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Closed manifolds

  1. May 2, 2005 #1
    1) General question :

    Let's take a usual line : it's a 1D manifold in 2D space. The line is closed if there are no border points. (circle, aso...)

    Let suppose a usual surface : it's a 2D manifold enbedded in 3D space.
    The surface is closed if there are no border line. (sphere, torus, aso...)

    Let's suppose a 3D manifold embedded in 4D space. The manifold (volume) is closed if there are no border surface ???

    How can one represent a closed volume which is not infinite ??

    The usual Ball (with it's interior) is not closed, because there is a border surface (the outer sphere). Neither a torus, neither any 3D shape I could imagine.

    2) Precise question :

    Let's assume a 4D manifold is given by F(w,x,y,z)=0, such that w=w(r,s,t), aso for x,y,z....

    Question : How do I know the described manifold is closed ?

    Any reference ?

    BTW : I ask this question in relation with the Poincare Conjecture.
     
  2. jcsd
  3. May 2, 2005 #2
    well, by "infinite" i think you mean: that the volume is finite yet the manifold is unbounded (closed), right? im not sure, but a 3d hyperbolic manifold would probably fit the bill since the hyperbolic plane (2d) can be projected into a fixed circle on the euclidean plane (closed manifold, yet finite area).

    well, the Generalized Stokes' Theorem of differential forms says that:

    [tex] \int _{\partial M} \alpha = \int _{M} d\alpha [/tex]


    so if our manifold is closed, then the boundary set of the manifold is empty:

    [tex] {\partial M} = \emptyset [/tex]


    so then:

    [tex] \int_{\partial M} \alpha = \int_{\emptyset} \alpha = 0 [/tex]


    by the other side of the equation then:

    [tex] \int _{M} d\alpha = 0 [/tex]


    since we know that [tex]M[/tex] is not empty, this implies that [tex] d\alpha = 0[/tex]. infact, a form whose exterior derivative is zero is known as a "closed form".

    so one way to test might be to take a differential form defined on your manifold, and take the exterior derivative and integrate it over your manifold. if the result is zero, then your manifold is closed.

    this is from De Rham cohomology, something that I'm only transiently familiar with. Here's the wikipedia page:

    http://en.wikipedia.org/wiki/De_Rham_cohomology

    maybe someone else can help me out here, but i think this is right..
     
    Last edited: May 2, 2005
  4. May 2, 2005 #3

    AKG

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    Note that a form and its exterior derivative are both forms, and that a both a manifold-with-boundary and its boundary are manifolds. So it seems you're arguing that if the integral of a form over the boundary, that the boundary is empty. This is not the case. If you consider the form w = 0, then this form on any surface will have an integral that's zero, closed or not.

    kleinwolf, if for every point x on the k-dimensional manifold M, there is a diffeomorphism f from an open set U in n-dimensional space to an open set V in n-dimensional space such that x is in U, and such that f(U n M) = {y in V : all but the first k co-ordinates of y are 0}, then M is a closed manifold. If M were not a closed manifold, suppose it were a manifold-with-boundary, then points x on the boundary would satisfy a similar property, but f(U n M) = {y in V : all but the first k co-ordinates of y are 0 and the kth co-ordinate of y is non-negative}, and f(x) would have kth component 0. A point x cannot satisfy both conditions, so if all points satisfy the first condition, then none of them satisfy the second condition, the one that makes them a boundary point. So, again, to check that a set is a closed manifold, show that all points satisfy the first condition. In a math class, this may be a question on a test. In a physics class, I don't think you'd even be expected to know this, you tell what the boundary is by looking at it, more or less.
     
  5. May 3, 2005 #4
    Ok, we assume I'm in the physics class, I get the exercise :

    w=(Lcos(c)+a)*sin(t)cos(p)

    x=(Lcos(c)+a)*sin(t)sin(p)

    y=(Lcos(c)+a)*cos(t)

    z=Sqrt[4(L+a)^2-(Lcos(c)+a)^2]

    with a>L>0 constants, c,t,p the free parameters.

    Then w^2+x^2+y^2+z^2=4(L+a)^2

    Which is the equation of a 3Sphere....starting from the fact that

    x^2+y^2=cste is a closed manifold (circle), x^2+y^2+z^2=cste is a closed manifold (sphere), I deduce the presented manifold is closed (3-sphere) ?
     
  6. May 3, 2005 #5
    Since we're talking about a hypersurface in Euclidean space, try to think of closed-ness from an analytical point of view: ask yourself: are there any sequences of points in the given space that converge to a point outside of the space.

