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14.22: pg 362 Lee's Smooth Manifold Let $X$ be a smooth vector field on $M$.

  1. If $w$ is a smooth differential form then $i_Xw$ is smooth.
    $$ i_X w:= (X \lrcorner w)_p =X_p \lrcorner w_p. $$

  2. $i_X:\Omega^k(M) \rightarrow \Omega^{k-1}(M)$ is linear over $C^\infty(M)$, corresponding to a smooth bundle homomoprhism $$ i_X: \wedge ^kT^*M \rightarrow \wedge^{k-1}T^*M. $$


I would really appreciate nice explanations for both aprts. My solution for 1. is ugly (below). For 2. how does linearity over $C^\infty(M)$ imply bundle homomorphism?


My thoughts for 2. are using lemma 14.13 and working in a local coordinate chart.

Lemma 14.13, pg 358: If $w \in \wedge^k(V^*) $ and $\eta \in \wedge^l(V^*)$, $$ i_v (w \wedge \eta) = (i_v w ) \wedge \eta + (-1)^k w \wedge (i_v \eta). $$

which I believe is what the solution wants.


May attempt for 1.

We work locally on a local chart $(U, (x^i))$ around $w$. We know $w = \sum w_I dx^I$, $X=X^i \frac{\partial}{\partial x^i} $, where $w_I, X^i$ are smooth on $U$. Thus, for $p \in U$, $v_2, \ldots, v_k \in T^kM$, \begin{align*} X_p \lrcorner w_p (v_2, \ldots, v_k) &= w_p(X_p, v_2, \ldots, v_k) \\ &= \sum_I w_I(p) \sum_i X^i (p)dx^I(\frac{\partial }{\partial x^i}, v_2, \ldots, v_k) \end{align*}

In this form, we can see the expression is clearly representable $\sum_{J} l_Jdx^J$ for $J$ of length $k-1$., such that $l_J$ is smooth on $U$.


EDIT:

Thanks for the comments below. The general theorem show that $i_X$ is then given by $i_X (v) = i_X (\tilde{v})(p)$ for some $\tilde{v} \in \Omega^k(M)$ where $\tilde{v}(p) = v$.

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    $\begingroup$ Your solution for 1 is fine. I wouldn't call working with coordinates ugly. In fact, in some cases, there is no other way. As for 2, it is a general (and important) fact that if $E,F\to M$ are vector bundles and $L:\Gamma(E)\to\Gamma(F)$ is linear over $C^\infty(M)$, then $L$ corresponds to a bundle morphism $E\to F$. Regardless, it follows from the definition that interior multiplication corresponds to a bundle morphism. $\endgroup$ – Amitai Yuval Oct 3 '18 at 10:14
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    $\begingroup$ As @AmitaiYuval said. The second part is true by Lemma 10.29 (Bundle Homomorphism Characterization Lemma) in Lee’s. $\endgroup$ – Sou Oct 3 '18 at 11:38
  • $\begingroup$ Thanks AmitaiYuval and Sou, I understand now. @Amitai do you want to copy/post your solution so the problem is resolved? $\endgroup$ – CL. Oct 5 '18 at 9:20

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