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Hilbert Spaces?

  1. Jul 27, 2009 #1
    I've been reading about them (briefly), and can't see any large difference between them and metric spaces or even euclidean spaces for that matter. What am I missing?

    I read a Hilbert Space is a complete inner product space. But a metric space is a complete space as well with the only difference in definition being the omission of the possibility to measure angles (in Rudin's "Principles of Analysis," and my definition of Hilbert space from Wikipedia...).
     
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  3. Jul 27, 2009 #2

    mathman

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    The usual definition of a Hilbert space is that the number of dimensions is infinite. Essentially it is infinite dimension analog of Euclidean space (finite dimensional). Both of them have the concept of inner product which leads to the definition of length. In both cases the length of a vector X is (X,X)1/2.
     
  4. Jul 27, 2009 #3
    A Hilbert space is, in particular, a linear space. A metric space need not be. A Hilbert space is, in particular, a metric space. But many metric spaces are not Hilbert spaces.
     
  5. Jul 28, 2009 #4
    Ok, those posts help clarify. I do gave some follow up questions though...

    Why the need for all these various spaces? I haven't studied all the math I eventually will and have not yet studied quantum physics either so maybe it's just naivity or lack of exposure but I can't figure out why we need this many different ways of describing space...Banach spaces, Hilbert spaces, metric spaces, Euclidean spaces, Vector spaces, Reimann spaces..... They all have slight differences but wouldn't it then just be useful to generalize all space? I mean this in a similar way to the way the complex number field generalizes (or at least, contains) the natural numbers, the integers, the rationals etc.
     
  6. Jul 28, 2009 #5
    For the same reason we have all those different kinds of numbers. I can prove unique factorization in the positive integers, but not in the complex numbers. Similarly some statements about special kinds of spaces don't hold about all kinds of space.

    Of the spaces you listed all except vector spaces are special cases of what we call topological spaces (which is studied in topology). Many results can be shown in general for topological spaces, but sometimes one needs additional constraints to obtain some results (for instance it's a quite nice result in metric spaces that a subset is compact if and only if it's totally bounded and complete).

    A vector space is not a space in the same sense as the others as there is no way to talk about "nearness" in an abstract vector space, but many vector spaces are actually normed vector spaces and these are topological spaces.
     
  7. Jul 28, 2009 #6
    Thanks, good post. In hindsight, I probably should have posted this in the Topology section but since I'm here...

    Is there any analogous space to the complex number field in that every other number field is a subfield or subset of the complex number field?
     
  8. Jul 28, 2009 #7

    HallsofIvy

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    No, a metric space is NOT necessarily a complete space. For example, the set of all rational numbers with the "standard" metric, d(x,y)= |x- y|, is a metric space that is NOT complete.

    Formally, a complete metric space is called a "Frechet space". A "Banach space" is a complete space with a norm defined. Given a norm, we can define a metric by d(x,y)= ||x- y||. Thus, every Banach space is a Frechet space but there exist Frechet spaces that are not Banach spaces. A "Hilbert space" is a complete space with an inner product defined. Given an inner product, we can define a norm by ||x||= [itex]\sqrt{<x, x>}[/itex] so every Hilbert space is a Banach space (and so a Frechet space) but there exist Banach spaces that are not Hilbert spaces.

    The most important examples are L1(A), the set of functions f(x), on set A, such that [itex]\int_A |f(x)|dx[/itex] is finite (and we define |f| to be that integral), and L2(A), the set of functions f(x), on set A, such that [itex]\int_A f^2(x) dx[/itex] is finite (and we define <f, g>= [itex]\int_A f(x)g(x)dx[/itex] which can be shown to be finite also. They are important because they also form vector spaces. In L1(A) we can define "unit vectors" and in L2(A) we can define "orthogonal vectors".
     
  9. Jul 28, 2009 #8

    Wow, alot of information to read about, thanks. And yes I had forgotten that not every metric space is complete, don't know why I typed that.
     
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