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

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.

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 $$1,\dots,m$$.