    Going back to the 3-sphere ([itex]x^2+y^2+z^2+w^2=1[/itex]), if you take any sequence of points on it and it converges to another point, that limit point has to have the same distance from the origin (0,0,0,0) as all of the other points in the sequence, so it must also be on the 3-sphere. Hence the 3-sphere is a closed manifold.

    You mentioned the ball in 4-space, which I'm taking to be the open set of points of distance less than 1 from the origin. Just consider a sequence where the x-value gets closer and closer to 1, say, [itex] x_n=1-{1\over n}[/itex] but the y, z, and w values remain 0. Each element of the sequence lives in the ball, but its limit doesn't (it lives on the sphere). Thus, the ball (under the Euclidean metric) is not closed.
     
  7. May 3, 2005 #6
    OK, so my terminology is wrong and misleading : What I mean is

    "Is the boundary of the above manifold empty ?"...

    Like the sphere has a vanishing boudary, but a disc has a circle as boundary....
    and a Ball has the outer sphere as boundary...

    (NB : For the Ball, there are closed and open balls, just take "a distance less or equal than 1")
     
  8. May 3, 2005 #7

    AKG

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    Yes, that works.
     
  9. May 3, 2005 #8
    No, I have a counterexample :

    [tex] x=cos(cos(t)\pi/2)\quad y=sin(cos(t)\pi/2)\quad t\in[-\infty,\infty] [/tex]

    Then we have x^2+y^2=1...but there are two boundary points because it's only half a circle...

    So I need a precise criterion to decide if the above 4D manifold has a boundary or not, because personally, I am not able to do this just by looking at it...whereas I can for the case here, because it's simpler.
     
  10. May 4, 2005 #9

    AKG

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    Oops, I see. Well then you have to use the criteria I gave you: show that for every point x in the manifold M, if it is a k-dimensional manifold in n-dimensional space, then there is a smooth diffeomorphism h from an open set U in n-dimensional space containing x to an open set V in n-dimensional space such that h(U n M) = {y in V : the last n-k co-ordinates are zero}. This doesn't look like it would be a fun task given the example you have. Since x,y,z,w are paramterized by c,p,t, that might give you a natural smooth diffeomorphism you can use, but you'll have to work that out.

    Alternatively, there is an equivalent condition you could check. Show that for every point x in M, if M is a k-dimensional manifold in n-dimensional space, then there is a one-to-one smooth function h from W, and open set in k-dimensional space, to n-dimensional space such that the rank of the Jacobian of h at each y in W is k, h(W) = U n M for some open set U in n-dimensional space containing x, and the inverse of h from h(W) to W is continuous.

    These conditions don't sound nice, but they're not so bad. The second condition just says that around any point in the k-dimensional manifold, you can make a co-ordinate system that is like k-dimensional space. The Earth is a sphere, but it seems flat to us because we see a lcoal region of it. So, we're at a point on the manifold (the crust of the earth) and we can make a co-ordinate system around this point (which is just like making a map, which is 2-d).

    It seems reasonable that since x² + y² = 1 gives us a 1-d manifold and x² + y² + z² = 1 gives us a 2 dimensional manifold, that your example should give us a three dimensional manifold. And we have a natural parameterization from 3-dimensional space to this manifold by c, p, and t. Let h be the function:

    h(c, p, t) = (x(c,p,t), y(c,p,t), z(c,p,t), w(c,p,t))

    The Jacobian at any point for h (note that the h does have to work for all points, it just has to work for a neighbourhood around a point x in the manifold, and there has be an h for each x, but it might work for all of them if we're lucky) is just:

    [ dx/dc dx/dp dx/dt ]
    [ dy/dc dy/dp dy/dt ]
    [ dz/dc dz/dp dz/dt ]
    [ dw/dc dw/dp dw/dt ]

    I'll leave it to you to find the rank of this matrix. You also need to check: is h 1-1? Can you pick two different points (c,p,t) and (d,q,s) such that h(c,p,t) = h(d,q,s)? You need to check that it is smooth, which shouldn't be too hard to show because the component functions of h are, for the most part, just addition or multiplication of constants or sinusoidal functions, both of which are smooth. It's obvious that this function will only map to some subset of M (never outside of M) so the criteria h(W) = U n M seems satisfiable. I think the implicit function theorem will tell you if the inverse is continuous, but this will be obvious if your function is smooth.

