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Nonzero exact 1-form implies foliation?

  1. Jul 20, 2011 #1

    ivl

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    Hi all,
    First of all I would like to apologize, as my knowledge of differential geometry is not as good as it should be (I only took an introductory course for physicists).

    Anyhow, here is a question for you. Any help is greatly appreciated!

    -I have a four dimensional manifold, whose properties are rather banal (for instance, all closed differential forms are also exact).
    -On this manifold, I am given a nowhere-vanishing 1-form (I call it "dt").
    -I would like to know if this nowhere-vanishing exact 1-form "dt" determines a foliation (the folia are t=const surfaces).

    An attempt to an answer: dt is an exact 1-form, and t is a function (which maps points in the manifold to real numbers). Now, if dt is nowhere zero, t is a monotonic function. The surfaces of t=const are thus folia which are monotonically labeled.

    Related topics: Frobenius theorem, space+time foliation
     
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  3. Jul 20, 2011 #2

    quasar987

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    Yes:

    There is a theorem called the regular value theorem (which is really just a straightfoward application of the implicit function theorem), which states that if f:M-->R is a smooth map and c is a regular value of f (i.e. df is non vanishing at every point of f-1(c)), then f-1(c) is an embedded submanifold of dimension dim(M)-1 (a "hypersurface").

    Hence, if df is nowhere vanishing on all of M, then f-1(c) is an embedded submanifold for every c in R, and exhausting every value of R exhaust every point of M and you have yourself a foliation of M by embedded hypersufaces.
     
  4. Jul 21, 2011 #3

    ivl

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    Thanks for the useful reply, quasar987!

    Let me just re-state what you said, so as to check if I understood.

    ====================
    Remark: An exact 1-form that does not vanish in any point of a manifold M determines a foliation of M with folia of dimension dim(M)-1.
    ====================
    Proof: Any exact 1-form df determines a function
    f:M --> R
    p |--> c=f(p)
    which is smooth and defined up to a constant. The image of f, labeled Im(f), is the set of all c belonging to R such that c=f(p) for some point p of M. The inverse image of a real number c, denoted f^{-1}(c), is the set of all points p of M which are mapped to the value c by the function f.
    Two crucial facts ought to be observed. Firstly, the manifold M is the domain of f, and thus trivially coincides with the inverse image of the full set Im(f). As a consequence, any point p in M can be labeled by a real number c by virtue of f(p)=c. Secondly, if the 1-form df is non-zero for all points of some f^{-1}(c), then the inverse image of a value c is an embedded sub-manifold of dimension dim(M)-1 [Regular Value Theorem]. Given that df is assumed to be nowhere vanishing, the inverse image of every c in Im(f) is an embedded sub-manifold. In conclusion, given an exact 1-form df which is always non-zero, all points of M can be labeled according to which embedded submanifold f^{-1}(c) they belong to. This provides a foliation, QED
    ======================

    Great! Just two further silly questions... any help, again, is greatly appreciated:
    -Is the direction of df the direction of increasing c?
    -Do you know any good (hopefully easy) reference for the regular value theorem?

    Thanks so much!
     
  5. Jul 21, 2011 #4

    quasar987

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    Excellent!

    This is a bit ambiguous as df is not a vector. But if you pick a chart (x_1,...,x_n) of M and consider f(x_1,..,x_n) as a map on R^n, and then identify the 1-form [itex]df = \sum\frac{\partial f}{\partial x_i}dx_i[/itex] with the vector field [itex]\left(\frac{\partial f}{\partial x_1},\ldots,\frac{\partial f}{\partial x_n}\right)[/itex], then of course this is just the gradient of f and it points in the direction of increasing c.

    It can be found in any book about smooth manifold theory. See for instance John M. Lee's Introduction to smooth manifold.

    Or you could try to prove it yourself if you know the definition of an embedded submanifold and the implicit function theorem. The implicit function theorem simply says that if f:R^n=R^k x R^m -->R^m is a smooth map and p is a regular value of f (Df(p) is surjective), then locally near p the level set f-1(f(p)) is the graph of some function (there exists an open rectangle O = U x V in R^k x R^m such that O n f-1(f(p)) = {(x,F(x))| x in U} for some smooth map F:U-->V). But graphs of maps and clearly submanifolds since projection onto the domain is a (global) chart. Doing this for each point in f-1(f(p)) proves that it is a submanifold of dimension k=n-m.
     
    Last edited: Jul 21, 2011
  6. Jul 28, 2011 #5

    ivl

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    Thanks a lot for the answer (and apology for the delay). The textbook you suggested seems to be useful -- well done, then!

    Unfortunately, I do not know anything about graph theory... so I cannot follow your proof. Bu never mind, I think I got my solution :-)

    Cheers!
     
  7. Jul 28, 2011 #6

    quasar987

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    The graph of a function f:A-->B is simply the set G={(x,f(x)|x in A} (i.e. that thing that you draw on a piece of paper when you draw a function R-->R). Nothing to do with graph theory.
     
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