1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Regular Values (Introduction to Smooth Manifolds)

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider the map [itex]\Phi[/itex] : ℝ4 [itex]\rightarrow[/itex] ℝ2

    defined by [itex]\Phi[/itex] (x,y,s,t)=(x2+y, yx2+y2+s2+t2+y)

    show that (0,1) is a regular value of [itex]\Phi[/itex] and that the level set [itex]\Phi^{-1}[/itex] is diffeomorphic to S2 (unit sphere)

    2. Relevant equations




    3. The attempt at a solution

    So I have the Jacobian

    D[itex]\Phi[/itex] = [2x 2y 0 0; 2x 2y+1 2s 2t]

    the reduced row echelon form of D[itex]\Phi[/itex] is a rank 2 matrix which implies that F is a smooth submersion.

    So I guess I'm a little confused on the definition: Let M and N be smooth manifolds

    If F:M→N is smooth and let p [itex]\in[/itex] M, then p is a regular point if D[itex]\Phi[/itex] at p is onto.

    D[itex]\Phi[/itex] is a rank 2 matrix which implies that D[itex]\Phi[/itex] is onto right?

    Another question that I have is that the pre-image of (0,1) under [itex]\Phi[/itex] is a subset of ℝ4 any hints on the diffeomorphism part?
     
  2. jcsd
  3. Mar 30, 2013 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I don't get the same Jacobian

    But ##D\Phi## is not a matrix at all. You should evaluate it in different points. You should ask for which ##(x,y,s,t)## that ##D\Phi_{(x,y,s,t)}## is surjective. This is not true for all ##(x,y,s,t)##. For example ##(0,0,0,0)##.

     
  4. Mar 30, 2013 #3
    Shoot!
    [itex]\Phi[/itex](x,y,s,t) = (x2+y, x2+y2+s2+t2+y)

    D[itex]\Phi[/itex] = [2x 1 0 0; 2x 2y+1 2s 2t]

    How is D[itex]\Phi[/itex] not a matrix? I'm confused on that?

    I'm going to switch notation up a bit

    If we have a smooth map F: M → N and let p [itex]\in[/itex] M, then p is a regular point if

    TpF is onto. In this case TpF is the total derivative which we used D[itex]\Phi[/itex].

    TpF : TpM → TF(p)N

    However I do see your point about D[itex]\Phi[/itex](0,0,0,0), we would have a rank 1 matrix in that case.
     
    Last edited: Mar 30, 2013
  5. Mar 30, 2013 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I still don't agree.

    It's not a matrix in the sense that it depends on specific values of ##(x,y,s,t)##. If you're given specific values, then it is a matrix.

    Can you classify all points such that ##D\Phi_{(x,y,s,t)}## is surjective?
     
  6. Mar 30, 2013 #5
    sorry my equation was wrong
     
  7. Mar 30, 2013 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    OK, I agree with the Jacobian now. But please don't edit any posts after you already received a reply. Just reply with the correct equation.

    So, can you now figure out for which ##(x,y,s,t)## that your Jacobian has rank 2?
     
  8. Mar 30, 2013 #7
    First sorry about that reply thing, I realized that once I made the edit, thought about re-editing so that I made my mistake, but then that's just too many re-edits, in any case I won't make that mistake again!

    so the point (0,1), the pre-image of (0,1) under [itex]\Phi[/itex] is a subset of ℝ4

    so we will look at the specific case:

    (x2+y,x2+y2+s2+t2+y) = (0,1)

    this means that

    x2+y = 0

    and that x2+y2+s2+t2+y = 1

    this reduces the second component to

    y2+s2+t2 = 1 (which is nice sine S2 : y2+s2+t2)

    [itex]\Phi^{-1}[/itex](0,1) = {(x,y,s,t)|(x2+y=0 and y2+s2+t2 = 1}

    so for all (a,b,c,d) [itex]\in[/itex] [itex]\Phi^{-1}[/itex](0,1)

    we have D[itex]\Phi[/itex](a,-a2,c,d) is a rank 2 matrix

    I think that's good now right?
     
  9. Mar 30, 2013 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    If it's clear to you why ##D\Phi(a,-a^2,c,d)## has rank 2, then it's good.
     
  10. Mar 30, 2013 #9
    for all p [itex]\in[/itex] [itex]\Phi^{-1}[/itex](0,1), p is a regular point, and it follows that (0,1) is a regular value. Thanks so much! That really helped a lot. I think I have the second part.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Regular Values (Introduction to Smooth Manifolds)
  1. A smooth manifold (Replies: 6)

  2. Smooth manifolds (Replies: 2)

Loading...