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Informal working notes, apparently drafts for Phil's tensor and wedge-product document, dated around March 2016. They compare notation with Sjamaar and Spivak, define the pullback φ* of forms from x-space to t-space, and derive vector and component transformations using the Jacobian matrix R. They also revisit axis-aligned and tangent base vectors, with marginal remarks flagging confusions and paradoxes to resolve.
AI-written summary; may contain errors. This description is approximate.
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Warning: Different authors use different symbols, sometimes within the same document, so things can be confusing. Here is a quick comparison :
us: φ: Rn → Rm and φ(t) = x left = n right = m n ≤ m
Sjamaar p 39 φ: Rn → Rm and φ(x) = y
Sjamaar p 73 ψ: Rn → RN and ψ(t) = x
Spivak p 90 f: Rn → Rm and f(p) = x
Recall from Section 2 above that a smooth piece V of a manifold M (surface of dimension n in Rm) can be generated by applying the 1-to-1 transformation x = φ(t) to a simple open region U of t-space. We then have a mapping:
φ : U Rn → V M Rm x = (x1,x2.....xm) = φ(t1,t2...tn) = φ(t) (10.6.1)
t-space x-space
One might say that in this mapping, a point t in U is "pushed forward" to a point x on manifold M.
A pullback takes a differential form α defined at point x on M (x-space), and "pulls it back" to a different differential form β = φ*(α) defined at point t in U (t-space). Here φ* is a pullback operator which is associated with the transformation x = φ(t) . The operator φ* acts on a k-form α to produce (it turns out) another k-form β, where α exists in x-space and β exists in t-space. One sometimes says that β is the pullback of α along φ.
****************
Comments
1. We are free to add more dimensions to t-space on the left of the above picture, to bring it up to Rm, and the figure would still describe the pullback of a 1-form (k = 1). As we know, the Rm space will support k-forms for any k ≤ m. Once cannot have k > m because then there would need to be k basis vectors λi wedged together for such a k-form, but such a wedge product vanishes if k > m because one can only produce m linearly independent basis vectors in Rm, see (8.2.6).
2. If one has Rm on the left for t-space, the mapping x = φ(t) cannot create a surface in x-space on the right that has dimension more than m. The surface on the right is supposed to be embedded in Rn so we always have in mind that n ≥ m.
3. Officially, the set U in t-space on the left can be any open set in t-space, not just the 1-cube we have shown (or the c-cube shown next). STOP. Then you cannot do what I have shown below because a curve in t-space is NOT an open set in t-space if t-space is dimension 2. I am confused. Does this really mean that we must have m = k ? I am confused indeed, manana! See page 37 Sjamaar for help.
**********
When t-space is Rn, it is possible to have k-forms defined on Rn for any k ≤ n. For example, in the above figure which concerns the 2-forms tλj ^ tλj and βt, we can also talk about the same 1-form addressed in the previous drawing, as indicated by the two red markings in (10.6.7). A 1-form on the left would for example be any linear combination of tλ1 and tλ2, and similarly on the right any linear combination of xλ1,xλ2 ... xλm.
Since our interest is going to be integrating forms over the manifold shown in the right, we only care about k-forms where k = n which for the above figure is the 2-form. When k = the maximal value m, the form is called a "volume form".
Our interest will be the case that k = n, which is the largest possible k-form that "fits"
******************8
*****************
φ*(xei) = ri = the ith row of the R matrix = something with n components = vector in t-space
So this then defines the "pullback" of a basis vector in x-space to a vector in t-space.
So I should be able to say
[φ*(xei)] = Σ
There exists a related "vector transformation" which exists entirely in t-space,
R(tei) = Σa=1n Rai (tea) i = 1,2..n (10.6.5)
STOP. We know that R(tei) = xei is a vector with m components. But any linear combination of (tea) can have only n components, so the above equation cannot be value.
Proof: Since the tei form a basis in t-space, and since R is a linear operator, we certainly know that R(tei) is some linear combination of the tei, and it turns out that the coefficients of that linear combination are as shown. This can be verified as follows by applying (teb) to both sides and showing the results are equal. Then since the teb form a complete set (being a basis), the two sides of (10.6.5) are equal:
left side: (teb) R(tei) = Σa=1n (teb)j [R(tei)]j // (2.2.5)
= Σa=1n (teb)j Σc=1n Rjc (tei)c // matrix algebra
= Σa=1n δbj Σc=1n Rjc δic = Rbi // (10.6.3)
right side: (teb) [Σa=1n Rai (tea)] = Σa=1n Rai [(teb) (tea)]
= Σa=1n Raiδba = Rbi . // (2.3.2) QED
To summarize the above, we have two transformations:
(xei)j = Σa=1n Rja(tei)a i = 1,2...n component xform (10.6.4a)
R(tei) = Σa=1n Rai (tea) i = 1,2...n vector xform (10.6.5)
As noted in (2.9.2) and elsewhere, one can move indices up and down in covariant equations as long as the "tilts" are maintained. The above equations can then be written for dual basis vectors,
(xei)j = Σa=1n Rja(tei)a i = 1,2...n component xform (10.6.4a)
or
xei = R(tei)
R(tei) = Σa=1n Rai (tea) i = 1,2...n vector xform (10.6.5)
Combining these we get
xei = Σj=1n Rji (tej) i = 1,2..n
φ*(xei) = Σj=1m Rij (tej)
where the second line shows my "target equation".
Although one could derive it mimicking the above, we already know from (10.6.5) that
R(tei) = Σa=1n Rai (tea) (10.6.5a)
and from
is the correct vector transformation for the dual basis vectors tei . In Dirac notation (2.11.a.1) we can write the above as
R |tei> = Σa=1n Rai |tea> (10.6.5b)
**************************************
I know that
xei = R(tei) = Σa=1n Rai (tea)
Recall now (2.11.d.9) which defines the transpose of a matrix in covariant notation
(MT)ab = Mba (RT)ab = Rba
(MT)ab = Mba (RT)ab = Rba
(MT)ab = Mba (RT)ab = Rba
(MT)ab = Mba (RT)ab = Rba , (2.11.d.9)
so that (RT)ai = Rai.
*********************
3.14.16
**********************************************************
Axis-Aligned Vectors and Tangent Base Vectors
In Tensor it is shown in Section 3.2 that e'n are axis-aligned basis vectors [ (e'n)i = δni] in x'-space which map into the "tangent base vectors" en in x-space according to en= Se'n and e'n = Ren . One writes en = ∂x/∂x'n so en is tangent to the "coordinate line" of a constant value x'n drawn in x-space, see Tensor (3.2.8), (1.13) and (3.4.3).
Then in Tensor Section 3.5 the inverse transformation is discussed, x = F-1(x'), and for that transformation the un are axis-aligned basis vectors [ (un)i = δni] in x-space which map into the "inverse tangent base vectors" u'n in x'-space according to u'n= Run and un= Su'n as in Tensor (3.5.3) In this case the inverse tangent base vector u'n is tangent to the inverse coordinate line in x'-space which correspond to a constant value of xn in x-space.
To maintain control of our notation, we first review the above two paragraphs and then make changes step by step:
1. Forward transformation from x-space to x'-space (Tensor 3.2 )
x' = F(x) Rik ≡ (∂x'i/∂xk) = ∂kx'i
Sik ≡ (∂xi/∂x'k) = ∂'kxi
e'n with (e'n)i = δni axis-aligned basis vectors in x'-space
en en= Se'n tangent base vector in x-space
2. Inverse transformation from x'-space to x-space (Tensor 3.5, R and S the same as above )
x = F-1(x') Rik ≡ (∂x'i/∂xk) = ∂kx'i
Sik ≡ (∂xi/∂x'k) = ∂'kxi
un with (un)i = δni axis-aligned basis vectors in x-space
u'n u'n= Run tangent base vector in x'-space
3. In the last case, for our purposes it makes more sense to redefine the R and S matrices this way to get
x = φ(x') Rik ≡ (∂xi/∂x'k) = ∂'kxi
Sik ≡ (∂x'i/∂xk) = ∂kx'i
un with (un)i = δni axis-aligned basis vectors in x-space
u'n u'n= Sun tangent base vector in x'-space
where the new R matrix naturally goes with (is the differential of) the transformation named φ = F-1. Comparing this new definition of R and S with the original definition one has R ↔ S. Finally,
4. To be compatible with Sjamaar we rename spaces x' → t and x' → x to get
x = φ(t) Rik ≡ (∂xi/∂x'k) = ∂'kxi
Sik ≡ (∂x'i/∂xk) = ∂kx'i
un with (un)i = δni axis-aligned basis vectors in x-space
u'n u'n= Sun tangent base vector in x'-space
but same R and S as above
un with (un)i = δni axis-aligned basis vectors in x-space
u'n u'n= Run tangent base vector in x'-space
If we were to rename spaces x' → t and x → x' the above would say
STOP. Tensor doc is simply unclear on this topic and needs a rewrite. Do that (but not inline) right now.
Then in Tensor Section 3.5 the inverse transformation is discussed, x = F-1(x'), and for that transformation the un are axis-aligned basis vectors [ (un)i = δni] in x-space which map into the "inverse tangent base vectors" u'n in x'-space according to u'n= Run and un= Su'n as in Tensor (3.5.3) In this case the inverse tangent base vector u'n is tangent to the inverse coordinate line in x'-space which correspond to a constant value of xn in x-space.
As we show below, it is this last mapping that will be of great interest. As just stated, it has
(un)i = δni axis aligned basis vectors in x-space
u'n tangent base vectors in x'-space
It will map axis-aligned basis vectors tui in t-space into tangent base vectors we shall call xei in x-space. These vectors xei for i = 1,2..n will be seen below to span that tangent space TxM at a point x on a manifold M. Thus we have a clear connection between the notion of tangent base vectors and the tangent space TxM.
About t-space and x-space
We now change x → t and x' → x to get instead the following Picture F,
BUT we then physically switch the two sides of Picture F to get this rendering new of Picture F which is the more traditional picture used in our current context,
(10.6.1)
Instead of being x' = F(x), the new transformation is x = φ(t). The new (2.1.2) R matrix is this
(xVa) = Σb=1n Rab (tVb) Rab ≡ (∂xa/∂tb) = ∂bxa ≡ (φ)ab ≡ (Dφ)ab
or
(xV) = R (tVb) dx = R dt . (10.6.2)
We adopt the notation that xA is something in x-space while tA is something in t-space.
In Chapter 2 (and in the underlying Tensor document) the two spaces had the same dimension, but now
t-space = Rn and x-space = Rm and we have in mind that m ≥ n. The R matrix is then in general no longer square, but is in fact an m x n matrix with m rows (first index a) and n columns (second index b). R is a "tall" matrix when m > n. We have chosen the dimension names m and n to be consistent with Spivak and Sjamaar.
******************* 3/15/16 *******************
Paradox 2 of 3/15/16 I will try to concentrate the paradox even more:
On the one hand, according to Argument 1:
1 [R tua]i // restrict interest to i = 1..n
2 = Σj=1n Rij(tua)j // transformation of a covariant vector from t-space to x-space
3,4 = Ria // (xua)i = δai
On the other hand, according to Argument 2:
[R tua]i = [xua]i
So that is better! There you are, big as day! So what have I assumed in the two arguments?
Arg 1: matrix multiplication by vector, and (tua)j = δaj
Arg 2: (xua) ≡ R (tua) definition of the (xua) and (xua)i = δai
I am almost ready to do a numerical example! In the above, I could say that (xua) ≡ R (tua) is true as the only definition. Then given these n (xua) vectors which span TxM I know I can always construct dual vectors (xua) such that <xub | xua> = δba , so I don't need the Task A argument.
**********************************
For the S matrix elements, start with this rewrite of (7.19.15),
en = S e'n // S = R-1
un = S u'n
or
|en> = S |e'n>
|un> = S |u'n> (7.19.19)
Matrix elements of S must be of the form <a | S | b'> = <a | S b'>. Again using the table (7.19.12) for scalar product evaluations, close on the left to get
<ei |en> = <ei | S | e'n> = g'in
<ui |en> = <ui | S | e'n> = Sin = Rni (7.19.20)
and similarly
<ei |un> = <ei | S | u'n> = Rin = Sni
<ui |un> = <ui | S | u'n> = gin . (7.19.21)
Here then is a summary of matrix elements of R and S, where it happens that the basis vector label is up on the left side and down on the right side,
<e'i | R | en> = g'in <ei | S | e'n> = g'in
<u'i | R | en> = Sin = Rni <ui | S | e'n> = Sin = Rni
<e'i | R | un> = Rin = Sni <ei | S | u'n> = Rin = Sni
<u'i | R | un> = gin <ui | S | u'n> = gin (7.19.22)
****************
Notice in (7.19.22) that the right-side expressions for the left column of equations is exactly the same as for the right column. The reason for this perhaps surprising fact can be explained by first reordering the right column while maintaining the order of the left column
<e'i | R | en> = g'in <ui | S | u'n> = gin
<u'i | R | en> = Sin = Rni <ei | S | u'n> = Rin = Sni
<e'i | R | un> = Rin = Sni <ui | S | e'n> = Sin = Rni
<u'i | R | un> = gin <ei | S | e'n> = g'in
Now the second column can be obtained from the first column using the earlier rules
g'↔ g R ↔ S en ↔ u'n e'n ↔ un en ↔ u'n e'n ↔ un . (7.18.2)
Basically one is just changing from the transformation x' = F(x) to the inverse transformation x = F-1(x'). For example, the tangent basis vectors for F are the en while those for F-1 are the u'n, and so on.
***************************
(10.6.2)
The properties of the vectors ea and e'a are summarized in Tensor (7.18.1), while those of ua and u'a are summarized in Tensor (7.18.3). For example, (ea)b = Rab and (u'a)b = Rba = Sab. It is shown in Tensor (7.19.10) that the u-basis (e'-basis) is the natural default basis for x-space (x'-space).
To be compatible with Sjamaar we now rename our two spaces and our transformation to obtain a new picture, Picture F, and a new set of basis vectors
x-space → t-space e → te
x'-space → x-space e' → xe
F → φ u → tu
x' = F(x) → x = φ(t) u' → xu
unprimed vector → vector with t subscript
primed vector → vector with x subscript
(10.6.3)
For this picture,
x = φ(t) transformation Rab ≡ (∂xa/∂tb) = ∂(t)bxa differentials
Sab ≡ (∂ta/∂xb) = ∂(x)bta
xea with (xea)b = δab axis-aligned basis vectors in x-space
tea (tea) = S (xea) tangent base vector in t-space
tua with (tua)b = δab axis-aligned basis vectors in t-space
xua (xua) = R(tua) tangent base vector in x-space (10.6.4)
For example, xua is a tangent base vector in the new x-space. We put the x subscript to the left of the vector name to keep it out of the way of the basis vector label a.
************************
Now we are going to "pull back" all m of the xui basis vectors in x-space to vectors in t-space which vectors will turn out to be the rows of the R matrix. Here we are not inverse-mapping the first n xui back to their corresponding tui in t-space, we are doing something different. We first define,
φ*T(xei) ≡ Σj=1n Rij (tuj) i = 1..m (10.6.14)
where φ*T is a pullback operator which maps from x-space to t-space, φ*T : Rm→ Rn. The reason for writing this as the transpose of operator φ* will become clear below.
The sum on the right of (10.6.9) is a linear combination of t-space basis vectors, it is not a sum of vector components as in (10.6.4), so φ*T(xui) is a vector in t-space, not in x-space.
equation numbers only good to this point!
