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Let $X$ be a Kronecker vector field on the two dimensional torus $\mathbb{T}^2$. Let $K$ be the space of all 1- forms $\alpha$ of class $C^1$ on $\mathbb{T}^2$ which satisfy $d\alpha=0,\;\alpha(X)=1$. Then $K$ is a convex closed subset of all $C^1$ 1-forms on $\mathbb{T}^2$.

Is $K$ a compact subset of the space of 1-forms with respect to $C^1$ topology? If the answer is affirmative. according to the Krein Millman theorem, what is a presise description of its extreme points of $K$?

Does the topological structure of $K$ depends on chosing the vector field $X$ tangent to our initial Kronecker foliation of torus? Does the topological structure of $K$ depend on the slope of our Kronecker foliation?

Motivation:

A motivation for this question is the following:

In this post and some other related linked posts we try to find a Riemannian metric compatible to orbits of a non vanishing vector fields. Choosing various metrics enable us to have different curvatuare functions. Possessing an appropriate curvature function is very essential for appllying the Gauss Bonnet theorem to the problem of limit cycles of vctor fields.(For counting them as closed geodesics). So this situation leads us to think about the diversity of closed differential 1-forms $\alpha$ with $\alpha(X)=1$. Under these conditions, in particular the propery of closed convexity of this set $K$. one is tempt to be curious about the presice description of possible extrem points of $K$.

Remark: For generalization of this question to $n$ dimensional space we should consider the spaceof all 1-form $\alpha$ with $i_X d\alpha=0,\;\alpha(X)=1$.

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  • $\begingroup$ What is a kronecker vector field? $\endgroup$
    – Thomas Rot
    Commented Nov 29, 2020 at 10:41
  • $\begingroup$ @ThomasRot Vector field $\partial_x+\alpha\partial_y$ where $\alpha$ is an irrational number. This planar vector field gives us a foliation of torus. Two different irational $\alpha, \beta$ may introduce topologically different foliation of $\mathbb{T}^2$. $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2020 at 10:45

1 Answer 1

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I don't think it is compact, but perhaps I miss a normalization condition?

Let $X=\partial_x+a\partial_y$, with $a$ irrational (doesn't actually matter for the following). Let $\alpha\in C$ (e.g. $\alpha=dx$) and let $\omega_\lambda=\lambda(a dx-dy)$, for $\lambda\in \mathbb R$. As $X$ lives in the kernel of $\omega_\lambda$, and $d\omega_\lambda=0$, then $\alpha+\omega_\lambda\in C$.

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  • $\begingroup$ Thank you for your attention and your very interesting answer. $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2020 at 14:41
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    $\begingroup$ But what about if we impose an upper bound on the $C^1$ norm of $\alpha$? $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2020 at 14:43
  • $\begingroup$ As you see in the motivation of the question, the only $\alpha$ I was aware in quadratic system was $d\theta$ but may be some other $\alpha$ would produce more appropriate curvature function. $\endgroup$
    – Ali Taghavi
    Commented Nov 29, 2020 at 14:49
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    $\begingroup$ I think you can write down differential equations for the space $B$ of $1$-forms $\beta$ satisfying $d\beta=0$ and $\beta(X)=0$. The space $C$ is $\alpha+B$. Perhaps you can find all solutions of $B$ explicitly? And then look at the unitball in $B$? $\endgroup$
    – Thomas Rot
    Commented Nov 29, 2020 at 14:55
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    $\begingroup$ B is a vector space, hence K is affine. so I don’t think so $\endgroup$
    – Thomas Rot
    Commented Nov 29, 2020 at 21:49

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