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If $X$ is a smooth vector field on $U$ and $\omega$ is a smooth differential form, then $i_{X}\omega$ is smooth

The above is an Exercise 14.22 in Lee's introduction to smooth manifolds.

Recall the definition of smooth differential form: A smooth differential form on $U$ is a map $\omega:U\to \Lambda^k(T^*U)$ such that $\omega_p\in \Lambda^k(T^*_pU)$ and $\omega_p = \sum_{I}'\omega_I(p)dx^I|_p$ where $w_I:U\to \Bbb R$ are $\textbf{smooth}$ functions.

Also, if a smooth vector field on $U$ is a map $X:U\to TU$ such that $X_p\in T_pU$ and $X_p = \sum_{i=1}^n X^i(p)\frac{\partial}{\partial x^i}|_p$ where $X^i:U\to\Bbb R$ are smooth functions.

Here, $(i_{X}\omega)_p:= i_{X_p}\omega_p$. So I need to transform this $i_{X_p}\omega_p$ to the above form to check whether the coefficient function is smooth or not. But how can I transform? Any help?

Edit: For simplicity, I changed my question to an open subset $U\subset\Bbb R^n$ not $M$.

Edit: If $X = \sum_{i=1}^n X^i \frac{\partial}{\partial x^i}$ and $\omega_p = \sum_{I}' \omega_I dx^I$ then by multilinearity, $i_Xw = w(X,...) = w(\sum_{j}X^j\frac{\partial}{\partial x^j},...) = \sum_{j}X^j\omega(\frac{\partial}{\partial x^j},...) = \sum_jX^j(\sum_I'\omega_I dx^I)(\frac{\partial}{\partial x^j},...) = \sum_j\sum_I'X^jw_I dx^I(\frac{\partial}{\partial x^j},...) =\sum_j\sum_I'X^jw_I (\frac{\partial}{\partial x^j}\lrcorner dx^I)$

one potato two potato
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  • $\iota_X\omega = \sum_I\sum_j X^j \omega_I \left(\iota_{\frac{\partial}{\partial x^j}}dx^I\right)$. So, now all you have to do is calculate the thing in brackets. Try this yourself first, using the Leibniz rule and carefully unwinding the definitions; after that I'd suggest you take a look at this answer where I talk about how to do computations. Note by the way, it doesn't matter whether $U$ is in $\Bbb{R}^n$ or $M$, the same proof holds word for word. – peek-a-boo Apr 20 '21 at 16:17
  • @peek-a-boo Thank you for your comment. But how $i_X\omega = \sum_{I}\sum_{j}X^j\omega_I(i_{\frac{\partial}{\partial x^j}}dx^I)$ come from? I saw your link and It says $(\sum_a X^a\frac{\partial}{\partial x^a})\lrcorner (\sum_{I}\alpha_Idx^I) = \sum_{\alpha,I}X^a\alpha_I(\frac{\partial}{\partial x^a}\lrcorner dx^I)$. Why is this ture? – one potato two potato Apr 20 '21 at 16:25
  • Can you tell me how is $\iota_X\omega$ defined (and for clarity, don't go all the way down to the "pointwise level" of $p\in M$)? – peek-a-boo Apr 20 '21 at 16:26
  • You mean $(X\lrcorner\omega)_p = X_p\lrcorner \omega_p$? – one potato two potato Apr 20 '21 at 16:34
  • I meant for you to write it without the $p$ there but nvm. You still haven't answered my question though: you just substituted one piece of notation for another. How is the tensor $(\iota_X\omega)p:= \iota{X_p}\omega_p\equiv X_p \lrcorner \omega_p$ defined? After all tensors are certain multilinear maps. So how is the action of this multilinear map defined? – peek-a-boo Apr 20 '21 at 16:36
  • you mean this? $i_{X_p}\omega_p(\omega_1,...,\omega_{k-1}) = \omega_p(X_p,\omega_1,...,\omega_{k-1})$ – one potato two potato Apr 20 '21 at 16:41
  • Well... almost the $\omega$'s should really be $X$'s. More succinctly, $\iota_X\omega = \omega(X,\cdots)$, which means we plug in the vector field $X$ into the first slot of $\omega$ (some people define it to be the last slot... but doesn't really matter here). Ok, so now if $X=\sum_j X^j \frac{\partial}{\partial x^j}$, what can you say using mulitlinearity of $\omega$? In case you haven't noticed yet, the answer to your original question is "by definition" which is why I'm asking you to say all of this out explicitly. – peek-a-boo Apr 20 '21 at 16:44
  • I just edit my post since it's too long to write in comment. – one potato two potato Apr 20 '21 at 16:55

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