Let $T$ be an $n$-dimensional area-minimising hypersurface in $\mathbf{R}^{n+1}$. If $T$ has bounded area growth in the sense that there is a constant $C > 0$ so that $\mathcal{H}^n(T \cap B\_R) \leq C R^n$ for all $R > 0$, then there are rigidity theorems for $T$. For example, when $n \leq 6$ then the work of Simons [1] implies that $T$ must be an $n$-dimensional plane. (In larger dimensions there are singular area-minimising hypercones.)


**Question.** Is there an example of an area-minimising hypersurface with unbounded growth? Could such an example exist in low dimensions, when $n \leq 6$? What about $n = 2$?


[1] James Simons. Minimal varieties in Riemannian manifolds. *Annals of Mathematics*, Second Series, Vol. 88, No. 1 (1968), pp. 62-105.



Area-minimising hypersurface with unbounded area growth
This is a straightforward comparison argument. Let $\omega\_n$ be the volume of $\partial B\_1\subset \mathbb{R}^{n+1}$.


For generic $R$, one has $\partial B\_R \cap T=\tau$ a smooth submanifold. By Alexander duality, there is a subset, $\Omega$, of $ \partial B\_R\setminus \tau$ so $\partial \Omega=\tau$. Clearly, $\mathcal{H}^n(\Omega)\leq \omega\_n R^n$. In fact, up to replacing $\Omega$ by it's complement one has $\mathcal{H}^n(\Omega)\leq \frac{1}{2}\omega\_n R^n$.


By the area minimization property,
$$\mathcal{H}^n(\Sigma\cap B\_R)\leq \mathcal{H}^n(\Omega)\leq \omega\_n R^n.$$


The monotonicity formula ensures the bound holds for all $R$.


In fact, this argument should work (using slicing) for an area minimizing integral $\mathbb{Z}\_2$ current.


