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Topology, lemiscate not an embedded submanifold

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that the image of the curve Let β: (-π,π) → ℝ2 be given by β(t) = (sin2t, sint)
    is not an embedded submanifold of ℝ2


    2. Relevant equations



    3. The attempt at a solution

    So I'm not too great with the topology. I do see that β'(t) = (2cos2t, cost) ≠ 0 for all t. So β is a regular curve and as a result β is an immersion.

    So here's what I have to work with.

    A subset S of M is an embedded submanifold of M if:
    - S is a smooth Manifold ( the image of (-π,π) under β is smooth)
    -the inclusion map S to M is a smooth embedding

    So this means that the topology of S (in our case β(-π,π)) is the subspace topology coming from M (in our case ℝ2)

    We declare a set U[itex]\subset[/itex]β(-π,π) is open if and only if there exists a set W[itex]\subset[/itex]ℝ2 (namely the open discs) such that

    U = W [itex]\cap[/itex] S

    So to be more specific let's take U = β(-π/6, π/6), this set includes the origin.

    Let W be the open unit disc B(0,1) which includes the origin. I've done a lot of reading where if we remove the origin we have trouble, not really seeing how W [itex]\cap[/itex] S will look like the letter X. Any thoughts?
     
    Last edited: Mar 30, 2013
  2. jcsd
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