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Submanifold diffeomorphic to sphere

  1. Jun 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the map ##\phi: \mathbb{R}^4 \rightarrow \mathbb{R}^2## defined by ##\phi(x,y,s,t) = (x^2 + y, x^2 + y^2 + s^2 + t^2 + y)##.

    Show that ##(0,1)## is a regular value of ##\phi##, and that the level set ##\phi^{-1}(0,1)## is diffeomorphic to ##\mathbb{S}^2##.

    2. Relevant equations


    3. The attempt at a solution

    I get two equations describing the level set:

    (1) ##x^2 + y = 0 \implies y = -x^2##
    (2) ##x^2 + y^2 + s^2 + t^2 + y = 1 \implies y^2 + s^2 + t^2 = 1##

    So ##\phi^{-1}(0,1) = \{(x,y,s,t) \in \mathbb{R}^4: y = -x^2 \hspace{0.1cm} \mathrm{ and } \hspace{0.1cm} y^2 + s^2 + t^2 = 1\}##.

    I need to show that ##d\phi(x,y,s,t)## is surjective for all ##(x,y,s,t) \in \phi^{-1}(0,1)##. I calculate:

    ##d\phi(x,y,s,t) = \begin{pmatrix} 2x & 1 & 0 & 0 \\ 2x & 2y + 1 & 2s & 2t \end{pmatrix}##

    It is easy to show that this matrix has rank ##2## for all ##(x,y,s,t) \in \phi^{-1}(0,1)## and so ##\phi^{-1}(0,1)## is an embedded submanifold of ##\mathbb{R}^4##. Quick question: Did I calculate the differential properly?

    Now, I need to show that this level set is diffeomorphic to the unit sphere. I can kind of see that it may be diffeomorphic to a spheroid, and I know I can show that the spheroid is diffeomorphic to the sphere. My only problem is coming up with this diffeomorphism from the level set onto the spheroid.

    I imagine I can define a map ##\psi: \phi^{-1}(0,1) \rightarrow S## by ##\psi(x,y,s,t) = (x,s,t)##, where ##S = \psi(\phi^{-1}(0,1))##. It is easy to see that this map is invertible and its inverse is given by ##\psi^{-1}(x,s,t) = (x, -x^2, s, t)## since ##y = -x^2##. My only trouble is showing that these maps are both smooth. Both the domain and codomain are submanifolds of ##\mathbb{R}^4## and ##\mathbb{R}^3##, respectively, so I need to express ##\psi## in appropriate local coordinates before differentiating, but coming up with these local coordinates seems like I am making things overly complicated. What would you suggest I do?
     
  2. jcsd
  3. Jun 10, 2017 #2

    andrewkirk

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    I think if we instead set ##\psi(x,y,s,t)=(y\ \mathrm{sign}(x),s,t)## we would have ##S## a perfect sphere, rather than just a spheroid.

    I can't see a way of avoiding local coordinates. Perhaps there's a theorem that can be used to shortcut it, but none pops to mind.

    Coordinate maps to use might be:
    for ##S##:
    ##\ (y,s,t)\mapsto (s,t)##, with inverse map ##(s,t)\mapsto (\pm\sqrt{1-s^2-t^2},s,t)##, if ##s^2+t^2\leq 0.9## (two disjoint coordinate patches), otherwise:
    ##(y,s,t)\mapsto (y,s)##, with inverse map ##(y,s)\mapsto (y,s,\pm\sqrt{1-s^2-y^2})##, if ##y^2+s^2\leq 0.9## (two disjoint coordinate patches), otherwise:
    ##(y,s,t)\mapsto (y,t)##, with inverse map ##(y,t)\mapsto (y,\pm\sqrt{1-y^2-t^2},t)##, (two disjoint coordinate patches)
    where the ##\pm## has a different sign for each of the two patches in a pair.

    and for the level set:
    ##\ (x,y,s,t)\mapsto (s,t)##, with inverse map ##(s,t)\mapsto (\pm(1-s^2-t^2)^\frac14,-\sqrt{1-s^2-t^2},s,t)##, if ##s^2+t^2\leq 0.9## (two disjoint coordinate patches), otherwise:
    ##(x,y,s,t)\mapsto (x,s)##, with inverse map ##(x,s)\mapsto (x,-x^2,s,\pm\sqrt{1-s^2-x^4})##, if ##y^2+s^2\leq 0.9## (two disjoint coordinate patches), otherwise:
    ##(x,y,s,t)\mapsto (x,t)##, with inverse map ##(x,t)\mapsto ((x,-x^2,\pm\sqrt{1-x^4-t^2},t)##.
     
  4. Jun 11, 2017 #3
    I can solve it like this:

    I will keep ##\psi## defined how it is in the original post. Let ##M = \phi^{-1}(0,1)##. I've already proven that ##M## is a submanifold and so the inclusion map ##\iota_M: M \rightarrow \mathbb{R}^4## is smooth. The projection map ##\pi: \mathbb{R}^4 \rightarrow \mathbb{R}^3## given by ##\pi(x,y,s,t) = (x,s,t)## is smooth. Because ##\psi = \pi \circ \iota_M##, then ##\psi## is smooth.

    We can prove the smoothness of ##\psi^{-1}## in a similar manner: Let ##\iota_S: S \hookrightarrow \mathbb{R}^3## be an inclusion map - it is smooth as ##S## is an embedded submanifold of ##\mathbb{R}^3##. Let ##\iota: \mathbb{R}^3 \rightarrow \mathbb{R}^4## be an inclusion map - it too is smooth. And finally let ##F: \mathbb{R}^4 \rightarrow M## be given by ##F(x,y,s,t) = (x,-x^2,s,t)##, which is obviously smooth. Then ##\psi^{-1} = F \circ \iota \circ \iota_S## is the composition of smooth maps and hence also smooth.

    Note: I know that ##S## is a submanifold of ##\mathbb{R}^3##, and hence ##\iota_S## is smooth, because ##S## is diffeomorphic to ##\mathbb{S}^2## (constructing a diffeomorphism is very easy).
     
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