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Let $M$ be a smooth differential manifold. Let ω be a $1$-form on $M$ and $X$ be a fixed vector field on $M$.

Then, in order to say "If $ω(X) = 0$, then $X = 0$ (i.e. $ω$ is injective?)",

what condition does $ω$ satisfy?

Antash Bob
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1 Answers1

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I believe this is not possible if $\dim M>1$. I say this because, at each point $p\in M$, $\omega_p\colon T_pM\to\mathbb{R}$ is a linear transformation, so you have $\dim(\ker\omega_p)\geq \dim M-1>0$. Therefore you can pick a non-vanishing vector field $X$ in some local neighborhood such that $\omega(X)=0$ in that neighborhood, and make it $0$ outside of it.

On the other hand, if $\dim M=1$, the necessary and sufficient condition that I can come up with is the set $\{p\in M\mid \omega_p=0\}$ containing no sets that are open in $M$ besides the empty set.

In fact, if $\dim M=1$, then $\omega(X)=0$ at a point $p\in M$ if and only if either $X_p=0$ or $\omega_p=0$. Another thing that will be useful to show this is that $M$ is, up to diffeomorphism, either $S^1$ or $\mathbb{R}$, and both have trivial tangent bundles. So picking vector fields and one-forms is the same as picking real-valued smooth functions on your manifold and your problem reduces to picking a smooth function $f\colon M\to \mathbb{R}$ such that, given any other smooth function $g\colon M\to \mathbb{R}$, then $fg=0\implies g=0$.

So if $V=f^{-1}(0)$ contains no non-empty sets that are open in $M$, then given a non-vanishing function $g\colon M\to \mathbb{R}$, $U=g^{-1}\left(\mathbb{R}\setminus\{0\}\right)$ is open so $fg$ is not zero in the non-empty set $U\setminus V$.

Conversely, if $V=f^{-1}(0)$ contains a non-empty set $U$ that is open in $M$, then $M\setminus U$ is closed in $M$ and the result follows from what is discussed here, by picking $g$ such that $g^{-1}(0)=M\setminus U$.

I hope this helps.

Paulo Mourão
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  • Thank you for answer my question. However, I think the my argument (If ω(X)=0 , then X=0) is satisfied even if dimM>1. – Antash Bob May 20 '19 at 13:35
  • For example, let f be a immersion from M to R. Then, df is injective i.e. If df(X)=0, then X=0. – Antash Bob May 20 '19 at 13:42
  • You can't have an immersion of $M$ into $\mathbb{R}$ if $\dim M>1$. This is general in the sense that if you have an immersion $f\colon N\to M$, then $\dim N\leq \dim M$. – Paulo Mourão May 20 '19 at 14:00
  • Sorry, it was clear that an immersion of M into R because injective maps don't given for n=dimM>1. By the way, if ω is R^{n+1}-valued 1-form, then the argument may be satisfied? – Antash Bob May 20 '19 at 14:18
  • $1$-forms are, by definition, maps $T_pM\to \mathbb{R}$ at each point $p$, so what you're saying isn't defined, I don't think. However, what you could do is take $n$ $1$-forms $\omega_1,...,\omega_n$ with independent kernels, i.e., $\ker(\omega_1)\cap...\cap\ker(\omega_n)={0}$. Then you would have $$\omega_1(X)=0, ...,\omega_n(X)=0\implies X=0$$ This is what is called a coframe on $M$. – Paulo Mourão May 20 '19 at 14:39
  • Keep in mind that this isn't always possible to do, however. – Paulo Mourão May 20 '19 at 14:49
  • Let ω be a R^{n+1}-values 1-from i.e. ω(X)=(ω1(X),...,ω{n+1}(X))', where ω_i(X) is a 1-form on M. Then, the argument may be satisfied iff ω_i(X) = 0 ? – Antash Bob May 20 '19 at 15:59
  • You have the right idea but that's usually written as taking $n+1$ $1$-forms $\omega_1,...,\omega_{n+1}$ and not an "$\mathbb{R}^{n+1}$-valued $1$-form", but this is just wording. That's not enough though, since they can be all the same: you could just pick the same $1$-form $n+1$ times and you'd be back at the original problem. A sufficient condition would be that they span $T_p^*M$ at each point $p$. – Paulo Mourão May 20 '19 at 16:07
  • Sorry, my definition of a R^{n+1}-values 1-from may be wrong. According to my text, a R^{n+1}-values 1-from is defined by a 1-form which take value on R^{n+1}. Please teach me the formula of a R^{n+1}-values 1-from. – Antash Bob May 21 '19 at 01:03
  • I created a chat to discuss this further if you want – Paulo Mourão May 21 '19 at 06:43
  • In the wikipedia which you taught me, the following is written:

    "Let V be a fixed vector space. A V-valued differential form of degree p is a differential form of degree p with values in the trivial bundle M × V. When V = R one recovers the definition of an ordinary differential form. "

    However, I can't understand this statement. So, please teach me the definition of a V-valued differential form of degree 1 with the some equations.

     – Antash Bob May 22 '19 at 05:48
  • Yes, I found the article and apparently it is a thing, but I've never seen it before, so I can't really help you – Paulo Mourão May 22 '19 at 08:32
  • Ohh... Thank you for previous responses! – Antash Bob May 22 '19 at 09:39