Lecture 2 described scalar fields on manifolds. We now turn to vectors on smooth manifolds. We must first talk about smooth paths in M.

Definition 3.1 A smooth path in the smooth manifold M is a smooth map defined on an open segment of the real line, r: (-a, a) M, where r is a vector-valued function with coordinates (y1, y2, . . ., ys). We say that r is a smooth path through m M if r(t0) = m for some t0.

We can specify a path in M at m by its coordinates:

y₁ = y₁(t)
y₂ = y₂(t)
    . . .
y_s = y_s(t),

where m is the point (y₁(t₀), y₂(t₀), . . . , y_s(t₀)). Equivalently, since the ambient and local coordinates are functions of each other, we can also express a path--at least that part of it inside a coordinate neighborhood--in terms of its local coordinates:

x¹ = x¹(t)
x² = x²(t)
    . . .
xⁿ = xⁿ(t).

Examples 3.2
(a) (borrowed from Example 1.7 in Lecture 1) Straight lines in E₃
(b) (from Example 1.7 in Lecture 1) Helix in a cylinder radius r embedded in E₃
(c) A smooth path in Sⁿ

Definition 3.3 A tangent vector at m M Er is a vector v in Er of the form v = y'(t0) for some path y = y(t) in M through m (with y(t0) = m.

Examples 3.4
(a) Let M be the surface y₃ = y₁² + y₂², which we paramaterize by

y₁ = x¹
y₂ = x²
y₃ = (x¹)² + (x²)²

This corresponds to the single chart (U=M; x¹, x²), where

x¹ = y₁   and
x² = y₂.

To specify a tangent vector, let us first specify a path in M, such as

y₁ = t sin t
y₂ = t cos t
y₃ = t²

(Check that the equation of the surface is satisfied.) This gives the path shown in the figure.
Now we obtain a tangent vector field along the path by taking the derivative:

dy1 dt

,

dy2 dt

,

dy3 dt

=

(t cos t + sin t,   - t sin t + cos t,   2t)

(To get actual tangent vectors at points in M, evaluate this at a fixed point t₀.)

Note We can also express the coordinates xⁱ in terms of t:

x¹ = y₁ = t sin t
x² = y₂ = t cos t,

giving

dx1 dt

,

dx2 dt

=

(t cos t + sin t,   -t sin t + cos t),

since xⁱ = y_i for this manifold. We also think of this as the tangent vector, given in terms of the local coordinates. A lot more will be said about the relationship between the above two forms of the tangent vector below.

Algebra of Tangent Vectors: Addition and Scalar Multiplication

The sum of two tangent vectors is, geometrically, also a tangent vector, and the same goes for scalar multiples of tangent vectors. However, we have defined tangent vectors using paths in M, and we cannot produce these new vectors by simply adding or scalar-multiplying the corresponding paths: if y = f(t) and y = g(t) are two paths through m Ž M where f(t₀) = g(t₀) = m, then adding them coordinate-wise need not produce a path in M. However, we can add these paths using some chart as follows.

Choose a chart x at m, with the property (for convenience) that x(m) = 0. Then the paths x(f(t)) and x(g(t)) (defined as in the note above) give two paths through the origin in coordinate space. Now we can add these paths or multiply them by a scalar without leaving coordinate space and then use the chart map to lift the result back up to M. In other words, define

(f+g)(t) = x^-1(x(f(t)) + x(g(t))
and(Âf)(t) = x^-1(Âx(f(t))).

The above constructions turn T_m into a vector space.

Let us return to the issue of the two ways of describing the coordinates of a tangent vector at a point m M: writing the path as y_i = y_i(t) we get the ambient coordinates of the tangent vector:

y'(t0) =

dy1 dt

, ... ,

dys dt

t=t0

Ambient coordinates

x'(t0) =

dx1 dt

, ... ,

dxn dt

t=t0

Local coordinates

Question In general, how are the dxⁱ/dt related to the dy_i/dt?

Answer By the chain rule,

dy1 dt = y1 x1 dx1 dt   +   y1 x2

dx2 dt

and similarly for dy₂/dt and dy₃/dt. Thus, we can recover the original three ambient vector coordinates from the local coordinates. In other words, the local vector coordinates completely specify the tangent vector.

