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I am reading Guillemin and Pollack's Differential Topology. For the proof on Page 164, I was not able to get through the last step.

$$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$$

Definition. 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: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

According to Daniel Robert-Nicoud's ansewr to $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, what is $\omega[f(x)]$?

Locally, differential form can be written as $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}$$ with $\alpha$ a smooth function. Then $$f^*\omega_\alpha(x) = (df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}].$$

We write $$\omega_\alpha(y) = \alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p},$$ $$\theta_\beta(y) = \beta(y)dx^{j_1}\wedge\ldots\wedge dx^{j_q}.$$

Hence, $$\omega \wedge \theta = (\alpha(y)dx^{i_1}\wedge\ldots\wedge dx^{i_p}) \wedge (\beta(y)dx^{j_1}\wedge\ldots\wedge dx^{j_q})$$

Following James S. Cook's very brilliant answer Pullback expanded form.

$$\omega \wedge \theta = \alpha(y) \beta(y) \sigma(I,J) dx^{k_1}\wedge\ldots\wedge dx^{k_{p+q}}$$

Rename $\gamma(y) = \alpha(y) \beta(y) \sigma(I,J)$, we get $$f^*(w \wedge \theta) = (df_x)^* [(\gamma \circ f)(x) dx^{k_1}\wedge\ldots\wedge dx^{k_{p+q}}].$$

But $$(f^*w) \wedge (f^* \theta) = ((df_x)^*[(\alpha\circ f)(x)dx^{i_1}\wedge\ldots\wedge dx^{i_p}])\wedge((df_x)^*[(\beta\circ f)(x)dx^{j_1}\wedge\ldots\wedge dx^{j_q}])$$

So here I got stuck - I don't really know how to move around $\alpha, \beta$ under $df^*$, to get close to the left hand side expression.

Thank you!~

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up vote 2 down vote accepted

I read your approach but I am giving another approach which might be helpful. Lets start with the definition. If $f:X\rightarrow Y$ is a map and $\omega$ is a $p$ form in $Y$ then its pull back is defined as $f^*(\omega)v=\omega (f_*v).,$ where $f_*v=v(f),$ for any $v\in T_pX.$ Therefore Expand (assuming $\omega$ is a p form and $\theta$ is a q form) the term (where $v_i$and $w_i$ are in $T_pX$ ) $f^*(\omega\wedge \theta)(v_1,\cdots ,v_p,w_1,\cdots ,w_q)=(\omega\wedge \theta)(f_*(v_1),\cdots ,f_*(v_p),f_*(w_1),\cdots ,f_*(w_q)).$ using the summation formula for wedge product. Which will be same as $\omega(f∗(v_1),⋯,f∗(v_p))\wedge \theta(f∗(w_1),⋯,f∗(w_q)).$ Which is (by definition) same as $f_*(\omega)\wedge f_*(\theta).$

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looks great, thanks tessellation. I wonder if my approach is wrong. – WishingFish Jul 31 '13 at 6:01
@WishingFish notice he is doing what I said in the other thread. That is what the $df_x^*$ intends. It might be good to look back at… or… where I focus more on the issue of how the pull-back replaces the coefficient with the composition of the old coefficient. – James S. Cook Jul 31 '13 at 6:08
It feels like physicists and geometers haven't met an agreement on many mathematical expressions yet, but I am really confused and frustrated by the different notation from the one I am introduced. Because I am constrained in the idea that if I want to use another definition, I shall prove its equivalence first... @JamesS.Cook – WishingFish Jul 31 '13 at 16:38
To be more specific, I am confused with tessellation's definition $f^*(\omega)v = \omega(f_*v)$ with my $f^*(\omega)v = (df_x)^*\omega[f(x)]$, @JamesS.Cook – WishingFish Jul 31 '13 at 16:59
This answer may be misleading...see here… – Eric Auld Dec 15 '13 at 20:54

By definition, $$f^*(w \wedge \theta) = (df_x)^*(\omega\wedge\theta)[f(x)].$$

Page 162: If $\omega$ is a $p$-form and $\theta$ is a $q$-form, then $p+q$ form $$(\omega \wedge \theta) (x) = \omega(x) \wedge \theta(x).$$

Therefore, $$(df_x)^*(\omega\wedge\theta)[f(x)] = (df_x)^*\omega[f(x)] \wedge \theta[f(x)].$$

Page 159: $A^*$ is linear and that $$A^*(T \wedge S) = A^*T \wedge A^*S.$$

Therefore, $$(df_x)^*\omega[f(x)] \wedge \theta[f(x)] = (df_x)^*\omega[f(x)] \wedge (df_x)^*\theta[f(x)].$$

That is exactly $$(f^*w) \wedge (f^* \theta).$$

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