## Lecture 3: Tangent Vectors and the Tangent Space

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)
. . .
ys = ys(t),

where m is the point (y1(t0), y2(t0), . . . , ys(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:

 dy1dt , dy2dt , dy3dt = (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

 dx1dt , dx2dt = (t cos t + sin t,   -t sin t + cos t),

since xi = yi 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))).
Taking their derivatives at the point t0 will, by the chain rule, produce the sum and scalar multiples of the corresponding tangent vectors. Since we can add and scalar-multiply tangent vectors

 Definition 3.5 If M is an n-dimensional manifold, and m M, then the tangent space at m is the set Tm of all tangent vectors at m.

The above constructions turn Tm 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 yi = yi(t) we get the ambient coordinates of the tangent vector:

 y'(t0) = dy1dt , ... , dysdt t=t0 Ambient coordinates
and, using some chart x at m, we get the local coordinates
 x'(t0) = dx1dt , ... , dxndt t=t0 Local coordinates

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

 dy1dt
= y1x1
 dx1dt
+    y1x2
 dx2dt

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:

Converting Between Local and Ambient Coordinates of a Tangent Vector

If the tangent vector V has ambient coordinates (v1, v2, . . . , vs) and local coordinates (v1, v2, . . . , vn), then they are related by the formulae

 vi = nk=1 yixk
vk
and
 vi = sk=1 xiyk
vk.

Note To obtain the coordinates of sums or scalar multiples of tangent vectors, simply take the corresponding sums and scalar multiples of the coordinates. In other words:

(v+w)i = vi + wi and (Âv)i = ÂvI
just as we would expect to do for ambient coordinates. (Why can we do this?)

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

 Einstein Summation Convention If an index appears twice in an expression, then summation over that index is implied.

Thus,

 nk=1 yixk
vk   becomes
yi

xk
vk  (because the index k repeats)
and
 sk=1 xiyk
vk   becomes
xi

yk
vk   (again because the index k repeats).

Examples 3.4 Contd.
(b) Take M = En, and let v be any vector in the usual sense with coordinates Œi. Choose x to be the usual chart xi = yi. 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
at t = 0. Thus this vector has local and ambient coordinates equal to each other, and equal to
 dxidt = i,
which are the same as the original coordinates. In other words, the tangent vectors are "the same" as ordinary vectors in En.

(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

 dy1dt , dy2dt , dy3dt = (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

 dx1dt , dx2dt = (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 yi(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 yi = yi(xj), giving the yi as functions of t.

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

 vj = dxjdt t=0
=
 1 if j = i 0 if j i
=  ij

ij 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 ij?

Question Using the Einstein summation convention, evaluate viij.

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

 vj = yjxk vk = yjxk ik = yjxi

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 = xjxi .

Its ambient coordinatres are given by

 j th coordinate = yjxi .

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 En correspond to the tangent vectors /x1, /x2, . . . , /xn. Therefore, the latter vectors are a basis of the tangent space Tm.

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 = xji 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
with 0 < xi < 2. Find the ambeint coordinates of the two orthogonal tangent vectors at a general point, and sketch the resulting vectors.

Last Updated: January, 2002