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

Vectors and Co-vectors

  1. Dec 8, 2007 #1
    Okay, So I have am elementary question to ask but it is of fundamental importance to me. First things first, I have been looking through the posts on "the difference between vectors and covectors'' and found them to be helpful. But not too conducive to the way I am trying to learn about them. The posts seem to revolve around tangent and cotangent spaces. Although I will eventually go on to use my definition of covectors and vectors to define natural bases on manifolds, I am trying to ascertain a ``stand alone'' version of the defintion of covectors and vectors.

    I have begun with good ol' reliable [itex]\mathbb{R}^3[/itex] for my vector space: Let us define a vector space [itex]V[/itex] such that:

    [tex]
    V=\mathbb{R}^3
    [/tex]

    [tex]V[/tex] is the set

    [tex]
    V=\{ v:v^i e_i=(v^1,v^2,v^3)^T | v^i \in \mathbb{R}^3 \}
    [/tex]

    with basis, say

    [tex]
    e_1=\left(
    \begin {array}{c}
    1 \\
    \noalign{\medskip}
    0 \\
    \noalign{\medskip}
    0 \\
    \end {array}
    \right)
    [/tex]

    [tex]
    e_2=\left(
    \begin {array}{c}
    0 \\
    \noalign{\medskip}
    1 \\
    \noalign{\medskip}
    0 \\
    \end {array}
    \right)
    [/tex]


    [tex]
    e_3=\left(
    \begin {array}{c}
    0 \\
    \noalign{\medskip}
    0 \\
    \noalign{\medskip}
    1 \\
    \end {array}
    \right),
    [/tex]

    To be more explicit, let me define what the vector is, say

    [tex]
    v=(1,2,-5)^T
    [/tex]

    So that [itex]v^1=1[/itex], [itex]v^2=2[/itex] and [itex]v^3=-5[/itex]. And so that:

    [tex]
    v=1\cdot e_1+2\cdot e_2-5\cdot e_3
    [/tex]

    Now define [itex]V^*[/itex], a space dual to [itex]V[/itex], by its elements [itex]f[/itex];
    [tex]
    V^*= \{ f: f=f_i e^i=(x,y,z) \}
    [/tex]

    so that [itex]f_1=x[/itex], [itex]f_2=y[/itex] and [itex]f_3=z[/itex].

    with (covariant) basis:


    [tex]
    e_1=\left(
    \begin {array}{ccc}
    1 & 0 & 0 \\
    \end {array}
    \right)
    [/tex]

    [tex]
    e_2=\left(
    \begin {array}{ccc}
    0 & 1 & 0 \\
    \end {array}
    \right)
    [/tex]

    [tex]
    e_3=\left(
    \begin {array}{ccc}
    0 & 0 & 1 \\
    \end {array}
    \right)
    [/tex]


    where it is demanded that

    [tex]e^i(e_j)=\delta^i_j[/tex].


    Further, if we want to know how the [itex]f \in V^*[/itex] acts on the [itex]v \in V[/itex], we must derive a relation:

    [tex]
    f(v)=f(v^i e_i)=v^i f(e_i)=v^i \delta^j_i f(e_j)
    [/tex]

    But by our previous demand we have:

    [tex]
    f(v)=v^i (e^j(e_i)) f(e_j)
    [/tex]

    By linearity we have:

    [tex]
    f(v)=v^i (e^j(e_i)) f(e_j)
    [/tex]

    Now [itex]v^i,f(e_j) \in \mathbb{R}[/itex] so we can just shift them around at will.

    [tex]
    f(v)=f(e_j) v^i (e^j(e_i))=f(e_j) v^i (e^j(e_i))
    [/tex]

    [tex]
    =(f(e_j) v^i e^j)(e_i)=(f(e_j) e^j)(v^i e_i)=(f(e_j) e^j)(v)
    [/tex]


    As this is true [itex]\forall v \in V[/itex] we must have:

    [tex]}
    f \equiv f(e_j) e^j
    [/tex]

    For notational purposes we define [itex]f_j=f(e_j)[/itex]. So that;
    [tex]
    f \equiv f_j e^j
    [/tex]


    So back to the problem at hand:

    [tex]
    f(v) = f_je^j(v)=f_1 e^1(v)+f_2 e^2(v)+f_3 e^3(v)
    [/tex]

    [tex]
    f(v) = f_je^j(1\cdot e_1+2\cdot e_2-5\cdot e_3)
    [/tex]

    [tex]
    =f_1 e^1(1\cdot e_1+2\cdot e_2-5\cdot e_3)+f_2 e^2(1\cdot e_1+2\cdot e_2-5\cdot e_3)+f_3 e^3(1\cdot e_1+2\cdot e_2-5\cdot e_3)
    [/tex]

    [tex]=f_1 e^1(1\cdot e_1)+f_2 e^2(2\cdot e_2)+f_3 e^3(-5\cdot e_3)[/tex]

    [itex]=f_1 (1)+f_2 (2 )+f_3 (-5)[/itex]

    And so

    [itex]f(v)=x (1)+y (2 )+z (-5)=x+2y-5z[/itex]

    So [tex]f(v)[/tex] is a plane.


