# pullback of wedge product of 1-forms

I would like some help understanding one of the equalities in the following proof:

Let $$f:\mathbb{R}^n\rightarrow \mathbb{R}^m$$ be a differentiable function. If $$\alpha^1,\cdots ,\alpha^k$$ are $$1$$-forms in $$\mathbb{R}^m$$, prove that $$f^*(\alpha^1\wedge\cdots \wedge\alpha^k)=f^*(\alpha^1)\wedge\cdots \wedge f^*(\alpha^k)$$

Proof:

\begin{align}f^*(\alpha^1\wedge\cdots \wedge\alpha^k)(v_1,\cdots ,v_k)&=(\alpha^1\wedge\cdots \wedge \alpha^k)(df(v_1),\cdots ,df(v_k))\\&=\det(\alpha^i(df(v_j))\\&=\det(f^*\alpha^i(v_j))\\ &=f^*(\alpha^1)\wedge\cdots \wedge f^*(\alpha^k)(v_1,\cdots ,v_k) \end{align}

My question is, why is the second equality true. I know that the wedge product can be defined using determinants, but if we are dealing with differentials what does $$\alpha^i(df(v_j))$$ actually mean and why would we write out that way?

The definition I am given for the wedge product is:

$$\alpha^1\wedge\cdots \wedge \alpha^k=\sum_{i_1<...

So as always in differential geometry, you have to keep track of what type your objects are.

Here notice that $$df$$ is a linear function $$\mathbb{R}^n \rightarrow \mathbb{R}^m$$ so it spits vectors of $$\mathbb{R}^m$$.

On the other hand $$\alpha^i$$ is a 1-forms so it takes a vector in $$\mathbb{R}^m$$ as argument. Thus it shoud be clear (at least formally) what $$\alpha^i(df(v_j))$$ means.

You can show (by induction for example, knowing that $$(\alpha \wedge \beta) (V_1,V_2)=\alpha(V_1)\beta(V_2)-\alpha(V_2)\beta(V_1)$$ for any vectors, that for any $$V_1, \ldots,V_k$$ in $$\mathbb{R}^m$$, $$\alpha^1 \wedge \ldots \wedge \alpha^k (V_1, \ldots ,V_k)=det(\alpha^i(V_j)).$$

(To me this was pretty much the definition of wedge products).

It really just comes from this result: Let $$\alpha^1,\cdots\alpha^k$$ be linear functionals on a vector space $$V$$, and $$v_1,\cdots v_k\in V.$$ Then, \begin{align*} (\alpha^1\wedge\cdots\wedge \alpha^k)(v_1,\cdots, v_k)&=A(\alpha^1\otimes\cdots\otimes\alpha^k)(v_1,\cdots, v_k)\\ &=\sum\limits_{\sigma\in S_k}(\text{sgn }\sigma)\alpha^1(v_{\sigma(1)})\cdots\alpha^k(v_{\sigma(k)})\\ &=\det [\alpha^i(v_j)] \end{align*} Here, $$A$$ is the alternating operator $$Af=\sum\limits_{\sigma\in S_k}(\text{sgn }\sigma)\sigma f,$$ and, by definition, $$f\wedge g=\frac{1}{k!\ell!}A(f\otimes g)$$ for a $$k$$-form $$f$$ and $$\ell$$-form $$g$$. By $$\sigma f$$, we mean $$(\sigma f)(v_1,\cdots, v_k)=f(v_{\sigma(1)},\cdots, v_{\sigma(k)}).$$
Also, since $$\alpha^i$$ is a one-form and $$df(v_j)$$ is a tangent vector, $$\alpha^i(df(v_j))$$ is perfectly well-defined.