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I'm thinking about the kernel of a differential one-form $\theta\in\Lambda^{1}(M)$:

$$ Ker(\theta):=\left\{X\in\mathfrak{X}(M) \;|\; \theta(X)=0\right\} $$

Now suppose $X\in Ker(\theta)$, then is $fX$, with $f$ a function on $M$, in $Ker(\theta)$?

Somewhat naively I think it is, in fact:

$$ \theta(fX)=f\theta(X)=0 $$

Is it right?

More specifically is $Ker(\theta)$ a real vector space or a real module over the functions on $M$?

My question arises because I need to understand the relationship between the foliation associated to $\theta$ and $Ker(\theta)$.

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A one-form is just a a global section of the cotangent bundle so it´s a module over the ring of smooth functions. Of course, you can see this as a real vector space, but then you lost the endomorphism of rings (of the module as an abelian group). In general, for a compact manifold, by Serrre-Swan theorem you can just say that this module is projective. –  user40276 Apr 11 at 11:38
    
I do not understand how your comment is related to the kernel $Ker(\theta)$ of a differential one-form, can you, please, be more explicit? Thank You in advance. –  Ilcapitano Apr 11 at 16:13
    
It is a real vector space and a module over smooth functions. –  user40276 Apr 12 at 2:51

1 Answer 1

up vote 2 down vote accepted

Yes, you are correct: a 1-form is $C^\infty(M)$ linear so if $\theta(X) = 0$ then $\theta(fX) = f\theta(X) = 0$ for all smooth functions $f$. This means that $\ker\theta$ is a module over smooth functions on $M$.

If $\theta$ is nowhere vanishing, then $\ker \theta$ will be a vector subbundle (i.e. distribution) of $TM$. If $\theta$ has some zeros then the rank will jump and so will not form a vector bundle but will be a sheaf of $C^\infty(M)$ modules.

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Thank You. One last thing,if I $\theta$ vanishes at some point $m\in M$ and I consider $\theta$ only in $\overline{M}:=M-\{m\}$ then will $ker(\theta)$ be a subbundle of $T\overline{M}$? I think it is so but I need a confirmation. –  Ilcapitano Apr 26 at 12:53
    
@Ilcapitano yes, that is correct (assuming of course $\theta$ vanishes only at $m$). –  Eric O. Korman Apr 26 at 13:41

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