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Edit: According to the comment of Prof. Bryant, I revise my question.

Is there a reasonable and non trivial (geometric) interpretation for the following quantity on a compact Riemannian manifold $(M,g)$ of dimension $3$

$$q=\sup_{\alpha}\int_M \alpha \wedge d\alpha $$

where $\sup$ is taken over all $1$-forms $\alpha$ of $g$-length $1$.

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    $\begingroup$ First, the Euler characteristic of any compact $3$-manifold without boundary is $0$, so this is no restriction. Second, if $\alpha(Y) = g(X,Y)$ for all vector fields $Y$, then $\alpha = X^\flat$, so there is only one such $\alpha$ for any given unit vector field $X$. Can you reformulate your question? Did you want to just require that $\alpha(X)=1$, which still allows for some variation in $\alpha$? Now the metric is irrelevant. It only depends on $X$. $\endgroup$
    – Robert Bryant
    Commented Sep 11, 2017 at 13:25
  • $\begingroup$ I am sorry for not paying attention to the Euler characteristic of odd dimensional manifold. Moreover, to every unit length vector field $X$ we associate a i-form $\alpha_X$ with $\alpha_X(Y)=<X,Y>$. Then we take $\sup$ over all possible unit length vector fields $X$. $\endgroup$
    – Ali Taghavi
    Commented Sep 11, 2017 at 13:36
  • $\begingroup$ @RobertBryant Now I revise my question. Could I clarify my question? $\endgroup$
    – Ali Taghavi
    Commented Sep 11, 2017 at 13:59
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    $\begingroup$ Your question is better now, but why not just say that the sup is to be taken over all $1$-forms $\alpha$ of $g$-length $1$? Thinking of the vector field $X$ doesn't seem to be relevant. $\endgroup$
    – Robert Bryant
    Commented Sep 11, 2017 at 16:39
  • $\begingroup$ @RobertBryant Thank you. I revise it again. $\endgroup$
    – Ali Taghavi
    Commented Sep 11, 2017 at 20:33

1 Answer 1

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There is no reason to believe that there is a supremum of this functional. For example, consider the $3$-torus $M = \mathbb{R}^3/\mathbb{Z}^3$ with the quotient metric and the unit $1$-forms $$ \alpha_n = \cos(2\pi n z)\,\mathrm{d} x - \sin(2\pi n z)\,\mathrm{d} y, $$ where $n$ is an integer, which are well-defined on $M$.

One finds by calculation that $$ \alpha_n\wedge\mathrm{d}\alpha_n = 2\pi n\, \mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\, $$ Hence $$ \int_M \alpha_n\wedge\mathrm{d}\alpha_n = 2\pi n, $$ so that the integral you describe can be made arbitrarily large (positive or negative) by appropriate choice of $\alpha$.

I expect that some similar construction could be made for any compact oriented $3$-manifold, showing that the functional is always unbounded (in either direction).

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    $\begingroup$ I honestly have admiration to your magnanimous and noble because while there are a lot of deeper question in MO but you have open and kind conduct to beginner or people as me whose questions are not deep. I observed this during my activity in MO. My best regard to you. $\endgroup$
    – Ali Taghavi
    Commented Sep 12, 2017 at 6:48

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