[proofplan]
The scalar [second variation formula](/theorems/2729) expresses $\delta^2\mathcal A[\phi]$ as the Dirichlet energy of $\phi$ minus the curvature potential term $q\phi^2$. The compact support of $\phi$ allows [integration by parts](/theorems/2098) on $\Sigma$ with no boundary contribution, converting the Dirichlet term into the negative Laplacian pairing. Substituting this identity and using the definition $L\phi=\Delta_\Sigma\phi+q\phi$ gives the operator form.
[/proofplan]
[step:Start from the scalar second variation formula with fixed sign conventions]
Since $\Sigma$ is two-sided, choose the globally defined smooth unit normal field $\nu: \Sigma \to TM|_\Sigma$. Let $A$ denote the second fundamental form of $\Sigma$ with respect to $\nu$, and let $q: \Sigma \to \mathbb{R}$ be the curvature potential defined, for each $p \in \Sigma$, by
\begin{align*}
q(p)=|A(p)|^2 + \operatorname{Ric}_g(\nu(p), \nu(p)).
\end{align*}
Let $\nabla^\Sigma: C^\infty(\Sigma) \to \Gamma(T\Sigma)$ denote the intrinsic gradient operator of the Riemannian manifold $(\Sigma,g|_\Sigma)$, and let $\operatorname{div}_\Sigma: \Gamma(T\Sigma) \to C^\infty(\Sigma)$ denote the intrinsic divergence operator. Define the Laplace-Beltrami operator $\Delta_\Sigma: C^\infty(\Sigma) \to C^\infty(\Sigma)$ by
\begin{align*}
\Delta_\Sigma\psi=\operatorname{div}_\Sigma(\nabla^\Sigma\psi).
\end{align*}
Define the Jacobi operator $L: C^\infty(\Sigma) \to C^\infty(\Sigma)$ by
\begin{align*}
L\psi=\Delta_\Sigma\psi+q\psi.
\end{align*}
The normal variation field determined by $\phi$ is the smooth section $V_\phi: \Sigma \to TM|_\Sigma$ defined, for each $p \in \Sigma$, by
\begin{align*}
V_\phi(p)=\phi(p)\nu(p).
\end{align*}
Because $\phi \in C_c^\infty(\Sigma)$ and $\nu$ is smooth, $V_\phi$ is smooth, normal to $\Sigma$, and compactly supported. Thus $V_\phi$ is an admissible compactly supported normal variation field for the [scalar second variation formula for two-sided minimal hypersurfaces](/page/Second%20Variation%20Formula). In the sign convention used here, that formula states that every compactly supported normal variation $V_\psi=\psi\nu$ with $\psi \in C_c^\infty(\Sigma)$ satisfies
\begin{align*}
\delta^2\mathcal A[\psi]
=
\int_\Sigma \left(|\nabla^\Sigma \psi|^2 - q\psi^2\right)\,d\mathcal H^m.
\end{align*}
Applying this formula with $\psi=\phi$ gives
\begin{align*}
\delta^2\mathcal A[\phi]
=
\int_\Sigma \left(|\nabla^\Sigma \phi|^2 - q\phi^2\right)\,d\mathcal H^m.
\end{align*}
Here the minimality hypothesis is used to eliminate the first variation term and put the second variation in this quadratic form.
[/step]
[step:Integrate the gradient term by parts using compact support]
Because $\phi \in C_c^\infty(\Sigma)$, the vector field $X: \Sigma \to T\Sigma$ defined, for each $p \in \Sigma$, by
\begin{align*}
X(p)=\phi(p)\nabla^\Sigma\phi(p)
\end{align*}
is smooth and compactly supported. The product rule for the intrinsic divergence gives
\begin{align*}
\operatorname{div}_\Sigma X
=
|\nabla^\Sigma \phi|^2 + \phi\,\Delta_\Sigma\phi.
\end{align*}
Integrating over $\Sigma$ and using the [divergence theorem](/theorems/3614) for compactly supported vector fields on $\Sigma$ yields
\begin{align*}
0
=
\int_\Sigma \operatorname{div}_\Sigma X\,d\mathcal H^m
=
\int_\Sigma |\nabla^\Sigma \phi|^2\,d\mathcal H^m
+
\int_\Sigma \phi\,\Delta_\Sigma\phi\,d\mathcal H^m.
