For the stated flat metric and orientation,
$$
*dx^i=(-1)^{i-1}
dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n.
$$
Because the metric coefficients and the coordinate one-forms are constant, the <Hodge Laplacian> acts coefficientwise:
$$
\Delta(\alpha_i\,dx^i)=(\Delta\alpha_i)\,dx^i.
$$
Thus a harmonic one-form $\alpha=\sum_i\alpha_i dx^i$ has harmonic coefficient functions. Every harmonic function on the compact connected torus is constant by the <maximum principle for harmonic functions>. Hence
$$
\mathcal H^1(T^n)
=\operatorname{span}_{\mathbb R}\{dx^1,\ldots,dx^n\}
\cong\mathbb R^n.
$$
The <Hodge decomposition theorem> now gives
$$
H^1_{\mathrm{dR}}(T^n)\cong\mathbb R^n.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 4 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-115/4/e/solution