# Fill in the hole for the proof for $f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$

My proof has a hole there, wonder if anyone can help me fill it in?

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

By definition,

$$f^*w(x) = (df_x)^*w[f(x)].$$

So I understand $w[f(x)]$ as the $p$-tensor eats a $p$-vector $f(x)$.

The main problem is, that I don't think the function $f$ whose range is a $p$-dimensional vector space can distribute into $w$ and $\theta$ respectively, as what I hope to reach after the question marks.

In order to make $w \otimes \theta$ makes sense on the left hand side, we want $w$ to eat an $m$-vector and $\theta$ to eat an $n$-vector, where $m+n=p$. Hence, how can each $w$ and $\theta$ eats a $p$-vector $f$?

\begin{eqnarray} f^*(w \wedge \theta) & = &(df_x)^* (w \wedge \theta) (f)\\ & = &(df_x)^* \operatorname{Alt}(w \otimes \theta) (f)\\ & = &(df_x)^* \frac{1}{p!}\sum_{\pi \in S_p}(-1)^\pi(w \otimes \theta)^\pi (f)\\ & = & ???\\ & = &\frac{1}{p!}\sum_{\pi \in S_p}(-1)^\pi((df_x)^*w(f) \otimes (df_x)^*\theta(f))^\pi\\ & = & \operatorname{Alt}[(df_x)^*w(f) \otimes (df_x)^*\theta(f)]\\ & = &(df_x)^*w(f) \wedge (df_x)^*\theta(f)\\ & = &(f^*w) \wedge (f^* \theta). \end{eqnarray}

• I'm confused, I would think $f$ is a function so $f^* \omega$ is a pull-back under $f$ of a form in the range of $f$ to a new form $f^* \omega$ in the domain of $f$. To understand it, I would evaluate on an appropriate tuple of vectors at a given point and work through the push-forward feeding into the range form... maybe I have a different problem in mind? Commented Jul 29, 2013 at 23:26
• Hi @JamesS.Cook, I think you are right, and I think my understanding to $w[f(x)]$ is wrong. How does $w[f(x)]$ work? Commented Jul 29, 2013 at 23:28
• I wrote up an answer to a question I think you want to ask... Commented Jul 29, 2013 at 23:49

Suppose $\omega$ and $\phi$ are one-forms on $N$ and $f: M \rightarrow N$ is a function. The form $\omega \wedge \phi$ is a two-form on $N$ and the pull-back under $f$ would be denoted $f^*(\omega \wedge \phi)$ it is a two-form on $M$. In particular, the definition can be made precise in terms of two vectors at $p \in M$. \begin{align} f^*(\omega \wedge \phi)(v,w) &= (\omega \wedge \phi) (d_pf(v),d_p(w)) \\ &= (\omega \otimes \phi- \phi \otimes \omega) (d_pf(v),d_p(w)) \\ &= (\omega \otimes \phi)(d_pf(v),d_p(w))- (\phi \otimes \omega) (d_pf(v),d_p(w)) \\ &= \omega (d_pf(v) \phi(d_p(w))- \phi (d_pf(v))\omega (d_p(w)) \\ &= (f^*\omega)(v) (f^*\phi)(w)- (f^*\phi)(v)(f^*\omega)(w) \\ &= (f^*\omega \otimes f^*\phi)(v,w)- (f^*\phi \otimes f^*\omega)(v,w) \\ &= (f^*\omega \otimes f^*\phi - f^*\phi \otimes f^*\omega)(v,w) \\ &= (f^*\omega \wedge f^*\phi)(v,w) \end{align} This holds for all $v,w \in T_pM$ hence the identity $f^*(\omega \wedge \phi) = f^*\omega \wedge f^*\phi$ holds. This argument can be generalized for $\omega$ a $k$-form and $\phi$ a $r$-form, you just have to sort through the antisymmetrization. Looks like you may already have those tools ready to use, so I hope this case suffices to get you started.
• Dear James - thank you. You look right, but you are not using the definition I was given... > $$f^*w(x) = (df_x)^*w[f(x)].$$ :-P Commented Jul 30, 2013 at 0:48
• @WishingFish I think I am... $(df_x)^*$ is the pull-back of the $df_x$ map. How does that work? Well, $(df_x)^*(w[f(x)])(v) = w[f(x)](df_x(v))$. The $f(x)$ in the $w[f(x)]$ means you are using the form $w$ at $f(x)$. I suppressed the point-dependence in my answer. Commented Jul 30, 2013 at 1:36
• Hi James, that is exactly where confuses me - I never get what $w[f(x)]$ really is. My guess is that $w$ is a $p$-tensor, and $f(x)$ fed it assuming $f$ evaluated at $x$ is a $p$-tuple vector. So what do you mean by "The $f(x)$ in the $w[f(x)]$ means you are using the form $w$ at $f(x)$"? I think you are right and I was wrong, so I am trying to get your point! I used a new question, here: math.stackexchange.com/questions/455322/definition-of-pullback Commented Jul 30, 2013 at 1:56
• @WishingFish right, I do believe that notation indicates using the form $w$ at the point $f(x)$. I'll look at your other question shortly... Commented Jul 30, 2013 at 4:51