[proofplan] We identify $C_{p,q}^{\mathrm{reg}}$ as the cone over the product link $L_{p,q}=S^p(\sqrt{p/(p+q)})\times S^q(\sqrt{q/(p+q)})\subset \mathbb{S}^{p+q+1}$. The chosen radii make the two families of principal curvatures cancel, so the link is minimal in the unit sphere, and therefore the cone is minimal in Euclidean space. For $p=q=3$, the link has $|A_L|^2=6$, so the second variation quadratic form on the cone reduces to a radial Hardy inequality with weight $r^6$. The sharp one-dimensional Hardy constant is $25/4$, which dominates $6$, giving nonnegativity for every compactly supported variation away from the vertex. [/proofplan]

[step:Describe the cone as the radial extension of its product link] Set $m:=p+q$. Define \begin{align*} a&:=\sqrt{\frac{p}{m}},& b&:=\sqrt{\frac{q}{m}}, \end{align*} and define the product submanifold \begin{align*} L_{p,q}:= S^p(a)\times S^q(b) \subset \mathbb{R}^{p+1}\times\mathbb{R}^{q+1}. \end{align*} Since $a^2+b^2=1$, we have $L_{p,q}\subset \mathbb{S}^{m+1}$. Define the smooth map $\Psi:(0,\infty)\times L_{p,q} \to C_{p,q}^{\mathrm{reg}}$ by \begin{align*} \Psi(r,z):=rz \quad \text{for } (r,z)\in (0,\infty)\times L_{p,q}. \end{align*} Then $\Psi$ is a smooth diffeomorphism. Indeed, if $z=(u,v)\in L_{p,q}$, then \begin{align*} q|u|^2-p|v|^2 = q\frac{p}{m}-p\frac{q}{m} = 0, \end{align*} so $rz\in C_{p,q}^{\mathrm{reg}}$. Conversely, if $(x,y)\in C_{p,q}^{\mathrm{reg}}$, set \begin{align*} r:=\sqrt{|x|^2+|y|^2}. \end{align*} Then $r>0$ and $z:=(x/r,y/r)\in \mathbb{S}^{m+1}$. The defining equation gives \begin{align*} q|x|^2=p|y|^2, \qquad |x|^2+|y|^2=r^2, \end{align*} hence \begin{align*} |x/r|^2=\frac{p}{m}, \qquad |y/r|^2=\frac{q}{m}. \end{align*} Thus $z\in L_{p,q}$ and $(x,y)=rz$. [/step]

[step:Compute the principal curvatures of the link in the unit sphere] Define the smooth unit normal field $\nu_L:L_{p,q}\to T\mathbb{S}^{m+1}\big|_{L_{p,q}}$ by \begin{align*} \nu_L(u,v):=\left(\frac{b}{a}u,-\frac{a}{b}v\right) \quad \text{for } (u,v)\in L_{p,q}. \end{align*} For $(u,v)\in L_{p,q}$, \begin{align*} |\nu_L(u,v)|^2 = \frac{b^2}{a^2}|u|^2+\frac{a^2}{b^2}|v|^2 = b^2+a^2 = 1, \end{align*} and \begin{align*} \nu_L(u,v)\cdot (u,v) = \frac{b}{a}|u|^2-\frac{a}{b}|v|^2 = ab-ab = 0. \end{align*} Thus $\nu_L$ is tangent to $\mathbb{S}^{m+1}$ and normal to $L_{p,q}$ inside $\mathbb{S}^{m+1}$. Let $A_L$ denote the shape operator of $L_{p,q}\subset \mathbb{S}^{m+1}$ with respect to $\nu_L$, using the convention $A_L(W)=-D_W\nu_L$, where $D$ is the Euclidean derivative. If $W=(\xi,0)$ is tangent to the $S^p(a)$ factor, then \begin{align*} D_W\nu_L=\frac{b}{a}(\xi,0), \qquad A_L(W)=-\frac{b}{a}W. \end{align*} If $W=(0,\eta)$ is tangent to the $S^q(b)$ factor, then \begin{align*} D_W\nu_L=-\frac{a}{b}(0,\eta), \qquad A_L(W)=\frac{a}{b}W. \end{align*} Therefore the principal curvatures of $L_{p,q}$ in $\mathbb{S}^{m+1}$ are \begin{align*} -\frac{b}{a}=-\sqrt{\frac{q}{p}} \quad\text{with multiplicity }p, \qquad \frac{a}{b}=\sqrt{\frac{p}{q}} \quad\text{with multiplicity }q. \end{align*} The mean curvature scalar of $L_{p,q}$ in $\mathbb{S}^{m+1}$ is the trace of $A_L$: \begin{align*} \operatorname{tr}A_L = -p\sqrt{\frac{q}{p}}+q\sqrt{\frac{p}{q}} = -\sqrt{pq}+\sqrt{pq} = 0. \end{align*} Thus $L_{p,q}$ is minimal in $\mathbb{S}^{m+1}$. [/step]

