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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 vectorvalued function with coordinates (y_{1}, y_{2}, . . ., y_{s}).
We say that r is a smooth path through m M if r(t^{0}) = m for some t_{0}. 
We can specify a path in M at m by its coordinates:
where m is the point (y_{1}(t_{0}), y_{2}(t_{0}), . . . , y_{s}(t_{0})). Equivalently, since the ambient and local coordinates are functions of each other, we can also express a pathat least that part of it inside a coordinate neighborhoodin terms of its local coordinates:
Examples 3.2
(a) (borrowed from Example 1.7 in Lecture 1) Straight lines in E_{3}
(b) (from Example 1.7 in Lecture 1) Helix in a cylinder radius r embedded in E_{3}
(c) A smooth path in S^{n}
Definition 3.3 A tangent vector at m M E_{r} is a vector v in E_{r} of the form v = y'(t_{0}) for some path y = y(t) in M through m (with y(t_{0}) = m. 
Examples 3.4
(a) Let M be the surface y_{3} = y_{1}^{2} + y_{2}^{2}, which we paramaterize by
This corresponds to the single chart (U=M; x^{1}, x^{2}), where
To specify a tangent vector, let us first specify a path in M, such as
(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:
dy_{1} dt  ,  dt  ,  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 t_{0}.)
Note We can also express the coordinates x^{i} in terms of t:
giving
dx_{1} dt  ,  dt  =  (t cos t + sin t, t sin t + cos t), 
since x^{i} = 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 scalarmultiplying the corresponding paths: if y = f(t) and y = g(t) are two paths through m Ž M where f(t_{0}) = g(t_{0}) = m, then adding them coordinatewise 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))Taking their derivatives at the point t_{0} will, by the chain rule, produce the sum and scalar multiples of the corresponding tangent vectors. Since we can add and scalarmultiply tangent vectors
and(Âf)(t) = x^{1}(Âx(f(t))).
Definition 3.5 If M is an ndimensional manifold, and m M, then the tangent space at m is the set T_{m} of all tangent vectors at m. 
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:
and, using some chart x at m, we get the local coordinates
y'(t_{0}) = dy_{1}
dt, ... , dy_{s}
dt_{t=t0} Ambient coordinates
x'(t_{0}) = dx_{1}
dt, ... , dx_{n}
dt_{t=t0} Local coordinates
Question In general, how are the dx^{i}/dt related to the dy_{i}/dt?
Answer By the chain rule,


and similarly for dy_{2}/dt and dy_{3}/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 viceversa:
Converting Between Local and Ambient Coordinates of a Tangent Vector
If the tangent vector V has ambient coordinates (v_{1}, v_{2}, . . . , v_{s}) and local coordinates (v^{1}, v^{2}, . . . , v^{n}), then they are related by the formulae
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} = v^{i} + w^{i} and (Âv)^{i} = Âv^{I}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,
 v^{k}  becomes  x^{k}  v^{k}  (because the index k repeats)  
and  
 v_{k}  becomes  y_{k}  v_{k}  (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 Œ^{i}. Choose x to be the usual chart x^{i} = y_{i}. If p = (p^{1}, p^{2}, . . . , p^{n}) is a point in M, then v is the derivative of the path
x^{1} = p^{1} + t^{1}at t = 0. Thus this vector has local and ambient coordinates equal to each other, and equal to
x^{2} = p^{2} + t^{2};
. . .
x^{n} = p^{n} + t^{n}
dt  =  ^{i}, 
(c) Let M = S^{2}, and the path in S^{2} given by
This is a path (circle) through m = (0, 0, 1) following the line of longitude x^{2} = 0, and has tangent vector
dy_{1} dt  ,  dt  ,  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
The derivative
dx_{1} dt  ,  dt  =  (1, 0) 
give the local coordinates (the coordinates of its image in coordinate Euclidean space).
(e) In general, if (U; x^{1}, x^{2}, . . . , x^{n}) is a coordinate system near m, then we can obtain paths y_{i}(t) by setting
x^{j}  = 
 , 
where the constants are chosen to make x^{i}(t_{0}) 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 = t_{0} is called /x^{i}. It has local coordinates
 = 
 =  _{i}^{j} 
_{i}^{j} is called the Kronecker Delta, and is defined by
_{i}^{j}  = 
 =  _{i}^{j}  . 
Question Which matrix has ij entry equal to _{i}^{j}?
Answer
Question Using the Einstein summation convention, evaluate v^{i}_{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):
v_{j}  =  x^{k}  v^{k}  =  x^{k} 
_{i}^{k}  =  x^{i} 
We call this vector /x^{i}. Summarizing,
Its ambient coordinatres are given by

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 ndimensional manifold, and let m M.
Proposition 3.6 (The Tangent Space)
There is a linear onetoone correspondence between tangent vectors at m and plain old vectors in E_{n}. In other words, the tangent space "looks like" E_{n}. 
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 ndimensional plane, and all ndimensional planes are just copies of ndimensional 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 11 correspondence works (a) with this definition, and (b) in arbitrary sdimensional space? That is why we really need the proof.
Note Under the onetoone correspondence in the proposition, the standard basis vectors in E_{n} correspond to the tangent vectors /x^{1}, /x^{2}, . . . , /x^{n}. 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 x^{i} at m with v^{i} = 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 S^{2}, where ¿ and ú are spherical polar coordinates ( = x^{1}, = x^{2}).
(b) Sketch these vectors at some point on the sphere.
3. Prove that
^{i}  =  ^{i} 
x^{j} 
4. Consider the torus T^{2} with the chart x given by
y_{1} = (a+b cos x^{1})cos x^{2}with 0 < x^{i} < 2. Find the ambeint coordinates of the two orthogonal tangent vectors at a general point, and sketch the resulting vectors.
y_{2} = (a+b cos x^{1})sin x^{2}
y_{3} = b sin x^{1}
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