Now take the contravariant s component of both sides of (10.6.9),
[φ*T(xei)]s = Σj=1n Rij (tuj)s i = 1,2...m s = 1,2...n
= Σj=1n Rij δjs // (10.6.3)
= Ris . (10.6.10)
Reversing the tilts as done earlier, we can rewrite (10.6.9) and (10.6.10) as
φ*T(xei) ≡ Σj=1n Rij (tuj) i = 1..m
[φ*T(xei)]s = Ris . (10.6.11)
Therefore we conclude from (10.6.10) and (10.6.11) that
Fact: The m t-space vectors φ*T(xui) for i =1 to m are the rows of the R** matrix.
The m t-space vectors φ*T(xui) for i =1 to m are the rows of the R** matrix. (10.6.12)
If the rows of R** [ or R**] are denoted ri [ or ri ] then we can write
R** = = and ri = φ*T(xui) .
R** = = and ri = φ*T(xui) . (10.6.13)
Recall that R is a tall matrix with m rows and n columns, so each ri [ or ri ] has n components.
Thus, the pullback of an x-space basis vector xui is nothing more than the corresponding row of the R matrix for the transformation x = φ(t) . Every basis vector xui for i = 1 to m has such a pullback.
This drawing represents the three mappings so far described,
(10.6.14)
Here we show the mappings for the basis vector tu1.
In Dirac notation one can write (10.6.9) and (10.6.11) as ket equations,
φ*T | xui> = Σj=1n Rij | tuj> i = 1,2..m
φ*T | xui> = Σj=1n Rij | tuj> . (10.6.15)
**********
If the columns of R** [or R**] are denoted ci [ or ci ] then one can write,
R** = [ c1, c2.....cn ] = [xu1, xu2 ....xun ] ci = xui
R** = [ c1, c2.....cn ] = [xu1, xu2 ....xun ] ci = xui . (10.6.12)
Recall that R is a tall matrix with m rows and n columns, so each ci [ or ci ] has m components.
Metric Tensor Comment: We have been quiet about the metric tensors tg and xg in the two spaces. Since we have used covariant notation, this entire section should be valid for any choice of metric tensors. For example
(tg)ij = (tui) (tuj) i,j = 1,2..n // (2.4.2) translated from Pic A to Pic F
(xg)ij = Σrs=1nRirRjs(tg)rs i,j = 1,2..m // (2.2.3) translated from Pic A to Pic F (10.6.13)
One normally selects tg in t-space (often Cartesian meaning (tg)ij = δi,j) and then the last line above gives xg in x-space. The last line just states that the metric tensor transforms as a rank-2 tensor. The object (tg)ij is n x n whereas (xg)ij is m x m since the R matrix has m rows and n columns.
ok to here 3.17.16
****************8
In Dirac notation one may write *** as <xei | R | tun> = Rin = Sni
|xui> = R|tui>
|tui> = R-1|xui> = RT |xui> = S |xui>
*********************
where now the operator φ* = R acts on the bra to its left.
(10.6.11)
corresponding bra (dual space) equations are
<xui | φ* = Σj=1n Rij <tuj | i = 1,2..m
<xui | φ* = Σj=1n Rij <tuj | . (10.6.16)
The bras in these equations are dual space vectors (rank-1 linear functionals) as discussed in Section 2.11.
The ket <xui | is an element of dual space (Rm)* while <tuj | inhabits (Rn)*. Note that Fig (10.6.14) displays only the spaces Rn and Rm and not the corresponding dual spaces.
************************88
In terms of the (2.11.c.2) basis functionals xλi = <xei| of the dual space (Rm)*, this second line of (10.6.16) may be written,
φ*(xλi) = Σj=1n Rij (tλj) = Σj=1n (Dφ)ij (tλj) i = 1,2..m (10.6.17a)
or in cosmetic notation of Section 10.1 above,
φ*(dxi) = Σj=1n Rij (dtj) = Σj=1n (Dφ)ij (dtj) . i = 1,2..m . (10.6.17b)
Since xλi = dxi is a very simple example of a differential form, we see in (10.6.17) our first example of "the pull back of a differential form". What is being "pulled back" is the dual vector <xei| = xλi ϵ (Rm)* . The pulled back vector is φ*(xλi) ϵ (Rn)*. The mapping is φ*: (Rm)*→ (Rn)*.
Noted added 3.19.16. I am having massive problems, so I need to tie down my claims here with claims in Sjamaar or Spivak. Do ANY of the above equations appear in Sjamaar? He does show this equation,
So I think this would match (10.6.17b) except for Sjamaar's constant changing of space names. In this section he is using α = ΣI fI dyI and I would then use page 37 to say (x,y)Sja → (t,x)me . Then what I have in (10.6.17b) is the special case α = dyi (him) or α = dxi(me). So I think I can say that my equation
(10.6.17b) has verification in Sjamaar. And then I think (10.6.17a) is also OK. BUT what I don't have is any connection between xλi and some <x*i| basis vector. Sja just avoids that kind of language. So basically above I just "took a guess" and assumed that <xui | φ* = Σj=1n Rij <tuj | and that this pullback acted on the tangent base vectors in x-space (rather than the axis-aligned vectors).
What happens if I change this assumption. Maybe this is the correct starting point:
<xei | φ* = Σj=1n Rij <tuj | i = 1,2..m
How would this affect my presentation above and below? I will use up this v4 doc by making edits below to see what happens with this assumption.
The result is that basically very little changes! I made all changes below in red!
**************8
The dual vectors (functionals) in the sense of xλi = <xei| then span the cotangent space of TxM.
But this is wrong with our changed approach! The vectors <xei| are axis aligned vectors in x-space and they do NOT span the tangent space TxM. It is the <xui| which span the cotangent space. This was not a critical fact so in blue it is removed from the presentation.
*****************
Why do I use < xu^I | instead of < xe^I | at this point?? In Chapter 2 I never said whether you should use the natural basis or the tangent base vectors basis, I just used there a generic basis e. But here I am making a decision without justification. What does Sjamaar do? He writes α = ΣIfIdyI on page 36 and this goes with the famous page 37 picture. But he never says that dyI = <???I | But in Chapter 7 he writes on page 87 that α = ΣI fI dxI. and on page 83 he has written dxi = <ei| and he shows them on p 83A as axis-aligned basis vectors!!!! So what I am doing above is WRONG, since in Picture F it is the xea which are the axis-aligned vectors!!!!! So I will now save this file off into a new v4 and edit from this point onwards. This was a suspicion I have had for a few days now.
****************** following is removed from Ch2 , both statements are really wrong!
The reciprocal base vectors are the tangent base vectors of the inverse transformation x = F-1(x'). The are also surface normals to the faces of the n-piped spanned by the en.
except e'n → en(e) = the e-space version of en . // this was above (2.6.7), seems wrong.
****************
One is used to the fact that RTR = 1 for a rotation in Cartesian space (real orthogonal), but in our covariant world where matrix multiplication is either "down tilt" or "up tilt", RTR = 1 is true for the differential matrix R of any transformation x' = F(x), whether R is a pure rotation or a more general matrix including stretch. This claim of course requires a covariant definition of RT which we take to be the following,
(MT)ab = Mba (RT)ab = Rba
(MT)ab = Mba (RT)ab = Rba
(MT)ab = Mba (RT)ab = Rba
(MT)ab = Mba (RT)ab = Rba , (2.11.d.9)
where M is any rank-2 tensor and R is the differential R matrix. The middle two lines show that the transpose of a "tilted" matrix is obtained not by swapping the indices, but by reflecting them in the vertical line between them, as is done in the inversion rule of (2.1.11). Thus, RT = R-1 and RRT = RTR= 1 with tilted matrix multiplication. Specifically,
(RTR)ac = (RT)abRbc = RbaRbc = δac = (1)ac (2.11.d.10)
where RbaRbc = δac is the orthogonality rule (2.1.9) #1.
******************************
This stuff removed from Section 10_6 v6.
In our discussion below we want to match the notations of Sjamaar and Spivak who replace our x and x' with their t and x. We need then to translate the above hard-won "kinematics package" into this new choice of coordinate systems using the following translation table,
x-space → t-space e → te u → tu g → tg
x'-space → x-space e' → xe u' → xu g' → xg
F → φ
x' = F(x) → x = φ(t)
unprimed vector → vector with t pre-subscript
primed vector → vector with x pre-subscript (10.6.3)
The resulting kinematics package for Picture F is then (F just because other letters are used in Tensor),
x = φ(t) xform Rab ≡ (∂xa/∂tb) = ∂(t)bxa R = (Dφ)
xV = R tV vector Sab ≡ (∂ta/∂xb) = ∂(x)bta
xea with (xea)b = δab axis-aligned basis vectors in x-space
tea tea = S xea tangent base vector in t-space
tua with (tua)b = δab axis-aligned basis vectors in t-space
xua xua = R tua tangent base vector in x-space
(xua)b = Rbc (tua)c
1= | xei> <xei| = | xei> <xei| = | xui> <xui| = | xui> <xui| completeness in x-space
1= | tei> <tei| = | tei> <tei| = | tui> <tui| = | tui> <tui| completeness in t-space
(tun)i = tui tun = <tui | tun > = tgin = xui xun = <xui | xun >
(ten)i = tui ten = <tui | ten > = Sin = Rni
(xen)i = xei xen = <xei | xen > = xgin = tei ten = <tei | ten >
(xun)i = xei xun = <xei | xun> = Rin = Sni
ten = xgni tei xen = xgni xei tun = tgni tui xun = tgni xui
ten = xgni tei xen = xgni xei tun = tgni tun xun = tgni xui
<ten | S | xei> = <xei | R | ten> = xgin
<ten | S | xui> = <xui | R | ten> = Sin = Rni
<tun | S | xei> = <xei | R | tun> = Rin = Sni
<tun | S | xui> = <xui | R | tun> = tgin
(10.6.4)
The final Picture F' is just a left/right reflection of Picture F, since we want to have t-space on the left in all the pictures below.
The basis vectors xua for a = 1,2..n will be seen below to span the tangent space TxM at a point x on a manifold M. Thus we have a clear connection between the notion of tangent base vectors and the tangent space TxM discussed earlier in Section 6.****.
Here is a more specific drawing of Picture F' for our application,
(10.6.5)
In Chapter 2 (and in the underlying Tensor document) the two spaces had the same dimension, but now
t-space = Rn and x-space = Rm and we have in mind that m ≥ n. The matrix Rab is then in general no longer square, but is in fact an m x n matrix with m rows (first index a) and n columns (second index b). R is a "tall" matrix when m > n. We have chosen the dimension names m and n to be consistent with Spivak and Sjamaar.
Question: What now happens to the equations quoted above
RRT = RTR = 1 SST = STS = 1 RS = SR = 1
RT = R-1 = S ST = S-1 = R . (2.11.g.3) (10.6.1)
when R is no longer square?
**********************
"Raising/lower components and maintaining tilts" as below (2.9.2) we can write a version of (10.6.7) for the dual basis vectors xui and tui,
xui = R tui |xui> = R |tui>
or
(xui)j = Σa=1n Rja (tui)a = Σa=1n Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.9)
Since (xui)j = Rji and (xui)j = Rji, we conclude from (10.6.7) and (10.6.9) that
Fact: The first n x-space basis vectors xui for i = 1 to n are the columns of R** .
The first n x-space basis vectors xui for i = 1 to n are the columns of R** . (10.6.10)
R** = [xu1, xu2 ....xun ] R** = [xu1, xu2 ....xun ]
That is to say, in (xui)j = Rji if we fix i and examine j = 1,2...m, we describe the column i of R**.
**************************
(a) x' = F(x) xform Rij ≡ (∂x'i/∂xj) = ∂jx'i R = (DF)
V' = R V vector Sij ≡ (∂xi/∂x'j) = ∂'jxi
(b) e'i with (e'i)j = δij axis-aligned basis vectors in x'-space
ei ei= Se'i tangent base vector in x-space (7.18.1)'
(c) ui with (ui)j = δij axis-aligned basis vectors in x-space
u'i u'i= Rui tangent base vector in x'-space (7.18.3)'
(u'i)j = Rjk (ui)k
(d) 1= | e'i> <e'i| = | e'i> <e'i| = | u'i> <u'i| = | u'i> <u'i| completeness in x'-space
1= | ei> <ei| = | ei> <ei| = | ui> <ui| = | ui> <ui| completeness in x-space
(e) (uj)i = ui uj = <ui | uj > = gij = u'i u'j = <u'i | u'j >
(ej)i = ui ej = <ui | ej > = Sij = Rji
(e'j)i = e'i e'j = <e'i | e'j > = gij = ei ej = <ei | ej >
(u'j)i = e'i u'j = <e'i | u'j > = Rij = Sji (7.19.12)'
(f) ei = g'ij ej e'i = g'ij e'j ui = gij uj u'i = gij u'j
ei = g'ij ej e'i = g'ij e'j ui = gij uj u'i = gij u'j (7.19.14)'
(g) <ej | S | e'i> = <e'i | R | ej> = g'ij
<ej | S | u'i> = <u'i | R | ej> = Sij = Rji
<uj | S | e'i> = <e'i | R | uj> = Rij = Sji
<uj | S | u'i> = <u'i | R | uj> = gij . (7.19.19)' (10.6.2)
***************
Above we have defined F, DF and DF as alternate names for matrix R because many authors (like Spivak) use this notation. In Tensor (E.4.4) we show that this is in fact a "reverse dyadic notation". Often (DF)ab is written unbolded (DF)ab so then R = (DF) with the idea that a matrix like R is normally not bolded.
V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a ≡ (F)ab ≡ (DF)ab ≡ (DF)ab (2.1.2)
dx'a = Rabdxb dx' = Rdx . (2.1.12) (10.6.1)
************************************8
Since we shall always be working in Λk we now drop the ^ subscript and change notation so now
λI ≡ λi ^ λi .....^ λi
dxI ≡ dxi ^ dxi .....^ dxi . (10.1.10)
Similarly we change T^ to be just T. The traditional names for differential forms are α, β and so on, so we take T^ → T → α and write our arbitrary differential form (1.8) now as
α = Σ'I fI(x) λI = Σ'I fI(x) dxI α ϵ Λk(V) V = x-space = Rn (10.1.11)
where fI is the more traditional name for AI. We have now taken V = Rn, Euclidean space, where the basis vectors ei = |ei> are independent of x, and so the λi = <ei| are also independent of x.
*****************************
But then it is useful to label the two forms [φ*α] and α by their respective space indicators (the first is a t-space k-form, the second
where
α = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I // modified (10.9.9)C
But then since α is a k-form in x-space and [φ*α] is a k-form in t-space, we label each form so that then α is αx and [φ*α] is [φ*α]t so then
[φ*α]t (v1,v2...vk) = αx(Rv1,Rv1...Rvk)
Replacing R by (D(t)φ) since Rij = (D(t)φ)ij = (∂φi(t)/∂tj) as in (10.9.17), this becomes
= αx((D(t)φ)v1,(D(t)φ)v1...(D(t)φ)vk) (10.9.27)
where we replace R by (D(t)φ) since Rij = (D(t)φ)ij = (∂φi(t)/∂tj) as in (10.9.17) .
