- #1
ConfusedMonkey
- 42
- 15
I have an issue with the definition of coordinate system in differential geometry vs the definition of coordinate system in linear algebra. The post is a bit long, but it's necessary so that I get my point across.
Let ##V## be an ##n##-dimensional normed space over the reals and equip ##V## with the topology that its norm induces. ##V## can be given a natural smooth structure, making it into a smooth ##n##-manifold. Now, let ##\{v_1, \dots, v_n\}## be an ordered basis for ##V##. Any ##p \in V## can be written as ##p = c^i v_i##, where we are using Einstein notation. This means that ##p## has the coordinate representation ##(c^1, \dots, c^n)##, relative to the given basis. This seems to define a coordinate system - not in the usual differential geometric sense, but if ##V = \mathbb{R}^n## then a basis gives us coordinate axes.
However, we also have the usual definition of a coordinate system about ##p##: The ordered pair ##(U, \varphi)## is a coordinate system about ##p## if ##U \ni p## and ##U## is open and ##\varphi## is a diffeomorphism onto some open subset of ##\mathbb{R}^n##. This allows us to naturally identify ##p## with ##(x^1(p), \dots, x^n(p))## where the ##x^i## are the local coordinates of ##\varphi##.
So it seems that the two definitions of coordinate systems above give us the same thing: a way to uniquely identify ##p## with a point of ##\mathbb{R}^n##, which is precisely what we want. However, the two definitions are not equivalent. Let me demonstrate:
Given that ##V## is a finite-dimensional vector space, I will make the usual identification of ##V## with ##T_p V## and just write ##V## instead. Similarly, even though ##p \in V##, it will also be used interchangeably as both an element of ##V## as well as ##T_p V##. Given a coordinate system ##(U, \varphi)##, it induces a coordinate basis at ##p##, and this is like a coordinate system in the first linear algebraic sense that I described. That's fine. In a differential geometric sense, ##p## is identified with ##(x^1(p), \dots, x^n(p))##. In a linear algebraic sense, we can write ##p = p^i \frac{\partial}{\partial x^i}## where ##p^i = p(x^i)##. The coordinate representation of ##p##, in the linear algebraic sense is then ##(p^1, \dots, p^n)## which is naturally identified with ##(x^1(p), \dots, x^n(p))##. So whether we are using the differential geometric or linear algebraic definition of coordinate system, we get the same identification ##p \leftrightarrow (x^1(p), \dots, x^n(p))##.
However, the two definitions gave the same identification only because we used a coordinate basis. From what I have previously read (I don't remember the source, but I am sure that you more knowledgeable posters will be aware of this), not every basis for ##V## is a coordinate basis. That is, there could be an ordered basis ##\{w_1, \dots, w_n\}## such that no coordinate chart induces it. This bothers me, because by giving an ordered basis ##\{w_1, \dots, w_n\}##, we indeed do have a coordinate system - every element of ##V## has a coordinate representation relative to the basis, BUT, this basis may not necessarily give rise to a coordinate chart. So now we have a coordinate system in one sense (the linear algebraic) but we do not have an equivalent coordinate system in the differential geometric sense. This bothers me a lot!
The differential geometric definition of coordinate system was conceived of for when there is no natural or useful linear algebraic definition of coordinate system: That is, for when we cannot identify a manifold with its tangent space. But in the case when the manifold is a finite dimensional normed space, we can identify the manifold with its tangent space (for example, ##\mathbb{R}^n \leftrightarrow T_p \mathbb{R}^n##), and so in this case, both definitions should be equivalent, i.e. give the same coordinate system, but they do not, as I just demonstrated. How do I reconcile this?
Let ##V## be an ##n##-dimensional normed space over the reals and equip ##V## with the topology that its norm induces. ##V## can be given a natural smooth structure, making it into a smooth ##n##-manifold. Now, let ##\{v_1, \dots, v_n\}## be an ordered basis for ##V##. Any ##p \in V## can be written as ##p = c^i v_i##, where we are using Einstein notation. This means that ##p## has the coordinate representation ##(c^1, \dots, c^n)##, relative to the given basis. This seems to define a coordinate system - not in the usual differential geometric sense, but if ##V = \mathbb{R}^n## then a basis gives us coordinate axes.
However, we also have the usual definition of a coordinate system about ##p##: The ordered pair ##(U, \varphi)## is a coordinate system about ##p## if ##U \ni p## and ##U## is open and ##\varphi## is a diffeomorphism onto some open subset of ##\mathbb{R}^n##. This allows us to naturally identify ##p## with ##(x^1(p), \dots, x^n(p))## where the ##x^i## are the local coordinates of ##\varphi##.
So it seems that the two definitions of coordinate systems above give us the same thing: a way to uniquely identify ##p## with a point of ##\mathbb{R}^n##, which is precisely what we want. However, the two definitions are not equivalent. Let me demonstrate:
Given that ##V## is a finite-dimensional vector space, I will make the usual identification of ##V## with ##T_p V## and just write ##V## instead. Similarly, even though ##p \in V##, it will also be used interchangeably as both an element of ##V## as well as ##T_p V##. Given a coordinate system ##(U, \varphi)##, it induces a coordinate basis at ##p##, and this is like a coordinate system in the first linear algebraic sense that I described. That's fine. In a differential geometric sense, ##p## is identified with ##(x^1(p), \dots, x^n(p))##. In a linear algebraic sense, we can write ##p = p^i \frac{\partial}{\partial x^i}## where ##p^i = p(x^i)##. The coordinate representation of ##p##, in the linear algebraic sense is then ##(p^1, \dots, p^n)## which is naturally identified with ##(x^1(p), \dots, x^n(p))##. So whether we are using the differential geometric or linear algebraic definition of coordinate system, we get the same identification ##p \leftrightarrow (x^1(p), \dots, x^n(p))##.
However, the two definitions gave the same identification only because we used a coordinate basis. From what I have previously read (I don't remember the source, but I am sure that you more knowledgeable posters will be aware of this), not every basis for ##V## is a coordinate basis. That is, there could be an ordered basis ##\{w_1, \dots, w_n\}## such that no coordinate chart induces it. This bothers me, because by giving an ordered basis ##\{w_1, \dots, w_n\}##, we indeed do have a coordinate system - every element of ##V## has a coordinate representation relative to the basis, BUT, this basis may not necessarily give rise to a coordinate chart. So now we have a coordinate system in one sense (the linear algebraic) but we do not have an equivalent coordinate system in the differential geometric sense. This bothers me a lot!
The differential geometric definition of coordinate system was conceived of for when there is no natural or useful linear algebraic definition of coordinate system: That is, for when we cannot identify a manifold with its tangent space. But in the case when the manifold is a finite dimensional normed space, we can identify the manifold with its tangent space (for example, ##\mathbb{R}^n \leftrightarrow T_p \mathbb{R}^n##), and so in this case, both definitions should be equivalent, i.e. give the same coordinate system, but they do not, as I just demonstrated. How do I reconcile this?