[proofplan] We first identify the carré du champ of the weighted Laplacian with the squared gradient norm, so that $\Gamma_2(f)$ has the concrete form $\frac{1}{2}L|\nabla f|^2-g(\nabla f,\nabla Lf)$. We then expand $L=\Delta-g(\nabla V,\nabla\cdot)$ in this expression. The Laplace-Beltrami part is exactly the classical Bochner identity, while the drift terms are computed by the product rule for the Levi-Civita connection and reduce to $\nabla^2V(\nabla f,\nabla f)$. Combining this drift contribution with $\operatorname{Ric}(\nabla f,\nabla f)$ gives the Bakry-Émery tensor. [/proofplan]

[step:Compute the carré du champ of the weighted Laplacian] Let $f,h\in C^\infty(M)$. The operator $L:C^\infty(M)\to C^\infty(M)$ is defined by $Lu=\Delta u-g(\nabla V,\nabla u)$ for $u\in C^\infty(M)$. By the product rule for the Laplace-Beltrami operator and the derivation property of the vector field $\nabla V$, \begin{align*} L(fh)=fLh+hLf+2g(\nabla f,\nabla h). \end{align*} Therefore the definition of $\Gamma$ gives \begin{align*} \Gamma(f,h)=g(\nabla f,\nabla h). \end{align*} In particular, \begin{align*} \Gamma(f)=|\nabla f|^2. \end{align*} [/step]

[step:Rewrite $\Gamma_2(f)$ in terms of $L|\nabla f|^2$ and $\nabla Lf$]Using the preceding identity with $h=f$ and with $h=Lf$, the definition of $\Gamma_2$ gives \begin{align*} \Gamma_2(f)=\frac{1}{2}L|\nabla f|^2-g(\nabla f,\nabla Lf). \end{align*} Expanding $Lf=\Delta f-g(\nabla V,\nabla f)$ and $L|\nabla f|^2=\Delta|\nabla f|^2-g(\nabla V,\nabla|\nabla f|^2)$, we obtain \begin{align*} \Gamma_2(f)=\frac{1}{2}\Delta|\nabla f|^2-g(\nabla f,\nabla\Delta f)-\frac{1}{2}g(\nabla V,\nabla|\nabla f|^2)+g(\nabla f,\nabla g(\nabla V,\nabla f)). \end{align*}[/step]

[guided]The definition of $\Gamma_2$ is algebraic, but for this operator it becomes a differential identity. Since we already proved $\Gamma(f,h)=g(\nabla f,\nabla h)$, we have $\Gamma(f)=|\nabla f|^2$ and \begin{align*} \Gamma(f,Lf)=g(\nabla f,\nabla Lf). \end{align*} Because $\Gamma_2(f)=\frac{1}{2}L\Gamma(f)-\Gamma(f,Lf)$ by symmetry in the diagonal case, this gives \begin{align*} \Gamma_2(f)=\frac{1}{2}L|\nabla f|^2-g(\nabla f,\nabla Lf). \end{align*} Now substitute the actual operator. The weighted Laplacian is the Laplace-Beltrami operator plus the first-order drift term $-g(\nabla V,\nabla\cdot)$. Hence \begin{align*} L|\nabla f|^2=\Delta|\nabla f|^2-g(\nabla V,\nabla|\nabla f|^2). \end{align*} Also, \begin{align*} Lf=\Delta f-g(\nabla V,\nabla f). \end{align*} Taking the gradient of this last identity and pairing with $\nabla f$ gives \begin{align*} g(\nabla f,\nabla Lf)=g(\nabla f,\nabla\Delta f)-g(\nabla f,\nabla g(\nabla V,\nabla f)). \end{align*} Substituting both formulas into the expression for $\Gamma_2(f)$ yields \begin{align*} \Gamma_2(f)=\frac{1}{2}\Delta|\nabla f|^2-g(\nabla f,\nabla\Delta f)-\frac{1}{2}g(\nabla V,\nabla|\nabla f|^2)+g(\nabla f,\nabla g(\nabla V,\nabla f)). \end{align*} This separates the calculation into the usual Bochner part and the new drift correction.[/guided]

[step:Apply the classical Bochner identity to the Laplace-Beltrami part] By the classical Bochner formula for the Laplace-Beltrami operator, applied to the smooth function $f\in C^\infty(M)$ with the same sign convention for $\Delta$, one has (citing a result not yet in the wiki: Classical Bochner formula) \begin{align*} \frac{1}{2}\Delta|\nabla f|^2-g(\nabla f,\nabla\Delta f)=|\nabla^2f|^2+\operatorname{Ric}(\nabla f,\nabla f). \end{align*} Thus \begin{align*} \Gamma_2(f)=|\nabla^2f|^2+\operatorname{Ric}(\nabla f,\nabla f)-\frac{1}{2}g(\nabla V,\nabla|\nabla f|^2)+g(\nabla f,\nabla g(\nabla V,\nabla f)). \end{align*} [/step]