Our original x-space k-form was
αx = Σ'I fI(x) dx^I (10.9.9)C
More traditionally this is just written with α instead of αx
α = Σ'I fI(x) dx^I (10.9.28)
so then (10.9.27) becomes
[φ*α] (v1,v2...vk) = α((Dφ)v1,(Dφ)v1...(Dφ)vk)
But since α on the right is an x-space differential form, it is tagged as αx , and since [φ*α] is a t-space differential form, it is tagged as [φ*α]t.
which in Dirac notation is
<αx | φ* | vn,vn...vn> = <αx | Rvn,Rvn...Rvn> . (10.9.28)
Writing the right side as <αx | R | vn,vn...vn> we get
<αx | φ*| vn,vn...vn> = <αx | R | vn,vn...vn> (10.9.29)
which serves as a check on the fact that φ* = R. Alternatively, one can regard the last three equations in reverse order as a very quick derivation of (10.9.27).
If we had written our original x-space differential form as α = Σ'I fI(x) dx^I (not αx),
[φ*αx] (vn,vn...vn) = αx(Rvn,Rvn...Rvn)
Comment: Note that <αx | ϵ Λk(Rm) (Rm)*k whereas | vn,vn...vn> ϵ (Rn)k but does not lie in the wedge space Lk(Rn).
We can rewrite (10.9.27) making these changes:
Replace R by (D(t)φ) since Rij = (D(t)φ)ij = (∂φi(t)/∂tj) as in (10.9.17)
Replace the dummy vector arguments vn by the simpler set of arguments vr
Replace α subscript x on the right side by φ(t) so the entire right side depends only on t
Write [φ*αx] as [φ*αx]t to emphasize [ ] is a t-space k-form (we called it βt in (10.9.15))
The result is
[φ*αx]t (v1,v2...vk) = αφ(t)((D(t)φ)v1,(D(t)φ)v2...(D(t)φ)vk) . (10.9.30)
Everything on the right side involves only t-space objects.
If we were to change variable names so that x = φ(t) became y = φ(x), the above would be written
[φ*αy]x (v1,v2...vk) = αφ(x)((D(x)φ)v1,(D(x)φ)v2...(D(x)φ)vk) . y = φ(x) (10.9.31)
where now we are pulling back the k-form αy from y-space to x-space where it becomes [φ*αy]x .
where we interpret the left side φ*(α)x as meaning [φ*αy]x with the y label on αy suppressed. Then the original y-space k-form would be written α = Σ'I fI dy^I which Sjamaar writes as α = ΣI fI dyI and this does appear on page 96. From our Dirac notation point of view, it seems slightly illogical to have a subscript on one α but not on the other α,
<αx | φ*| vn,vn...vn> = <αx | R | vn,vn...vn> = <αx | Rvn,Rvn...Rvn> (10.9.28,29)
[φ*αx] (vn,vn...vn) = αx(Rvn,Rvn...Rvn) (10.9.27)
****************************************
I would be looking for a theorem of this general type where RIJ factors and xJ does not.
AltI [ ΣJ RIJ xJ] = ΣJ AltJ[RIJ] [AltJ (xJ) ]
or
ΣJ AltJ[RIJ xJ] = ΣJ AltJ[RIJ] AltJ [xJ]
From appendix C I do know that for any tensors of any ranks, where ai can be a hat or no hat
Alt[(T1)(T2) ...... (TN)] = Alt[(T1)a(T2)a ...... (TN)a] (C.4.17)
where each ai can independently be a blank or can be a ^. As an example
Alt[(T1)(T2)] = Alt[(T1)^(T2)^ ] (*)
Suppose (T1)IJ = RIJ and (T2)J = xJ // tensors of different sizes
Then
{Alt[(T1)(T2)]}IJJ = AltI [ RIJ xJ]
The right side of (*) would then be
Alt[Alt(T1)Alt(T2) ]IJJ = Alt
**************************
I am missing an important theorem, will try to figure it out right here. From Section 10.1
ΣITIλ^I = Σ'IAIλ^I A = k!Alt(T)
So can write
ΣITIλ^I = Σ'I k!Alt(T)I λ^I = Σ'I k! AltI(TI) λ^I
Would this apply to TI = RJI where we treat J as a bystander index? Then would have
ΣI RJI λ^I = Σ'I k! AltI(RJI ) λ^I = Σ'I det(RJI ) λ^I
Maybe generalize this to say
ΣI RJI X^I = Σ'I k! AltI(RJI ) X^I = Σ'I det(RJI ) X^I
where X^I is ANY hat product . Try to prove this directly
ΣI RJI AltI(XI) = Σ'I det(RJI ) AltI(XI) = Σ'I k! AltI(RJI ) AltI(XI)
swap I and J to state this hoped-for theorem as
ΣJ RIJ AltJ(XJ) = Σ'J det(RIJ ) AltJ(XJ) = Σ'J k! AltJ(RIJ ) AltJ(XJ)
I did this basically above, but let's do it again Sam,
************
In Section 7.4 it is proven that these two tensor expansions are the same
T = Σii....i Tii....i (ei ei ..... ei) . (7.4.1)
T^ = Σi<i<....<i Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.5)
provided one sets Aii...i = k! [Alt(T)]ii...i . In the proof, the ei are certain basis vectors, but they play a bystander role and the proof would be valid for an arbitrary set of vectors xi
Σii....i Tii....i (xi xi ..... xi) .
= Σi<i<....<i Aii...i (xi ^ xi ^ .... ^ xi)
where Aii...i = k! [Alt(T)]ii...i = k! AltI(Tii...i) .
In multiindex notation, this theorem states that
ΣI TI xI = Σ'I k! AltI(TI) x^I
Suppose we set TI = ZIJ where J is a bystander multiindex. Then the above says
ΣI ZIJ xI = Σ'I k! AltI (ZIJ) x^I
If ZIJ happens to have the form Zij Zij ..... Zij then we know from (A.8.30) that
AltI (ZIJ) = (1/k!) det(ZIJ)
and then our theorem says
ΣI ZIJ xI = Σ'I det(ZIJ) x^I
This shows how a symmetric sum involving a tensor product xI becomes an ordered sum involving the wedge product x^I : the coefficients ZIJ turn into det(ZIJ) . Since our matrix RIJ has the factored form,
RIJ = Rij Rij ..... Rij
we can write
ΣI RIJ xI = Σ'I det(RIJ) x^I
In this theorem xI is some tensor product vector xI = xi xi .... xi . But the vector x^I is just a linear combination of such tensor product vectors, so we could apply the above theorem to x^I to get
ΣI RIJ x^I = Σ'I det(RIJ) x^^I = Σ'I det(RIJ) x^I
Here we take x^^I = x^I which says [Alt(Alt(x))]I = Alt(x)I as per ***. If this were really true, I would have shown that
ΣI RIJ xI = Σ'I det(RIJ) x^I = ΣI RIJ x^I
which implies that
ΣI RIJ xI = ΣI RIJ x^I
I don't believe this. Try n = 2:
ΣI RIJ xI = Σii' RijRi'j' xi xi' =?= Σii' RijRi'j' (1/2)[ xi xi' - xi' xi ]
= Σii' RijRi'j' (1/2) xi xi' - Σii' RijRi'j' (1/2) xi' xi
= Σii' RijRi'j' (1/2) xi xi' - Σii' Ri'jRij' (1/2) xi xi'
= Σii' (1/2)[ RijRi'j' - Ri'jRij'] xi xi'
= (1/2) ΣI det(RIJ) xI
and this is NOT the same as ΣI RIJ xI . So my theorem is only this:
ΣI RIJ xI = Σ'I det(RIJ) x^I
OK, but earlier I claim that
k! φ*(dx^I) = ΣJ det(RIJ) dtJ
k! φ*(dx^I) = ΣJ det(RIJ) dt^J
k! φ*(dx^I) = ΣJ k! RIJ dt^J
which is to say
ΣJ det(RIJ) dtJ = ΣJ det(RIJ) dt^J = ΣJ k! RIJ dt^J
and in my generic notation I think this would say
ΣJ det(RIJ) xJ = ΣJ det(RIJ) x^J = ΣJ k! RIJ x^J
So I then have two different theorems:
Σ'I det(RIJ) x^I = ΣI RIJ xI
ΣJ det(RIJ) xJ = ΣJ det(RIJ) x^J = ΣJ k! RIJ x^J
Let's recheck this theorem
ΣI RIJ xI = ΣI RIJ x^I
RHS = ΣI RIJ AltI [xI]
= ΣI RIJ (1/k!) ΣP(-1)P xP(I)
= (1/k!) ΣP(-1)P ΣI RIJ xP(I)
= (1/k!) ΣP(-1)P ΣI RP(I)P(J) xP(I)
= (1/k!) ΣP(-1)P ΣI RIP(J) xI
= ΣI [ (1/k!) ΣP(-1)P RIP(J)] xI
= ΣI [ AltJ(RIJ) ] xI
= ΣI [ (1/k!) det(RIJ) ] xI
************************************************
Fact : ΣJ det(TIJ) xJ = ΣJ det(TIJ) x^J = ΣJ k! TIJ x^J if T has factored form.
1 2 3
Proof 2 = 1
ΣJ det(TIJ) x^J = ΣJ k! AltI(TIJ) AltJ(xJ) // (A.8.30) and (7.4.3)
= ΣJ k! [(1/k!) ΣP(-1)PTP(I)J ] [(1/k!) ΣQ(-1)Q xQ(J) ] // (A.5.3c) used twice
= (1/k!)ΣP(-1)PΣQ(-1)Q [ΣJ TP(I)J xQ(J))] // reorder
= (1/k!)ΣP(-1)PΣQ(-1)Q [ΣJ TIP(J) xQP(J))] // (A.1.20) that ΣJ fJ = ΣJ fP(J)
= (1/k!)ΣP(-1)P ΣJ TIP(J) [ΣQ(-1)Q xQP(J))] // reorder
= (1/k!) [ΣP(-1)P ΣJ TIP(J)] [ΣQ(-1)Q xQ(J))] // (A,1.3) rearrangement theorem
= k! [(1/k!)ΣP(-1)P ΣJ TIP(J)] [(1/k!) ΣQ(-1)Q xQ(J))] // rewrite
= k! AltJ(TIJ) Alt(xJ) // rewrite
dxi dxi = Σj< j det dtj dtj calculus ???? (10.8.24)
This is wrong but I will fill it in soon, have to do the Sjamaar measure stuff.
************************************
I am bothered by (10.9.7). How can these two results be the same
AltI [ ΣJ RIJ <tuJ | ] = ΣJ (1/k!) det(RIJ) <tuJ | (1)
AltI [ ΣJ RIJ <tuJ | ] = ΣJ RIJ AltJ <tuJ | (2)
Check for a 1-form
Σj Rij <tuj | = Σj Rij <tuj | ok
Check for a 2-form
LHS(1) = (1/2) [ Σjj'Rij Ri'j' <tuj | <tuj'| - Σjj'Ri'j Rij' <tuj | <tuj'| ]
= Σjj' { (1/2)[Rij Ri'j'- i'j Rij'] } <tuj | <tuj'| = Σjj' { (1/2) det(RIJ) <tuJ |
So (1) seems to be correct. Now check (2) :
RHS(2) = Σjj'Rij Ri'j' (1/2) [ <tuj | <tuj'| - <tuj' | <tuj| ]/2
= Σjj'Rij Ri'j' (1/2) <tuj | <tuj'| - Σjj'Rij Ri'j' (1/2) <tuj' | <tuj|
= Σjj'Rij Ri'j' (1/2) <tuj | <tuj'| - Σj'jRij' Ri'j (1/2) <tuj | <tuj'|
= Σjj' { (1/2) [ Rij Ri'j' - Rij' Ri'j ] } <tuj | <tuj'|
OK, for 2-forms the claims (1) and (2) also agree.
Question: Is there a third result having this form?
AltI [ ΣJ RIJ <tuJ | ] = ΣJ (1/k!) det(RIJ) <tu^J| (3)
Try testing this in the general case. Idea of pre-antisym. So
RHS(3) = ΣJ (1/k!) det(RIJ) AltJ (<tuJ | ) ?
= ΣJ AltI(RIJ) AltJ (<tuJ| )
My main weapon is to write this out in permutation notation
= (1/k!)2 ΣJ ΣP(-1)P [ RP(I)J] [ ΣQ(-1)Q (<tuQ(J)| ]
= (1/k!)2 ΣP(-1)P ΣQ(-1)Q [ΣJ RP(I)J <tuQ(J)| ]
= (1/k!)2 ΣP(-1)P ΣQ(-1)Q [ΣJ RP(I)P(J) <tuQP(J)| ]
= (1/k!)2 ΣP(-1)P ΣQ(-1)Q [ΣJ RIJ <tuQP(J)| ]
= (1/k!)2 ΣJ RIJ ΣP(-1)P ΣQ(-1)Q [ <tuQP(J)| ]
= (1/k!)2 ΣJ RIJ ΣP(-1)P ΣQ(-1)Q [ <tuQ(J)| ] // rearrangement theorem
= ΣJ RIJ (1/k!) ΣQ(-1)Q [ <tuQ(J)| ]
= ΣJ RIJ AltJ [ <tuJ| ]
= ΣJ RIJ <tu^J| // but this is not LHS (3) !
So I have shown that shown that (3) is in fact correct! To summarize, here then is what I know
AltI [ ΣJ RIJ <tuJ | ] = ΣJ (1/k!) det(RIJ) <tuJ | (1)
AltI [ ΣJ RIJ <tuJ | ] = ΣJ RIJ <tu^J | (2)
AltI [ ΣJ RIJ <tuJ | ] = ΣJ (1/k!) det(RIJ) <tu^J| (3)
This then gives three versions of an identity of interest
< xe^I | φ* = ΣJ (1/k!) det(RIJ) <tuJ |
< xe^I | φ* = ΣJ (1/k!) det(RIJ) <tu^J|
< xe^I | φ* = ΣJ RIJ <tu^J |
In cosmetic notation
φ*(dx^I) = ΣJ (1/k!) det(RIJ) dtJ
φ*(dx^I) = ΣJ (1/k!) det(RIJ) dt^J
φ*(dx^I) = ΣJ RIJ dt^J
But none of these seem to provide what I am looking for. Sjamaar has dφi differential objects which I don't seem to have. I think in my variables he would be writing, since x = φ(t),
dxi = dφi = (∂φi/∂tj) dtj // old Sja notes p 40 A
Are these regular calculus things, or are they red things? Well, think of φi(t) as a function = 1 0-form. Then dφi is the differential of a 0-form and we have a formula for doing that which is this
. So you can have the above as either black or red. So write also as
dφi = (∂φi/∂tj) dtj // old Sja notes p 40 A
which is just a special case of the dα general differentiation rule. Good. So we can either or ^ these together
dφi dφi' = (∂φi/∂tj) dtj (∂φi'/∂tj') dtj'
But for functions the can be removed and we then have
dφi dφi' = (∂φi/∂tj)(∂φi'/∂tj') dtj dtj' = RijRi'j' dtj dtj'
But you could also hat them together if you wanted to get
dφi ^ dφi' = (∂φi/∂tj)(∂φi'/∂tj') dtj ^ dtj' = RijRi'j' dtj ^ dtj'
So in general this becomes
dφ^I = ΣJ RIJ dt^J
Using my new Fact (A.8.35) this can be rewritten
dφ^I = Σ'J det(RIJ) dt^J
But since φ(t) = x, this is the same as
dx^I = ΣJ RIJ dt^J or dx^I = Σ'J det(RIJ) dt^J
or
xλ^I = ΣJ RIJ tλ^J
or
<xe^I | = ΣJ RIJ <tu^J |
I do have in table (10.8) that
<xei| R = Rij<tuj|
and I do have in
[<(T1)^| ^ < (T2)^| ^...^ < (TN)^|] Q = <(T1)^|Q ^ < (T2)^|Q ^...^ < (TN)^|Q (8.9.d.15)
which in this application would be
<xe^I | R = <xe^i1|R ^ <xe^i2|R ^...^ < <xe^iN|R
= ΣJ Ri1j1 Ri2j2 .....RiNjN <tuj1 | ^<tuj2 | ^ ... ^ <tujN |
= ΣJ RIJ <tu^J |
so the pieces are hanging together. So I really do have this result
***********************
= ( * * *)
**************************
***************************************************************************
Exercise: Verify that em(e) = Rem shown in the first line of (2.6.7) is consistent with (em(e))n = (en em) in the last line of (2.5.4):
(2.3.4) (2.1.9) (2.3.2)
em(e) = Rem (em(e))n = Rnj(em)j = RnjRmj = δnm = (en em) QED .
