Upper bound on volume growth of area minimizers

Question

Let $M^n$ be a complete simply connected Riemannian manifold with $\operatorname{sec}_M \leq 0$ (i.e. a Hadamard manifold) and assume that there is a constant $a \geq 0$ such that $\operatorname{sec}_M \geq -a^2$. Do you know whether this implies an upper bound on the volume growth of area minimizing submanifolds i.e. whether there exists a function $f: [0,\infty) \rightarrow [0,\infty)$ such that for any area minizing submanifold $\Sigma^k \subseteq M$ and any $p\in M$ it holds that $\mathcal{H}^k(B(p,r)\cap \Sigma) \leq f(r)$ for every $r\geq 0$. Here $\mathcal{H}^k$ is the $k$-dimensional Hausdorff measure.

I would be very grateful for a reference or a counterexample in the case when $M = \mathbb{R}^n$.

Answer 1

Consider the complex curve $w = z^k$ in $\mathbb{C}^2\simeq\mathbb{R}^4$, which is calibrated and therefore area-minimizing. The area of the part of this curve that lies inside the polydisk $\max\{|z|,|w|\}\le 1$ is $(k{+}1)\pi$. (The reason is that, because the standard Kähler form $\Omega = \tfrac{i}{2}(\mathrm{d}z\wedge\mathrm{d}\overline{z} + \mathrm{d}w\wedge\mathrm{d}\overline{w})$ calibrates this curve, the total area of this part of the curve is the sum of the areas of the projections, counted with multiplicities, onto the $z$- and $w$-axes. The projection onto the $z$-axis is $1$-to-$1$ onto the unit disk and the projection onto the $w$-axis is $k$-to-$1$ onto the unit disk (except for $w=0$).

Thus, there is no upper bound on the area of an area-minimizing surface inside a fixed ball in $\mathbb{R}^4$.