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I am reading Guillemin and Pollack's Differential Topology Page 163:

If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

So my question is, what is $\omega[f(x)]$?

I have been trying to make this question self-contained, but here is the whole background: Definition of pullback. In that question, I have shown $(df_x)^*T(v_1, \dots, v_p) = T \circ df_x (v_1, \dots, v_p)$ and $\omega$ is $(df_x)^*T$. But I want the missing piece about what that $\omega[f(x)]$ is equal to, to make sense of the definition of $f^* \omega (x)$.

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WishingFish
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2 Answers2

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$\omega[f(x)]$ is the value of the form $\omega$ at the point $f(x)$. It is a vector in the $p$-th exterior power of the cotangent space at $f(x)$. (Remember that a $p$-form is a smooth function assigning to each point of a manifold such a vector.) I guess you were just confused by the use of square brackets, which were, I think, intended just to make it easier to read; round brackets would have been OK too, and it seems they didn't cause you any problems with the entirely analogous expresson $f^*\omega(x)$ on the left side of the equation.

Andreas Blass
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  • Thank you so much Andreas. However, as I updated, I am looking for some expression making $\omega[f(x)]$ computable, because right on the next page asks me to prove $f^(w \wedge \theta) = (f^w) \wedge (f^* \theta)$. – WishingFish Jul 30 '13 at 22:49
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You can write your differential form (at least locally) as a sum of "pure forms", i.e. forms of the type $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}$$ with $\alpha$ a smooth function. Then $$f^*\omega_\alpha(x) = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}]$$ In other words, $\omega[f(x)]$ means that you precompose the "weight functions" of the pure forms with $f$.

Daniel Robert-Nicoud
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  • Oh! So you are saying, $\omega$ is defined to be $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}.$$

    Hence $$\omega[f(x)] = (\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}.$$

    Therefore, $$f^\omega(x) = (df_x)^\omega[f(x)] = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}].$$

     – WishingFish Jul 31 '13 at 01:46
  • And an application of your theorem here: http://math.stackexchange.com/questions/456086/pullback-expanded-form hope you can help me see if I got it correctly? :) – WishingFish Jul 31 '13 at 03:46
  • @WishingFish Yes, you got it correctly, except for the little detail that locally a $p$-form can be written as a sum of pure forms. If you take the wedge of two forms (written as sums of pure forms), then the wedge distributes over the sum. – Daniel Robert-Nicoud Jul 31 '13 at 07:51
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    I hope I got it correctly - so it should be, $$f^\omega_\alpha(x) = (df_x)^ \sum_{1 \leq i_1 < \cdots < i_k \leq \dim X}[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}]?$$ Thank you! – WishingFish Jul 31 '13 at 17:49