Manifolds – Part 2

by Octavian on Sunday, January 10, 2010

— 1. Some objects on the manifolds —

— 1.1. Curves and functions —

When I was in the beginner physics class, I often analyzed the track of a particle on a plane that moves under some potential. Ordinarily, the curve on the plane can be parameterized using one parameter, say ${t}$ , such that every points on the curve can be marked with some definite value of this parameter ${t}$ . We also can make an arbitrary choice to place the zero of ${t}$ ; nature will not know which point you identify with the ${t = 0}$ . If I make a coordinate system on this plane (just imagine for simplicity that this is a two-dimensional plane and we use Cartesian system for the coordinate), then a point on the curve has the coordinate

$\displaystyle (x,y) = (x(t), y(t)) \ \ \ \ \ (1)$

such that if you have a single number ${t}$ , then you will have two numbers, one for ${x}$ and one for ${y}$ , and it gives you the precise position of the point on the curve (this is why we often say that curve is one-dimensional). You can make an arbitrary curve on this plane which touches every parts of this plane, and by principal, you always can identify its points using this single parameter ${t}$ .

However, this advantage to use the powerful parameter ${t}$ to mark points on the curve cannot hold peacefully if we make a generalization from the plane to a manifold. The problem is, we cannot make a global coordinate system which covers all parts of manifold. Instead, we need to define some local coordinate systems, provided that we still can transform each other using continuous mapping. Therefore, the principal that we can identify the points on the curve by using coordinate ${(x,y) = (x(t), y(t))}$ doesn’t make sense anymore, since two neighborhoods on the manifold have different convention about the values of ${x}$ and ${y}$ .

To solve this problem, we need to make two mappings. First, consider the mapping ${c: [a,b] \in {\mathbb R} \rightarrow M}$ such that ${c(s) = c(t)}$ whenever ${s = t \in {\mathbb R}}$ . We can imagine that this is the curve which connects two points on the manifold, namely ${p = c(a)}$ and ${q = c(b)}$ , and it is a simple one; there is no intersection of this curve with itself (this assumption is just for the sake of simplicity). Of course you don’t need to have finite numbers for ${a}$ and ${b}$ , the two queens of pain ( ${-\infty}$ and ${\infty}$ ) will also work. While the first mapping is the one which maps ${{\mathbb R}}$ to ${M}$ , the second mapping is ${\phi: M \rightarrow {\mathbb R}^n}$ , i.e. the mapping that serves as the coordinates marker for each points on the manifold. Therefore, we can construct the composition mapping ${\phi \circ c: [a,b] \rightarrow {\mathbb R}^n : t \mapsto \{x^1, \ldots, x^n \}}$ , such that from a single parameter ${t}$ you will get ${n}$ values of coordinates.

If the part of ${{\mathbb R}}$ is mapped to ${M}$ and produces what we intuitively call as curve, then the mapping from ${M}$ to ${{\mathbb R}}$ makes a very different object: this is simply the ordinary function defined on a manifold. Concretely, if we have a function ${f: M \rightarrow {\mathbb R}}$ , and take a point ${p \in M}$ with the coordinate ${\phi(p)}$ , then we will have the composition mapping ${f \circ \phi^{-1}: {\mathbb R}^n \rightarrow {\mathbb R}}$ . In physics, we often encounter this composition mapping ${f \circ \phi^{-1}}$ , which relates the coordinates of points on the space with a value in each points. Actually, this physicist’s view will not make any trouble if we can define a global coordinate system on the manifold, but if we just could make a system of local coordinates, then we need to distinguish the role of ${f}$ , the function from points of the manifold to some real numbers, and the role of ${\phi}$ , the attachment of points to a local coordinate system.

— 1.2. Vector and tensor fields —

Suppose we have an ${n}$ -dimensional manifold ${M}$ , and a neighborhood ${U \subset M}$ which contains ${p \in M}$ . Then, consider another point ${q \in U}$ near ${p}$ such that the coordinate of ${q}$ differs with ${p}$ by a small number ${\varepsilon^\mu}$ , i.e.

$\displaystyle y^\mu = x^\mu + \varepsilon^\mu \ \ \ \ \ (2)$

where ${\{y^\mu\}}$ and ${\{x^\mu\}}$ are the coordinates of ${q}$ and ${p}$ respectively. Then, if we have a function defined over the manifold, ${f:M \rightarrow {\mathbb R}}$ , and if the value of this function at ${p}$ is ${f(p)}$ , then its value at ${q}$ can be expanded using Taylor series as

$\displaystyle f(q) = f(p) + \varepsilon^\mu \frac{\partial f}{\partial x^\mu} \vert_p + \cdots \ \ \ \ \ (3)$

From equation above, we can see that the difference between ${f(q)}$ and ${f(p)}$ in the first order is ${\varepsilon^\mu \frac{\partial f}{\partial x^\mu}}$ , the directional derivative of the function ${f}$ (i.e. the derivative of ${f}$ on the direction along the curve which connects ${p}$ and ${q}$ ). We can guess that this directional derivative does not depend on the coordinate system we use; for if we are given a function ${f}$ on a manifold ${M}$ , the value of directional derivative of ${f}$ at a certain point on ${M}$ is coordinate independent. Then, we can also guess that

$\displaystyle \varepsilon^\mu \frac{\partial}{\partial x^\mu} \ \ \ \ \ (4)$

is the tangential vector of the curve which connects points ${p}$ and ${q}$ , by setting ${\lim \varepsilon \rightarrow 0}$ .

