The <wedge product of differential forms> is the alternating tensor product
$$
(\alpha\wedge\beta)(v_1,\ldots,v_{p+q})
=\frac1{p!q!}\sum_{\sigma\in S_{p+q}}\operatorname{sgn}(\sigma)
\alpha(v_{\sigma(1)},\ldots,v_{\sigma(p)})
\beta(v_{\sigma(p+1)},\ldots,v_{\sigma(p+q)}).
$$
For nonzero $\eta\in\Lambda^{n-1}((\mathbb R^n)^*)$, choose a volume form $\mu$. There is a nonzero vector $v$ with $\eta=\iota_v\mu$. Extend $v$ to a basis and use its dual coframe; then $\eta$ is a scalar multiple of $e^2\wedge\cdots\wedge e^n$, hence equals $\xi\wedge\eta_0$. The zero form is immediate.

To integrate a top form on a compact oriented $n$-manifold, choose a finite oriented atlas and a subordinate <partition of unity>; integrate each compactly supported coordinate expression and sum. A smooth map pulls forms back by
$$
(\phi^*\alpha)_x(v_1,\ldots,v_p)=
\alpha_{\phi(x)}(d\phi_xv_1,\ldots,d\phi_xv_p).
$$
If $F^*\omega=\omega$ for a nowhere-zero top form, $F$ preserves its orientation. The change-of-variables theorem gives
$$
\int_M(h\circ F)\omega
=\int_MF^*(h\omega)
=\int_Mh\omega.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 115 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-115/2/a/solution