0$ such that, if $0\leq\Jg\leq \Jg_0$, then there exists $\rho_1 >0$ such that, for any cube $\Lambda\subset \zz^d$, \begin{equation}\label{4.13} W_1^\Jg\geq \rho_1\;. \end{equation} \end{proposition} \begin{remark}\label{Remark4.2}:\\ {\rm This proof could probably work also for other boundary conditions (periodic, free,...). But the condition that $|\Jg|$ is smaller than some $\Jg_0$ may depend on the choice of these boundary conditions. This will be analyzed in \cite{He5n}.\\ Let us also observe that the condition on the sign of $\Jg$ is unimportant in the proof. Sokal's approach is more related to ferromagnetic systems.} \end{remark} \section{The case $\Jg$ small~: an $1$-dimensional approach } In the preceding section, we were suggesting that in the highly non convex case there was a problem to find a strictly positive lower bound uniformly with respect to the parameters $x_k$ ($k\sim j$) for the operator $w_j^{(1;\Jg)}$ seen as a family of operators on $L^2(\rz)$. We presented in Section~\ref{Section2} one way which was efficient in the case when the non convexity was not too strong. In Section~\ref{Section4}, we explained the trick used by Bach-Jecko-Sj\"ostrand \cite{BaJeSj} consisting in the use of the non diagonal terms for controlling the perturbation and also eliminating any assumption of weak convexity.\\ The study in this section will present another way, less perturbative in spirit, for treating the difficulty to have uniform control with respect to $\Jg$. \\ The problem is indeed the following. We want to analyze \begin{equation}\label{5.1} w^{(1;\Jg)}(t,\frac{d}{dt}):= - \frac{d^2}{d t^2} + \frac 14 \left(\phi'(t) + 2\Jg d\; t - 2 \Jg \sum_{0\sim k} x_k\right)^2 + \frac 12 (\phi''(t) + 2\Jg\; d)\;. \end{equation} The criterion given in Section~\ref{Section2} is (when taking $\epsilon =1$) is actually related to the assumption that the operator $$ p^{(1;\Jg)}(t,\frac{d}{dt}):=w^{(1;\Jg)}(t,\frac{d}{dt}) - 2\Jg d\;, $$ that is, \begin{equation}\label{5.2} p^{(1;\Jg)}(t,\frac{d}{dt}):= - \frac{d^2}{d t^2} + \frac 14 \left( \phi'(t) + 2\Jg d\; t -2 \Jg \sum_{0\sim k} x_k\right)^2 + \frac 12 (\phi''(t) - 2 \Jg d)\;, \end{equation} is uniformly stricty positive (with respect to $x_k$ and $\Jg$).\\ The previous criteria correspond to taking the lower bound obtained by forgetting the term $ \frac 14 \left( \phi'(t) + 2\Jg d\; t -2 \Jg \sum_{0\sim k} x_k\right)^2$ and comparing with the lowest eigenvalue of $s(t,\frac{d}{dt})$ but this works only when this last operator is strictly positive. This is not satisfactory for $\Jg$ small. In this regim, it is sufficient (the two operators differ from $ 2 \Jg d$), to consider $w^{(1;\Jg)}(t, \frac{d}{dt})$ and, changing slightly the various parameters, to analyze the following family depending on $\Jg$ and $\alpha$: \begin{equation}\label{5.3} w(t,\frac{d}{dt}, \alpha, \Jg):= - \frac{d^2}{d t^2} + \frac 14\left ( \phi'(t) + 2\Jg d\; t -\alpha\right )^2 + \frac 12 (\phi''(t) + 2\Jg d)\;, \end{equation} with \begin{equation}\label{5.4} \alpha= -2 \Jg \sum_{0\sim k} x_k\;. \end{equation} We will then analyze the uniform strict positivity of this family with respect to $(\alpha,\Jg) \in \rz\times [-\Jg_0,+\Jg_0]$ which is better because we can forget the relation (\ref{5.4}). For any fixed pair $(\alpha,\Jg)$, the operator is strictly positive by Proposition \ref{Proposition2.1} and a standard perturbation argument permits to show (under assumptions $(h1)$, $(h2)$ and with $\delta\geq 1$) that the lowest eigenvalue of $w(t,\frac{d}{dt}, \alpha, \Jg)$ is continuous with respect to $\Jg$ and $\alpha$. \\ So it is a problem of uniformity as $\alpha = \pm \infty$. Let us treat our model where $\phi(t)= \phi^{TM}$ using a semi-classical approach. Changing of $\nu$, we can forget $\Jg$ although remembering that our argument has to be locally uniform in $\nu$.