Being the vector, we can view the ${\varepsilon^\mu}$ in equation above as its component, and ${\frac{\partial}{\partial x^\mu}}$ as the basis. This choice of basis vectors is called the coordinate basis, since it depends on the orientation of axis of coordinate system we make (well, remember that we can construct the basis vectors independently without noticing the axis orientation of the coordinate system being used). Therefore, this basis vectors replace our old basis vectors ${\hat{i}}$ , ${\hat{j}}$ and ${\hat{k}}$ of the Cartesian coordinate system.

One thing to remember for this construction of vector is about the comparison of two vectors. On the plane, we can say that two vectors on two different points are identic if they are parallel and their components are the same. On manifold, the first statement still holds (that two vectors are the same if their directions are the same), but the second statement fails. We cannot compare the components of two vectors on two different points because the coordinate systems we use are generally different at those points. Although the vectors are parallel, their components are different. Mathematically, if we have two parallel vectors ${A = A^\mu \frac{\partial}{\partial x^\mu}}$ at ${a \in M}$ and ${B = B^\mu \frac{\partial}{\partial y^\mu}}$ at ${b \in M}$ , then

$\displaystyle A = A^\mu \frac{\partial}{\partial x^\mu} = B^\mu \frac{\partial}{\partial y^\mu} = B \ \ \ \ \ (5)$

but generally ${A^\mu \neq B^\mu}$ . Since the transformation between these two basis vectors is

$\displaystyle \frac{\partial}{\partial y^\mu} = \frac{\partial x^\nu}{\partial y^\mu} \frac{\partial}{\partial x^\nu} \ \ \ \ \ (6)$

then the equation which relates the components of ${A}$ and ${B}$ is

$\displaystyle A^\nu = \frac{\partial x^\nu}{\partial y^\mu} B^\mu \ \ \ \ \ (7)$

The equations above are the mathematical versions of two statements I said previously.

Take a specific point ${p \in M}$ . If there is a curve through ${p}$ , there is a vector tangential to this curve at ${p}$ (sounds like “if there is a will, there is a way” huh?). Therefore, if we can draw all possibilities of curves through ${p}$ , we can draw all vectors which are tangential to manifold ${M}$ at ${p}$ . The vector space which is spanned by the set of tangential vectors at ${p}$ is denoted as ${T_pM}$ , and this vector space coincides with the locally flat space which we can create around ${p}$ . Of course, the tangential vector space at ${q}$ , ${T_qM}$ , is generally different with ${T_pM}$ . They are same only if our manifold is flat.

We also can create another object on manifold which we will name as the dual vector. Shortly, the dual vector (or, people often call it as one-form) ${\omega}$ can be defined as the mapping that takes a vector ${V}$ into a real number ${\omega(V)}$ . The dual vector ${\omega}$ can be stated explicitly as

$\displaystyle \omega = \omega_\mu dx^\mu \ \ \ \ \ (8)$

where the ${\omega_\mu}$ is the components of ${\omega}$ and ${dx^\mu}$ is its basis. What is the advantage of making this definition? Well, the statement that ${\omega}$ is a mapping from vector to a real number can be translated to the statement that there exists inner product between the dual vector and the vector, which has a real number as the result, i.e.

$\displaystyle \langle \omega, V \rangle = \omega(V) = \text{real number} \ \ \ \ \ (9)$

By using the coordinate representation, we will have

$\displaystyle \begin{array}{rcl} \langle \omega, V \rangle &=& \langle \omega_\mu dx^\mu, V^\nu \frac{\partial}{\partial x^\nu} \rangle = \omega_\mu V^\nu \langle dx^\mu, \frac{\partial}{\partial x^\nu} \rangle \\ & \equiv & \omega_\mu V^\nu \frac{\partial x^\mu}{\partial x^\nu} = \omega_\mu V^\nu \delta^\mu_\nu = \omega_\mu V^\mu \end{array}$

The space which is spanned by the dual vectors at ${p}$ is denoted as ${T_p^\ast M}$ , and it is called the dual space of ${T_pM}$ . Please note that the inner product we defined above is between the dual vector and the vector, not between two vectors. Don’t be trapped by the ordinary inner product between two vectors; this ordinary inner product actually can give us the length of vector. Here, there is no such definition for the length of something and distance between two points. What we only have is the components of the vector, not its length. The discussion about the length of vector and the distance between two points on the manifold is appropriate only if we have a metric on the manifold. What is metric? We will discuss it next.

