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Schwarz Inequality

  1. Apr 25, 2007 #1
    I'm thoroughly confused by the Schwarz inequality.

    In many books on analysis, the author defines the Euclidean distance and then shows that the distance function is a metric, ie the Schwarz inequality holds. This isn't very satisfying as I feel it masks the properties of R^n that generate this relation.

    In reading on Wikipedia, I came across the topological idea of a uniform space. Although I don't quite understand the characterization of a uniform space, it satisfied me to read that we could salvage some parts of analysis without resorting to a real-valued distance function. We can still formalize the notion of "relative closeness".

    From a simple algebraic viewpoint, one can easily define an absolute value in an ordered ring. This order relation satisfies the triangle inequality, and I also believe it satisfies the Schwarz inequality. So is every ordered ring a "metric space" where the standard metric takes on values from the ring?

    This makes sense until we look at the complex plane. It's not an ordered field (or ring) but we can still define a absolute value. This work because we identify the complex plane with R^2... but that brings me back the original question, what properties of R^n allow us to prove the Schwarz inequality?

    If this seems like a half-baked post, it is... I'm having a hard time even formulating my question and wanted to see if somebody could direct me to further reading on these ideas (my home library is coming up short).
  2. jcsd
  3. Apr 25, 2007 #2


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    Um, well it might help you to know that the C-S inequality in R^n is a special case of the C-S inequality that holds in any inner-product space. That is, although there might be more abstract generalization of the C-S inequality, the only property/structure you need to prove the case in R^n is that of an inner product space, hence it holds in any inner product space.

    An inner product space is a vector space together with a function of two variables mapping into some field satisfying a few axioms. So it seems that a ring is not sufficient for some reason I have no intention of investigating a 5h42 AM; it has to be a field.
  4. Apr 28, 2007 #3
    Ah... right an inner product space.

    I'll tell you what really motivates these questions. I'm having trouble sorting out the topological and algebraic properties of R^n that generate the differentiable structure. I found an article that talks about some of the ideas that interest me:

    Geometry, calculus and Zil'ber's conjecture

    What field does this fall under? I'm very much interested in logic, but also differential geometry.

  5. Apr 29, 2007 #4


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    There isn't going to be a unique construction.

    Anyways, the ordering relation isn't necessary to do differential geometry. The simplest case is:

    Let k be an algebraically closed field.
    Let [itex]K = k(t_1, t_2, \ldots, t_n)[/itex] be the field of rational functions over k in n variables.

    Then the module [itex]\Omega^1_{K/k}[/itex] of differential forms of K over k is abstract K-module generated by the formal basis elements [itex]dt_i[/itex]. The derivation [itex]d:K \to \Omega^1_{K/k}[/itex] is uniquely determined by the sum rule, the product rule, and the constraints [itex]d(t_i) = dt_i[/itex].

    We obtain the cotangent space at any point P in k^n by further imposing the relation that [itex]f dt_i = g dt_i[/itex] whenever f(P) = g(P).
    Last edited: Apr 29, 2007
  6. Apr 29, 2007 #5


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    those of us inalgebraic geometry, where that construction arises, do not consider ourselves to be doing diffrential geometry, even we are taking derivatives. when we use the phraSE differential geometry we mean there is also a length and angle measure present. so curvature is an essential ingredient of differential geometry.

    i.e. not aLL CIRCLEs SHOULD BE ISOMORPHIC IN DIFFERENTIAL GEOMETRY, as they are in algebraic geometry.
  7. Apr 29, 2007 #6


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    Ah! I never really formed a mental distinction between differential geometry and differential topology.

    My point still stands, I think -- we can still give an algebraic construction of the cotangent sheaf, and thus construct the tensor algebra over our variety. From there, we can do (the algebraic analog of) anything we want, such as study the geometry arising from a metric tensor. On algebraic varieties, though, I only ever hear people talking about differential k-forms, and occasionaly tangent vectors. Is there a serious flaw in the theory of the tensor algebra over an algebraic variety, or is it just not useful for the questions often asked?
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