my book on geometric analysis states that given two sets $ X_0, X_1 \subset \mathbb{R}^n$ and a diffeomorphism $\varphi$ between them, the diagram
is commutative. $\Omega(X_i)$ are the differential forms on $X_i$ and $\Theta_i:X_i \rightarrow \Omega(X_i)$ are the riesz isomorphisms defined by $\theta_i v(w)= (v | w)$.
Trying to verify this relation i get that the differential of $\varphi$ has to be self-adjoint as a necessary condition, that is $\partial_j \varphi^k $ = $\partial_k (\varphi^{-1})^j$. With matrices this means the transpose is the inverse, but something like $x\mapsto \lambda x$ would already be a counterexample. I get this out when checking for $((\varphi^{-1})^*\Theta_0)(e_j)(f_k)=(\Theta_1\varphi_*)(e_j)(f_k)$ (with $e_j, f_k$ basis vectors, because of linearity this is equivalent to the statement).
Am i wrong or could this be true? If im wrong, could someone post the correct calculation for me?
This statement is only true for isometries $\varphi:(X_0,g_0)\rightarrow (X_1,g_1)$ where $g_0,g_1$ are non-degenerate inner products on $X_0, X_1$ corresponding to $\theta_0,\theta_1$. Since $\varphi$ is a diffeomorphism, we have an isomorphism $(d\varphi)^{-1}_*:\mathcal{C}^\infty(X_1,\mathbb{R^n})\rightarrow\mathcal{C}^\infty(X_0,\mathbb{R^n})$ and the statement reads as follows \begin{align*} (d\varphi^{-1})^*\circ \theta_0\circ d\varphi^{-1}_*=\theta_1. \end{align*} Take any vector field $V_1\in \mathcal{C}^\infty(X_1,\mathbb{R^n})$. Then by computation we have \begin{align*} (d\varphi^{-1})^*\circ \theta_0\circ d\varphi^{-1}_*(V_1)=(d\varphi^{-1})^*\circ g_1((d\varphi^{-1})(V_1),\cdot)=g_1((d\varphi^{-1})(V_1),(d\varphi^{-1})(\cdot))\overset{!}{=}g_0(V_1,\cdot). \end{align*} That is the defining condition for $\varphi$ to be an isometry.