\end{align*}
Therefore
\begin{align*}
\int_\Sigma |\nabla^\Sigma \phi|^2\,d\mathcal H^m
=
-\int_\Sigma \phi\,\Delta_\Sigma\phi\,d\mathcal H^m.
\end{align*}
[guided]
The goal is to replace the energy term $\int_\Sigma |\nabla^\Sigma \phi|^2\,d\mathcal H^m$ by an expression involving $\Delta_\Sigma\phi$, because the Jacobi operator is defined using $\Delta_\Sigma$. Here $\nabla^\Sigma: C^\infty(\Sigma) \to \Gamma(T\Sigma)$ is the intrinsic gradient operator, $\operatorname{div}_\Sigma: \Gamma(T\Sigma) \to C^\infty(\Sigma)$ is the intrinsic divergence operator, and $\Delta_\Sigma=\operatorname{div}_\Sigma\nabla^\Sigma$ is the Laplace-Beltrami operator. Define the smooth tangent vector field $X: \Sigma \to T\Sigma$ by the rule
\begin{align*}
X(p)=\phi(p)\nabla^\Sigma\phi(p)
\end{align*}
for each $p \in \Sigma$. This vector field is compactly supported because $\phi$ is compactly supported. That compact support is the reason no boundary term appears in the [integration by parts](/theorems/210) identity.
Using the product rule for the intrinsic divergence on $\Sigma$, we compute
\begin{align*}
\operatorname{div}_\Sigma X
=
\operatorname{div}_\Sigma(\phi\nabla^\Sigma\phi)
=
|\nabla^\Sigma\phi|^2+\phi\,\operatorname{div}_\Sigma(\nabla^\Sigma\phi).
\end{align*}
Since $\Delta_\Sigma=\operatorname{div}_\Sigma\nabla^\Sigma$, this becomes
\begin{align*}
\operatorname{div}_\Sigma X
=
|\nabla^\Sigma\phi|^2+\phi\,\Delta_\Sigma\phi.
\end{align*}
The [divergence theorem](/theorems/2754) for compactly supported vector fields on $\Sigma$ gives
\begin{align*}
\int_\Sigma \operatorname{div}_\Sigma X\,d\mathcal H^m=0.
\end{align*}
Substituting the displayed formula for $\operatorname{div}_\Sigma X$ gives
\begin{align*}
0
=
\int_\Sigma |\nabla^\Sigma \phi|^2\,d\mathcal H^m
+
\int_\Sigma \phi\,\Delta_\Sigma\phi\,d\mathcal H^m.
\end{align*}
Rearranging this equality yields the integration by parts identity
\begin{align*}
\int_\Sigma |\nabla^\Sigma \phi|^2\,d\mathcal H^m
=
-\int_\Sigma \phi\,\Delta_\Sigma\phi\,d\mathcal H^m.
\end{align*}
[/guided]
[/step]
[step:Substitute the integration by parts identity into the second variation]
Substituting the preceding identity into the scalar second variation formula gives the sum of the negative Laplacian pairing and the negative potential term:
\begin{align*}
\delta^2\mathcal A[\phi]= -\int_\Sigma \phi\,\Delta_\Sigma\phi\,d\mathcal H^m - \int_\Sigma q\phi^2\,d\mathcal H^m.
\end{align*}
Combining the two integrals over the same [measure space](/page/Measure%20Space) $(\Sigma,\mathcal H^m)$ and factoring the common factor $-\phi$ inside the integrand gives
\begin{align*}
\delta^2\mathcal A[\phi]= -\int_\Sigma \phi\left(\Delta_\Sigma\phi+q\phi\right)\,d\mathcal H^m.
\end{align*}
By the definition of the Jacobi operator $L: C^\infty(\Sigma) \to C^\infty(\Sigma)$ fixed at the start of the proof, we have
\begin{align*}
L\phi=\Delta_\Sigma\phi+q\phi.
\end{align*}
Hence
\begin{align*}
\delta^2\mathcal A[\phi]
=
-\int_\Sigma \phi\,L\phi\,d\mathcal H^m,
\end{align*}
which is the claimed operator form of the second variation.
[/step]
