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Let $\omega $ be a $1$-form on $M$. For $f:M\to \mathbb R$, and $X\in \mathfrak{X}(M)$, Does the following true?

$f. i_X \omega = i_{f.x}\omega$

According to me proof is the following: Please point out the mistake in the following and if possible please make comment on above expression. Thanks.

For $p\in M$, we have $$f.i_X\omega|_p= f(p). \omega_p(X_p)= \omega(f(p)X_p)= i_{fX}\omega|_p$$

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When the hyphen in "$1$-form" is INSIDE the $\TeX$ tags, then it looks like a minus sign instead of a hyphen, thus: $1-$ form. (I fixed it.) – Michael Hardy Sep 10 '12 at 17:25
thanks.... i will take care in future.. – Junu Sep 10 '12 at 20:23
up vote 5 down vote accepted

The fact is true and your proof is true (although you should add a $p$ subscript after $\omega$ in the third equality of your proof).

You have used the $\mathbb{R}$-linearity of $\omega_p$, which is an element of the real vector space $T^\star_pM$, but you could have also used the $\mathcal{C}^\infty(M)$-linearity of $\omega$, which is an element of the $\mathcal{C}^\infty(M)$-module of sections of $T^\star M$ (the cotangent bundle). You have a $1$-form $\omega$ so applying to it the interior product (insertion operator) yields the smooth function $\omega(X)$: $$i_X\omega=\omega(X)$$ (there is only one slot to fill since we have a $1$-form). Now $i_{fX}\omega=\omega(fX)$ and remember that a differential form is $\mathcal{C}^\infty(M)$-multilinear alternate so you can take out the $f$ which yields: $$i_{fX}\omega=\omega(fX)=f\omega(X)=f\cdot i_X\omega$$.

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thanks for the answer... :) – Junu Sep 10 '12 at 15:34
@Junu: You are welcome :) – Benjamin Sep 10 '12 at 15:39

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