    You'll have to go through this more rigorously. I would suggest doing some examples with basic manifolds in lower-dimensional spaces, to get the hang of it. Or, alternatively, for this example specifically you could check to see that every point (x,y,z,w) satisfying x² + ... + w² = 1 is described by the equations you were given. If so, then you know that this is the 3-sphere. In the second example (the simpler one with two boundary points) you can certainly pick a pair (x,y) one the 1-sphere that is not described by x = cos(cos(t)pi/2), y=sin(cos(t)pi/2). Take an arbitrary point on the 1-sphere and show either that it must be described by your equations for x,y,z,w or show that it need not be. So my alternative suggestion is to practice this kind of thing, so you'd still be solving the problem in a "looking at it" kind of way, but a more rigorous one. While the second alternative might be simpler, it is only useful if you already have in mind what the manifold will look like. In this case, it is either the 3-sphere, or a portion of it with a hole in it somewhere, and thus it either has a boundary or is not a manifold at all (suppose it just has two points missing, then it's not a manifold, nor a manifold-with-boundary, nor mainfold-with-corners, etc.). However, if it is a weird shape, then you'll probably just have to prove it analytically. It depends on what you physics class is like, but I can't imagine a physics class requiring that kind of analysis.
     
  11. May 4, 2005 #10
    It's very nice from you to give me some help. I think the criterion of "no boundary" is : "For every point x in the manifold, there exist an open domain contained in the manifold, which contains x".

    In the case of the half circle, the domain containing the end in the manifold is not open, but closed where the end is.

    I'm not in a physics class, but I try to get around the Poincaré Conjecture, which I think is believed to have been proved by Perelman in 2003 :

    "every closed simply connected three-manifold is homeomorphic to the three-sphere"

    Since I have a physics education, and not math, I have absolutely no idea how to prove this is true. However, I was trying to find a counterexample.

    I'll explain you my idea : take a 3D domain which is a sphere with a spheric hole inside (like cheese hole).....Then it's clear that this domain is simply connected. Note here that in a 3D domain, one could define another connexity, and instead of closed path, taking closed 2D manifolds, in order to detect chees holes precisely.

    a) This domain is given the first three coordinates I gave before.

    Then I add an extra 4 coordinate that makes the domain a certain parametrization of the 3Sphere, hence a 3D manifold in 4D space....the problems are : show that the manifold has no boundary and is compact (I think compact is clear in this case, since all the coordinates are periodic and that the manifold is finite ??).

    b) Hence It's a 3D manifold in 4D space, which is simply connected, So I need to prove it has no boundary.

    c) Then the question is : how to prove it is homeomorphic to the 3Sphere...I already know it is a certain parametrization contained in the 3Sphere...but remains to prove homeomorphy ( I still have to learn what it is, but I think it means : there exist a countinuous bijection, whose inverse is continuous from the 3Sphere to the manifold)

    The point here is : I think precisely because it is a contained in the 3Sphere, then it has a boundary.....Take the anologous in 2D : take a domain contained in the surface of a sphere, then it has forcedly a boundary....

    So my basic Idea was wrong I think, but I want to prove it.
     
  12. May 4, 2005 #11
    No, what I'm saying is that if the integral over the boundary is zero due to the manifold having an empty boundary, then the exterior derivative of every form defined over the manifold must be zero.

    Here is a good link:

    http://mathworld.wolfram.com/StokesTheorem.html

    "When M is a compact manifold without boundary, then the formula holds with the right hand side zero."

    For example, simply by knowing that [tex]\partial\partial{M} = \emptyset[/tex] you can derive the fact that [tex]d^2 \omega = 0[/tex] from the GST.

    The fact that the boundary is empty, and the fact that the exterior derivative of an exact form is zero, are sides of the same coin. For example:

    [tex]\int_M d\alpha = \int_{\partial{M}} \alpha [/tex]

    let [tex]\alpha = d\omega[/tex]:

    [tex]\int_M d\alpha = \int_M d(d\omega) = \int_{\partial{M}} d\omega = \int_{\partial\partial{M}} \omega = \int_\emptyset \omega = 0[/tex]

    so [tex] \partial\partial{M} = \emptyset \implies d^2 \omega = 0[/tex]

    now, let [tex]N = \partial{M}[/tex], then:

    [tex]\int_N d\omega = \int_{\partial{N}} \omega = \int_{\partial(\partial{M})} \omega = \int_{\partial{M}} d\omega = \int_M d(d\omega) = 0[/tex]

    which says that [tex]d^2\omega = 0 \implies \partial\partial{M} = \emptyset[/tex]

    therefore, showing that the exterior derivative of every form over a manifold is zero, that is [tex]d\alpha = 0[/tex] (a closed form) is equivalent to saying that the boundary is empty. This sounds exactly like what the OP wants to show.
     
