There is a proof I am currently working trough, where I don't really understand the use of the Poincaré inequality. The preliminaries which lead to my question are pretty long. I will try to explain the question giving all the details needed but also making it as short as possible.
The setup is: We have a Robertson Walker spacetime $V= I \times S$, where $I=[t_0, \infty) \subset \mathbb{R}$ with metric $$ g=-dt^2+R^2 \sigma $$ where $\sigma$ is a Riemannian metric on $S$ and $R$ a smooth function on $I$.
The wave equation is defined as
$$ \text{tr}_g (\nabla^2 u)=0$$
where $u:V \rightarrow M$ and $(M,h)$ is a Riemannian manifold.
We can also define energies (here $S_t=\{t\} \times S$):
$$e_t:= \int\limits_{S_t} \vert d_t u \vert^2+ R^{-2} \vert du \vert^2 \ dx$$
and
$$e^1_t:= \int\limits_{S_t} \vert dd_t u \vert^2+ R^{-2} \vert d^2u \vert^2 \ dx$$
Now, given initial values $\varphi, \psi$, where $u(t_0,.)=\varphi, \partial_t u(t_0,.)=\psi$, we obtain initial valied for the energies, $e_0, e^1_0$.
We define
$$z_0=R(t_0) \sqrt{e_0^1}$$
We further define
$$ M(a, z_0)= \int\limits_{z_0}^{\infty} \dfrac{1}{a+b\xi^3 } \ d \xi $$
where $a,b$ are some values depending on $e_0, R(t_0)$ and the curvatures of $S$ and $M$, in particular it depends on the initial data.
One can show that a wave map solution exists, if for the given initial data it holds that
$$M(a, z_0) > \int\limits_{t_0}^{\infty} R^{-1}(\tau) \ d \tau$$
Now there is a corollary stating that for given $R$, there exists an open set of initial values in $H^1 \times H^1$, s.t. there exists a wave map solution. To prove that, one defines
$$ N_{\lambda}= \int\limits_{\lambda}^{\infty} \dfrac{1}{1+z^3} \ dz$$
One then shows that it holds
$$M(a, z_0) > a^{-2/3} b^{-1/3}N_{\lambda_0} $$ where $\lambda_0 = \sqrt{\dfrac{e_0^1}{e_0}}$
From the definition of $a,b$ one can also show that $a^{-2/3} b^{-1/3} \geq C \dfrac{1}{e_0}$ , so $a^{-2/3} b^{-1/3} $ will be greater than any number for $e_0$ small enough.
Now my question is about the last part of the proof. There, it says
"Suppose, therefore, the ratio $\dfrac{e_0^1}{e_0}$ to be equal to some number $\lambda_0 \geq 0$, then $M(a, z_0)$ will be greater than any required number if $e_0$ is small enough. There exists an open set of data satisfying the indicated required conditions, obtained by first choosing $\lambda_0$ greater than some constant linked with the Poincaré inequality of the manifold $(S, \sigma)$."
Here, I don't really know how to use this inequality. If I could have some sort of inequality
$$\Vert u \Vert_{L^2} \leq C_p \Vert \nabla u \Vert_{L^2}$$
This means we could somehow estimate
$$\dfrac{e_0}{e_0^1} \leq C_p$$
And therefore
$$\lambda_0 \geq 1/C_p$$ so $$ N_{\lambda_0} \leq N_{1/C_p} $$
Then we would get
$$M(a, z_0) > a^{-2/3} b^{-1/3}N_{\lambda_0} \geq C \dfrac{1}{e_0} N_{1/C_p} $$
And now choosing $e_0$ small enough, $M(a, z_0)$ would get arbitrarily large.
Is my idea here correct?
And if yes, how can I do that on arbitrarily complete $S$?