    Right so my questions are:


    1. What does [itex]f(v)[/itex] being a plane mean?

    2. I know that [itex]f(e_j)=f_j [/itex] is just notation, and that it's form may be deduced from the given expression for [itex]f[/itex] and the fact that the bases of [itex]V[/itex] and [itex]V^*[/itex] abide [itex]e^i(e_j)=\delta^i_j[/itex] but what does [itex]f(e_j)[/itex] mean? Is it just [itex]f[/itex] acting on the basis elements of [itex]V[/itex]? I mean, If i try to work out what [itex]f(e_k)[/itex] is from [tex]f(v)=f(e_j)e^j(v)[/itex], we just get a cyclic definition [itex]f(e_k)=f(e_j)e^j(e_k)=f(e_j)\delta^k_j=f(e_k)[/itex]. And if so, I am finding it hard to define, say [itex]f(e_3)\equiv f_3=z[/itex]. I mean would this be a valid description:

    [itex]f(e_i)[/itex] is "all of [itex]f[/itex]" acting on the i-th basis component of the corresponding vector space. It is defined by producing the i-th component of the covector [itex]f[/itex]

    If anyone can clarify I'd be ever so grateful.

    3. The form of [itex]f[/itex] I chose, relates to some sort of cartesian projection I think. Could someone shed some light on the situation.


    Cheers,



    edit: adjusted as requested.
     
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 8, 2007 #2

    Chris Hillman

    User Avatar
    Science Advisor

    As a courtesy to other PF readers, I'd encourage you to learn how to take advantage of the latex markup feature in VB, the software used in PF; see this primer. After reading a few lines from that page, you should have no trouble reformatting your document.
     
  4. Dec 8, 2007 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If V is a vector space over R, then it's dual space [itex]V^*[/itex] is, by definition, the space of all linear transformations from V to R. (also called linear functionals)

    If we are studying a particular vector space, then it is conventional to call an element of its dual space a "covector".


    If we choose a basis for V, then there is a natural choice of basis for [itex]V^*[/itex]; we call it the dual basis. And, generally, we write the coordinates of a vector as a column array and the coordinates of a covector as a row array. The reason is as follows:

    Let [itex]v \in V[/itex] and [itex]\omega \in V^*[/itex].
    Let [itex][v][/itex] be the coordinate representation of v with respect to our chosen basis, and similarly for [itex][\omega][/itex].

    Then, we have:
    [tex][\omega(v)] = [\omega][v][/tex]

    (The left hand side is the 1x1 matrix containing the number [itex]\omega(v)[/itex]. The right hand side is the product from matrix algebra, which yields a 1x1 matrix)
     
    Last edited: Dec 8, 2007
  5. Dec 8, 2007 #4
    >> If V is a vector space over R, then it's dual space [itex]V^*[/itex] is, by definition, the space of all linear transformations from V to R. (also called linear functionals)

    Okay

    >> If we are studying a particular vector space, then it is conventional to call an element of its dual space a "covector".

    Okay


    >> If we choose a basis for V, then there is a natural choice of basis for [itex]V^*[/itex]; we call it the dual basis. And, generally, we write the coordinates of a vector as a column array and the coordinates of a covector as a row array.

    Uh huh

    >> The reason is as follows:

    Let [itex]v \in V[/itex] and [itex]\omega \in V^*[/itex].
    Let [itex][v][/itex] be the coordinate representation of v with respect to our chosen basis, and similarly for [itex][\omega][/itex].

    Then, we have:
    [tex][\omega(v)] = [\omega][v][/tex]

    (The left hand side is the 1x1 matrix containing the number [itex]\omega(v)[/itex]. The right hand side is the product from matrix algebra, which yields a 1x1 matrix)


    Okay

    All that makes sense. It has given me some strength to know I am on the right track. However, my questions are still open, as far as my mind can stretch presently.
     
    Last edited: Dec 8, 2007
  6. Dec 23, 2007 #5
    For your first question, i give you an intutive sense of what 'the plane' means.
    Assuming we are unknown of what shape the earth has. But by some experiments, we derive a function f(v), which return the potential energy for an object, where v could be any point on earth. Then f(v)=0 is where all points with zero potential energy. As you obtain that f(v)=0 is a plane, we may say the earth is flat.
    The above explanation may not be strict. If i made any mistakes, please tell me.
     
  7. Dec 25, 2007 #6
    Not just arrows sitting at an origin are examples of vectors. There are other objects that also form vector spaces. Take a set of equally spaced lines in the plane. Enumerate the lines.

    Can you think of what it means to:


    • Multiply such a set of lines with a scalar? (easy)

    • Add two such sets of lines together? (slightly more complicated)
    What does the resulting set of lines look like in each case?

    Also, think a little about linear coordinate systems. Not in the way of basis vectors, but rather in the form of a parallelogramic grid. What does it mean to read off the coordinates of a vector from such a grid?
     
    Last edited: Dec 25, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Vectors and Co-vectors
  1. Vector or not vector? (Replies: 42)

  2. Are 4-vectors vectors? (Replies: 1)

  3. Parallel vectors (Replies: 3)

  4. Derivatives of vectors (Replies: 11)

Loading...