$\textbf{Theorem 2.9}$ Let $(M,g)$ be a smooth, compact Riemannian $n$-manifold.
(i) For any integers $j \geq 0$ and $m \geq 1$, any real number $q \geq 1$, and any real number $p$ such that $1 \leq p < \frac{nq}{n - mq}$, the embedding of $H_{j+m}^q(M)$ in $H_j^p(M)$ is compact. In particular, for any $q \in [1,n)$ real and any $p \geq 1$ such that $\frac{1}{p} > \frac{1}{q} - \frac{1}{n}$, the embedding of $H_1^q(M)$ in $L^p(M)$ is compact.
$\textbf{Proof.}$ Since $M$ is compact, $M$ can be covered by a finite number of charts $$(\Omega_s,\varphi_s)_{s=1,\cdots,N}$$ such that for any $s$ the components $g_{ij}^s$ of $g$ in $(\Omega_s,\varphi_s)$ satisfy $$\frac{1}{2} \delta_{ij} \leq g_{ij}^s \leq 2 \delta_{ij}$$ as bilinear forms.
Let $(\eta_s)$ be a smooth partition of unity subordinate to the covering $(\Omega_s)$. Given $(u_m)$ a bounded sequence in $H_1^q(M)$, and for any $s$, we let $$u_m^s = (\eta_s u_m) \circ \varphi_s^{-1}$$
Clearly, $(u_m^s)$ is a bounded sequence in $H_{0,1}^q(\varphi_s(\Omega_s))$ for any $s$. By lemma $2.5$ one then gets that a subsequence $(u_m^s)$ of $(u_m^s)$ is a Cauchy sequence in $L^p(\varphi_s(\Omega_s))$. Let $(u_m)$ be a a subsequence of $(u_m)$ chosen that for any $s$, $(u_m^s)$ is a Cauchy sequence in $L^p(\varphi_s(\Omega_s))$. Coming back to the inequalities satisfied by the $g_{ij}^s$'s, one easily gets that for any $s$, $(\eta_s u_m)$ is a Cauchy sequence in $L^p(M)$. But for any $m_1$ and $m_2$, $$||u_{m_2} - u_{m_1}||_p \leq \sum_{s=1}^N ||\eta_s u_{m_2} - \eta_s u_{m_1}||_p$$ where $||\cdot||_p$ stands for the $L^p$-norm. Hence, $(u_m)$ is a Cauchy sequence in $L^p(M)$. This proves the result. $\square$
I would like to know how the inequalities satisfied by the $g_{ij}^s$'s are used in the proof. I know that the inequalities satisfied by the $g_{ij}^s$'s in the sense of bilinear forms is in the sense of what is proved here, that the norm of $\nabla^k u$ is given by
$$|\nabla^k u| = g^{i_1j_1} \cdots g^{i_kj_k} (\nabla^k u)_{i_1\cdots i_k} (\nabla^k u)_{j_1\cdots j_k}$$
and that the norm of $L^p(M)$ is defined by
$$||u||_p = \left( \int_M |u|^p dv(g) \right)^{\frac{1}{p}}$$
I can not see how the inequalities satisfied by the $g_{ij}^s$'s, ensure that for any $s$, $(\eta_s u_m)$ is a Cauchy sequence in $L^p(M)$ once that the $L^p$-norm over $M$ does not depend of the metric except for the volume element, but the volume element will be bounded because $g_{ij}^s$ are smooth on $\Omega_s$ and we can suppose that they are bounded in these domains because we can work with a subdomain $\Omega_s'$ compactly contained in $\Omega_s$ to ensure that $g_{ij}^s$ are bounded in $\Omega_s'$, we can restrict the charts to $\Omega_s'$ to work with $\Omega_s'$, then I really can not see how the inequalities satisfied by the $g_{ij}^s$'s, ensure that for any $s$, $(\eta_s u_m)$ is a Cauchy sequence in $L^p(M)$.
Thanks in advance!
