- #1
cathalcummins
- 46
- 0
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.
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: