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 $T^*(dx_1\wedge\dots\wedge dx_n) = (\det T)dx_1\wedge\dots\wedge dx_n$.)


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