Question How are the local coordinates of a given tangent vector for one chart related to those for another?

Answer Again, we use the chain rule. The formula

di dt = i xj

dxj dt

In other words, a tangent vector through a point m in M is a collection of n numbers (local coordinates) Vⁱ = dxⁱ/dt (specified for each chart x at m) where the quantities for one chart are related to those for another according to the formula

v

i

=

i xj

vj

This leads to the following definition.

Definition 4.1 A contravariant vector at m M is a collection vi of n quantities (defined for each chart at m) which transform according to the formula v   i   = i xj vj It follows that contravariant vectors "are" just tangent vectors: the contravariant vector vi corresponds to the tangent vector given by v = vi xi so we shall henceforth refer to tangent vectors and contravariant vectors. A contravariant vector field V on M associates with each chart x a collection of n smooth real-valued coordinate functions Vi of the n variables (x1, x2, . . . , xn), such that evaluating Vi at any point gives a vector at that point. Further, the domain of the Vi is the whole of the range of x. Similarly, a contravariant vector field V on U M is defined in the same way, but its domain is restricted to x(U).

The tranformation rule for all contravariant vector fields is therefore given as follows.

where now the Vⁱ are functions of the associated coordinates (x¹, x², . . . , xⁿ), and similarly for the barred coordinates. Note that the transformation rule is only valid on the intersection of the images of x and .

Notes 4.2
1.The above formula is reminiscent of matrix multiplication: In fact, let be the matrix whose ij th entry is ⁱ/x^j, then the above equation becomes, in matrix form:

V

=

D

V.

where we think of V and as column vectors.

2. By "transform," we mean that the above relationship holds between the coordinate functions Vⁱ of the xⁱ associated with the chart x, and the functions ⁱ of the ⁱ, associated with the chart .

3. Note the formal symbol cancellation: if we cancel the 's, the x's, and the superscripts on the right, we are left with the symbols on the left!

4. From the proof of 3.6, we saw that, if V is any smooth contravariant vector field on M, then

V = Vi

xi

.

Examples 4.3
(a) Take M = E_n, and let F be any tangent vector field in the usual sense with coordinates Fⁱ. If p = (p¹, p², . . . , pⁿ) is a point in M, then F is the derivative of the path

x¹ = p¹ + tF¹
x² = p² + tF²;
. . .
xⁿ = pⁿ + tFⁿ

at t = 0. Thus this vector has coordinate functions

dxi dt

=

Fi,

which are the same as the original coordinates. In other words, the tangent vectors are "the same" as ordinary vectors in E_n.

(b) An Important Local Vector Field Recall from Example 3.4 (e) the definition of the vectors /xⁱ: At each point m in a manifold M, we have the n vectors /x¹, /x², . . . , /xⁿ, where the typical vector /xⁱ was obtained by taking the derivative of the path:

xi

=

vector obtained by differentiating the path   xj   =

t + const. if j = i const. if j i

,

where the constants are chosen to make xⁱ(t₀) correspond to m for some t₀. This gave

xi

j

=

ij

.

Now, there is nothing to stop us from defining n different vector fields /x¹, /x², . . . , /xⁿ, in exactly the same way: at each point in the coordinate neighborhood of the chart x, associate the vector above.

Note: /xⁱ is a field, and not the i th coordinate of a field. Its jth local coordinate under the chart x is given by _i^j = x^j/xⁱ at every point in the image of x.

Question Since the coordinates do not depend on x, does it mean that the vector field is constant?
Answer No. Remember that a tangent filed is a field on (part of) a manifold, and as such, it is not, in general, constant. The only thing that is constant are its coordinates under the specific chart x. The corresponding coordinates under another chart are ^j/xⁱ (which are not constant in general).

Question What does the vector field /xⁱ look like?
Answer Click here to see.