Now, after we successfully construct the vectors and dual vectors, we can make a straightforward generalization to an object called tensor. If we define the dual vector as a mapping from vector to real number, then we define the tensor of type ${(r,s)}$ as a mapping from ${r}$ dual vectors and ${s}$ vectors to a real number. Then, we can say that the vector is a tensor of type ${(1,0)}$ (because if we have a vector, then we need one dual vector to make a real number) and the dual vector is a tensor of type ${(0,1)}$ . If the coordinate representation of vector ${V}$ is ${V = V^\mu \frac{\partial}{\partial x^\mu}}$ and dual vector ${\omega}$ is ${\omega = \omega_\mu dx^\mu}$ , then the ${(r,s)}$ -tensor ${T}$ is

$\displaystyle T = {T^{\mu_1 \cdots \mu_r}}_{\nu_1 \cdots \nu_s} \frac{\partial}{\partial x^{\mu_1}} \otimes \cdots \otimes \frac{\partial}{\partial x^{\mu_r}} \otimes dx^{\nu_1} \otimes \cdots \otimes dx^{\nu_s} \ \ \ \ \ (10)$

such that if we have ${r}$ dual vectors ${\omega_1, \ldots, \omega_r}$ and ${s}$ vectors ${V_1, \ldots \, V_s}$ , the mapping ${T}$ is

$\displaystyle T(\omega_1, \ldots, \omega_r, V_1, \ldots \, V_s) = {T^{\mu_1 \cdots \mu_r}}_{\nu_1 \cdots \nu_s} {\omega_1}_{\mu_1}, \ldots, {\omega_r}_{\mu_r} {V_1}^{\nu_1}, \ldots \, {V_s}^{\nu_s} \ \ \ \ \ (11)$

Last but not least, if we have a tensor ${T}$ on the manifold ${M}$ , and we make two different coordinate systems ${\{x^\mu\}}$ and ${\{y^\mu\}}$ on the neighborhood ${U}$ of point ${p}$ , then

$\displaystyle \begin{array}{rcl} T &=& {T^{\mu_1 \cdots \mu_r}}_{\nu_1 \cdots \nu_s} \frac{\partial}{\partial x^{\mu_1}} \otimes \cdots \otimes \frac{\partial}{\partial x^{\mu_r}} \otimes dx^{\nu_1} \otimes \cdots \otimes dx^{\nu_s} \\ &=& {T^{\alpha_1 \cdots \alpha_r}}_{\beta_1 \cdots \beta_s} \frac{\partial}{\partial y^{\alpha_1}} \otimes \cdots \otimes \frac{\partial}{\partial y^{\alpha_r}} \otimes dy^{\beta_1} \otimes \cdots \otimes dy^{\beta_s} \end{array}$

such that the transformation between the components of tensor ${T}$ is

$\displaystyle {T^{\mu_1 \cdots \mu_r}}_{\nu_1 \cdots \nu_s} = \frac{\partial x^{\mu_1}}{\partial y^{\alpha_1}} \cdots \frac{\partial x^{\mu_r}}{\partial y^{\alpha_r}} \frac{\partial y^{\beta_1}}{\partial x^{\nu_1}} \cdots \frac{\partial y^{\beta_s}}{\partial x^{\nu_s}} {T^{\alpha_1 \cdots \alpha_r}}_{\beta_1 \cdots \beta_s} \ \ \ \ \ (12)$

Any indexed object on the manifold which transforms like the equation above is a tensor. It’s why we commonly use the coordinate transformation, which is simple and easy to use, as the defining method to distinguish a tensor, not its original definition as a mapping from a number of vectors and a number of dual vectors to a real number.

From → Differential Geometry, The Theory of Space and Gravitation

8 Comments →

Pradisty permalink

kak, kita di GR kan baru belajar manifold, nanti ada kaitannya sama black hole, event horizon & singularity ga?

Reply
- Octavian permalink
  
  ya nanti kita juga belajar itu kalau waktunya cukup
  
  Reply
  - WV7VS permalink
    
    Lanjutkan!
juanmarqz permalink

Hi Andy, a quizz: how do you called the convention of sub- and super-indexing tensor’s components that you are using here above in (10), (11) and (12) formulas?… maybe Einstein-Penrose?

Reply
- Octavian permalink
  
  Wow it’s really hard quiz juanmarqz, because I never take it too seriously before. What I know that if we use this rule then everything is so easy. Since tensor is the mapping of some vectors and some dual vectors then the tensor itself must be the direct product of each basis which spans each vector spaces and dual vector spaces. I just wonder, is there another convention to write a tensor? Interesting!
  
  Reply
juanmarqz permalink

Agree, the reason why this way is usefulness and my propose came from
http://en.wikipedia.org/wiki/Abstract_index_notation... how do you see this?
greets from Mexico : )

Reply
juanmarqz permalink

http://en.wikipedia.org/wiki/Abstract_index_notation

Reply
Octavian permalink

Very interesting! When I studied tensor first, I thought that the index notation is like that way because it’s the only simple way that keeps every calculation about tensor becomes efficient. And at that time I also felt that my thought was such an obvious and no one ever took it too seriously. Well the Penrose graphical notation of tensor (http://en.wikipedia.org/wiki/Penrose_graphical_notation) is very nice (although catastrophic) ^^

Reply

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Manifolds – Part 2

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