  13. May 4, 2005 #12

    AKG

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    Have you even shown that [itex]\alpha[/itex] is exact?
    I don't see how your proof does anything of this. How could every form on a manifold be closed? I have no idea how the parts of your proof relate to other parts of the proof at all, or how your statements follow, or what any of this has to do with anything being discussed. When "the right hand side of the formula is zero" that doesn't mean that the integrand is zero, it means that the whole integral is zero. The fact that d²w = 0 holds for, if I'm not mistaken, any differential form, and has nothing to do with the manifold it's defined on.
     
  14. May 5, 2005 #13
    Well, I'm not familiar with exterior derivatives, but there are closed form that are not exact : [tex] d\theta [/tex] is closed.

    However [tex] \theta\neq d\eta \forall \eta [/tex]...

    But what I think quetzalcoatl wants to say is that :

    [tex] \int_M d^2\omega=\int_{\partial M}d\omega=\int_{\partial^2 M}\omega=0 [/tex]

    Hence : the integration of an exact form over the boundary of every manifold is zero....but this does not indicate that the boundary is empty.

    On ther other hand, quetzalcoatl showed that :

    [tex] \partial M=\emptyset \Rightarrow \int_Mdw=0 [/tex] : if a manifold has no boundary, then the integration of every exact form on the manifold is zero....

    So then quetzalcoatl reversed the theorrm by extending (without saying) the exact forms to ALL forms, using the reverse implication and hence deduce the boundary was zero....but this is not allowed, since there are forms that are not exact....

    What was asked is a condition of the form : [tex] P\Rightarrow \partial M=\emptyset [/tex]

    And not the opposite way...

    Remark :

    Interpretation of a 3D manifold :

    let's assume : w=f1(a,b,c), x=f2(a,b,c), y=f3(a,b,c), z=f4(a,b,c) a 3D manifold in 4D space, with free parameters (a,b,c)...

    Then we can look at the reverse transformation :

    a=g1(w,x,y,z), b=g2(w,x,y,z), c=g3(w,x,y,z)

    If you interprete w as the physical time t, then the above equations represent a time-dependent transformation of R^3 into itself......

    Maybe this could help the imagination
     
    Last edited: May 5, 2005
  15. May 5, 2005 #14
    kleinwolf, you are correct.

    I had mistakenly thought that the converse was true. last night i realized that this is not the case when i consulted a text.
     
  16. May 5, 2005 #15
    quetzalcoatl,

    thank your idea of using Stokes theorem I found a way to check if the boundary of a 3D manifold is empty (and thanks to David Bachman's work presented in the next thread in this same "Tensor Analysis" part :

    The (non-linear) differential form to compute the Area of a 2D manifold is :

    [tex] Area=\sqrt{\sum_{i>j}(dx_i\wedge dx_j)^2}[/tex]

    In my problem it is the boundary of a 3D manifold (which is 2-dimensional). The problem is we don't know the boundary, so we can use Stoke's theorem :

    [tex] S=\int_{\partial M}Area=\int_Md(Area) [/tex]

    Hence it just suffice to compute the exterior derivative of Area with the coordinates as function of the free parameters, and then integrate over the domain of the defintion of the free parameters.

    If this is 0, then the boundary vanishes, without actually knowing the boundary....
    (I tried in particular for the sphere, it worked)

    Seems correct ???
     
  17. May 5, 2005 #16
    kleinwolf, congratulations! i never occured to me to take the area formula and integrate over the manifold, this makes perfect sense. (the elegance of differential forms at its best :)
     
  18. May 7, 2005 #17
    So that in fact your assumption was right :

    you affirmed : [tex] \int_Mdw=0\forall w\Rightarrow \partial M=\emptyset [/tex]

    This is of course true...just take in particular [tex] w=Area [/tex]....but why test on other forms then ?
     
    Last edited: May 8, 2005
  19. May 7, 2005 #18
    yes, the problem was precisely that...i had erroneously extended this to the exterior derivative of ALL forms on the manifold, which is incorrect, and is certaintly incorrect in the case that you have described. there may be some circumstances of cohomology where this is true, but i dont know enough about that to say.
     
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