Note The chain rule as used above shows us how to convert local coordinates to ambient coordinates and vice-versa:

From now on, we shall omit the summation signs, and use the Einstein Summation Convention:

Thus,

	n k=1 yi xk	vk	becomes	yi xk	vk	(because the index k repeats)
and
	s k=1 xi yk	vk	becomes	xi yk	vk	(again because the index k repeats).

Examples 3.4 Contd.
(b) Take M = E_n, and let v be any vector in the usual sense with coordinates Œⁱ. Choose x to be the usual chart xⁱ = y_i. If p = (p¹, p², . . . , pⁿ) is a point in M, then v is the derivative of the path

x¹ = p¹ + t¹
x² = p² + t²;
    . . .
xⁿ = pⁿ + tⁿ

dxi dt

=

i,

(c) Let M = S², and the path in S² given by

y₁ = sin t
y₂ = 0
y₃ = cos t

This is a path (circle) through m = (0, 0, 1) following the line of longitude x² = 0, and has tangent vector

dy1 dt

,

dy2 dt

,

dy3 dt

=

(cost,   0,   -sint)

=

(1, 0, 0) at the point m.

(c) We can also use the local coordinates to describe a path; for instance, the path in part (b) can be described using spherical polar coordinates by

x¹ = t
x² = 0

The derivative

dx1 dt

,

dx2 dt

=

(1, 0)

give the local coordinates (the coordinates of its image in coordinate Euclidean space).

(e) In general, if (U; x¹, x², . . . , xⁿ) is a coordinate system near m, then we can obtain paths y_i(t) by setting

xj

=

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

,

where the constants are chosen to make xⁱ(t₀) correspond to m. (The paths in parts (c) and (d) are examples of this.) To view this as a path in M, we just apply the parametric equations y_i = y_i(x^j), giving the y_i as functions of t.

The associated tangent vector at the point where t = t₀ is called /xⁱ. It has local coordinates

vj = dxj dt t=0

=

1   if j = i 0 if j i

=

ij

_i^j is called the Kronecker Delta, and is defined by

ij

=

1   if j = i 0 if j i

=

ij

.

Question Which matrix has ij entry equal to _i^j?
Answer

Question Using the Einstein summation convention, evaluate vⁱ_i^j.
Answer

We can now get the ambient coordinates by the above conversion formula (we are using the Einstein summation convention from this point on):

vj

=

yj xk

vk

=

yj xk

ik

=

yj xi

We call this vector /xⁱ. Summarizing,

Definition of xi xi is the vector whose local coordinates are given by j th coordinate = xi j     = ij = xj xi . Its ambient coordinatres are given by j th coordinate = yj xi . What do these strange vectors look like?

Now that we have a better feel for local and ambeinet coordinates of vectors, let us state some more "general nonsense": Let M be an n-dimensional manifold, and let m M.

Proposition 3.6 (The Tangent Space) There is a linear one-to-one correspondence between tangent vectors at m and plain old vectors in En. In other words, the tangent space "looks like" En.

Click here for a proof (which will also explain why local coordinates are better than ambient ones).

Question Wait a minute! Isn't that obvious from the picture? The tangent space is just an n-dimensional plane, and all n-dimensional planes are just copies of n-dimensional space!

Answer Geometrically, that seems true -- at least in three dimensions. But remember, we have defined the tangent space at m M as the set of tangent vectors to paths through m. How can we be so sure that this 1-1 correspondence works (a) with this definition, and (b) in arbitrary s-dimensional space? That is why we really need the proof.

Note Under the one-to-one correspondence in the proposition, the standard basis vectors in E_n correspond to the tangent vectors /x¹, /x², . . . , /xⁿ. Therefore, the latter vectors are a basis of the tangent space T_m.

Exercise Set 3

1. Suppose that v is a tangent vector at m Ž M with the property that there exists a local coordinate system xⁱ at m with vⁱ = 0 for every i. Show that v has zero coordinates in every coefficient system, and that, in fact, v = 0.

2. (a) Calculate the ambient coordinates of the vectors / and / at a general point on S², where ¿ and ú are spherical polar coordinates ( = x¹, = x²).
(b) Sketch these vectors at some point on the sphere.

3. Prove that

i

=

xj i

xj

4. Consider the torus T² with the chart x given by

y₁ = (a+b cos x¹)cos x²
y₂ = (a+b cos x¹)sin x²
y₃ = b sin x¹