We can apply the four equations shown on the right of (2.6.5) sequentially to the basis vectors V = um, um, em, em to obtain a set of 16 equations. The very first equation would be un um = umn. One can then look up the dot product in (2.4.2) to find that un um = δnm and then one gets the result that umn = δnm . The next equation is un um = (um)n and we look up this dot product to find un um = gnm and so (um)n = gnm. Rather than do all these calculations, since the dot products are already listed in (2.5.4), we can just read off the 16 results we want. This then produces the rightmost column of equations in (2.5.4) above, which we transcribe here,
(um)n = δmn (um)n = g(u)mn (em)n = Rmn (em)n = Rmn
(um)n = gmn (um)n = δmn (em)n = Rmn (em)n = Rmn
(um(e))n = Rnm (um(e))n = Rnm (em(e))n = δmn (em(e))n = g(e)mn
(um(e))n = Rnm (um(e))n = Rnm (em(e))n = g(e)mn (em(e))n = δmn .
(2.6.8)
We have replaced g'→g(e) and have made cosmetic changes such as δnm = δmn as well as gmn = gnm since all metric tensors are symmetric.
The set (2.6.8) gives the contravariant (up) and covariant (down) components of all eight basis vectors: um, um, em, em and um(e), um(e), em(e), em(e).
******************************
2.5 Expansion of vectors onto basis vectors
A vector V can be expanded onto an arbitrary complete basis bn as follows:
V = Σn Vn bn (2.5.1)
Any such basis has a dual basis bn such that ( see Tensor ...)
bm bn = δmn (2.5.2)
Comment: By definition any basis bn is complete, so "complete basis" is redundant, but we like to emphasize the fact of completeness which also means the basis vectors are linearly independent and that in turn means that the dual basis bm exists.
Dotting both sides of (2.5.1) with bm shows that
Vn = bn V = <bn | V> (2.5.3)
where the dot product is also written in Dirac notation to be described later. This leads to a fact which is obvious but can be quite confusing in the midst of calculations:
Fact: The meaning of the component index n on Vn is dependent on the basis vectors bn which appear in the expansion for which Vn is the coefficient.
Just looking at Vn one cannot tell what basis was used for the expansion. To be precise one could write
V = Σn [V]n(b) bn
[V]n(b) = bn V = <bn | V> (2.5.4)
where the superscript n has a marker (b) indicating the meaning of the superscript. Then in this case the component index n is a "bn type component index".
A similar issue will arise later when we expand rank-2 tensors. Suppose bn and cn are two bases for vector space V so we could expand rank-2 tensor M as follows:
M = Σmn Mmn bm cn
bm cn M = Mmn (2.5.5)
where the reader is asked for the moment to ignore the and symbols to be defined later. The point here is that the meaning of the indices on Mmn is unclear if only stares only at Mmn. This could be clarified as above by writing Mmn = Mm(b)n(c). For many indices, this is obviously a very clumsy notation, although it would be precise. We shall in fact use this notation below but in the following manner
Vn(b) is written as [V(b)]n
Mm(b)n(c) is written as [M(b,c)]mn (2.5.6)
where the order of the markers matches the order of the indices. Then we would have
M = Σmn [M(b,c)]mn bm cn
bm cn M = [M(b,c)]mn (2.5.7)
The same markers serve for any position of the indices, for example
M = Σmn [M(b,c)]mn bm cn
bm cn M = [M(b,c)]mn (2.5.8)
If both bases are the same, one marker is sufficient
M = Σmn [M(b)]mn bm bn
bm bn M = [M(b)]mn (2.5.9)
In most of what follows below, we shall be considering the basis en to be "an arbitrary basis" like bn above. Although we have used en to refer to the tangent base vectors of a transformation, we imagine that for any arbitrary basis en we can find a transformation for which the en are the tangent base vectors. We shall then be writing
V = Σn [V(e)]n en [V(e)]n = en V = <en | V>
M = Σmn [M(e)]mn em en em en M = [M(e)]mn (2.5.10)
When it is clear that we are always using this en basis exclusively, we will suppress the marker (e) and write simply
V = Σn Vn en Vn = en V = <en | V>
M = Σmn Mmn em en em en M = Mmn (2.5.11)
with the implication that all component indices are of the en type.
This whole matter seems simple enough, but here is an example of the seeming paradoxes that can arise.
Given the axis-aligned basis vectors un and the tangent base vectors en we first expand
V = Σn An en (2.5.12)
where An is initially unknown. Then
An = en V = e'n V' = (e'n)iV'i = δniV'i = V'i (2.5.13)
and we end up with
V = Σn V'i en where V'i = RijVi (2.5.14)
and this seems to conflict with (2.5.11).
OUCH!! How do I explain this??? Is tensor doc wobbling??? Consider
e'n V' = (e'n)iV'i
What "type" of component indices are these?? The expansion is
V' = Σn V'i (e'n)i V'i = e'n V'
Here the detailed notation would be this
V' = Σn [V'(e')]n (e'n) [V'(e')]n = e'n V'
Apply S to both sides to get
V = Σn [V']n(e') en could abbreviate as Σn V'n en
which compare to my earlier
V = Σn [V ]n(e) en could abbreviate as Σn Vn en
and those abbreviations then lead to a contradiction. Consider again the meaning of
V'n = RnmVm
Which "component types" are implied here on each side? Even simple questions are hard! My two defining expansions are
V = ΣnVn un expanded on axis-aligned basis vectors in x-space
Vn = un V = [V(u)]n
V' = ΣnV'n e'n expanded on axis-aligned basis vectors in x'-space
V'n = e'n V = [V(e')]n
So the components of V' = RV are
[V(e')]n = Rnm [V(u)]n
so in each space, the meaning of index n corresponds to an expansion on the axis-aligned basis vectors in that space.
Then apply R to the first equation and you get
RV = ΣnVn Run
or
V' = ΣnVn u'n
Not a conflict. But then I am running two different conventions at the same time which seems dangerous.
Convention #1: Used in tensor doc
V'n = RnmVm meaning Vn = [V(u)]n and V'n = [V(e')]n
V = ΣnVn un
V' = ΣnV'n e'n
Convention #2: Used for expansions
V = Σn Vn en meaning Vn = [V(e)]n
Say it again Sam:
Convention #1: V = ΣnVn un Vn = [V(u)]n = un V
Convention #2: V = ΣnVn en Vn = [V(e)]n = en V
Where are each of these conventions used?
Convention #1 Use: V'n = RnmVm meaning [V(u)]n and V'n = [V(e')]n
Convention #1 Use: all tensor expansions on en .
Things are then internally inconsistent if you use these tensor expansions
V = Σn Vn en
M = Σmn Mmn em en
To be consistent, you MUST say this instead
V = Σn [V(e)]n en
M = Σmn [M(e)]mn em en
Then what happens when you "dot from the left " ?
em V = [V(e)]m consistent
This would make my general tensor expansion be
T = Σii....i [T(e)]ii....i (ei ei ..... ei ) . T = ΣI[T(e)]I eI (2.11.f.2)
Maybe do it like this
T = Σii....i eTii....i (ei ei ..... ei ) . T = ΣI eTI eI (2.11.f.2)
M = Σmn eMmn em en
V = Σn eVn en
But the transform rule is then
[V(e')]n = Rnm [V(u)]m
e'V'n = Rnm uVm
2.5 Expansion of vectors onto the u and e basis vectors in x-space
A vector V can be expanded onto the bases defined above as follows (implied sum on n),
V = Vn un where un V = Vn
V = Vn un where un V = Vn
V = V'n en where en V = V'n
V = V'n en where en V = V'n . (7.13.10)' (2.5.1)
This reveals the interesting fact that the coefficients for expansions on the u basis vectors are the x-space components of the vector V, whereas the coefficients for expansions on the e basis vectors are the x'-space components of the vector (recall V'n = RnmVm ). The projection equations on the right all arise from the duality relations of the vector pairs, en em = δnm and un um = δnm . To verify that V'n are the expansion coefficients on the third line, consider
en V = en (Vk uk) = Vk (en uk) = Vk Rnk = RnkVk = V'n (2.5.2)
where we used the first line of (2.5.1), then the first line of (2.4.5) and then (2.1.2).
Inserting the dot products shown in (2.5.1) into the expansions of (2.5.1) gives.
V = (un V) un
V = (un V) un
V = (en V) en
V = (en V) en . (2.5.3)
We now write the above four expansions for the cases V = um, um, em, em . Please ignore the rightmost column of equations for now, they will be referenced later.
um = (un um) un = δnm un = um // not very interesting (um)n = δnm
um = (un um) un = gnm un // a fact already known in (2.4.3) (um)n = gnm
um = (en um) en = Rnm en (um(e))n = Rnm
um = (en um) en = Rnm en (um(e))n = Rnm
um = (un um) un = gnm un // a fact already known in (2.4.3) (um)n = gnm
um = (un um) un = δnm un = um // not very interesting (um)n = δnm
um = (en um) en = Rnm en (um(e))n = Rnm
um = (en um) en = Rnm en (um(e))n = Rnm
em = (un em) un = Rmn un (em)n = Rmn
em = (un em) un = Rmn un (em)n = Rmn
em = (en em) en = δnm en = em // not very interesting (em(e))n = δnm
em = (en em) en = g'nm en // a fact already known in (2.3.3) (em(e))n = g(e)nm
em = (un em) un = Rmn un (em)n = Rmn
em = (un em) un = Rmn un (em)n = Rmn
em = (en em) en = g'nm en // a fact already known in (2.3.3) (em(e))n = g(e)nm
em = (en em) en = δnm en = em // not very interesting (em(e))n = δnm
why do care about these equations? Go to the original doc and see how this is referenced other than the next one set of equations. Seems like a lot of ink for not much payoff.
(2.5.4)
The new information these equations provide is this:
um = Rnm en em = Rmn un
um = Rnm en em = Rmn un
um = Rnm en em = Rmn un
um = Rnm en em = Rmn un . (2.5.5)
In each pair, the second equation can be obtained from the first by reversing the tilt of the summation index n.
These equations show that the matrix R in its various forms is the "basis change matrix" which relates the u and e basis vectors. Each u basis vector is a certain linear combination of the e basis vectors, and vice versa, and the R matrix provides the coefficients for those linear combinations.
As noted in the Comment at the end of Section 2.3, one can regard the above equations regarding en to apply to an arbitrary set of basis vectors en(x), with corresponding dual en(x).
2.7 Default Vector Component Conventions
2.7 A change in notation : Picture E
To clarify the way expansions (2.5.1) work, we shall now cosmetically modify Picture A noted above so that x'-space becomes e-space (E is used because B,C,D are already used up in Tensor , and because E matches e)
(2.1.1)
(2.7.1)
As a first stage we do this heavy-handed conversion as follows
X → X(u) where X is any x-space object
X' → X(e) where X' is any x'-space object . (2.7.2)
Of particular interest are the expansions of vectors in x-space and in x'-space. We have shown the table for x-space expansions in (2.5.1) above, while that for x'-space vectors is given in Tensor (7.13.11)' . These expansions are easy to verify, so we won't do that here. V is a vector in x-space, while V' is the corresponding vector in x'-space according to V' = RV.
V = Σn Vn un where un V = Vn = [V]n
V = Σn Vn un where un V = Vn = [V]n
V = Σn V'n en where en V = V'n
V = Σn V'n en where en V = V'n
V' = Σn V'n e'n where e'n V' = V'n = [V']n
V' = Σn V'n e'n where e'n V' = V'n = [V']n
V' = Σn Vn u'n where u'n V' = Vn
V' = Σn Vn u'n where u'n V' = Vn (2.7.3)
Notice that each set of coefficients appears twice on the right, once for V and once for V'. This duplication arise because a b = a' b' for any pair of vectors in either space. For example,
Vn = un V = u'n V' appearing in lines 1 and 7 (2.7.4)
The rightmost column just shows that [V]n = Vn and [V']n = V'n so the notation is consistent.
Consider then
[V]n = un V = the component of vector V on the axis-aligned un basis vector.
[V']n = e'n V' = the component of vector V' one the axis-aligned e'n basis vector. (2.7.5)
Therefore,
Fact: The meaning of a component index on a vector is that it indicates the projection of that vector onto the relevant axis-aligned basis vector. (2.7.6)
Converting now our vector expansions (2.7.3) from Picture A to Picture F we find,
V(u) = Σn V(u)n un(u) where un(u) V(u) = V(u)n ≡ [V(u)]n
V(u) = Σn V(u)n un(u) where un(u) V(u) = V(u)n ≡ [V(u)]n
V(u) = Σn V(e)n en(u) where en(u) V(u) = V(e)n
V(u) = Σn V(e)n en(u) where en(u) V(u) = V(e)n
V(e) = Σn V(e)n en(e) where en(e) V(e) = V(e)n ≡ [V(e)]n
V(e) = Σn V(e)n en(e) where en(e) V(e) = V(e)n ≡ [V(e)]n
V(e) = Σn V(u)n un(e) where un(e) V(e) = V(u)n
V(e) = Σn V(u)n un(e) where un(e) V(e) = V(u)n (2.7.7)
We have made the following changes (shown only for labels down)
un → un(u) V → V(u) g → g(u) x' = F(x) → x(e) = F(x(u))
u'n → un(e) V' → V(e) g' → g(e)
en → en(u) Vn = [V]n → V(u)n = [V(u)]n
e'n → en(e) V'n = [V']n → V(e)n = [V(e)]n (2.7.8)
The notation is cumbersome but is helpful with higher rank tensors. From (2.6.6) we have
(um(u))n = g(u)mn = δmn (em(u))n = Rmn
(um(e))n = Rnm (em(e))n = g(e)nm = δnm (2.7.9)
Each of these 4 equations generates 4 equations when labels and indices are raised and lowered, so
(um(u))n = δmn (um(u))n = g(u)mn (em(u))n = Rmn (em(u))n = Rmn
(um(u))n = g(u)mn (um(u))n = δmn (em(u))n = Rmn (em(u))n = Rmn
(um(e))n = Rnm (um(e))n = Rnm (em(e))n = δmn (em(e))n = g(e)mn
(um(e))n = Rnm (um(e))n = Rnm (em(e))n = g(e)mn (em(e))n = δmn (2.7.10)
Any metric tensor is symmetric regardless of the index positions, so various forms are possible above.
Notice in (2.7.10) [ col 4 row 4] that (em(e))n = δmn . According to the Comment at the end Section 2.3, this equation applies for an arbitrary set of basis functions em(x). The equation is eminently reasonable. Expanding em = [ Σn (em(e))n en one sees that one must have (em(e))n = δmn. Expanding a basis vector on its own basis vectors yields a single term in the sum. The same applies to (um(u))n = δmn .
