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Homeomorphism (defining a chart)

  1. Mar 23, 2010 #1
    hello,
    im trying to get a homeomorphism between a n-dim vector space and R^n that is independent of the basis. (im actually defining a C \infinity structure on V)
    since i want a homeomorphism, i know i should define a topology on my vector space, which is the norm, since that would be independent of the basis.
    but i cant move forward from there.
    any ideas?
    btw im fine with a homeom. btw a 1-dim vector space and R since i can write v in V as v1 + v2 + ... + vn where each vi is a vector in 1-dim vector space.
    thank you.
     
  2. jcsd
  3. Mar 24, 2010 #2

    quasar987

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    Why do you insist that the homeo be basis-independant if your goal is simply to be a smooth structure on V?
     
  4. Mar 24, 2010 #3
    well because this is the question, to find a basis independent homeomorphism, ofcourse, since the basis- dependent one is trivial,...
    im working on sth actually, i'll post it up when it's done, take your opinion about it.
    still open though for suggestions.
     
  5. Mar 24, 2010 #4

    quasar987

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    What is the precise statement of the question you're working on?

    If you're trying to construct a basis independant smooth structure on V, this does not imply finding a basis-independant homeo. from V to R^n (which I would be surprised is possible).
     
  6. Mar 24, 2010 #5
    Maybe I'm misunderstanding what you mean by basis-independent, but couldn't you just choose some basis [itex]e_1,\ldots,e_n \in V[/itex] beforehand and define the homeomorphism in terms of that in the obvious way, and then show that it is actually independent of the basis?

    For instance in the 1-dimensional case couldn't you just choose a non-zero element [itex]e \in V[/itex] and then define [itex]f : V \to R^1[/itex] by [itex]f(\lambda e) = \lambda[/itex] for [itex]\lambda \in \mathbb{R}[/itex]. Then given another non-zero element [itex]e' \in V[/itex] if [itex]xe' = ye[/itex], then you would have,
    [tex]f(xe') =f(ye)[/tex]

    I'm sorry if I'm being obtuse, but can you state what it means for a homeomorphism to be basis-independent because I feel like I'm missing something?

    The only other definition I could imagine would be is that if [itex]\{e_1,\ldots,e_n\}[/itex] and [itex]\{e_1',\ldots,e_n'\}[/itex] are bases of V, then [itex]f(\sum x_i e_i) = f(\sum x_i e_i')[/itex] for all [itex]x_1,\ldots,x_n \in \mathbb{R}[/itex]. However this is clearly impossible since given a basis {e} of a 1-dimensional vector space, {2e} is also a basis, but different. Yet we would have f(e) = f(2e) which contradicts that f is injective.
     
  7. Mar 24, 2010 #6

    quasar987

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    "Basis-independant" here mean that there is a "natural recipe" to assign to a generic vector space V a homeomorhism btw V ans R^n, without picking out a specific basis for V and R^n.

    In your 1-dimensional example, the first step in the recipe for defining f it to chose a nonzero element e in V (i.e. a basis). This disqualifies your recipe as being "natural" or "basis-independant".
     
  8. Mar 25, 2010 #7
    ok, im failing to do that,..
    the question im working on is from spivak, volume 1, 3rd edition:
    let V be an n-dim vector space (n: finite) over R.
    define a C^infinity structure on V and a homeomorphism between TV and VxV which is independent of the choice of bases.

    maybe im not grasping the question right, i dont know because everything im doing, im getting a tiopology on V that is basis independent but my homeomorphism is not.

    rasmhop, what if f(xe') = f(ye), so you're saying that x = y, and thus basis independent?
    mmm,...

    quasar987, you said sth interesting, you think im thinking of this the wrong way?
     
  9. Mar 25, 2010 #8
    rasmhop, while ur idea works for 1 -dim vector space, it will fail for the n-dim case right, so i guess we need sth else.
     
  10. Mar 25, 2010 #9

    quasar987

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    I'm sorry, this isn't exactly right. You need a recipe that may involve a choice of basis at some point, but the final result is independant of the choice of basis.

    In your example, you chose a nonzero element e and define f: V-->R by f(xe)=x. This recipe for constructing a homeo V<-->R is not basis independent because for a different nonzero element e', the map f':V-->R defined by f'(xe')=x is not the same as the map f. (If v=xe=ye' is any nonzero element of V, then f(v)=x, f'(v)=y, with x different from y).

    An example of a basis independent construction is the following: Let R^n have its usual topology. Chose a basis v_1,..,.v_n of V and define an isomorphism T:V-->R^n by T(v_1)=e1, ..., T(V_n)=e_n. Define a topology on V by setting [itex]\mathcal{O}\subset V[/itex] open iff [itex]T(\mathcal{O})[/itex] is open in R^n. It is easy to show that this indeed defines a topology and that it is independent of the isomorphism T (which amount to say it is independent of the choice of basis v_1,..,v_n). In fact, it is the so-called "norm topology" on V.
     
    Last edited: Mar 25, 2010
  11. Mar 25, 2010 #10

    quasar987

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    So, like I suspected, you are never actually required to construct a basis independant homeo. between V and R^n. You just need to construct a basis independant smooth structure on V.

    So put the norm topology on V, chose a basis v_1,...,v_n of V and define an isomorphism T:V-->R^n by T(v_1)=e_1,...,T(v_n)=e_n. Show it is an homeomorphism. Hence it is a global chart (V,T) for V which makes V into a smooth manifold with smooth structure defined as the one containing your global chart.
    Now, to show that the construction of this smooth structure is basis independent, chose another basis w_i and define an iso S:V-->R^n by the same recipe, i.e. S(w_1)=e_1,...,S(w_n)=e_n. This defines another global chart (V,S) of V and it induces another smooth structure on V, namely, the one containing the chart (V,S). You need to show that the smooth structure induce by (V,S) is the same as the one induced by (V,T). This only means that you need to show that the charts (V,T) and (V,S) are smoothly compatible. That is to say, the transition functions T o S^-1 and S o T^-1 are both smooth.
     
  12. Mar 26, 2010 #11
    thank you quasar987,
    thing is, professor is telling me that i can find a homeomorphism between a 1-dim vector space and R, that has nth to do with bases,..
    so pick a non zero vector in V, map it to a point in R.
    then i guess i cannot think of another element in V to be a linear combination of the chosen vector bc that is inducing a basis.
    but whatever can i map it to?
     
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