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I Coordinate system vs ordered basis

  1. May 27, 2017 #1
    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?
     
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  3. May 27, 2017 #2

    fresh_42

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    To be honest, I don't get your point. You have ##\mathbb{R}^n## as manifold, which you call ##V##, ##\mathbb{R}^n## as tangent space ##T_p(V)## and ##\mathbb{R}^n## as Euclidean space where the charts are from. So all of your basis are those of ##\mathbb{R}^n## and you get different representations: ##(c^1,\ldots , c^n)\, , \,(x^1(p),\ldots ,x^n(p))\, , \,(p^1,\ldots ,p^n)## and whatever the coordinates according to ##\{w_1,\ldots ,w_n\}## are. The basis ##\{\frac{\partial}{\partial x^1},\ldots ,\frac{\partial}{\partial x^n}\}## is a orthogonal basis, the others any. Fine until here. Then you write
    What should this mean? What is a coordinate basis? And "I have previously read" is a miserable source.
    What does "induces it" mean? If you have a basis of a vector space, then the coordinates are the components according to this basis. If you have a manifold, then basis means either one of the tangent space or one of the chart map. The rest are basis transformations, as you always have ##\mathbb{R}^n## everywhere in your example.

    The only interesting point is, that the basis vectors of the tangent space above are the directional derivatives of the standard basis in ##\mathbb{R}^n## which makes the notation a bit different than usual.
    A basis gives rise to a chart map? A chart maps a part of the manifold on a part of a Euclidean space which has a basis. What did you mean by the first basis you mentioned? If you have a different basis in the tangent space than for your charts, you also have a basis transformation. So what? If you have a basis in your manifold, which only makes sense in an example as the one above, where the manifold itself is a Euclidean space, then you can have again a different basis than the one for the charts. However, since all vector spaces here: ##V,T_p(V),\mathbb{R}^n## are the same, I would chose only one basis for all of them: ##(x^i) \textrm{ in }V \, , \,(\frac{\partial}{\partial x^i}) \textrm{ in }T_p(V)\, , \,(\mathfrak{e}^i)\textrm{ in }\mathbb{R}^n## are noted differently but are basically all the same.
     
  4. May 27, 2017 #3
    A coordinate basis is a basis for the tangent space which is induced by the coordinate chart. For example, if ##(U, \varphi)## is a coordinate chart for a smooth manifold ##M## containing a point ##p##, where ##\varphi = (x^1, \dots, x^n)##, then the coordinate basis for ##T_p M## induced by this chart is ##\{\frac{\partial}{\partial x^i}, \dots, \frac{\partial}{\partial x^n}\}##.

    Here is a source. See post #6: https://www.physicsforums.com/threads/classifications-of-basis-vectors.785676/

    Also, I asked the same question here: https://math.stackexchange.com/questions/2299345/coordinate-system-vs-ordered-basis
    and was told: "At a single, fixed point, any basis comes from a coordinate chart. But, yes, if you look at a (smooth) choice of basis throughout an open set, it'll only come from a coordinate chart if the pairwise Lie brackets ##[w_i,w_j]## all vanish."

    Anyway, my initial post may have been too long winded, obscuring my point. Let me try again.

    If we are in a finite dimensional vector space, then any ordered basis ##\{v_1, \dots, v_n\}## is a coordinate system, because for any ##p \in V## we can write ##p = c^i v_i## and so the coordinate representation of ##p## is ##(c^1, \dots, c^n)##.

    Now suppose we are in a general topological space. In general, it isn't a vector space and so we cannot just impose a coordinate system on the manifold by using a basis. However, what we can do is equip the topological space with local coordinate charts, making it into a manifold.

    So we have two definitions of coordinate systems. One that uses basis vectors, and one that uses coordinate charts. But, there are cases, such as ##\mathbb{R}^n## where we could use both definitions of "coordinate systems". We could use basis vectors to define a coordinate system or we could use smooth coordinate charts. However, the two definitions are not equivalent because while every coordinate chart gives a basis for ##\mathbb{R}^n## (we are still using the identification ##\mathbb{R}^n \leftrightarrow T_p \mathbb{R}^n##), not every basis for ##\mathbb{R}^n## has a corresponding coordinate chart. So if I wanted a coordinate chart on ##\mathbb{R}^n##, which definition do I use? That is what I am confused about - the fact that there are slightly differing notions for "coordinate system" in the case of ##\mathbb{R}^n##, or in the slightly more general case where the manifold is a finite-dimensional normed space.
     
    Last edited: May 27, 2017
  5. May 28, 2017 #4

    Orodruin

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    You seem to have several confusions. Let us try to clear them up.
    An ordered basis is not a coordinate system. However, it defines a coordinate system on the vector space where the coordinates are taken to be the components of a vector with respect to the ordered basis. Note that this will give you an affine coordinate system, which is only a subset of the possible coordinate systems. There is no reason to expect an arbitrary coordinate system to be affine.

    You need a topological space that is locally homeomorphic to ##\mathbb R^n##. Just any topological space will not do.

    As I said earlier, your first definition defines an affine coordinate system, it is not the most general type of coordinate system.

    You are contradicting yourself here. Every vector basis in a vector space does give a corresponding affine coordinate system. Note that the set of basis vectors for a (non-affine) coordinate system changes from point to point. For example, in spherical coordinates the position vector is ##r\vec e_r## everywhere because ##\vec e_r## depends on the position.

    There is only one definition of a coordinate chart, it needs to locally uniquely specify a point in the manifold. An affine coordinate system is a type of coordinate system (which includes the subset of Cartesian coordinates when the basis is orthonormal), not all coordinate systems are affine.
     
  6. May 28, 2017 #5
    I guess I will just have to live with the fact that not every ordered basis for Euclidean space will be a coordinate basis, and whether I use a linear algebraic or differential geometric coordinate system will depend on the case at hand.

    EDIT: I posted this at the same time you replied, Orodruin. I will read your post then reply to it some time tomorrow as it is very late where I am.
     
  7. May 28, 2017 #6
    I'm not sure whether it's appropriate entering into this discussion, but I'm following this thread since the opener started it. I have a question.
    What you mean here? e.g. in ##\mathbb{R}^3##, take a point ##p## with coordinates ##(x,y,z)##; if a vector has components ##(x,y,z)##, then the ordered basis ##(1,0,0), (0,1,0),(0,0,1)## defines an affine coordinate system at ##p##?
     
  8. May 28, 2017 #7

    Orodruin

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    Given an ordered basis (it does not need to be orthonormal) ##\vec w_i## in a vector space, you can use the ##n##-tuple of numbers ##\zeta^i## as coordinates such that an element in the vector space is written ##\vec v = \zeta^i w_i##. The basis itself is a basis, not a coordinate system.
     
  9. May 28, 2017 #8
    @Orodruin

    Yes, a basis is a basis and only defines a coordinate system, I was speaking loosely, which is not a good thing in mathematics. I have realized where I was going wrong. A basis ##\{v_i\}## for ##V## determines a smooth global chart, ##(V, \varphi)## on ##V## and this chart determines a global frame of coordinate vector fields on ##TV##. However, the key is that not every frame will be determined by a coordinate chart.
     
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