Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Non-coordinate basis for vector fields

  1. Sep 16, 2012 #1
    Im trying to read some mathematical physics and have problems with the understanding of vector fields. Th questions are regarding the explanations in the book "Geometrical methods of mathematical physics"..
    The author, Bernard Schutz, writes:

    "Given a coordinate system x^i, it is often convenient to adopt {∂/∂x^i} as a basis for vector fields."

    This seems like a good basis for the tangent space to a given point (which can be described by our coordinates) but the vector space of our vector fields on M is an infinite dimensional space.. Im not really sure what he means by the set {∂/∂x^i}?? The elements doesnt even seem to be vector fields.

    He writes that noncoordinate basis is a basis that can not be expressed by any coordinate system.. but when he shows that two vector fields not necessary commutes he actually expands them in the coordinate system! Im not really sure what he wants to show. (page 44.)

    I am really thankful for your effort!

    All the best!
  2. jcsd
  3. Sep 16, 2012 #2


    User Avatar
    Science Advisor
    Gold Member

    {∂/∂x^i} which runs from i=1 to D (where D is the dimension of your manifold M) forms a set of basis vectors, at each point, for the tangent space (also dimension D) at each point that the coordinate system is valid for.

    In other words, one should always evaluate each basis vector (differential operator) at a specific point on the manifold. And that set forms a basis for the vector space that IS the tangent space.
  4. Sep 16, 2012 #3
    First of all, you must realize that the situation is local. If we claim that the [itex]x^i[/itex] are a coordinate system, then this must be interpreted locally. That is, there is an open subset U of our manifold M such that [itex]x^i:U\rightarrow \mathbb{R}[/itex] form a coordinate system. As a consequence, the [itex]\frac{\partial}{\partial x^i}[/itex] are a vector field on U and not a global vector field on M.

    So, when we say that [itex]\frac{\partial}{\partial x^i}[/itex] are a basis for the vector fields, then this must be interpreted locally on an open subset U. Now, what do we mean with a basis. First of all, we should interpret it as follows: for each p in U, we have that [itex]\frac{\partial}{\partial x^i}\vert_p[/itex] is a basis of the tangent space. This is equivalent to saying that there exists smooth functions [itex]f_i:U\rightarrow \mathbb{R}[/itex] such that [itex]\sum_i f^i\frac{\partial}{\partial x^i}[/itex].

    Also, if we expand a vector field in a coordinate basis, then we do so locally. That is, we write [itex]X=\sum_i X^i\frac{\partial}{\partial x^i}[/itex] locally on a set U. This is not in general globally.
  5. Sep 16, 2012 #4
    Ah I see where the misstake was! I was expecting a basis for a vector space where the coefficients where scalars.. I.e. some kind of infinite set of basis vector fields that can span the space of all continuous vector fields.

    I see we are using smooth functions as "coefficients". He is not trying to construct the vector space, \Tau(M), Im thinkin of right? :)
  6. Sep 19, 2012 #5
    The confusion arises due to the the fact that in index-free notation of vector fields, tangent vectors are no longer "vectors in the conventional sense" but rather operators on smooth functions. Of course as already mentioned, tangent vectors refer only to a single neighborhood of the manifold, where their space exists tangentially.
  7. Sep 19, 2012 #6


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The coordinate vector fields form a finite dimensional basis - using simple scalars - for the tangent space at each point. The same is true for any set of vector fields that are lineally independent at each point of their domain.

    They do not form a finite dimensional vector space basis for the vector fields over a domain. They form a basis for a finitely generated module over the ring of smooth functions on the domain.

    Over an entire manifold, there may be no basis for the vector fields.
    For instance on the sphere, every vector field must have at least one zero so no two vector fields can be everywhere linearly independent.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook