Let $\Sigma^n \subseteq \mathbb{R}^{n+1}$ be a minimal complete hypersurface. Let's consider $\bar{\Sigma} := \Sigma \times \mathbb{R}^k \subseteq \mathbb{R}^{n+k +1}$. Clearly $\bar{\Sigma}$ is a minimal hypersurface as well.
Moreover it is easy to check that if $\Sigma$ is stable then $\bar{\Sigma}$ is stable as well. Is it true the converse? Namely, if $\bar{\Sigma}$ is stable, is it true that $\Sigma$ is stable?
Yes, this is true.
I will prove it for $k=1$, and the result clearly follows by induction. If $\Sigma$ is unstable, then there is some compactly supported function $\varphi(x)$ with $\mathscr{Q}_\Sigma(\varphi,\varphi) < -\delta <0$ (where $\mathscr{Q}$ is the second variation operator). Consider a function $\chi(y)$ which is $1$ on $[-R,R]$ and cuts off to $0$ at $\pm 2R$. Then (since $\nabla \varphi \cdot \nabla \chi =0$) $$ \mathscr{Q}_{\bar \Sigma}(\chi \varphi,\chi\varphi) = \int_\mathbb{R} |\nabla \chi|^2 \int_{\Sigma}\varphi^2 + \mathscr{Q}_\Sigma(\varphi,\varphi) \int_{\mathbb{R}} \chi^2 $$ Note that $$ \int_\mathbb{R} |\nabla \chi|^2 \leq C R^{-1} $$ and $$ \int_{\mathbb{R}} \chi^2 \geq c R $$ Hence, we have $$ \mathscr{Q}_{\bar \Sigma}(\chi \varphi,\chi\varphi) \leq CR^{-1} \int_{\Sigma}\varphi^2 - c\delta R, $$ which is negative for $R$ large. Thus, if $\Sigma$ is unstable, then $\bar \Sigma$ is also unstable.