[step:Pass minimality from the link to the cone] Under the parametrization $\Psi:(0,\infty)\times L_{p,q}\to C_{p,q}^{\mathrm{reg}}$, the induced metric on $C_{p,q}^{\mathrm{reg}}$ is \begin{align*} g_C = d\mathcal{L}^1(r)^2+r^2 g_L, \end{align*} where $g_L$ is the induced metric on $L_{p,q}$. The radial direction has zero normal curvature, and each principal curvature $\kappa$ of $L_{p,q}\subset \mathbb{S}^{m+1}$ contributes the cone principal curvature $\kappa/r$ on $C_{p,q}^{\mathrm{reg}}$. Hence the Euclidean mean curvature scalar of the cone is \begin{align*} H_C(r,z) = \frac{1}{r}\operatorname{tr}A_L(z) = 0 \end{align*} for every $(r,z)\in (0,\infty)\times L_{p,q}$. Therefore $C_{p,q}^{\mathrm{reg}}$ is a minimal hypersurface of $\mathbb{R}^{m+2}$. [/step]

[step:Write the second variation of the Simons cone in radial and angular variables]Now assume $p=q=3$. Then $m=6$, $a=b=1/\sqrt{2}$, and \begin{align*} L:=L_{3,3} = S^3(1/\sqrt{2})\times S^3(1/\sqrt{2}) \subset \mathbb{S}^7. \end{align*} From the principal curvature computation above, the principal curvatures of $L\subset\mathbb{S}^7$ are $-1$ with multiplicity $3$ and $1$ with multiplicity $3$. Hence \begin{align*} |A_L|^2=3\cdot 1^2+3\cdot 1^2=6. \end{align*} The cone $C:=C_{3,3}^{\mathrm{reg}}$ has dimension $7$, and its squared second fundamental form is \begin{align*} |A_C(r,z)|^2=\frac{|A_L(z)|^2}{r^2}=\frac{6}{r^2}. \end{align*} Let $\phi\in C_c^\infty(C)$, and define its pullback $\widetilde{\phi}:(0,\infty)\times L\to \mathbb{R}$ under $\Psi$ by \begin{align*} \widetilde{\phi}(r,z):=\phi(rz) \quad \text{for } (r,z)\in (0,\infty)\times L. \end{align*} The [second variation formula](/theorems/2729) for a Euclidean minimal hypersurface gives \begin{align*} \delta^2 A_C[\phi\nu] = \int_C \left( |\nabla_C\phi|^2-|A_C|^2\phi^2 \right) \,d\mathcal{H}^7. \end{align*} Here $\nabla_C\phi$ is the intrinsic gradient on $C$, and $\mathcal{H}^7$ is the $7$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $C$. In the warped product coordinates $(r,z)$, \begin{align*} |\nabla_C\phi|^2 = |\partial_r\widetilde{\phi}|^2+\frac{1}{r^2}|\nabla_L\widetilde{\phi}|^2, \end{align*} and the Hausdorff measure decomposes as \begin{align*} d\mathcal{H}^7(rz) = r^6\,d\mathcal{L}^1(r)\,d\mathcal{H}^6_L(z), \end{align*} where $\mathcal{H}^6_L$ is the induced $6$-dimensional Hausdorff measure on $L$. Therefore \begin{align*} \delta^2 A_C[\phi\nu] = \int_0^\infty\int_L \left( |\partial_r\widetilde{\phi}(r,z)|^2 + \frac{1}{r^2}|\nabla_L\widetilde{\phi}(r,z)|^2 - \frac{6}{r^2}\widetilde{\phi}(r,z)^2 \right) r^6 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r). \end{align*} Since $|\nabla_L\widetilde{\phi}|^2\geq 0$, it is enough to prove \begin{align*} \int_0^\infty\int_L |\partial_r\widetilde{\phi}(r,z)|^2 r^6 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r) \geq 6 \int_0^\infty\int_L \widetilde{\phi}(r,z)^2 r^4 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r). \end{align*}[/step]

