# Pullback of differential form

Given an open $$U\subset\mathbf{R}^n$$, we define the $$k$$-forms $$\Omega^k(U):=C^{\infty}(U,\Lambda^k((\mathbf{R}^n)^*))$$ (smooth functions from $$U$$ to the $$k$$-linear alternate forms).

Given a smooth $$\phi:\mathbf{R}^n\supset U\longrightarrow V\subset \mathbf{R}^m$$ and a $$k$$-form $$\alpha\in\Omega^k(V)$$, we can 'pull back' the $$k$$-form using $$\phi$$, to obtain a $$k$$-form $$\phi^*\alpha\in \Omega^k(U)$$, defined by

$$\phi^*\alpha(x)(v_1,\ldots,v_k)=\alpha(\phi(x))(d_x\phi(v_1),\ldots,d_x\phi(v_k)),$$ for all $$x\in U$$ and $$v_1,\ldots,v_k\in \mathbf{R}^n$$. ($$d_x\phi:T_xU\to T_{\phi(x)}V$$ is the tangent map.)

I hope my understanding of the pull-back so far is correct.

Now I am stuck on an easy example, which sets $$k=n=m$$ and asks to compute $$\phi^*\alpha$$ for $$\alpha=f\cdot dx_1\wedge \ldots \wedge dx_n$$ the volume form ($$f:V\to \mathbf{R}$$ smooth).

My try: let $$x\in U$$ and $$v_1,\ldots,v_n\in\mathbf{R}^n$$,

\begin{align*} (\phi^*\alpha)(x)(v_1,\ldots,v_n) &= \alpha(\phi(x))(d_x\phi(v_1),\ldots,d_x\phi(v_n)) \\ &= f(\phi(x))\cdot dx_1\wedge \ldots\wedge dx_n(\phi(x))(d_x\phi(v_1),\ldots,d_x\phi(v_n)) \end{align*}

I don't know what to do from this point on. I know the answer should be $$f(\phi(x))\det(d_x\phi)dx_1\wedge\ldots\wedge dx_n$$.

If $$T\colon\Bbb R^n\to\Bbb R^n$$ is a linear map, show that $$dx_1\wedge\dots\wedge dx_n(Tv_1,\dots,Tv_n) = (\det T) dx_1\wedge\dots\wedge dx_n(v_1,\dots,v_n).$$ (This is showing that $$T^*(dx_1\wedge\dots\wedge dx_n) = (\det T)dx_1\wedge\dots\wedge dx_n$$.)