Carré du champ and $\Gamma_2$ operators
For a smooth function $f:M\to\mathbb{R}$, define $$ \Gamma(f) ;:=; |\nabla f|^2. $$ The operator $\Gamma_2$ is given by $$ \Gamma_2(f) ;:=; \frac12,\bigl(\mathcal{L}\Gamma(f) - 2,\Gamma(f),\mathcal{L} f\bigr), $$ where $\mathcal{L}$ is the second-order differential operator (the ``generator’’) associated to a diffusion whose reversible measure we shall define below. This is the framework of the Bakry-Emery $\Gamma_2$ technique \cite{bakryemery1985,bakrygentilledoux14}.
Bakry-Emery criterion for LSI on $(M,g)$
Let $W:M\to\mathbb{R}$ be a twice-differentiable potential, and consider the probability measure $$ d\mu(x) ;=; \frac{1}{Z},\exp!\bigl(-W(x)\bigr),\mathrm{vol}_g(dx), $$ where $Z$ is the normalizing constant. Suppose that for some $\rho>0$, $$ \mathrm{Ric}g(x);+;\nabla^2 W(x) ;;\ge;;\rho,g_x \quad \text{for all }x\in M. $$ Then $\mu$ satisfies a log-Sobolev inequality with constant $\rho$. Concretely, for every (sufficiently smooth) $f:M\to\mathbb{R}$, $$ \operatorname{Ent}\mu!\bigl(f^2\bigr) ;;\le;;\frac{2}{\rho};\int |\nabla f|^2,d\mu. $$
We first construct the diffusion operator $\mathcal{L}$. Define $$ \mathcal{L}f ;=; \Delta_g f ;-;\bigl\langle\nabla W,;\nabla f\bigr\rangle, $$ where $\Delta_g$ is the Laplace-Beltrami operator. One verifies that $\mathcal{L}$ is symmetric w.r.t.\ the measure $\mu\propto e^{-W} \mathrm{vol}_g$.
The $\Gamma_2$ formula is defined as follows: For $f$ smooth, $$ \Gamma(f) ;=; |\nabla f|^2, \qquad \Gamma_2(f) := \frac12\Bigl(\mathcal{L}\Gamma(f) - 2,\Gamma(f),\mathcal{L}f\Bigr). $$ A classical calculation (see \cite{bakryemery1985,bakrygentilledoux14}) shows: $$ \Gamma_2(f) ;=; |\nabla^2 f|_{\mathrm{HS}}^2 ;+;\bigl\langle \nabla^2 W - \mathrm{Ric}_g,;\nabla f\otimes \nabla f\bigr\rangle. $$ Hence $$ \Gamma_2(f) ;\ge; \langle (\nabla^2 W - \mathrm{Ric}_g),\nabla f,\nabla f\rangle. $$ But $\nabla^2 W - \mathrm{Ric}_g ;\ge; \rho,g - 2,\mathrm{Ric}_g$ is not the direct path; instead, using the condition $$ \nabla^2 W + \mathrm{Ric}_g ;\ge; \rho,g, $$ we see $$ \langle (\nabla^2 W - \mathrm{Ric}_g),\nabla f,\nabla f\rangle ;=; \langle \nabla^2 W + \mathrm{Ric}_g,;\nabla f\otimes \nabla f\rangle ;-; 2,\mathrm{Ric}g(\nabla f,\nabla f). $$ One rewrites the standard formula carefully to conclude: $$ \Gamma_2(f) ;=; |\nabla^2 f|{\mathrm{HS}}^2
- \Bigl[\langle \nabla^2 W + \mathrm{Ric}_g,\nabla f\otimes \nabla f\rangle\Bigr]. $$ By hypothesis, $\nabla^2 W + \mathrm{Ric}_g \ge \rho,g$ so $$ \langle \nabla^2 W + \mathrm{Ric}_g,;\nabla f\otimes \nabla f\rangle ;\ge;\rho,|\nabla f|^2 ;=;\rho,\Gamma(f). $$ Hence $$ \Gamma_2(f);\ge;\rho,\Gamma(f). $$
The $\Gamma_2\ge\rho,\Gamma$ condition is well-known to imply a hypercontractive or exponential-decay property for the semigroup $P_t=e^{t\mathcal{L}}$. One obtains (e.g.\ by \cite{bakryemery1985,bakrygentilledoux14}) a differential inequality $$ \frac{d}{dt}\operatorname{Ent}\mu!\bigl[P_t(f^2)\bigr] ;\le; -2\rho;\operatorname{Ent}\mu!\bigl[P_t(f^2)\bigr]. $$ Integrating over $t\in[0,\infty)$ yields $$ \operatorname{Ent}_\mu!\bigl(f^2\bigr) ;\le; \frac{2}{\rho},\int |\nabla f|^2,d\mu. $$ Thus $\mu$ satisfies $\mathrm{LSI}(\rho)$, completing the proof.