I'm looking to prove that the pullback of a smooth function distributes over wedge product, i.e. $$\varphi^*(\omega \wedge \eta) = \varphi^* \omega \wedge \varphi^* \eta. $$

Here the question is answered. This is an excerpt from the answer:

$f^*(\omega\wedge \theta)(v_1,\cdots ,v_p,w_1,\cdots ,w_q)=(\omega\wedge \theta)(f_*(v_1),\cdots ,f_*(v_p),f_*(w_1),\cdots ,f_*(w_q)).$ 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 $f_*(\omega)\wedge f_*(\theta).$

The part I have trouble with is the third line. $\omega(f∗(v_1),⋯,f∗(v_p))$ and $\theta(f∗(w_1),⋯,f∗(w_q))$ are real numbers, so how can we wedge them? There might be some abuse of notation here I'm not familiar with.

When I try to show $\varphi^*(\omega \wedge \eta) = \varphi^* \omega \wedge \varphi^* \eta $, my attempt breaks down when I try to simplify $(\varphi^*\omega \wedge \varphi^* \eta)(v_1, \dotsc, v_{p+q})$. Should I try to work with decomposable forms?

  • $\begingroup$ The wedge product of two $0$-forms (aka scalars or functions, depends on whether you look at a single point or not), is just the ordinary (pointwise) product. $\endgroup$ – Daniel Fischer Dec 15 '13 at 20:45
  • $\begingroup$ @DanielFischer A ha, and where do we get $$(\omega \wedge \theta)(f_*(v_1), \dotsc, f_*(v_{p+q}))=\omega(f_*(v_1), \dotsc, f_*(v_p))\wedge \theta(f_*(w_{1}), \dotsc, f_*(w_q))?$$ $\endgroup$ – Eric Auld Dec 15 '13 at 20:48
  • $\begingroup$ I don't think we do. $$(dx\wedge dy)\left(\frac{\partial}{\partial y},\frac{\partial}{\partial x}\right) \neq dx\left(\frac{\partial}{\partial y}\right) \cdot dy\left(\frac{\partial}{\partial x}\right).$$ We must permute and sum. $\endgroup$ – Daniel Fischer Dec 15 '13 at 20:53
  • $\begingroup$ OK, good. At least I'm not thinking about it wrong. How would you show $$(\omega \wedge \eta) (dF(v_1), \dotsc, dF(v_{k+l})) = (F^*\omega \wedge F^*\eta) (v_1, \dotsc, v_{k+l})?$$ Everywhere I'm seeing has the proof omitted, as if it's trivial. Should I perhaps try so show on decomposable forms $(\omega_1 \wedge \dotsb \wedge \omega_k) \wedge (\eta_1 \wedge \dotsb \wedge \eta_l)$? But that doesn't seems promising because I don't know how to take $\varphi^*((\omega_1 \wedge \dotsb \wedge \omega_k) \wedge (\eta_1 \wedge \dotsb \wedge \eta_l))$. $\endgroup$ – Eric Auld Dec 15 '13 at 20:59
  • $\begingroup$ @DanielFischer Oh, I think I'm seeing it. We have to permute and sum, but on each term, they are the same. $\endgroup$ – Eric Auld Dec 15 '13 at 21:08

The answer mentioned should have noted that we can say: \begin{equation} \begin{aligned} \varphi^*(\omega)(v_1, \dotsc, v_k)&\varphi^*(\eta)(v_{k+1}, \dotsc, v_{k+l}) \\&= \omega(dF^*(v_1), \dotsc, dF^*(v_k))\eta(dF^*(v_{k+1}), \dotsc, dF^*(v_{k+l})) \end{aligned} \end{equation} and we can extend this termwise result over the $\text{Alt}$ map.


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