\\ So we have finally to verify the following \begin{lemma}:\\ There exists $\tau >0$ such that for $|\alpha| \geq\tau$ the operator \begin{equation} s(t,\frac{d}{dt}, \alpha):= - \frac{d^2}{d t^2} + \frac 14(\frac \lambda 3 t^3 + \nu t -\alpha )^2 + \frac 12 (\lambda t^2 +\nu)\end{equation} satisfies \begin{equation} s(t,\frac{d}{dt}, \alpha)\geq\frac 1\tau\; |\alpha|^{\frac 23}\;. \end{equation} \end{lemma} The problem is symmetric in $\alpha$. It is consequently sufficient to treat the case $\alpha >0$.\\ Let us introduce the scaling $t= \alpha^\frac 13 y$. We then obtain the following inequality \begin{equation} q(y,\frac{d}{dy}, \alpha):= - \alpha^{-\frac 23}\frac{d^2}{d y^2} + \frac {\alpha^2}{4} (\frac \lambda 3 y^3 + \nu \alpha^{-\frac 23} y - 1 )^2 + \frac 12 \alpha^\frac 23 (\lambda y^2 +\alpha^{-\frac 23}\nu)\geq\frac 1\tau |\alpha|^\frac 23\; \end{equation} to be proved. In order to get a more semi-classical form we can write it as \begin{equation} q(y,\frac{d}{dy}, \alpha):=\alpha^2 {\tilde q}(y,h\frac{d}{dy};h;\mu)\;, \end{equation} with \begin{equation} {\tilde q}(y,h\frac{d}{dy};h;\mu):=\left( - h^2 \frac{d^2}{d y^2} + \frac {1}{4} (\frac \lambda 3 y^3 + \mu y - 1 )^2 + \frac 12 h (\lambda y^2 +\mu)\right)\;, \end{equation} where \begin{equation} h = \alpha^{-\frac 4 3}\;;\; \mu = \nu \alpha^{- \frac 23}= \Og (h^\frac 12)\;. \end{equation} This is a non degenerate one-well semi-classical problem near the point $y=1$ and this leads, for the operator ${\tilde q}(y,h\frac{d}{dy};h;\mu)$, to a lowest eigenvalue which is asymptotically equal to $ h \lambda$. We finally have obtained a behavior like $\lambda \alpha^\frac 23$ for the lowest eigenvalue of $q (y,\frac{d}{dy}, \alpha)$ as $\alpha \ar +\infty$ uniformly with respect to $\nu$ and consequently of $s(t,\frac{d}{dt},\alpha)$. This shows the lemma.\\ We have consequently given an alternative proof of Proposition \ref{Proposition4.1}, in the case of the toy model $\phi=\phi^{TM}$. \begin{remark}\label{remark5.2}:\\ {\rm We shall discuss the uniformity of the argument with respect to other boundary conditions in \cite{He5n}. The price to pay is that the estimates given in this section will be much more implicit.} \end{remark} \section{Conclusion} As we have seen, the lowest eigenvalue of the Witten Laplacian plays an important role. A natural question is the understanding of the phase transition through this approach. The answer is not clear. Let us just observe here that the results obtained on the transition of phase by other methods like infrared estimates or Peierls argument (see \cite{GlJa} for a presentation of these methods) will give an upper bound on the splitting between the two lowest eigenvalues of the Witten Laplacian on the functions or an upper bound of the smallest eigenvalue of the Witten Laplacian on the $1$-forms.\\ We recall that in the infrared estimates (see \cite{He6} for a recent presentation and a bibliography) one gets (but under the condition $d\geq 2$), a uniform\footnote{with respect to $\Lambda$} lower bound $\delta_0>0$ of the mean value of the variance $\Cov(f,f)$ of the function $x\mapsto f(x)=\frac{1}{|\Lambda|} (\sum_{\j\in \Lambda} x_j)$ with respect to the measure $\exp -\Phi(X)\; dX$ and this gives an upper bound of the smallest eigenvalue, and consequently of the splitting, by $\frac{1}{\delta_0 |\Lambda|}$.\\ \noindent {\bf Acknowledgements~:} \\ We are grateful to V.~Bach, T.~Jecko and J.~Sj\"ostrand for useful discussions (and in particular for communicating before publication the results of \cite{BaJeSj}). Many thanks to T.~Bodineau for discussions around possible applications to logarithmic Sobolev inequalities and to J.~Johnsen for useful comments on various versions of these notes.\\ \noindent We thank also the Fields Institute in Toronto where one part of this work was done and the European Union which partially supported this research through the TMR Programme FMRX-CT 960001 of the European Commission - Network Postdoctoral training programme in partial differential equations and application in quantum mechanics-.\\ \begin{thebibliography}{99} \bibitem [{AA}]{AA} A.V.~Antoniouk, A.V.~Antoniouk~: \newblock Decay of correlations and uniqueness of Gibbs lattice systems with nonquadratic interaction. \newblock {\it J. Math. 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