Let's now review the paradox (2.1) presented at the start of Chapter 2, but in our "improved" notation. We first restate the paradox in covariant notation:
Paradox. Let v = Σnvnen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em)n en. This implies that (em)n = δmn. Thus, the only possible basis vectors em allowed in the universe are axis-aligned unit vectors. (2.7.11)
The paradox arises because the expansion as stated is ambiguous and in our Picture E notation it is in fact wrong. The expansion should be em(e) = Σn(em(e))n en(e) and it is true that (em(e))n = δmn. This does not put any restriction on (em(u))n = Rmn where Rmn can be a set of arbitrary numbers.
Exercise: Explain why two of these equations are valid and two are not:
Picture F Picture A
1 em(u) = Σn(em(u))n en(u) // invalid em = Σn(em)n en
Rmn
2 em(u) = Σn(em(u))n un(u) // valid em = Σn(em)n un
3 em(e) = Σn(em(e))n en(e) // valid e'm = Σn(e'm)n e'n
δmn
4 em(e) = Σn(em(e))n un(e) // invalid e'm = Σn(e'm)n u'n (2.7.12)
This relates to Fact (2.7.6) which we replicate here
Fact: The meaning of a component index on a vector is that it indicates the projection of that vector onto the relevant axis-aligned basis vector. (2.7.6)
Because of our desire shown in (2.6.1) to have u'n = R un and e'n = R en, it turns out that the un are the axis-aligned basis vectors in x-space, but the e'n are axis-aligned in x'-space, see below (2.6.2). In equation 1 above, the vector em(u) is an x-space vector, so its component (em(u))n is the coefficient in an expansion over x-space axis-aligned vectors which are the un as in equation 2, not the en as in equation 1. We leave the explanation of 3 and 4 to the reader.
Exercise: Verify that equation 2 above is valid using entries in (2.7.10) :
2 (em(u))i =?= Σn (em(u))n (un(u))i
Rmi =?= Σn Rmn δni = Rmi so 2 is valid (2.7.13)
Conclusion: One must be careful when dealing with tensors and their components to understand the space to which a tensor belongs, and to understand the meaning of the component indices.
We now restate the transformation rules for rank-1 and rank-2 tensors in our (2.7.1) Picture E context:
[V(e)]a = Rab [V(u)]b (2.1.5)
[M(e)]ab = Raa' Rbb' [M(u)]a'b' . (2.1.7) line 1 (2.7.14)
If the above tensors are tensor fields, one then has x(e) = F(x(u)) so
[V(e)(x(e))]a = Rab [V(u)(x(u))]b
[M(e)(x(e))]ab = Raa' Rbb' [M(u)(x(u))]a'b' (2.7.15)
Comment. The ambiguity of the meaning of a vector component is made very clear in the Dirac notation discussed below in Section 2.11. In that notation one writes for example,
(um)n = (um(u))n = <un | um > = the vector | um > projected onto the <un| basis element
(um(e))n = <en | um > = the vector | um > projected onto the <en| basis element (2.7.16)
ok to here but please review first thing
***************** above put here 4/3/16 at 8 PM
Item (e) has multiple arguments which are different points in the space of x. Here F must be linear, otherwise one would have to deal with Rij(x) ≠ Rij(y).
Item (f) is is the transformation of a scalar field, whether or not F is linear.
Item (g) has those multiple arguments like item (e), so F must be linear.
Item (h) describes the transformation of a tensor field which has multiple arguments each of which is itself a vector field. This type of transformation is viable for linear or non-linear F. In the latter case, despite the multiple arguments, only the coordinate x is involved.
If F(x) is non- linear, then Rab(x) depends on x and even if Tabcd is independent of x, T'a'b'c'd' does depend on x, so one would have
T'a'b'c'd'(x') = Ra'a(x)Rb'b(x)Rc'c(x)Rd'd(x)Tabcd
The dependence of T' on x' = F(x) is entirely induced by the transformation.
Item (d) requires that F be linear because, for a non-linear F, the linearization Rij(x) near x would differ
from Rij(y) near y. When F is linear, Rij is the same everywhere. There can be any number of coordinate arguments, where x' = F(x), y' = F(y), and so on, but only one F .
Item
*******
make a change of notation, and then apply the machinery to the simple situation of a change of basis. This then provides a bulletproof way to understand expansions of vectors and other objects onto basis vectors. This helps one avoid paradoxes like the following:
Paradox. Let v = Σnvnen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em)n en. This implies that (em)n = δm,n. Thus, the only possible basis vectors em allowed in the universe are axis-aligned unit vectors! (2.1)
This paradox involves confusion about the meaning of a basis vector. Hopefully this paradox will motivate the reader to trudge through the following presentation of covariant notation.
************ above from Chapter 2
As for bases, one can consider (ab)ij = aibj of (2.8.9) in several situations depending on the spaces in which the two vectors lie. Here are two examples ( the basis must of course match on the two sides )
[(ab)(e)]ij = a(e)ib(e)j
[(ab)(e,u)]ij = a(e)ib(u)j . (2.8.14)
As an example of the second equation, one could write
[(enem)(e,u)]ij = (en(e))i (em(u))j = δni Rmj (2.8.15)
where the last expression comes from (2.7.4).
**********8
M = Σab [M(e)]ab eaeb . (2.10.2)
As a check, take a tensor component of both sides of (2.10.2) in the ei basis :
[M(e)]ij = { Σab [M(e)]ab eaeb}(e)ij
= Σab [M(e)]ab (eaeb)(e)ij
= Σab [M(e)]ab (ea(e))i (eb(e))j // (2.8.9)
= Σab [M(e)]ab δaiδbj // (2.7.4) [ col 3 row 3]
= [M(e)]ij (2.10.3)
so things are consistent.
*************
Exercise: Consider these two expansions of the rank-2 tensor M,
M = Σab[M(e)]ab eaeb
M = Σab[M(u)]ab uaub . (2.10.8)
How are the coefficients in these expansions related?
Method 1: Use the result in (2.5.5) that em = Rmiui. Then
M = Σab [M(e)]ab eaeb
= Σab [M(e)]ab Σij(Raiui)(Rbjuj)
= Σij {Σab[M(e)]abRaiRbj} uiuj // now do dummy i↔a and j↔b
= Σab { Σij[M(e)]ijRiaRjb} uaub . (2.10.9)
Comparing with the second line of (2.10.8) one sees that
[M(u)]ab = ΣijRiaRjb[M(e)]ij . (2.10.10)
and this then shows how the (u) coefficients in (2.10.8) are related to the (e) coefficients.
Method 2: Consider the rule for the transformation of a rank-2 tensor as stated in (2.1.6) but converted to Picture E:
[M(e)]ab = Σa'b'Raa' Rbb' [M(u)]a'b' . (2.10.11)
This at once shows how the (e) coefficients in (2.10.8) are related to the (u) coefficients.
We can invert this equation using the R matrix orthogonality rules (2.1.8), but instead we just use the inversion rule stated in (2.1.11)
***********
= Mab uaub
M = Σab M'ab eaeb
= Σab M'ab Σij(Raiui)(Rbjuj)
= Σij {ΣabM'abRaiRbj} uiuj // now do dummy i↔a and j↔b
= Σab { ΣijM'ijRiaRjb} uaub . (2.10.9)
Comparing with the second line of (2.10.8) one sees that
[M(u)]ab = ΣijRiaRjb[M(e)]ij . (2.10.10)
and this then shows how the (u) coefficients in (2.10.8) are related to the (e) coefficients.
Method 2: Consider the rule for the transformation of a rank-2 tensor as stated in (2.1.6) but converted to Picture E:
[M(e)]ab = Σa'b'Raa' Rbb' [M(u)]a'b' . (2.10.11)
This at once shows how the (e) coefficients in (2.10.8) are related to the (u) coefficients.
We can invert this equation using the R matrix orthogonality rules (2.1.8), but instead we just use the inversion rule stated in (2.1.11)
**
[M(u)]ab = Σa'b'Ra'a Rb'b [M(e)]a'b' = ΣijRiaRjb[M(e)]ij (2.10.13)
*********************************8
New Definition of a Component Index
Up to this point the meaning of a general tensor component index has been defined in terms of the meaning of a component index on a vector, namely,
vi ≡ ui v = <ui|v> v = Σi viui
where the ui are the axis-aligned vectors of (2.4.1). In a more general notation one could write this as
v(u)i ≡ ui v = <ui|v> v = Σi v(u)iui
where v(u)i means the component is taken in the u basis.
In the upcoming chapters, we want to work with general arbitrary basis vectors we call en which may or may not be axis-aligned vectors. We can certainly then write
v(e)i ≡ ei v = <ei|v> v = Σi v(e)iei
The connection between these two kinds of vector components is given by
v(e)i = <ei|v> = <ei|uj><uj|v> = Rij v(u)j .
v(u)i = Rji v(e)j // inversion rule on the above line
The cost of adding a descriptive (e) superscript to every vector index is quite high, so starting here we change the meaning of vi to be that of the above line
vi ≡ v(e)i ≡ ei v = <ei|v> v = Σi v(e)iei = Σiviei
One large implication of this change is that all Chapter 2 equations have to be "processed" according to the rule
vi → Rji vj
For example
(en)i = Rni → (en)i = Rji Rni
which we could write in a more detailed notation as [v(u)]i , meaning vi is a component of a vector in the ui basis
v = Σn viui
Recall (2.7.5) which said,
Fact: The meaning of a component index on a vector is that it indicates the projection of that vector onto the axis-aligned basis vector in the space of the vector. (2.7.5)
All of Chapter 2 up to this point has used the above definition where vi ≡ ui v = <ui|v> .
Starting right here, and continuing for many chapters, we change the meaning of a component as follows. We allow that en is some arbitrary basis for vector space V, and the component of a vector is now
vi = ei v = <ei|v> = [v(e)]i
Now vi is the component of v in the e basis, not the axis-aligned u basis
************************************
We normally use |ei> as our go-to basis in V, and, due to the orthogonality relation noted above, we like to use <ei | as our go-to basis in V*.
************************************
Consider then
Vn = [V]n = un V = the component of vector V on the axis-aligned un basis vector.
V'n = [V']n = e'n V' = the component of vector V' on the axis-aligned e'n basis vector. (2.7.4)
We arrive then at the following fact which should be clearly stated though it might be considered obvious:
Fact: The meaning of a component index on a vector is that it indicates the projection of that vector onto the axis-aligned basis vector in the space of the vector. (2.7.5)
This "meaning" applies in fact to all the index components on all equations which have so far appeared in Chapter 2. As a special example, one can write
x = Σn xn un x' = Σn x'n e'n Rij = (2.7.6)
so the i index on Rij is associated with e'i while the j index with uj. Later in Dirac notation we will write
Rij = <e'i | R | uj> (2.7.7)
which makes this fact more explicit.
***************88
(d) Transformations of tensor functions
We pause to comment on the way the objects discussed above "transform" under the kinds of transformations discussed in Chapter 2. For example, there we said that v'i = Rijvj described the transformation of a contravariant vector v where R is the linearization of some underlying transformation x' = F(x). The subject of transformations is always potentially confusing and we now describe three kinds of transformations we shall call A,B and C.
transformation A tensors like v and Tij change, basis vectors like ei do not change
transformation B tensors do not change, basis vectors do change
transformation C tensors and basis vectors change together (2.11.d.1)
Our interest is only in transformation A which is often called an "active" transformation. For example, an apparatus described by some set of tensors is rotated, while the observing system (including basis vectors) is not rotated. In a very simple case, one would start with a vector V and actively rotate it into a rotated vector V' = RV where the coordinate system does not move:
Active Transformation A Passive Transformation B (2.11.d.2)
For transformation A, the basis vectors do not move, so objects like ui = |ui>. |ui>, <ui| and <ui| are vectors in the sense that they are objects having n components, but under transformation A they are "non transforming vectors" which act more like scalars. For example,
Vi = <ui| V> // before transformation A
V'i = <ui| V'> // after transformation A : V moved, ui did not move . (2.11.d.3)
In the tensor expansion notation, transformation A works this way for a rank-1 and rank-2 tensor,
V = ΣaVa ua |V> = ΣaVa |ua>
V' = ΣaV'a ua |V'> = ΣaV'a |ua> V'a = RabVb (2.11.d.4)
T = ΣabTab uaub |T> = ΣabTab |ua> |ub>
T' = ΣabT'ab uaub |T'> = ΣabT'ab |ua> |ub> T'ab = RacRbdTcd . (2.11.d.5)
u1 u2 V V'1 V'2
Basically, V and V' are just two different vectors, as the above Figure shows, and T and T' are two different rank-2 tensors.
____________________________
Very confused, resume here tomorrow. Figure it out, then decide whether to delete the discussion.
I am confused now. I would say that Transformation B then works this way (Just apply the R operator)
V = ΣaVa ua |V> = ΣaVa |ua>
V' = ΣaVa u'a |V'> = ΣaVa |u'a> V'a = RabVb (2.11.d.4)
T = ΣabTab uaub |T> = ΣabTab |ua> |ub>
T' = ΣabTab u'au'b |T'> = ΣabTab |u'a> |u'b> (2.11.d.5)
The second case aligns with my (2.7.1) table of possible expansions. You get the second equation just by applying R to the first equation. So I would say this passive transformation B is more in line with what I am writing about! You are moving
Problems here! I see that the numbers in the u-basis change from Va to V'a from (2.11.d.3).
****************
(g) The Covariant Transpose
Whereas the matrix transpose of a matrix Mab would be (MT)ab = Mba (swap the rows and columns), it is the covariant transpose (MT)ab = Mba that is significant in covariant notation. We quote from Tensor where M is a general rank-2 tensor while R and S are the transformation "differentials" of (2.1.2),
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . (7.9.3)' (2.11.g.1)
Equations in any column can be obtained by lowering one or both indices in the top equation, so that the covariant transform MT is a rank-2 tensor if M is a rank-2 tensor.
For all-up or all-down indices, the two kinds of transposes are the same: (MT)ab = (MT)ab = Mba.
The transpose always has the indices reflected in a vertical line between the indices.
This subject is discussed in Tensor Section 7.9 where all claims are proved. We quote some of the conclusions:
det(M) = det(MT) = det(MT) (7.9.7)' (2.11.g.2)
RRT = RTR = 1 SST = STS = 1 RS = SR = 1
RT = R-1 = S ST = S-1 = R . (7.9.8)' (2.11.g.3)
v' = (Mv) = (M)v (v')T = (Mv)T = vT MT
|v'> = |Mv> = M|v> <v'| = <Mv| = <v|MT . (2.11.g.4)
<a | M | b> = <b| MT |a> . (7.9.17)' (2.11.g.5)
Here is a proof of the last item above, a good exercise in Dirac notation (implied sums as usual),
<a | M | b> ≡ <a | M b> = a (Mb) = ai(Mb)i = ai[ Mijbj] = ai Mij bj
= bj Mij ai = bj (MT)ji ai = bj [MTa]j = b (MTa) = <b| MTa> = <b| MT |a> .
Matrix products (like RRT or SR) are always in either up-tilt or down-tilt form, as for example in (2.1.3). Note that both S and R are "covariant real orthogonal" (RT = R-1) even when S and R are not rotation matrices. Equations like RTR = 1 are just the orthogonality rules (2.19), for example,
(RTR)ac = (RT)abRbc = RbaRbc = δac = (1)ac . // (2.1.9) #1
Another exercise is to show that wv is a scalar under any transformation x' = F(x) ,
w'v' = <w' | v'> = <Rw|Rv> = <w|RTR|v> = <w|1|v> = <w|v> = wv . (2.11.g.6)
*****************
2.11 Dual Spaces and Tensor Functions
We denote dual-space vectors and tensors by Greek or script font letters.
