the lambda i puzzle REVIEWED

The λi Puzzle PhL 4.7.16 5.16.16: Here I nailed down the right notation and interpretation for λi = <ei| and related equations. Before this it was all hazy and Sjamaar just added to the haze. I think now it is in a very good place. I am snagged on this issue and the related "vector vs scalar" issue. This λi has to be properly treated early in Section 2.11 because it then forms the basis of the entire dual part of the doc. In particular, it forms the basis of the differential forms section where I want to have a λi object in both spaces. I have made many attempts to resolve these issues and it just refuses to resolve, so I will make more attempts right here in this little doc. Plan A. The subject first comes up in Section 2.11 (c) where I say λn ≡ <un| . Comment: I think of | un> as being in x-space, and <un| being in dual x-space. Question 1: Does definition λn ≡ <un| imply the existence of an object λ'n in dual x'-space? Consider this earlier stated fact u'n = R un or | u'n> = | R un> This certainly says that <u'n | = < R un| but how does that help me for question 1? So put question 1 on hold. Question 2: Do the objects λn for n = 1..N form a vector or transform as a vector? I have asked this before. Each λn = <un| is itself a dual vector, so the objects λn for n = 1..N are really a set of dual vectors, they do not form a vector of scalar values like Vn does. So the answer to question 2 is no, the λn object is not a vector where λn are components of that vector. I think this is just a bad impression encouraged by the look of the notation. If the objects λn did form a single vector, you could then reasonably say λ'n = Rnmλm and then the answer to Question 1 would be that yes, there is an implied (forced to exist) λ'n object in dual x'-space. Question 3. Is it reasonable to say λn ≡ <un| λ'n = <u'n| and then this is an implied meaning for the object λ'n in dual x'-space? Since λn ≡ <un| is a definition of an object in dual x-space, I don't think you are forced to say λ'n ≡ <u'n| in dual x'-space. It seems that you might be allowed to say λ'n ≡ <e'n| in x'-space if you wanted. Then you would argue that the motivation for the two definitions is that you are using axis-aligned basis vectors in each case. The key thing is that both λn and λ'n are just definitions , nothing more. Question 4. I like to say in (2.11.c.11) that α(v) = α v is a scalar and so α'(v') = α' v' = α v = α(v) . (2.11.c.11) Here α is a functional = dual vector just like λn . If you can say the above for general α, then that seems to imply that you should be able to restate the above replacing α by λn. What would that look like?? Answer: Well, we have α = Σiαiλi so we could select αi = δin to get α = λn. Then α v = αivi = δinvi = vn. So far then we would have that: α(v) = α v → λn(v) = vn but beyond that is hazy. Question 5. If you write α'(v') then you must have some definition for α' . What is that definition? Answer: What we do have is α = Σiαiλi being the definition of α. Based on a negative answer to Question 1, I would say that this definition of α does not imply α' = Σiα'iλ'i because we don't really yet have any definition of λ'i in x'-space. So if when you write α' = Σiα'iλ'i you mean this to be some kind of statement in dual x'-space, then such a statement has no meaning since λ'i has no definition there (unless I manually make one). Therefore α' does not exist, so writing α'(v') has no meaning. You could make the definition α'(v') ≡ α' v' = α v = α(v) but that hardly seems useful to do. Conclusion: I am inclined to remove that equation (2.11.c.11) above since it has no meaning. Question 6. I claim that α(v) = <α|v> = α v is a "scalar function". Since α and v are bona fide vectors, you certainly could say α v = α' v' where the primed vectors exist in x'-space in the usual way. Does this somehow imply that there exists some α'(v') = α' v' ?? Answer: What we do know is this: α(v) = Σiαiλi(v) = α v = α' v' You are searching for some parallel statement α'(v') = Σiα'iλ'i(v') = α' v' = α v Suppose you were to make this definition of the λ'i object, λ'i(v') ≡ v'i We know from (2.7.1) line 5 that V' = Σn V'n e'n. Therefore we know that v' = Σn v'n e'n if v' is a vector in x'-space. In order to have λ'i(v') ≡ v'i we would (I think) have to say λ'i ≡ <e'i| Let's see if this works right: λ'i(v') = <e'i| v'> = v'i !!! So if we make the definition λ'i ≡ <e'i| , then we get λ'i(v') ≡ v'i and then the objects α'(v') = Σiα'iλ'i(v') and α' = Σiα'iλ'i are both well-defined. Then we certainly can say α'(v') = Σiα'iλ'i(v') = Σiα'iv'i = α' v' = α v = α(v) Conclusion: This is the very first time in Section 2.11 that I had any reason to define λ'i ≡ <e'i| in dual x'-space while at the same time defining λi ≡ <ui| in x-space. This has two advantages: (1) it makes things look nice in that α'(v') exists and α'(v') = α' v' = α v = α(v) (2) it makes Chapter 10 work !!!!! It is essential for Chapter 10 as in v9 : 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) I will now pause here and try to edit up a new section 2.11 (c) . // I did this, and I now have more questions. Question 7. I have space V and the dual space V* . Are these associated only with x-space, or do they somehow also apply to x'-space? Or are there other spaces called V' and V'* ? Answer: For now I have decided that there are four spaces V, V*. V' and V'* and all have the same dimension and all have a 1 to 1, and I wrote this up in a comment below (2.11.c.17). We shall see if this causes trouble. I may move this comment to a more prominent position.