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A Question about de Rham's Theorem

  1. Dec 27, 2012 #1

    lavinia

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    Suppose there is a closed differential form whose periods lie inside an additive subgroup of the real numbers e.g the integers or the rational numbers.

    Does this form determine a cohomology class with coefficients in this subgroup? I think this follows from the universal coefficient theorem.

    Conversely given a non-torsion cohomology class (on a smooth manifold)with coefficients in one of the subgroups does there correspond a closed differential form with the same periods?
     
    Last edited: Dec 27, 2012
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  3. Dec 28, 2012 #2

    Bacle2

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    What do you mean by periods?
     
  4. Dec 28, 2012 #3

    lavinia

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    The periods are the values taken on homology cycles.

    The more I think about this, it seems that both cases follow from the Universal Coefficient Theorem and de Rham's theorem. Right?
     
  5. Dec 28, 2012 #4
    I'm a bit rusty on de Rham cohomology so excuse my overly long answer which was mostly just to convince myself, but now I feel I might as well leave the details in there. It seems like (for smooth manifolds at least) it follows from the universal coefficient theorem and deRham's theorem as you say.

    By deRham's theorem we have an isomorphism [itex]\theta : H_{deRham}^k(M) \to H^k(M,\mathbb{R})[/itex] (maybe reduced singular cohomology in dimension 0; I don't really recall the details and doubt we really care about dimension 0 anyway). By the universal coefficient theorem we have a surjection
    [tex]H^k(X,A) \xrightarrow{\tau_A} Hom(H_k(X),A) \to 0[/tex]
    which is natural in A and with kernel [itex]Ext(H_{k-1}(X),A)[/itex].

    If I understand your definition of periods right, then to say that the periods of a closed differential form [itex]\omega[/itex] lies in a subgroup [itex]A \subset \mathbb{R}[/itex] is simply to say that the image of [itex][\omega] \in H_{deRham}^k(X)[/itex] along
    [tex]H_{deRham}^k(X) \xrightarrow{\theta} H^k(X,\mathbb{R}) \xrightarrow{\tau_\mathbb{R}} Hom(H_k(X),\mathbb{R})[/tex]
    can be expressed as the image of an element [itex]\alpha_{[\omega]} \in Hom(H_k(X),A)[/itex] along the map [itex]Hom(H_k(X),A) \to Hom(H_k(X),\mathbb{R})[/itex] induced by the inclusion [itex]i : A \to \mathbb{R}[/itex].

    By naturality we have a commutative diagram
    [tex]H^k(X,A) \xrightarrow{\tau_A} Hom(H_k(X),A) \to 0[/tex]
    [tex]H^k(X,\mathbb{R}) \xrightarrow{\tau_\mathbb{R}} Hom(H_k(X),\mathbb{R}) \to 0[/tex]
    where the two rows are supposed to be connected by the morphisms induced by the inclusion [itex]i : A \to \mathbb{R}[/itex].

    By the exactness of the top row we get some [itex]c \in H^k(X,A)[/itex] with [itex]\tau_A(c) =\alpha_{[\omega]}[/itex], but then by the commutativity of the diagram we get
    [tex]\tau_\mathbb{R}(i_*c) = i_*(\alpha_{[\omega]}) = \tau_\mathbb{R}(\theta([\omega]))[/tex]
    but [itex]\tau_\mathbb{R}[/itex] is an isomorphism by the universal coefficient theorem ([itex]Ext(H_{k-1}(X),\mathbb{R})=0[/itex] because [itex]\mathbb{R}[/itex] is a field) so
    [tex]\theta([\omega]) = i_*c[/tex]
    This in particular shows that [itex]\omega[/itex] determines a cohomology class with coefficients in the subgroup A.

    The second statement follows similarily by the commutativity of the diagram above because if we start with [itex]c \in H^k(X,A)[/itex], we then get [itex]i_*c \in H^k(X,\mathbb{R})[/itex] which is represented by some closed differential form [itex]\omega[/itex]. However
    [tex]\tau_\mathbb{R}(i_*c) = i_*(\tau_A(c))[/tex]
    so the periods of [itex]i_*c = \theta([\omega])[/itex] lie in A.
     
  6. Dec 30, 2012 #5

    mathwonk

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    you seem to be saying that a form which determines a Q valued linear function on homology should be equivalent to a cohomology class with Q coefficients. Don't all such linear functions occur that way by duality between homology and cohomology? I am not so sure about Z valued functions, since integral cohomology can have torsion, hence a cohomology class is not "determined by" its values as a functional on homology. In that case I would guess there could be more than one cohomology class with those "periods".
     
  7. Jan 1, 2013 #6

    lavinia

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    Happy New Year Mathwonk.

    My thought was that a closed differential form with say integer periods many not take on integer values on all simplices in a smooth triangulation. So technically it does not represent an integer cochain even though it takes on integer values on cycles. So why would there be an integer cocycle that is cohomologous to it?

    The universal coefficient theorem says that the non-torsion part of the integer cohomology is just the homomorphisms of the integer homology into the integers. The differential form with integer periods certainly does this so it must be identifiable with an integer cochain.
     
  8. Jan 1, 2013 #7

    mathwonk

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    happy new year to you lavinia. i was being picky and quibbling with the use of the word "determines" which implies the integer cochain would be unique. i argued not against existence but against uniqueness.
     
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