Lecture 8: Covariant Differentiation

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 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 along some path xi = xi(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 non-constant, 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 = jxh Xh,
one has, by the product rule,

 djdt = 2jxkxh Xh dxkdt + jxh dXhdt ...     (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 E3 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 n-manifold with metric g, and let V be a vector in Es,. The projection V of V onto Tm has (local) coordinates given by (V)i = gik(V./xk), where [gij] is the matrix inverse of [gij], and gij = (/xi).(/xj) as usual.

Proof We can represent V as a sum,

V = V + W,
where W is the component of V normal to Tm. Now write /xk as ek, and write
V = a1e1 + ... + anen,
where the ai are the desired local coordinates. Then
 V = V + W = a1e1 + ... + anen + W
and so
 V.e1 = a1e1.e1 + ... + anen.e1 + 0 V.e2 = a1e1.e2 + ... + anen.e2 ... V.en = a1e1.en + ... + anen.en =
which we can write in matrix form as
[V.ei] = [ai]g**
whence
[ai] = [V.ei]g**.
Finally, since g** is symmetric, we can transpose everything in sight to get
[ai] = g**[V.ei],
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.

xp
[gqr] =  xp ysxq ysxr
=  2ysxpxq ysxr + 2ysxrxp ysxq
or, using "comma notation" (that is, R,p denotes partial derivative with respect to xp.),
gqr,p = ys,pq ys,r + ys,rp ys,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.
gqr,p=  ys,pq ys,r
+  ys,rp ys,q
(Original formula)
grp,q=  ys,qr ys,p
+  ys,pq ys,r
gpq,r=  ys,rp ys,q
+  ys,qr ys,p
Note that each term on the right occurs twice altogether as shown by the colors. This permits us to solve for the yellow term ys,pq ys,r by adding the first two equations and subtracting the third:
 ys,pq ys,r = 12 [ gqr,p + grp,q - gpq,r ].

Definition 8.2 Christoffel Symbols

We make the following definitions.

 [pq, r]
=  12 [ gqr,p + grp,q - gpq,r ].
Christoffel Symbols of the First Kind
 ip q
=  gir [pq, r]
Christoffel Symbols of the Second Kind

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

 [hk, l] = [ri, j] rxh ixk jxl + ij 2ixhxk jxl

Transformation Law for Christoffel Symbols of the Second Kind

 ph k
=  tr i
xp

t
r

xh
i

xk
+
xp

t
2t

xhxk

(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)

 DXidt = dXidt + ip q Xp dxqdt

Proof By definition,

 DXdt = dXdt
which, by the lemma, has local coordinates given by
 DXidt = gir dXdt . xr .
To evaluate the term in parentheses, we use ambeint coordinates. dX/dt has ambient coordinates
 ddt Xp ysxp = dXpdt ysxp + Xp 2ysxpxq dxqdt .
Thus, dotting with /xk = ys/xr gives
 dXpdt ysxp ysxr + Xp 2ysxpxq ysxr dxqdt
=  dXpdt gpr + Xp[pq, r] dxqdt
Finally,
DXi

dt
=  gir dXdt . xr
=  gir dXpdt gpr + Xp[pq, r] dxqdt
=  ip dXpdt + ip q Xp dxqdt (Defn of Christoffel symbols of the 2nd Kind)
=  dXidt + ip q Xp dxqdt
as required.

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) 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 8.4 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 8

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. 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 DXp/dt transform as a contravariant vector.

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

7. 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.

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

 DXidt = Xi|k dxkdt .

9. 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.

10. (a) What is |i if is a scalar field?
(b) Give a definition of the "contravariant" derivative, Xa|b of Xa with respect to xb, and show that Xa|b = 0 if and only if Xa|b = 0.

Last Updated: January, 2002