[guided]The purpose of this step is to translate geometric stability into an explicit inequality. The cone is minimal, so the second variation of area for a compactly supported normal variation $V=\phi\nu$ is governed by the Jacobi quadratic form \begin{align*} \delta^2 A_C[\phi\nu] = \int_C \left( |\nabla_C\phi|^2-|A_C|^2\phi^2 \right) \,d\mathcal{H}^7. \end{align*} The negative term is the possible source of instability: curvature tries to make the quadratic form negative. For the Simons cone, the link \begin{align*} L= S^3(1/\sqrt{2})\times S^3(1/\sqrt{2}) \end{align*} has six nonzero principal curvatures, namely three equal to $1$ and three equal to $-1$. Thus \begin{align*} |A_L|^2=3\cdot 1^2+3\cdot (-1)^2=6. \end{align*} When passing from the link to the cone, each angular principal curvature scales by $1/r$, while the radial direction contributes principal curvature $0$. Hence \begin{align*} |A_C(r,z)|^2=\frac{|A_L(z)|^2}{r^2}=\frac{6}{r^2}. \end{align*} Now write the variation function in cone coordinates. Define the map $\widetilde{\phi}:(0,\infty)\times L\to \mathbb{R}$ by \begin{align*} \widetilde{\phi}(r,z):=\phi(rz) \quad \text{for } (r,z)\in (0,\infty)\times L. \end{align*} Because $\phi$ is compactly supported away from the vertex, $\widetilde{\phi}$ is smooth and compactly supported in $[\varepsilon,R]\times L$ for some numbers $0<\varepsilon<R<\infty$. This compact support justifies all integrations by parts and all weighted integrals below. The metric on the cone is \begin{align*} g_C=d\mathcal{L}^1(r)^2+r^2g_L. \end{align*} Therefore the intrinsic gradient decomposes into radial and angular parts: \begin{align*} |\nabla_C\phi|^2 = |\partial_r\widetilde{\phi}|^2+\frac{1}{r^2}|\nabla_L\widetilde{\phi}|^2. \end{align*} The $7$-dimensional Hausdorff measure on the cone decomposes as \begin{align*} d\mathcal{H}^7(rz) = r^6\,d\mathcal{L}^1(r)\,d\mathcal{H}^6_L(z), \end{align*} because the link has dimension $6$. Substituting these identities into the second variation formula gives \begin{align*} \delta^2 A_C[\phi\nu] = \int_0^\infty\int_L \left( |\partial_r\widetilde{\phi}(r,z)|^2 + \frac{1}{r^2}|\nabla_L\widetilde{\phi}(r,z)|^2 - \frac{6}{r^2}\widetilde{\phi}(r,z)^2 \right) r^6 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r). \end{align*} The angular term is nonnegative, so it can only help stability. Thus the entire problem reduces to proving that the radial energy controls the negative curvature term: \begin{align*} \int_0^\infty\int_L |\partial_r\widetilde{\phi}(r,z)|^2 r^6 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r) \geq 6 \int_0^\infty\int_L \widetilde{\phi}(r,z)^2 r^4 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r). \end{align*} This is exactly where the dimension of the Simons cone enters: the radial weight is $r^6$, and the corresponding Hardy constant is large enough to dominate the curvature constant $6$.[/guided]

