section 2.11 on dual spaces
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A section from Phil's Wedge World tensor document, marked as installed and not to be edited. It defines V* as the space of linear functionals on V and builds the dual basis λi from the reciprocal vectors ei. It expands general vectors and rank-2 and rank-k tensors in V*, and shows how their components transform covariantly. Evaluating the tensors on vectors gives scalar fields.
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2.11 The dual space V*
In our treatment of the dual space side of things, we shall attempt to use Greek or script letters for all dual space vectors and tensors encountered. Vectors in V will continue to be represented by lower-case Latin letters.
Just as we used V as our prototype vector in space V, here we use V as our prototype vector in space V* .
The dual space V* is by definition the space of linear functionals over V. If V ϵ V*, we can then write
V : V → K V(v) = k ϵ K (2.11.1)
where K is any field such as the reals. Since V is a linear functional, V(v) is a linear function.
In normal calculus, if f: V → R, one refers to f as a function, and f(v) as that function evaluated at some point in V, though loosely speaking f(v) is also called a function. To emphasize the distinction, we shall refer to f as a "functional" and f(x) as a "function".
The basis λi
Given the n vectors {ei} which are a basis for V, one can find another set of n basis vectors {ei} in V such that,
ei ej = δij . (2.3.2) (2.11.2)
These are the same ei discussed in Section 2.3. Section 2.7 gave an example of computing the ei from the ei. We can then define a basis {λi) for V* as a set of linear functionals λi such that
λi(v) = ei v . // = (ei)a va (2.11.3)
Notice that i is a label, not a component. Functional λi is manifestly linear since
λi(kv) = kλi(v) and λi(v + v') = λi(v) + λi(v') k ϵ K . (2.11.4)
Here v and v' are vectors in V. As noted earlier, the definition of linearity is often combined into a single statement,
λi(k1v + k2v') = k1λi(v) + k2λi(v'), k1, k2 ϵ K (2.11.5)
The vectors ei are sometimes called "covectors" or "reciprocal vectors" or "dual vectors" which form a "dual basis". From (4.2.3) one then has
λi(ej) = ei ej = δij . (2.11.6)
Therefore,
λi(v) = λi(Σj vj ej) = Σj vjλi(ej) = Σj vjδij = vi . (2.11.7)
A component of λi is given by
(λi)j = (ei(e))j = δij (2.11.8)
where we use (2.6.4) col 4 row 4.
General vector in V*
A general linear functional V in V* can be written as a linear combinations of the basis functionals λi,
V = ΣiViλi = general vector in V* V(v) = ΣiViλi(v) V: V → K (2.11.9)
where on the right we show the corresponding function V(v). Evaluating at v = ej,
V(ej) = ΣiViλi(ej) = ΣiViδij = Vj . (2.11.10)
Consider now,
V(v) = ΣiViλi(v) = ΣiVi vi = V v = ΣiVivi (2.11.11)
Imagine a version of Picture A where we transform from x-space with basis vectors ei to some x'-space with basis vectors e'i. Under such a transformation, a dot product like V v transforms as a scalar.
Whereas v and V transform as vectors, the function V(v) transforms as a scalar field, just as in (2.2.6). That is to say,
V(v) = V v = V' v' = V'(v') (2.11.12)
Rank-2 tensor in V*V*
We want this object λiλj to be a linear functional over the space V*V* such that
λiλj: V*xV* → K . (2.11.13)
The natural way to accomplish this desire is to write
(λiλj)(v1,v2) = λi(v1)λj(v2) = scalar * scalar = scalar ϵ K (2.11.14)
Object (λiλj) is bilinear in its arguments because λi(v1) is linear in its argument.
A rank-2 tensor M in V*V* has the following expansion, analogous to (2.10.2)
M = Σab [M(e)]ab λaλb . (2.11.15)
and we do the verification similar to (2.10.3),
[M(e)]ij = Σab [M(e)]ab (λaλb)ij = Σab [M(e)]ab (λa)i (λb)j (2.11.16)
= Σab [M(e)]ab δaiδbj = [M(e)]ij
The components [M(e)]ab transform as a covariant rank-2 tensor,
[M'(e')]ij = Σab [M(e)]ab [λ'aλ'b]ij = Σab [M(e)]ab (λ'a)i(λ'b)j
= Σab [M(e)]ab [Ric (λa)c] [Rjd (λa)d]
= Σab [M(e)]ab [Ric δac] [Rjd δbd ]
= Σab[M(e)]ab Ria Rjb
= ΣabRiaRjb [M(e)]ab (2.11.17)
We evaluate the tensor M at (v1,v2) to get
M(v1,v2) = Σab [M(e)]ab (λaλb)(v1,v2) = Σab [M(e)]ab λa(v1) λb(v2) (2.11.18)
M(ei,ej) = Σab [M(e)]ab λa(ei) λb(ej) = Σab [M(e)]ab δaiδbj = [M(e)]ij (2.11.19)
One can rewrite M(v1,v2) as
M(v1,v2) = Σab [M(e)]ab λa(v1) λb(v2) = Σab [M(e)]ab (v1)a(v2)b (2.11.20)
Since this is a fully contracted outer product of three tensors (Section 3.1), we see that M(v1,v2) transforms as a scalar field over V* x V*,
M'(v'1,v'2) = M(v1,v2) (2.11.21)
Some authors refer to the function M(v1,v2) as a 2-tensor, but it is really the functional M which is the rank-2 tensor.
As a special case, consider M = α β . Then we get a result similar to (3.1.2),
(α β)ij = Σab αaβb (λaλb)ij = Σab αaβb λaiλbj = Σab αaβb δaiδbj
= αi βj (2.11.22)
and
(α β)(v1,v2) = Σabαaβb (v1)a(v2)b = [Σaαa (v1)a][Σbαb (v1)b]
= α(v1)β(v2) (2.11.23)
Again, α β is called a 2-tensor even though (α β)(v1,v2) is a scalar.
Rank-k tensor in V*V* .....
A rank-k tensor in V*V*.... can then be expanded as in (2.10.14),
M = Σabc... [M(e)]abc... (λaλbλc ...) (2.11.24)
and M(v1,v2, ....vk) is a scalar field over V*V* ... ,
M'(v'1,v'2, ....v'k) = M(v1,v2, ....vk) (2.11.25)
For the special case where M = α βδ..... ,
(α βδ.....)ijk.... = αi βjδk....... (2.11.26)
(α βδ.....)(v1,v2, ...vk) = α(v1)β(v2)δ(v3)...... (2.11.27)