(c) Patching Together Local Vector Fields
The vector field in the above example has the disadvantage that is local. We can "extend" it to the whole of M by making it zero near the boundary of the coordinate patch, as follows. If m M and x is any chart of M, lat x(m) = y and let D be a disc or some radius r centered at y entirely contained in the image of x. Now define a vector field on the whole of M by

w(p) = xj e-R2 if p is in D 0 otherwise

where

R =	|x(p) - y| r - |x(p) - y|)

The fact that the local coordinates vary smoothly with p M now follows from the fact that all the partial derivatives of all orders vanish as you leave the domain of x. Note that this field agrees with /xⁱ at the point m.

(d) (Based on Example 3.2(c)) Take M = Sⁿ, with stereographic projection given by the two charts discussed earlier (Example 2.3(f) in Lecture 2). Consider the circulating vector field on Sⁿ defined at the point y = (y₁, y₂, . . . , y_n, y_n+1) by the paths

t (y₁cost - y₂sint, y₁sint + y₂cost, y₃, ... , y_n+1).

(For fixed y = (y₁, y₂, . . . , y_n, y_n+1) this defines a path at the point y -- see Example 3.2(c).) This is a circulating field in the y₁y₂-plane. (See the figure. Note: the length of the tangent vector at a given point equals the radius of the latitutde circle on which it sits.)

Question What are its local coordinates under the two charts x and associated with stereographic projection?

Answer We saw in Lecture 2 that

x1 = y1 1-yn + 1 = y1cos t - y2sin t 1-yn + 1	so	V1	=	dx1 dt	=	- y1sin t + y2cos t 1-yn + 1	=	-x2
x2 = y2 1-yn + 1 = y1sin t + y2cos t 1-yn + 1	so	V2	=	dx2 dt	=	y1cos t - y2sin t 1-yn + 1	=	x1
x3 = y3 1-yn + 1	so	V3	=	dx3 dt	=	0
...
xn = yn 1-yn + 1	so	Vn	=	dxn dt	=	0

1 = y1 1+yn + 1 = y1cos t - y2sin t 1+yn + 1	so	1	=	d1 dt	=	- y1sin t + y2cos t 1+yn + 1	=	-2
2 = y2 1+yn + 1 = y1sin t + y2cos t 1+yn + 1	so	2	=	d2 dt	=	y1cos t - y2sin t 1+yn + 1	=	1
3 = y3 1+yn + 1	so	3	=	d3 dt	=	0
...
n = yn 1+yn + 1	so	n	=	dn dt	=	0

Thus, the local coordinates are given by

V = [-x²,  x¹,  0,  0, ... , 0] ,     and
= [-²,  ¹,  0,  0, ... , 0]

Question I don't believe that they transform according to the transformation rule for contravariant vectors!

Answer They do. Click here for the interesting details.

We now look at the gradient. If is a smooth scalar field on M, and if x is a chart, then we obtain the locally defined vector field /xⁱ. By the chain rule, these functions transform as follows:

i

=

xj

xj i

,

or, writing C_j = /x^j and _i = /ⁱ,

i

=

Cj

xj i

.

This leads to the following definition.

Notes 4.5

1. If D is the matrix whose ij th entry is xⁱ/^j, then the above equation becomes, in matrix form:

C

=

CD

where now we think of C and as row vectors.

2. Note that

(D)ij

=

xi k

k xj

=

xi xj

=

ji

,

and similarly for D. Thus, and D are inverses of each other.

3. Note again the formal symbol cancellation: if we cancel the 's, the x's, and the superscripts on the right, we are left with the symbols on the left!

4. Guide to memory: In the contravariant objects, the barred x goes on top; in covariant vectors, on the bottom. In both cases, the non-barred indices match.

Note From now on, all scalar and vector fields are assumed smooth.

Question Geometrically, a contravariant vector is a vector that is tangent to the manifold. How do we think of a covariant vector?

Answer The key to the answer is this:

Click here for a proof

Examples 4.8
(a) Let M = S¹ with the charts:

x = arg(z), = arg(-z)

discussed in Lecture 2. There, we saw that the change-of-coordinate maps, are given by

x

=

+ if < - if >

=

+x if x < -x if x >

,

with     /x = x/ = 1,

so that the change-of-coordinates do nothing. It follows that functions C and specify a covariant vector field iff C = . (Then they are automatically a contravariant field as well.) For example, let

C(x) = 1 = ().

This field circulates around S¹. On the other hand, we could define

C(x) = sin x   and   () = - sin = sin x.

This field is illustrated in the following figure.

(The length of the vector at the point eⁱ is given by sin .)

(b) Let be a scalar field. Its ambient gradient, grad , is given by

grad =

y1 y2 ... ys

F(V+W) = F(V) + F(W)
F(V) = F(V).

The coordinates of the corresponding covariant vector field are

F(/xi)	=	(/xi).grad
	=	dy1 dt dy2 dt ... dys dt . y1 y2 ... ys
	=	xi	,

which is the example that first motivated the definition.

(c) Generalizing (b), let be any smooth vector field in E_s defined on an open set containing M itself. Then the operation of dotting with is a linear function from smooth tangent fields on M to smooth scalar fields. Thus, by the proof of Proposition 4.7, it is a cotangent field on M with local coordinates given by applying the linear function to the canonical charts /xⁱ:

Ci

=

xi

.

The gradient is an example of this, since we are taking

=

in the preceding example.

Note that dotting with depends only on the tangent component of . This leads us to the (very important!) next example.

(d) If V is any tangent (contravariant) field, then we can appeal to (c) above and obtain an associated covariant field. The coordinates of this field are not the same as those of V. To find them, we write:

V

=

Vi

xi

(See Note 4.2 (4).)

Hence, the local coordinates are

Cj

=

xj

.

V

=

Vi

xj

.

xi

Question The vectors /xⁱ are mutually orthogonal, so that the last dot product is just _ij, right?

Answer Wrong! The tangent vectors /xⁱ are not necessarily orthogonal in general (look at the picture of these fields from earlier in this lecture) , so the dot products don't behave as simply as we might suspect.

Instead, we can define certain functions g_ij by

gij

=

xi

.

xj

so that

C_j = g_ijVⁱ

gives the correct relation between the coordinates of a covariant vector and the corresponding contravariant vector field. (Note how the indices cancel to leave us with a lowered index...) We shall see the quantities g_ij again presently. One last thing:

Definition 4.9 If V and W are contravariant (or covariant) vector fields on M, and if is a real number, we can define new fields V+W and V by       (V + W)i = Vi + Wi and   (V)i = Vi. It is easily verified that the resulting quantities are again contravariant (or covariant) fields.

These operations turn the set of all smooth contravariant (or covariant) fields on M into a vector space. Note that we cannot expect to obtain a vector field by adding a covariant field to a contravariant field.

1. Suppose that X^j is a contravariant vector field on the manifold M with the following property: at every point m of M, there exists a local coordinate system xⁱ at m with X^j(x¹, x², . . . , xⁿ) = 0. Show that Xⁱ is identically zero in any coordinate system.

2. Give and example of a contravariant vector field that is not covariant. Justify your claim.

3. Verify the following claim If V and W are contravariant (or covariant) vector fields on M, and if is a real number, then V+W and V are again contravariant (or covariant) vector fields on M.

4. Verify the following claim in the proof of Proposition 4.7: If C_i is covariant and V^j is contravariant, then C_kV^k is a scalar. 5. Let : SⁿE₁ be the scalar field defined by (p₁, p₂, . . . , p_n+1) = p_n+1.
(a) Express as a function of the xⁱ and as a function of the ^j.
(b) Calculate C_i = / xⁱ and _j = /^j.
(c) Verify that C_i and C-_j transform according to the covariant vector transformation rules.

6. Is it true that the quantities xⁱ themselves form a contravariant vector field? Prove or give a counterexample.

7. Prove that and in Proposition 4.7 are inverse functions.

8. Prove: Every covariant vector field is of the type given in Example 4.8(d). That is, obtained from the dot product with some contrravariant field.