[step:Apply the weighted Hardy inequality to dominate the curvature term] For each fixed $z\in L$, define the map $f_z:(0,\infty)\to\mathbb{R}$ by \begin{align*} f_z(r):=\widetilde{\phi}(r,z) \quad \text{for } r\in (0,\infty). \end{align*} Since $\widetilde{\phi}$ is compactly supported away from $r=0$, each $f_z$ belongs to $C_c^\infty(0,\infty)$. We prove the weighted Hardy inequality \begin{align*} \int_0^\infty |f_z'(r)|^2 r^6\,d\mathcal{L}^1(r) \geq \frac{25}{4} \int_0^\infty f_z(r)^2 r^4\,d\mathcal{L}^1(r). \end{align*} Integrating by parts with no boundary term, because $f_z$ is compactly supported in $(0,\infty)$, gives \begin{align*} \int_0^\infty f_z(r)^2 r^4\,d\mathcal{L}^1(r) &= -\frac{2}{5} \int_0^\infty f_z(r)f_z'(r) r^5\,d\mathcal{L}^1(r). \end{align*} Taking absolute values and applying the [Cauchy-Schwarz inequality](/theorems/432) in the [measure space](/page/Measure%20Space) $((0,\infty),\mathcal{B}(0,\infty),\mathcal{L}^1)$ to the functions $r\mapsto f_z'(r)r^3$ and $r\mapsto f_z(r)r^2$ yields \begin{align*} \int_0^\infty f_z(r)^2 r^4\,d\mathcal{L}^1(r) &\leq \frac{2}{5} \left( \int_0^\infty |f_z'(r)|^2 r^6\,d\mathcal{L}^1(r) \right)^{1/2} \left( \int_0^\infty f_z(r)^2 r^4\,d\mathcal{L}^1(r) \right)^{1/2}. \end{align*} If the rightmost weighted $L^2$ norm of $f_z$ is zero, the desired inequality is immediate. Otherwise, dividing by its square root gives \begin{align*} \left( \int_0^\infty f_z(r)^2 r^4\,d\mathcal{L}^1(r) \right)^{1/2} &\leq \frac{2}{5} \left( \int_0^\infty |f_z'(r)|^2 r^6\,d\mathcal{L}^1(r) \right)^{1/2}. \end{align*} Squaring proves the claimed Hardy inequality. Integrating this inequality over $L$ with respect to $\mathcal{H}^6_L$ gives \begin{align*} \int_0^\infty\int_L |\partial_r\widetilde{\phi}(r,z)|^2 r^6 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r) \geq \frac{25}{4} \int_0^\infty\int_L \widetilde{\phi}(r,z)^2 r^4 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r). \end{align*} Since $25/4>6$, we have \begin{align*} \int_0^\infty\int_L |\partial_r\widetilde{\phi}(r,z)|^2 r^6 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r) - 6 \int_0^\infty\int_L \widetilde{\phi}(r,z)^2 r^4 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r) \geq \frac{1}{4} \int_0^\infty\int_L \widetilde{\phi}(r,z)^2 r^4 \,d\mathcal{H}^6_L(z)\,d\mathcal{L}^1(r) \geq 0. \end{align*} Returning to the second variation formula and keeping the nonnegative angular term, we obtain \begin{align*} \delta^2 A_C[\phi\nu]\geq 0. \end{align*} Thus $C_{3,3}$ is stable for every compactly supported smooth normal variation away from the origin. [/step]