The dual space V* is by definition the space of linear functionals over V. If α ϵ V*, we can then write
α : V → K α(v) = k ϵ K (2.11.1)
where K is any field (but we always use the reals). Since α is a linear functional, α(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(v) as a "function".
Comments: Much of the rest of this section will be repeated in later Chapters. We have found that the notations involved can be a major stumbling block, and feel it is important to exercise the notation in many ways to make the reader (and author) feel comfortable with it. As with most endeavors, it is a matter of practice. We also try to explain why certain notations are used.
(a) The Dual Space V* in Matrix and Dirac Notation
For every column vector v in V, there exists a row vector vT such that (vT)i = vi. For example, for N=2,
v = = |v> vT = (a, b) = <v| (2.11.a.1)
Here we have snuck in the Dirac bra-ket notation where the ket |v> is a column vector and the bra <v| is the corresponding row vector. The notation vT means that the row vector is the Transpose of the column vector.
We now have multiple ways to write the dot (inner, scalar) products of Section 2.9 :
v v' = vTv' = (a, b) = aa'+bb' = <v | v' > . (2.11.a.2)
Because our vectors have real components, the above can also be written
v' v = v'Tv = (a', b') = aa'+bb' = <v' | v > . (2.11.a.3)
We mention real components only because in the Dirac notation one has <a|b> = <b|a>* where * means complex conjugation, so if this scalar product is real, then <a|b> = <b|a> .
We regard v or |v> as being a vector in the vector space V, while vT or <v| (the row vector) is a vector in the dual space V*. This is really a simple concept. Sometimes the dual-space vector vT = <v| is referred to as the covector of v = | v > .
Suppose αT = <α | is a vector in the dual space V* . One can regard this dual-space vector <α | as being a functional which acts on vectors in the space V. Then,
α = <α | = functional
α(v) = <α | v> = α v = function = a scalar number (2.11.a.4)
α : V → K . // K = any real field, such as the real numbers
Just as the space of column vectors V is a linear space (a vector space), so also the dual space of row vectors V* is a linear space, so we know that the functional <α | is a linear functional. That in turn implies that the function α(v) is a linear function, which we now show directly:
α(sv) = α (sv) = s ( α v ) = s α(v)
α(v+v') = α (v+v') = α v + α v' = α(v) + α(v') . (2.11.a.5)
(b) Functional notation
We have now a slight notational conundrum. We like to write a scalar-valued function F(v) in non-bold font, whereas a vector-valued function would be F(v). Thus we have written α(v) above with a non-bold α, since α(v) is a scalar-valued function. On the other hand, α(v) is really a function of the vector α, so it seems misleading to refer to it as α(v), and we ought to call it α(v) so then α(v) = <α | v> has everything bolded on both sides. But then the functional would have to be called α = <α | . But this contradicts our notation earlier that α is a vector, αT is the transpose, and we should write αT = <α |. If we use α = <α | then we avoid that contradiction. This is what authors end up doing, writing a functional as a scalar entity which for us means an unbolded entity. A possible solution would be to say,
fα = <α | = functional
fα(v) = <α | v> = α v = function (2.11.b.1)
where f is non-bold, and the subscript label α is bold, but then we have introduced a new symbol f which seems superfluous. So the conclusion is this: α = <α| = functional, α(v) = <α | v> = function, and one must understand that α(v) is a function of the vector quantity α. Obviously there is a unique functional α(v) for each vector α in V (and thus for each αT in V*). The spaces V and V* have the same dimension n and are isomorphic to each other in the sense just noted.
(c) Basis vectors for the dual space V*
Now recall that our axis-aligned x-space basis vectors ui have dual basis vectors ui where ui uj = δij which is the idea of orthogonality in the covariant world (which might be non-Cartesian). In our notations above,
ui uj = (ui)T uj = <ui | uj> = δij
= ui uj = (ui)T uj = <ui | uj> = δij = δi,j . (2.11.c.1)
Since ui and ui are in general different column vectors in V, (ui)T and (ui)T are different row vectors in the dual space V*.
Just as the column vectors |ui> and |ui> form two distinct bases for V, the row vectors <ui | and <ui | form two distinct bases for V*. Certainly dim(V) = dim(V*).
Definition of λi
Above we discussed α = <α| as a vector functional, and α(v) = <α|v> as the corresponding scalar function. Whereas <α| is some general vector in V*, we now consider in its place a basis vector <ui| in V*. With what notation shall we represent this functional? In analogy with α and α(v) we could use ui and ui(v) where the ui is unbolded to indicate a scalar function. Or we could use fu = <ui| and fu(v) = <ui| v> . The first notation is not uncommon (see wiki dual space where ui = ei), while the latter notation is unpleasant. Other common notations are v*i(v) or e*i(v) which for us would be u*i(v).
We shall use the following notation,
λi ≡ <ui| basis functional in V* // λi = (ui)T
so
λi(v) = <ui|v> basis function in V*f // λi(v) = (ui)Tv . (2.11.c.2)
The λ is unbolded, consistent with α(v). λ is a Greek letter consistent with our plan to use Greek or script letters for dual space objects. The index on λi is up, matching the index on ui in <ui|. Notice that
λi(uj) = <ui|uj> = δij . (2.11.c.3)
We think of λi ≡ <ui| as being in the dual space V* while λi(v) = <ui|v> lies in a directly corresponding space of functions which we call V*f.
Comment: For the dual space basis functionals Sjamaar used symbol λi in his 2006 notes (p 84), but changed to βi in his 2015 update (p 91). Spivak uses φi (p 76). Wiki (dual basis) uses basis vectors vi instead of ei so their λi is called vi. Wiki (dual space) uses ei while Lang Algebra uses fi (p 143). There seems to be no standard notation as in physics where F = ma is universally recognized as Newton's Second Law which would be hard to identify if written G = nb. Probably ui or ui (unbolded) is the most logical choice if the V basis vectors are ui, but it is so easy to confuse functional ui with the vector ui (especially when we drop our bolding of vectors starting in Chapter 3) that we shall stick with λi.
Eq. (2.7.1) line 3 gives the expansion of a vector v onto the ui
v = Σi vi ui where vi = ui v
or
|v> = Σi vi |ui> where vi = <ui| v > = ui v . (2.11.c.4)
Notice therefore that
λi(v) = <ui|v> = ui v = vi . (2.11.c.5)
The function λi(v) is sometimes called "the ith coordinate function" since it projects out the ith component the vector v. As summarized in (2.7.17), since each dot product ui v is a scalar, the functions λi(v) i = 1..N transform as scalars despite the fact that the values of these scalars are the components of the vector vi.
Transposing (2.11.c.4) produces a vector functional expansion in V*,
vT = Σi vi (ui)T or <v| = Σi vi <ui| = Σi vi <ui| . (2.11.c.6)
Using Greek letters for dual space objects we write this as
α = <α| = Σn αi <ui| = Σn αiλi . (2.11.c.7)
Then,
α = Σiαiλi functional (2.11.c.8)
α(v) = Σiαiλi(v) = Σiαivi = α v function (2.11.c.9)
or in bra-ket notation,
<α| = Σiαi<ui| functional
α(v) = Σiαi<ui|v> =Σiαivi α v = <α |v> function (2.11.c.10)
and we replicate the result (2.11.b.1).
Definition of λ'i
We have defined λi ≡ <ui| as a notation for a certain basis functional in dual x-space. We would like to somehow define an object λ'i which is a basis functional in dual x'-space. How should this be done?
One might intuitively feel that one should set λ'i ≡ <u'i| . Or one might think that once λi is defined as above, then the meaning of λ'i is forced upon us by some equation like λ'i = Rijλj . Both these notions are not what we want to do. We are not forced to say λ'i ≡ <u'i| just because λi ≡ <ui| since we are making two separate definitions. And λ'i = Rijλj is complete nonsense for the following reason. The N functionals λi for i = 1..N are each vectors in V*, so {λi} is a set of vectors, not a set of numbers, whereas when one tries to write λ'i = Rijλj one is implying that λj is a set of numbers which form a vector.
Recall that the ui are the "axis-aligned" basis vectors in x-space since (ui)j = (ui)j = δi,j.
Recall that the e'i are the "axis-aligned" basis vectors in x'-space since (e'i)j = (e'i)j = δi,j.
This suggests that the proper definition of λ'i is the following:
λ'i ≡ <e'i| (2.11.c.11)
One then finds that, for v' a vector in x'-space,
λ'i(v') ≡ <e'i| v'> = v'i (2.11.c.12)
which is then analogous to
λi(v) ≡ <ui| v'> = vi (2.11.c.5)
In both cases then λi and λ'i are the "ith coordinate functions", projecting out the ith coordinate from a vector.
Since v'i = Rijvj we can certainly write
λ'i(v') = Rij λj(v) (2.11.c.13)
as a statement relating two vectors of scalars. Notice this does not say λ'i = Rijλj which we already noted above does not even make sense. If we display the fact that Rij in general is Rij(x) then
λ'i(v') = Rij(x) λj(v) (2.11.c.14)
Since this does not fit into any of the molds shown in (2.1.16), one cannot quite claim that λj(v) transforms as a vector field, but the transformation is similar.
We can study (2.11.c.13) in Dirac notation as follows (see below for Dirac notation details).
λ'i(v') = <e'i|v'> = <ei|v> = <ei| 1 | v> = <ei|uj >< uj| v> = ei uj λj(v) = Rijλj(v)
(2.11.c.15)
where the last step comes from (2.4.3).
Once we have a functional λ'i, we can define a general rank-1 functional α' in dual x'-space as follows:
α' = Σiα'iλ'i functional in V'* (2.11.c.16)
α'(v') = Σiα'iλ'i(v') = Σiα'iv'i = α' v' function in V'*f (2.11.c.17)
It then follows that
α'(v') = α' v' = α v = α(v) (2.11.c.18)
and in some sense one could say that α(v) transforms as a scalar field, where v plays the role normally occupied by the position vector x. On the other hand, the vector |v> and the dual vector (functional) α = <α| transform as vectors and so α is a vector functional.
We now define α(v) to be a "rank-1 tensor function". Spivak would call it a "1-tensor". We have this seeming contradiction that α(v) is a rank-1 tensor function, yet that function transforms as a rank-0 scalar. The rank-1 description really applies to the functional α = <α| which is in fact a vector and transforms as a vector. When this is closed with the ket |v> one obtains the scalar object α(v) = <α | v>. (2.11.c.19)
Vector space names: V, V*, V*f and V', V'*, V'*f (2.11.c.20)
Here we have associated vector space names V and V* with x-space in Picture A (2.1.1), while V' and V'* are associated with x'-space. All these spaces have the same dimension N and all are isomorphic. There is a 1-to-1 relationship between V and dual space V* as noted above, and there is a 1-to-1 relationship between V and V' since for every vector v in x-space there is a unique corresponding vector v' = Rv in x'-space. We refer to V* as dual x-space and V'* as dual x'-space. Associated with the dual space V* of functionals is the space V*f of corresponding functions, and similarly for V'* and V'*f.
ok to here 9AMThurs 4.7
Section (d) has been removed.
(d) Rank-2 functionals and tensor functions
A rank-2 tensor may be represented as
T = Σab Tab ua ub (2.11.d.1)
V = Σa Va ua
where on the second line for comparison we show a general rank-1 tensor (vector). In Dirac notation, we write
| ua, ub> ≡ |ua> |ub> ↔ ua ub (2.11.d.2)
which represents any of the n2 basis vectors of the tensor product space V2 = VV. We could write this as | ua, ub>2 ≡ |ua>1 |ub>1 to distinguish the fact that some kets are in V1 and others in V2, but the contents of the ket usually make it obvious to which vector space a ket belongs. In Dirac notation, the tensor T is written
|T> = ΣabTab |ua> |ub> = ΣabTab | ua, ub> (2.11.d.3)
and this is a general element of the space V2. The corresponding rank-2 linear functional in the dual space V*2 is given by
<T| = ΣabTab <ua | <ub| = ΣabTab < ua, ub| . (2.11.d.4)
This is done in analogy with the vector case
|V> = Σa Va |ua>
<V| = Σa Va<ua | (2.11.d.5)
where we are careful to have the index "tilt" have the form of a contraction, even though we are not really contracting indices on a tensor. The rank-2 functional <T| is linear in both V* spaces of V*V*, so it is called a bilinear functional. If we let (subscript 1 and 2 are labels of two vectors, not components of v )
| v1, v2> ≡ |v1> |v2> (2.11.d.6)
represent an arbitrary (but pure) element of V2 = VV, then we may construct
T = <T| rank-2 tensor functional
T(v1,v2) = <T | v1, v2> rank-2 tensor function (a Spivak "2-tensor") . (2.11.d.7)
It follows that
T(v1,v2) = <T | v1, v2> = ΣabTab < ua, ub| v1, v2>
= ΣabTab <ua| v1> <ub| v2>
= ΣabTab (v1)a (v2)b (2.11.d.8)
where we have used the fact that the scalar product for elements of V*2 with elements of V2 is the product of two V*-with-V scalar products, as seen for example in (2.9.13). In the last line above we see that the tensor function T(v1,v2) is the contraction of a rank-2 tensor with two rank-1 tensors, and so is a scalar. Thus,
T'(v'1,v'2) = T(v1,v2) (2.11.d.9)
and a rank-2 tensor function transforms as a "scalar field of two arguments". The "rank-2" description applies to the tensor functional <T| , and when this is closed with an element of V2 the result is a scalar.
Note from (2.11.d.8) and (2.4.1) that (ui)a = δia ,
T(ui,uj) = ΣabTab (ui)a (uj)b = ΣabTab δia δjb = Tij (2.11.d.10)
so the tensor function evaluated at the basis vectors gives a corresponding element of the tensor.
Using λi = < ui| as defined above in (2.11.c.2), we can rewrite (2.11.d.4)
<T| = ΣabTab <ua | <ub|
as rank-2 tensor functional
T = ΣabTab λa λb (2.11.d.11)
which is in analogy to
<α | = Σa αa <ua|
rank-1 tensor functional
α = Σa αa λa . (2.11.d.12)
We continue to use script or Greek fonts to represent functionals, such as α and T.
Taking the special case of a rank-2 functional which is just λa λb we construct the following rank-2 tensor function,
(λa λb)(v1,v2) = < ua, ub| v1, v2> = <ua| v1> <ub| v2> = (v1)a (v2)b
= λa(v1) λb(v2) . (2.11.d.13)
This function is manifestly linear in both arguments, since λa(v1) is linear, so it is a bilinear function. For example,
(λa λb)(v1+v1',v2) = (v1+v1')a (v2)b = (v1)a (v2)b + (v'1)a (v2)b
= (λa λb)(v1,v2) + (λa λb)(v'1,v2) . (2.11.d.14)
Whereas <T| shown above is a general rank-2 tensor functional, we can consider the special case of a pure rank-2 functional formed from two vector functionals α = <α| and β = <β| . In that case one finds,
(α β) = <α| <β| = <α, β | rank-2 functional
(α β)(v1,v2) = <α, β | v1, v2>
= <α | v1> <β | v2> = α(v1)β(v2) rank-2 tensor function
(α β)(ui,uj) = α(ui)β(uj) = αiβj = (α β)ij rank-2 tensor (2.11.d.15)
where the very last item is αiβj expressed in the tensor product notation of (2.8.9). Once again, evaluation of a tensor function at two basis vectors creates an element of the tensor.