[step:Reduce the drift correction to the Hessian of $V$]Define the drift correction $D_f:M\to\mathbb R$ by \begin{align*} D_f=-\frac{1}{2}g(\nabla V,\nabla|\nabla f|^2)+g(\nabla f,\nabla g(\nabla V,\nabla f)). \end{align*} At an arbitrary point $p\in M$, use the Levi-Civita connection $\nabla$ and write $X=\nabla f$ and $Y=\nabla V$ as smooth vector fields near $p$. Metric compatibility gives \begin{align*} \frac{1}{2}Y(|X|^2)=g(\nabla_Y X,X). \end{align*} The product rule for the function $g(Y,X):M\to\mathbb R$ gives \begin{align*} X(g(Y,X))=g(\nabla_XY,X)+g(Y,\nabla_X X). \end{align*} Since $X=\nabla f$ is a gradient field, its covariant derivative is symmetric: for all vector fields $A$ and $B$, \begin{align*} g(\nabla_A X,B)=\nabla^2f(A,B)=\nabla^2f(B,A)=g(\nabla_B X,A). \end{align*} Taking $A=Y$ and $B=X$ gives \begin{align*} g(\nabla_YX,X)=g(\nabla_XX,Y)=g(Y,\nabla_XX). \end{align*} Therefore, at $p$, \begin{align*} D_f(p)=-g(\nabla_YX,X)(p)+g(\nabla_XY,X)(p)+g(Y,\nabla_XX)(p)=g(\nabla_XY,X)(p). \end{align*} By the definition of the Hessian of $V$, \begin{align*} g(\nabla_XY,X)=\nabla^2V(X,X)=\nabla^2V(\nabla f,\nabla f). \end{align*} Since $p\in M$ was arbitrary, \begin{align*} D_f=\nabla^2V(\nabla f,\nabla f). \end{align*}[/step]

[guided]The only term left to compute is the part caused by replacing $\Delta$ with $L$. Define \begin{align*} D_f:M\to\mathbb R \end{align*} by \begin{align*} D_f=-\frac{1}{2}g(\nabla V,\nabla|\nabla f|^2)+g(\nabla f,\nabla g(\nabla V,\nabla f)). \end{align*} We want to show that this equals $\nabla^2V(\nabla f,\nabla f)$. Fix a point $p\in M$. Set $X=\nabla f$ and $Y=\nabla V$ as smooth vector fields on a neighbourhood of $p$. First compute the derivative of the squared norm of $X$ in the direction $Y$. The Levi-Civita connection is metric-compatible, so for the function $|X|^2=g(X,X)$, \begin{align*} Y(|X|^2)=2g(\nabla_YX,X). \end{align*} Thus \begin{align*} \frac{1}{2}g(\nabla V,\nabla|\nabla f|^2)=\frac{1}{2}Y(|X|^2)=g(\nabla_YX,X). \end{align*} Next compute the derivative of the function $g(Y,X)$ in the direction $X$. Again using metric compatibility and the product rule, \begin{align*} X(g(Y,X))=g(\nabla_XY,X)+g(Y,\nabla_XX). \end{align*} In the original notation this is exactly \begin{align*} g(\nabla f,\nabla g(\nabla V,\nabla f))=g(\nabla_XY,X)+g(Y,\nabla_XX). \end{align*} Now compare the two terms involving the Hessian of $f$. Since $X=\nabla f$, the covariant derivative of $X$ is the Hessian of $f$: for vector fields $A$ and $B$, \begin{align*} g(\nabla_AX,B)=\nabla^2f(A,B). \end{align*} The Hessian of a smooth real-valued function is symmetric because the Levi-Civita connection is torsion-free. Therefore \begin{align*} g(\nabla_YX,X)=\nabla^2f(Y,X)=\nabla^2f(X,Y)=g(\nabla_XX,Y). \end{align*} By symmetry of the metric, $g(\nabla_XX,Y)=g(Y,\nabla_XX)$. Hence the two $f$-Hessian terms cancel: \begin{align*} D_f=-g(\nabla_YX,X)+g(\nabla_XY,X)+g(Y,\nabla_XX)=g(\nabla_XY,X). \end{align*} Finally, because $Y=\nabla V$, the Hessian of $V$ satisfies \begin{align*} \nabla^2V(X,X)=g(\nabla_X\nabla V,X)=g(\nabla_XY,X). \end{align*} Substituting $X=\nabla f$ gives \begin{align*} D_f=\nabla^2V(\nabla f,\nabla f). \end{align*}[/guided]

[step:Combine the Ricci and Hessian terms into the Bakry-Émery tensor] Substituting the drift computation into the expression for $\Gamma_2(f)$ gives \begin{align*} \Gamma_2(f)=|\nabla^2f|^2+\operatorname{Ric}(\nabla f,\nabla f)+\nabla^2V(\nabla f,\nabla f). \end{align*} By the definition $\operatorname{Ric}_V=\operatorname{Ric}+\nabla^2V$, this is \begin{align*} \Gamma_2(f)=|\nabla^2f|^2+\operatorname{Ric}_V(\nabla f,\nabla f). \end{align*} This proves the claimed Bochner formula for the weighted Laplacian. [/step]