbols of the second kind (called a connection) show that jik - kij is always a type (1, 2) tensor (called the associated torsion tensor).

(c) If aij and gij are any two symmetric non-degenerate type (0, 2) tensor fields with associated Christoffel symbols jika and jikg respectively. Show that jika - jikg 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 Yi is a covariant vector, then DYp = dYp - piq Yi dxq. are the components of a covariant vector field.

3. (See Rund, pp. 72-73) Covariant Differential of a Tensor Field We can again use the same analysis to obtain, for a type (1, 1) tensor, DThp = dThp + phqTrpdxq - piq Thi dxq.

4. Show that, if Xi is any vector field on En, then its ordinary partial derivatives agree with Xp|k.

5. Show that, if Xi and Yj are any two (contravariant) vector fields on M, then

(Xi + Yi)|k = Xi|k + Yi|k
(XiYj)|k = Xi|kYj + XiYj|k.

6. Show that, if C is a path on M, then

 DXidt = Xi|k dxkdt .

7. Show that, if X and Y are vector fields, then

 ddt X, Y = DXdt , Y + X DYdt ,
where the big D's denote covariant differentiation.

Last Updated: September, 1998