Comment on vertical bars in the Dirac Notation
Let |a> be a vector in V, and <b| a vector in the dual space V*. Notice that
<b| |a> = <b||a> = <b|a> . (2.11.d.16)
The official notation for the scalar product is <b | a> not <b || a> so one replaces the || with | . The same replacement is made for example doing a scalar product between elements of V*2 and V2
<a| <b| |c> |d> = <a,b||c,d> = <a,b | c,d>
or
<a| <b| |c> |d> = ( <a| |c> ) ( <b| |d> ) = <a|c><b|d> . (2.11.d.17)
(e) Rank-k functionals and tensor functions
It is a simple matter to generalize from k = 2 to k = k, so the vector space is Vk and the dual space is V*k,
Vk ≡ VxVx....xV k factors // Cartesian product of k spaces
Vk ≡ VV....V k factors // tensor product of k vector spaces
V*k ≡ V*V*....V* k factors // tensor product of k dual spaces . (2.11.e.1)
We then have as a most general element of Vk (a rank-k tensor),
T = Σii....i Tii....i (ui ui ..... ui ) . T = ΣITI uI (2.11.e.2)
with
(ui ui ..... ui) = |ui> |ui >..... |ui> = | ui, ui ....., ui >k (2.11.e.3)
= | ui, ui ....., ui > . uI ≡ ui ui ..... ui
On the right in red we show our equations expressed in the multi-index notation introduced in (2.10.17-22). The letter Z which appears below is used to represent the set of integers 1,2...k.
Then the rank-k tensor T in Vk is represented in Dirac notation as
|T> = Σii....i Tii....i | ui, ui ....., ui > . |T> = ΣITI |uI> (2.11.e.4)
The rank-k tensor functional <T| of V*k is then
<T | = Σii....i Tii....i < ui, ui ....., ui | <T| = ΣITI <uI|
or (2.11.e.5)
T = Σii....i Tii....i λi λi ..... λi . T = ΣITI λI
A general pure element of Vk is specified by
|v1, v2, ...vk> = |v1> |v2> .... |vk> . |vZ> = |v1> |v2> .... |vk> (2.11.e.6)
The corresponding rank-k tensor function is given by
T(v1, v2, ...vk) = <T | v1, v2, ...vk>
= Σii....i Tii....i < ui, ui, ..., ui | v1, v2, ...vk>
= Σii....i Tii....i < ui|v1>< ui|v2> .... < ui|vk> (2.11.e.7)
= Σii....i Tii....i (v1)i (v2)i....(vk)i T(vZ) = ΣITI (vZ)I
This shows that the rank-k tensor function is a linear combination of the products of the argument components weighted by the components of the corresponding rank-k tensor. Since this is the contraction of a rank-k tensor with k rank-1 tensors, the result transforms as a scalar, so then
T'(v'1, v'2, ...v'k) = T(v1, v2, ...vk) . T'(v'Z) = T(vZ) (2.11.e.8)
That is to say, the rank-k tensor function transforms as a scalar field, where the term "rank-k" is associated with the functional T = <T| which is an element of the dual space V*k . Finally we see that
T(uj,uj, .... uj) = <T | uj,uj, .... uj >
= Σii....i Tii....i (uj)i (uj)i....(uj)i
= Tjj....j . T(uJ) = TJ (2.11.e.9)
From (2.11.e.7) one sees that the tensor function T(v1, v2, ...vk) is manifestly k-multilinear, which is the generalization of linear for k = 1 and bilinear for k = 2.
Once can construct a rank-k tensor functional purely from the dual basis vectors,
(λiλi ... λi) = <ui| <ui| ... <ui| rank-k tensor functional λI = <uI| (2.11.e.10)
(λiλi ... λi)(v1,v2....vk) = λi(v1)λi(v2)....λi(vk)
= (v1)i(v2)i ... (vk)i rank-k tensor function λI(vZ) = (vZ)I (2.11.e.11)
(λiλi ... λi)(uj,uj, .... uj) = λi(uj)λi(uj)....λi(uj)
= (uj)i(uj)i ... (uj)i
= δjiδji ... δji evaluated at basis vectors . λI(uJ) = δIJ (2.11.e.12)
As an alternative to the most general rank-k tensor functional T and the all-basis-vector rank-k tensor functional (λiλi ... λi), one can consider a "pure" rank-k tensor functional constructed from k dual vectors which we shall call <αi| . In this case we find,
<α1, α2....αk| = <α1| <α2| .... <αk|
pure rank-k tensor functional
= α1 α2... αk = (α1α2...αk) (2.11.e.13)
(α1α2...αk)(v1, v2, ...vk) = α1(v1)α2(v2) ....αk(vk)
= (α1v1)(α2v2) ....(αkvk) pure rank-k tensor function (2.11.e.14)
(α1α2...αk)(uj,uj, .... uj) = α1(uj)α2(uj) ....αk(uj)
= (α1 uj) (α2 uj) ... (αk uj) evaluated at ur
= (α1)j(α2)j...(αk)j = (α1α2...αk)jj... j outer product notation (2.11.e.15)
Hopefully after this long slog, the following paragraph makes some sense to the reader:
A rank-k tensor function is the bra-ket closure (inner product) of a rank-k dual tensor functional <T| of V*k with a pure rank-k non-dual tensor |v1,v2...vk> of Vk such that T(v1,v2,...vk) = <T|v1,v2...vk>. The tensor function is k-multilinear in its arguments, and transforms as a scalar field with k vector arguments. When the rank-k tensor function is evaluated at the basis vectors ur, it replicates the non-dual rank-k tensor with which is it associated, which is to say, T(uj,uj, .... uj) = Tjj....j . Spivak on page 75 refers to a rank-k tensor function as a "k-tensor". (2.11.e.16)
As we shall see later, the motivation for using tensor functions is their crashingly simple description of the tensor product of an arbitrary rank-k tensor with an arbitrary rank-k' tensor to produce a rank-(k+k') tensor :
k<T | v1,v2...vk>k k'< S| vk+1,vk+2...vk+k'>k'
= [ k<T k'<S| ] [|v1,v2...vk>k |vk+1,vk+2...vk+k'>k']
= k+k'<TS | v1,v2...vk+k'>k+k' (2.11.e.17)
or
T(v1,v2,...vk) S(vk+1,vk+2,...vk+k') = (TS)(v1,v2 .... vk+k') . (2.11.e.18)
This equation appears below as (6.6.13) and also appears in Spivak page 75.
As noted by Benn and Tucker page 2, the relationship between the vector space Vk and the dual vector space V*k is a reciprocal one. One could, as they say, perversely regard V*k as the starting vector space and then Vk would be the dual space of V*k. This amounts to swapping bra ↔ ket in the Dirac notation outlined above. Instead of having a functional α(v) = <α|v>, one would have a functional v(α) = <v|α>. We find that things are hard enough to understand without doing this "perverse" swapping of things right off the bat as they do. They refer to a rank-k tensor as a tensor of degree k, while other authors refer to rank as the order of a tensor. We us the term rank and promise not to confuse it with the different notion of the rank of a matrix which is the number of linearly independent rows or columns, or with various other meanings of the word "rank" in mathematics.
(f) The Covariant Transpose
Whereas the matrix transpose of a matrix Mab would be (MT)ab = Mba (swap the rows and columns), it is the covariant transpose (MT)ab = Mba that is significant in covariant notation. We quote from Tensor where M is a general rank-2 tensor while R and S are the "differentials" of (2.1.2),
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba
(MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . (7.9.3)' (2.11.f.1)
Equations in any column can be obtained by lowering one or both indices in the top equation, so that the covariant transform MT is a rank-2 tensor if M is a rank-2 tensor.
For all-up or all-down indices, the two kinds of transposes are the same: (MT)ab = (MT)ab = Mba.
The transpose always has the indices reflected in a vertical line between the indices.
This subject is discussed in Tensor Section 7.9 where all claims are proved. We quote some of the conclusions:
det(M) = det(MT) = det(MT) (7.9.7)' (2.11.f.2)
RRT = RTR = 1 SST = STS = 1 RS = SR = 1
RT = R-1 = S ST = S-1 = R . (7.9.8)' (2.11.f.3)
(g) Linear Dirac Space Operators
Consider these three ways of writing the same real number, where M is a matrix sandwiched between vector b on the right and transpose vector a on the left,
aT (Mb) M acts to the right ( * * *) [ ]
(aTM)b M acts to the left, and note that (aTM) = (MTa)T [ (* * *) ]
aTM b can think of M acting either to the right or to the left. (2.11.g.1)
In writing these equations, one normally thinks of M as being a matrix
Mij = (M[u])ij or M = M[u] .
By default, the matrix elements are taken in the axis-aligned ui basis on both left and right (and this applies to all indices as discussed at the end of Section 2.4) so that
(ui)TM (uj) = Σa,b (ui)a Mab (uj)b = Σa,b δia Mab δjb = Mij // = (M[u])ij. (2.11.g.2)
One could, however, do this in some other basis, for example,
(ei) TM (ej) = Σa,b (ei)a Mab (ej)b = Σa,b Ria Mab Rjb = (RMRT)ij // = (M[e])ij (2.11.g.3)
and the result is a completely different matrix. In this case the matrices are related by a covariant similarity transformation by R
M[e] = R M[u]RT . (2.11.g.4)
It is useful to think of the object M as being a basis-independent abstract linear operator which, when sandwiched between certain basis vectors, has certain matrix elements. Different types of basis vectors yield different matrices. We could even have mixed basis elements,
(ui) TM (ej) = Σa,b (ui)aMab(ej)b = Σa,b δiaMabRjb = (MRT)ij // = (M[u,e])ij (2.11.g.5)
so in this case we get
M[u,e] = MRT . (2.11.g.6)
The abstract operator M only becomes a matrix when it is properly sandwiched.
This notion of thinking of the object M as a basis-independent linear operator becomes more pronounced in the Dirac notation. We restate the above equations as follows, all of which evaluate to the same real number,
<a| ( M |b>) M acts to the right = <a | Mb >
(<a|M ) |b> M acts to the left, and note that <a|M = <MTa | = <MTa | b >
<a| M |b> can think of M acting either to the right or to the left. (2.11.g.7)
The space between the vertical bars is inhabited by abstract linear operators like M. The matrix elements shown above are then
<ui | M | uj> = (M[u])ij = Mij
<ei | M | ej> = (M[e])ij = (RMRT)ij
<ui | M | ej> = (M[u,e])ij = (MRT)ij (2.11.g.8)
To emphasize this notion of abstract operator, we shall write the operator in a different font, so M is a matrix and M is a Dirac-space operator, and then
<a| M |b> = <a| ( M |b>) = <a |M b> = a scalar product of two vectors
<a| M |b> = (<a|M ) |b> = <MTa | b > = a scalar product of two vectors
<ui | M | uj> = (M[u])ij = Mij etc . (2.11.g.9)
Here then is a review of the matrix and Dirac notations,
a' = (Ma) = (M)a (b')T = (Mb)T = bT MT matrix notation
|a'> = |Ma> = M|a> <b'| = <Mb| = <b|MT . Dirac notation (2.11.g.10)
Then consider the following claim
Fact: <a | M | b> = <b| MT | a> (2.11.g.11)
where both M and MT are the names of abstract linear operators.
Proof:
<a | M | b> ≡ <a | M b> = a (Mb) = ai(Mb)i = ai[ Mijbj] = ai Mij bj
= bj Mij ai = bj (MT)ji ai = bj [MTa]j = b (MTa) = <b| MTa> = <b| MT |a> .
Operator M is defined by its action on an arbitrary ket vector M | b> = | M b>
Operator MT is defined by its action on an arbitrary ket vector MT | b> = | MT b>
Notice in the proof that the covariant transpose MT is the correct transpose to use since Mij = (MT)ji.
Exercise: Show that wv is a scalar under any transformation x' = F(x) :
w'v' = <w' | v'> = <Rw|Rv> = <w|RTR|v> = <w| 1 |v> = <w|v> = wv . (2.11.g.12)
In this example R is a matrix, whereas R is the corresponding Dirac space operator. The statement
RTR = 1 (2.11.g.13)
is the operator version of our (2.11.f.3) matrix statement
RTR = 1 (2.11.g.14)
which we verify as follows,
(RTR)ac = (RT)abRbc = Rba Rbc = δac // (2.1.9) #1 (2.11.g.15)
and which is valid for any transformation differential matrix Rij.
One may represent a Dirac operator M in various ways
M = Σij | ui> Mij <uj|
= Σij | ei> [M[e]] ij <ej|
= Σij | ui> [M[u,e]] ij <ej| (2.11.g.16)
as can be verified by closing with the appropriate basis vectors. For example, for the last line above,
<ua | M | eb> = <ua | { Σij | ui> [M[u,e]] ij <ej|} | eb>
= Σij <ua | ui> [M[u,e]] ij <ej| eb>
= Σij δai [M[u,e]] ij δjb
= [M[u,e]] ab . (2.11.g.17)
When M = 1 we find
1 = Σij | ui> δij <uj| = Σi | ui><ui| (2.11.g.18)
which is just a statement that the | ui> basis is complete.
Finally, we make a comparison between the abstract Dirac operator M and the abstract rank-2 "vector" M,
M = Σij | ui> Mij <uj| // Dirac operator
M = Σij Mij ui uj // (2.8.10), "vector" in vector space V2
or
|M> = Σij Mij | ui> | uj> . (2.11.g.19)
The first object M is an operator in the Dirac Hilbert Space V.
The second object M or |M> is a vector in the tensor product space V V.
M and M are completely different objects, though they both involve the same matrix elements Mij. In each case, we can project out those matrix elements in an appropriate fashion:
<ua | M | ub> = <ua | { Σij | ui> Mij <uj|} | ub> = Mab
[< ua| < ub| ] | M > = [< ua| < ub| ] Σij Mij | ui> | uj> = Mab . (2.11.g.20)
The above discussion is presented implicitly for a square matrix M, but only small adjustments are needed for it to apply to a non-square matrix. In this case, in aTM b one thinks of vectors a and b as having different dimensions. Perhaps b lies in x-space which is Rn while a lies in x'-space which is Rm with m > n, and then Mij is an m x n matrix. The x'-space V' has n basis vectors |u'i> while the x-space V has m basis vectors |ui>. Then one would have, for example,
<ui | M | u'j> = Mij
M = Σi=1m Σj=1n | ui> Mij <u'j|
|M> = Σi=1m Σj=1n Mij | ui> | u'j>
1' = Σi | u'i><u'i| completeness in V'
1 = Σi | ui><ui| completeness in V (2.11.g.21)
This is exactly the situation we shall encounter in Chapter 9 where the matrix R is an m x n matrix.
We shall not use a script font to represent Dirac space operators, but one should keep in mind that any non-scalar object which is seen to be operating to the left on a bra, or to the right on a ket, or which is found between the two vertical bars in < | | > is a Dirac space operator, regardless of the font used to represent it. Such operator objects are not matrices, but their matrix elements form matrices.
*******
Warning: For certain notational reasons, we have this unfortunate fact:
In Section 10.2 the tangent vectors at point x on M are called xei .
In Section 10.6 the tangent vectors at point x' on M are called u'i (this section).
In Section 10.8 the tangent vectors at point x on M are called xui .
