Formula for the Pullback of an $k$ form to $\mathbb{R}^{k}$

Question

1. $\phi^* \left( \sum_{i = 1}^n f_i dx_i \right) = \sum_{i = 1}^n f_i \circ \phi \frac{\partial \phi^i}{\partial t} dt$

2. Let $\omega = \sum_{i = 1}^n f_i x_1 \cdots \hat{x_i} \cdots x_n$ be a $n-1$ for m on $\mathbb{R}^n$ (an element of $\Omega^{n-1}( \mathbb{R}^n)$. Let $\phi : \mathbb{R}^{n-1} \rightarrow \mathbb{R}^{n}$ be a smooth map. Then
   $$ \phi^* (\omega) = \text{det} \left( \begin{bmatrix}
   f_1 \circ \phi & \cdots & f_n \circ \phi \\
   \frac{\partial \phi^1}{\partial x_1} & \cdots & \frac{\partial \phi^1}{\partial x_{n-1} } \\
   \vdots & & \vdots\\
   \frac{\partial \phi^n}{\partial x_1} & \cdots & \frac{\partial \phi^n}{\partial x_{n-1} } \\
   \end{bmatrix} \right) $$

3. Let $\phi : \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a map. The pullback of $f dx_1 \cdots dx_n \in \Omega^n ( \mathbb{R}^n)$ by $\phi$ is $f \circ \phi \text{det}(D(\phi))$.
   $$ \phi^* (\omega) = f \circ \phi \ \text{det} \left( \begin{bmatrix}
   \frac{\partial \phi^1}{\partial x_1} & \cdots & \frac{\partial \phi^1}{\partial x_{n} } \\
   \vdots & & \vdots\\
   \frac{\partial \phi^n}{\partial x_1} & \cdots & \frac{\partial \phi^n}{\partial x_{n} } \\
   \end{bmatrix} \right) $$

Can we generalize these formulas to find a formula for the pullback of a $k$ form in $\Omega^k (\mathbb{R}^n)$ to $\mathbb{R}^k$? I want a formula involving only multilinear operations that take $\frac{\partial \phi^i}{\partial x_j}$ and $f_i \circ \phi$ as inputs.

Answer 1

If I understood your question correctly, we can discuss first how to pullback a single wedge of $dx$s, then generalize by linearity. Let $\omega = dx_I = dx_{i_1}\wedge \dots \wedge dx_{i_k}$ and $\phi(u_1,\dots,u_m):\mathbb{R}^m\rightarrow\mathbb{R}^n$ the smooth map, then

$$\phi^*\omega =\bigwedge_{1\leq j\leq k} \sum_{q_j=1}^m \frac{\partial\phi_{i_j}}{\partial u_{q_j}}du_{q_j}=\sum_{1\leq q_1,\dots,q_k\leq m} \prod_{j=1}^k \frac{\partial\phi_{i_j}}{\partial u_{q_j}}du_{q_1}\wedge \dots du_{q_k}$$

Now for every $Q=(q_1,\dots,q_k)$ if two $q$s are the same we get $0$. Otherwize, there exists a unique increasing multi-index $J$ and a permutation $\sigma$ such that $\sigma(J)=Q$. If we sum according to $J$s and $\sigma$s, we get

$$\sum_{J}\sum_{\sigma \in S_k} \prod_{j=1}^k \frac{\partial\phi_{i_j}}{\partial u_{q_{\sigma(j)}}}du_{q_{\sigma(1)}}\wedge \dots du_{q_{\sigma(k)}} =\sum_{J}\sum_{\sigma \in S_k} \prod_{j=1}^k \frac{\partial\phi_{i_j}}{\partial u_{q_{\sigma(j)}}}\text{sgn}\sigma \ du_J$$ Note that by the permutation property of determinant we get the determinant created by the $i_1,\dots,i_k$ rows and $j_1,\dots,j_k$ columns of $D\phi$. We can write this as

$$\sum_J \det \frac{\partial \phi_I}{\partial u_J} du_J$$ Remember $J$ runs over all increasing multi-indexes of length $k$ from $1,\dots,m$.