Manifolds, tensors and fields

Discussion in 'Physics & Math' started by QuarkHead, Nov 11, 2014.

  1. Trippy ALEA IACTA EST Staff Member

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    Moderator Note:
    Twelve posts have been moved as being off-topic, trolling, or responses to posts that met those criteria.

    Missing posts can be found here: URL↑
     
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  3. QuarkHead Remedial Math Student Valued Senior Member

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    I am getting tired of repeating myself, and I am sure no member here has any more energy to read my detailed critique of your last that I have for writing.

    However
    ALL? No. it was defined by me as being \(\pi_j \circ h = x^j\) It hardly comes close to being all differentiable real functions of \(U\), And a coordinate (a single one, say) is just a real number, so I called this number as a coordinate on \(U\)
    You are aware that our manifold \(M\) starts life as a topological space - i.e. a set whose elements are subsets of some point set? My \mapsto is perfectly valid

    How can this be true? Since as you say later
    and since for each homeomorphism its inverse is assumed to be unique, I cannot see how this can be true

    You then go on to define a projection operator, and a set of elements \(\{x^j\}\) such that \(x^j= \pi_j \circ h\) but don't use them, rather continue to insist that a coordinate transformation is adequately written as, say, \(h^{-1} \circ h'\) as a mapping \(p \mapsto q\).

    I really don't agree - a lesser mortal might appeal to the authority of many texts here, but for now I am losing heart
     
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  5. rpenner Fully Wired Valued Senior Member

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    I shall abandon my conversation after this reply. In my opinion, you are abusing symbols you define differently in your head than in your writing and aren't being pedantic so much as testy. I think this might arise from you wanting to talk about SOMETHING not in my experience, but I'm having a hard time of seeing where you are going because your notation is unclear to me, and in cases I have highlighted, apparently contradictory.

    Since h is just one of the many-many possible mappings from U to \(R^n\), I stand by my description. Note here you use function composition so the bare symbol \(x^j\) is a function of points in the manifold that happen to be in subset U. In other places you refer to it as a real number when someone who did follow the source material would know you are talking of it's value at a particular point.

    I didn't write it was. I wrote that it was IN that set. The set of all functions from U to \(\mathbb{R}\) can be written in terms of set exponetiation: \(\mathbb{R}^{U}\) and \(x^j\) is somewhere in there.

    Even if you are using shorthand notation where \(x^j\) is both the function and the particular value for \(p \in U\), this latter quantity that you sometimes write \(x^j\) and I insist on \(x^j(p)\) is not IN U, but rather \(\mathbb{R}\). This forms the core of my objection that I started post #20 with.

    That's not what you said in the OP where you wrote
    While a topological space has open sets, it undoubtedly also has points. Under the mathematical foundations that I am familiar with, a manifold is yet another example of a "set with Structure".

    http://math.ucr.edu/home/baez/qg-spring2004/discussion.html
    http://www.goodmath.org/blog/2014/04/02/manifold-are-the-manifolds/

    A topological space is a set T and a second set \(\tau\) of open sets that gives the topological space its structure. Being regular, Hausdorff or a manifold are properties that restrict the general concept of topological space to a mathematically tractable playground.

    If both \(U \in M\) and \(U' \in M\) are true then \((U \cap U') \subset M\). If \((U \cap U') \neq \emptyset\) then it follows there is a mapping of \(H = \{ h(p) : p \in ( U \cap U') \} \subset \mathbb{R}^n\) to \(H' = \{ h'(q) : q \in ( U \cap U') \} \subset \mathbb{R}^n\) and that mapping is \(h' \circ h^{-1}\). The mapping \(\pi^j \circ h' \circ h^{-1}\) then is a mapping from H to \(\mathbb{R}\) and the line you complain about is exploring notation that can be used to express that because I have no idea what else you meant to say when you wrote \(x'^1 = x'^1(x^1)\) in post #17.

    I have tried to be crystal clear in distinguishing \((h^{-1} \circ h') : \{ q \in U' : \exists p \in U \; h(p) = h'(q) \} \subset M \to \{ p \in U : \exists q \in U' \; h(p) = h'(q) \} \subset M\) from \((h \circ h'^{-1}) : H' \subset \mathbb{R}^n \to H \subset \mathbb{R}^n\) and their inverses. For manifolds with a translation symmetry (like flat models of Euclidean and Lorentizan space time, or my "patronizing" spherical Earth) the former can give rise to useful properties of the manifold (Poincaré symmetry and the like). The latter can be used for change of variable in calculus or Poincaré transforms.

    By all means, if your notation is standard in some text, please cite that text.
     
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  7. QuarkHead Remedial Math Student Valued Senior Member

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    OK, I want to pick this thread off the floor, so to speak. But first a personal statement.......

    rpenner accused me of "being testy". This is correct and I apologize to him and any others attempting to follow this thread.

    Also, the difference between rpenner and myself on this subject is not huge - indeed, we only disagree on how coordinates are defined. However, in order to continue I need to stipulate the following........

    For the point \(p \in M\) (here \(M\) is a "nice" n-manifold, defined inter alia) we associate real numbers \(x^1,x^2,.....,x^n\) and call these the coordinates for (or do I mean of?) this point.

    For what may (or may not, depending on you guys) follow, I insist that the real number \(x^i\) above can also be thought of as functions. Otherwise nothing that follows will make any sense.

    My first attempt to convince you will seem totally preposterous, but nonetheless pedantically valid. Suppose a function \(f::\mathbb{R} \to \mathbb{R}\) with \(f(a) = 2a\) for all \(a \in \mathbb{R}\). By this simple equality I can say that "2" is the function that takes any real number to the set of even numbers. Nobody ever would say this, of course ,nonetheless there is no reason not to.

    Of more immediate relevance perhaps, consider the following. Suppose \(V\) a vector space with \(v,\,\,w \in V\). Take an arbitrary linear transformation \(T:V \to V\) such that \(Tv = w\). But the axioms of linear algebra mandate there is some \( \alpha \in \mathbb{F}\) (this being the field over which our vector space is defined) such that \(\alpha v = w\). Again, our field element (which may very easily be a real number) acts as a linear transformation, which is of course a "sort of" function.

    Oh dear, I hadn't expected this to be so lengthy. I guess what I really wanted to go on to say will have to wait
     

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