**********8
Fact: R is a linear operator acting to the left on elements of Λ1(R). That is to say, show that
(s1α'+s2β')R = s1(α'R) + s2( β'R) (10.7.9)
Proof: This fact seems intuitively obvious from the fact that R is associated with the matrix R and matrix algebra is linear. But here is a proof requiring no intuition. Define these four 1-forms:
α' ≡ Σi=1n Ai <e'i| α ≡ Σi=1n Ai <ei|
β' ≡ Σi=1n Bi <e'i| β ≡ Σi=1n Bi <ei| (10.7.10)
Then
α'R = [ Σi=1n Ai<e'i| ] R = Σi=1n Ai <e'i|R // bra space Λ1 is a linear space
= Σi=1n Ai<ei| = α // (10.7.7) that <ei| = <e'i|R
A similar result holds for β so,
α'R = α
β'R = β (10.7.11)
Then,
(s1α'+s2β')R = [ s1Σi=1n Ai <e'i| + s2Σi=1n Bi <e'i| ] | R // (10.7.10) left
= [ s1Σi=1n Ai + s2Σi=1n Bi] <e'i| R // bra space is linear
= [ s1Σi=1n Ai + s2Σi=1n Bi] <ei| // (10.7.7) that <ei| = <e'i|R
= [ s1Σi=1n Ai<ei| + s2Σi=1n Bi<ei|]
= [ s1α + s2β]
= s1(α'R) + s2( β'R) QED
***********************
Fact: R is a linear operator acting on Λk(Rm)
Proof: Define
α' ≡ ΣiAI(x')<e'I| ϵ Λk(Rm) a 1-form in dual x'-space
β' ≡ ΣiBI(x')<e'I| ϵ Λk(Rm) a 1-form in dual x'-space
α ≡ ΣiAI(x')<e'I| ϵ Λk(Rm) a 1-form in dual x'-space
β' ≡ ΣiBI(x')<e'I| ϵ Λk(Rm) a 1-form in dual x'-space
**********8
<u'^I|R = <u'i|R ^ <u'i|R ^ .....^ <u'i| R // (8.9.d15
= <ui| ^ <ui| ^ .....^ <ui| // (10.7.6)
= <u^I| (10.7.12)
On the other hand,
*******************
Proof: Let
α' = Σ'I aI(x)λ'^I
β' = Σ'I bI(x)λ'^I
Then
F*( s1α' + s2β') = F*( s1Σ'I aI(x)λ'^I + s2Σ'I bI(x)λ'^I)
= F*( [ Σ'I s1aI(x) + Σ'I s2bI(x)]λ'^I)
= { [ Σ'I s1aI(x) + Σ'I s2bI(x)] <e'^I| }R
= [ Σ'I s1aI(x) + Σ'I s2bI(x)] {<e'^I| }R // the bra space is linear
= [ Σ'I s1aI(x) + Σ'I s2bI(x)] ΣJ RIJ <u^J| // (10.7.12)
= [ Σ'I s1aI(x) + Σ'I s2bI(x)] ΣJ RIJ <u^J| // (10.7.12)
s1F*(α') + s2F*(β') = s1F*(Σ'I aI(x)λ'^I) + s2F*(Σ'I bI(x)λ'^I)
= {s1 Σ'I aI(x)<e'^I| } R + {s2 Σ'I bI(x)<e'^I| } R // def of F* twice
= {s1 Σ'I aI(x)<e'^I| + {s2 Σ'I bI(x)<e'^I| } R // bra space is linear
= {s1 Σ'I aI(x) + s2 Σ'I bI(x)} <e'^I| R // bra space is linear
= F*( [ Σ'I s1aI(x) + Σ'I s2bI(x)]λ'^I)
s1F*(α') + s2F*(β') = s1 <α' | R + s2 <β' | R // (10.7.20) twice
= [ s1 <α' | + s2 <β' | ] R // the bra space Λk is a linear space
= < s1α' + s2β' | R
F*( s1α' + s2β') = [ < s1α' + s2β' | ]R // (10.7.20)
= s1F*(Σ'I aI(x)λ'^I) + s2F*(Σ'I bI(x)λ'^I)
= F*(Σ'I aI(x)λ'^I)
Proof for k = 0
F*( s1α(x') + s2β(x') ) = s1α(x') + s2β(x') // (10.7.22) item 1
= s1 F*(α(x')) + s2F*(β(x')) // (10.7.22) item 1 used twice QED
Proof for k = 1 Define
α' ≡ Σiαi<e'i| ϵ Λ1(Rm) a 1-form in dual x'-space α ϵ V = Rm
α ≡ Σiαi<ei| ϵ Λ1(Rn) a 1-form in dual x-space
β' ≡ Σiβi<e'i| ϵ Λ1(Rm) a 1-form in dual x'-space
β ≡ Σiβi<ei| ϵ Λ1(Rn) a 1-form in dual x-space
Then
F*(α') = [α'R] = Σiαi<e'i|R = Σiαi<ei| = α
F*(β') = [β'R] = Σiβi<e'i|R = Σiβi<ei| = β
Next, consider
s1α' + s2β' = <s1Σiαie'i| + <s2Σiβie'i|
= [ s1Σiαi + s2Σiβi ] <e'i| // since Λ1(Rm) is a linear space
Then
F*( s1α' + s2β') = (s1α' + s2β') R
= [ s1Σiαi + s2Σiβi ] <e'i| R // from just above
= [ s1Σiαi + s2Σiβi ] <ei| // (10.7.7)
= s1Σiαi<ei| + s2Σiβi<ei| // since Λ1(Rn) is a linear space
= s1α + s2β // ***
= s1F*(α') + s2F*(β') QED
In pullback function notation this says
F*(s1α' + s2β') = s1F*(α') + s2F*(β')
***********************
***********************************************
Consider now
F*(f(x')λ'^I) = <f(x')e'^I|R
= f(x') <e'^I|R // the space Λk(Rm) is linear since it is a vector space
= f(F(x)) <e'^I|R // get rid of references to x'-space
= F*(f(x')) F*(λ'^I) // using the ** and ** above
= F*(f(x')) ^ F*(λ'^I) // ^ is same as * when one item is a scalar
The formal (and trivial) last step here arises from the fact that
Λ0 ^ Λk = Λ0 Λk = Λk
f(x) ^ α = f(x) α = f(x) α α ϵ Λk
The only reason for the last formal step is to allow us to formally say
F*(0-form * k-form) = F*(0-form) ^ F*(k-form) k ≥ 1
Then if the k-form is also a 0-form we get the even more elaborately trivial result
F*(0-form times 0-form) = F*(0-form) ^ F*(0-form)
or
F*[ f(x') g(x')] = F*[f(x')] ^ F*[g(x')] = f(F(x)) ^ g(F(x)) = f(F(x)) g(F(x))
since Λ0 ^ Λ0 = Λ0 = K (scalars).
Consider now the action of F* on the most general α form in dual x'-space,
αx' = Σ'I fI(x')λ'^I
Then
F*(αx') = F*(Σ'I fI(x')λ'^I ) =
= F*(f(x')) F*(λ'^I) = F*(f(x')) ^ F*(λ'^I)
The action of a scalar function f(x') times a basis vector k-form λ'^I is this
F*( f(x')λ'^I ) = f(x') F*( λ'^I) = f(F(x)) F*( λ'^I)
which
In the context of differential forms, the operator R is sometimes given the new name F*, where F* ≡ R. This F* is called the pullback operator. The F in F* refers to the F in the transformation x' = F(x), so F* is just the "differential" R operator of this transformation. So,
<e'i| F* = <e'i| R = Rij<uj|
<u'i| F* = <u'i| R = <ui| // <ui| = <RTu'i| . (10.7.9)
Recall from Section 2.11 our special notation for basis functionals in a dual space,
λi = <ei| basis functional // λi = (ei)T
λi(v) = <ei|v> = vi basis function // λi(v) = (ei)Tv (2.11.c.2)
where there the ei were axis-aligned basis vectors. Here we write,
λi ≡ <ui| = basis functional in dual x-space i = 1..n // ui axis-aligned
λ'i ≡ <e'i| = basis functional in dual x'-space i = 1..m // e'i axis-aligned . (10.7.12)
From (10.7.8) , (10.7.9) and (10.7.12) one has
<e'i| F* = λ'iF* = <e'i| R = λ'iR = Rij<uj| = Rij λi (10.7.13)
It is convenient (and more traditional) to define a pullback function named F* which acts on a bra <a'| in x'-space in this manner
F*(<a'| ) ≡ <a' | F* = <a' | R . F* ≡ R (10.7.14)
The equation
F*(<e'i| ) = F*( λ'i ) = <e'i| F* = Rij<uj| = Rij λi (10.7.15)
shows clearly that the pullback function F* is mapping from dual x'-space back to dual x-space, so
F* : Λ1(Rm) → Λ1(Rn) . (10.7.16)
*****************
F*=1 and G*=1 and the Fact has nothing to say, see (10.7.22) item 1.
In Dirac notation we denote spaces by subscripts on the bras (dual space vectors)
F*(G*α) = <G*α |
x<α| // start with t-space k-form
x<α| F* = x'<β| // apply F* to get an x'-space k-form
x'<β| G* = y<κ| // then apply G* to get a y-space k-form
Combining these two steps to get
(t<α| F*)G* = y<κ| . // normally written t<α| φ*ψ*
Instead of taking two steps, do it in one step using Φ
t<α| Φ* = y<κ| . // apply Φ* to get a y-space form directly from t<α|
Since both methods give the same y<κ| conclude that
(t<α| F*)G* = t<α| Φ* = t<α| (G o F)*
or in function notation,
G*(F*(α)) = (G o F)*(α) QED
A Chapter 1 style category diagram for this scenario would be
******************8
Notice that, using x-space completeness 1 = |uj><uj| from the package (10.6.a.1) item (d),
<e'i| R = <e'i| R [1] = <e'i| R |uj><uj| = Rij<uj| = <ei| (10.7.8)
so that operator R pulls back the axis-aligned dual basis vector <e'i| in dual x'-space to a certain linear combination of the basis vectors in dual x-space.
********************88
The ordered sum form of a k-form pullback
It was shown in (10.7.13) that
αx' = Σ'I fI(x')λ'^I
F*(αx') = ΣJ GJ(x) λ^J where GJ(x) ≡ Σ'I fI(F(x)) RIJ . (10.8.1)
Recall that ΣJ is the redundant symmetric sum where each basis element occurs k! times. We prefer to state this result using the ordered sum Σ'J where each basis element occurs only once. To this end, we write
F*(αx') = ΣJ [Σ'I fI(F(x)) RIJ] λ^J // install GJ(x)
= Σ'I fI(F(x)) [ΣJ RIJ λ^J] // reorder
= Σ'I fI(F(x)) [Σ'J det(RIJ) λ^J ] // (A.8.36)
= Σ'J [Σ'I fI(F(x)) det(RIJ)] λ^J // reorder
= Σ'J gJ(x) λ^J (10.8.2)
where
gJ(x) ≡ Σ'I fI(F(x)) det(RIJ) . (10.8.3)
We then obtain this result for the pullback of a k-form,
αx' = Σ'I fI(x')λ'^I ϵ Λ'k // k-form in dual x'-space
F*(αx') = Σ'J Σ'I fI(F(x)) det(RIJ) λ^J // (10.8.2)
= Σ'J Σ'I fI(F(x)) det(RIJ) dx^J // cosmetic
= Σ'J gJ(x) dx^J // (10.8.3)
≡ βx ϵ Λk // definition (10.8.4)
where we make up a simple name βx for the pulled back form.
Comment: It might be logical to use symbol αx in place of βx, but we retain βx just to emphasize that the differential forms αx' and βx = F*(αx') are totally different forms living in totally different spaces. The symbols αx' and αx are so similar, it seems one could just replace x' by x to get from one to the other.
*******************8
Given x' = F(x) and x'i = Fi(x) one can write the following calculus differential,
dFi = (∂Fi/∂xj) dxj = (∂x'i/∂xj) dxj = Rijdxj
or
dFi = Rijdxj . (10.8.5)
Mimicking this last equation we define a new differential 1-form dFi in dual x-space as follows
dFi ≡ Rij λj = F*( λ'i) // (10.7.19) item 5
Then consider
F*( λ'^I) = ΣJ RIJ λ^J // (10.7.19) item 4
= RijRij ...Rij λj ^ λj ^ .... ^ λj // deconstruct multiindex
= ( Rijλj) ^ (Rijλj) ^ ... ^ (Rijλj) // regroup
= F*( λ'i) ^ F*( λ'i) ^ ... ^ F*( λ'i) // (10.7.19) item 5
= dFi ^ dFi ^ ... ^ dFi // (10.8.6)
≡ dF^I . // definition (10.8.7)
Therefore
dF^I = ΣJ RIJ λ^J // (10.8.7) **
= Σ'J det(RIJ) λ^J // (A.8.36) from **
This can be installed into (10.8.2) second line to get this alternate form for the pullback of αx' :
dFi ≡ Rij dxj
= Rij λj // cosmetic
= F*( λ'i) // (10.7.19) item 5
= F*( dx'i) . // cosmetic (10.8.6)
Then consider,
F*( dx'^I) = F*( λ'^I) // cosmetic
= ΣJ RIJ λ^J // (10.7.19) item 4
= RijRij ...Rij λj ^ λj ^ .... ^ λj // deconstruct multiindex
= ( Rijλj) ^ (Rijλj) ^ ... ^ (Rijλj) // regroup
= F*( λ'i) ^ F*( λ'i) ^ ... ^ F*( λ'i) // (10.7.19) item 5
= dFi ^ dFi ^ ... ^ dFi // (10.8.6)
≡ dF^I . // definition (10.8.7)
Therefore
dF^I = ΣJ RIJ λ^J // (10.8.7) **
= ΣJ RIJ dx^J // cosmetic
= Σ'J det(RIJ) λ^J // (A.8.36) from **
= Σ'J det(RIJ) dx^J // cosmetic (10.8.8)
This can be installed into (10.8.2) second line to get this alternate form for the pullback of αx' :
αx' = Σ'I fI(x')λ'^I ϵ Λ'k
F*( αx') = Σ'I fI(F(x)) ΣJ RIJ λ^J
= Σ'I fI(F(x)) dF^I (10.8.9)
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We will have shown that dα" = dα if we can show that
ΣI [∂jTI(x)] λ^I = Σ'I [∂jAI(x)] λ^I ? (10.3.9)
where TI and AI are related by (10.1.4), AI(x) = k!Alt(TI(x)). But
AI(x) = k!AltI(TI(x)) [∂jAI(x)] = k! AltI [∂jTI(x)]
CI(j) = k! AltI(DI(j)) (10.3.10)
where CI(j) ≡ ∂jAI(x) and DI(j) ≡ ∂jTI(x) and j is regarded as a fixed label. But (10.3.11) is exactly the relationship which says that a form β = ΣI DI(j)λ^I can also be written β = Σ'I CI(j)λ^I. Therefore we conclude that ΣI DI(j)λ^I = Σ'I CI(j)λ^I so (10.3.9) is true. QED
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Recall now that one can always write a k-form with either a symmetric or an ordered sum:
α = ΣI TI(x) λ^I = Σ'I AI(x) λ^I where AI = k! AltI(TI) . (10.3.8)
This is shown in (10.1.1-3) and also in (A.8.34). If one is given TI, then the corresponding AI can be computed as shown. If one is instead given AI, one can select the following corresponding TI,
TI = or TI = AI θ(I = ordered) . // θ(bool) = 1 if true else 0 (10.3.9)
Then
ΣI TI(x) λ^I = ΣI AI(x) θ(I = ordered) λ^I = Σ'I AI(x)λ^I . (10.3.10)
Notice in Σ'I AI(x) λ^I that the object AI(x) is only "sensed" for ordered I values.
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