Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homomorphic normed linear spaces

  1. Nov 14, 2006 #1
    Hi all experts,
    I've just visitted another Maths forum and picked up two interesting questions:
    Question 1. Let H and K be homomorphic normed linear spaces. Is it necessary that H and K have the same dimension if both H and K are finite-dimension ? Is there possible a homomorphism between an infinite-dimension normed space and a finite-dimension one ?
    Question 2. If f is a homomorphism between two normed linear spaces, is f necessary uniformly continuous ?
    Hope to get your ideas.
     
  2. jcsd
  3. Nov 14, 2006 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Did you menan to say "isomorphic"?


    Have you given much thought to these questions?
     
  4. Nov 14, 2006 #3
    By saying "H and K is homomorphic", I mean that there is an one-to-one mapping f from H to K such that both f and f^{-1} (the inverse of f) are continous. Perhaps, this concept should be called "isomorphic" and sorry if I made a mistake.

    If H=R and K=R^2, I think R and R^2 cannot be holomorphic because if we remove a point of R, such as x, then R\{x} is no longer a connected set, but R^2\{f(x)} is still a connected set. Here, f is an one-to-one from R to R^2 such that f and f^{-1} are continuos.
     
  5. Nov 15, 2006 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A continuous bijection with a continuous inverse is an homeomorphism.
     
  6. Nov 15, 2006 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    A one-to-one homomorphism between 2 algebraic structures is an isomorphism.


    Daniel.
     
  7. Nov 15, 2006 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    No it isn't. The inclusion of Z into Q or R (as additive groups) is a one to one homomorphism. It is not an isomorphism.
     
  8. Nov 15, 2006 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    So, I think the phrase needed here is "homeomorphic isomorphism" or perhaps "isomorphic homeomorphism"!
     
  9. Nov 15, 2006 #8

    StatusX

    User Avatar
    Homework Helper

    To show R^n is not isomorphic to R^m as a vector space for n[itex]\neq[/itex]m, you can just show that isomorphism preserves dimension (ie, show a basis is mapped to a basis). Showing they are not homeomorphic as topological spaces is much more difficult. It can be done using cut points as in post 3 for n=1, but above this you need more advanced tools from algebraic topology. And what do you mean by uniformly continuous?
     
  10. Nov 15, 2006 #9

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    real normed linear spaces of finite dimension are probably always isomorphic to R^n. then they are determined by their homeomorphism type i would guess. a proof tends to involve homology theory.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Homomorphic normed linear spaces
  1. Normed space (Replies: 5)

Loading...