Let $(M^{n+k},g)$ be a Riemannian manifold. Call a surface $\Sigma^n \subset M$ *calibrated* if there is a closed $n$-form $\omega$ defined on a neighbourhood $U \subset M$ of $\Sigma$ so that $\omega \lvert \Sigma = \mathrm{vol}\_\Sigma$ and for any $p \in U$ and $n$-tuples $(X\_1,\dots,X\_n) \in T\_p M$ of orthonormal vectors $\omega(X\_1,\dots,X\_n) \leq 1$. (This is slightly different from the usual definition, where usually $\omega$ is defined on $M$.) A simple argument shows that a calibrated surface $\Sigma$ is *area-minimising* in the neighbourhood $U$, and a small perturbation of $\Sigma'$ of $\Sigma$ will have $\mathrm{Area}(\Sigma') \geq \mathrm{Area}(\Sigma)$. In particular a calibrated surface is *minimal*, that is stationary for the area functional, and has mean curvature $H\_\Sigma = 0$.


There are many examples of calibrated area-minimising surfaces:


1. linear subspaces of $\mathbf{R}^n$,
2. minimal graphs of $u: \Omega \subset \mathbf{R}^n \to \mathbf{R}$, where $\Omega$ is an open domain in $\mathbf{R}^n$,
3. special Lagrangian submanifolds $\Sigma \subset M$ in Calabi-Yau manifolds, that is Lagrangian submanifolds so that $\mathrm{Im} \, \Omega \lvert \Sigma = 0$ where $\Omega$ is the holomorphic volume form,
4. holomorphic subvarieties of $\mathbf{C}^n$,
5. area-minimising cones with an isolated singularity at the origin, for example the Simons cone $\mathbf{C}\_S = \{ (X,Y) \in \mathbf{R}^n \times \mathbf{R}^n \mid \lvert X \rvert = \lvert Y \rvert \}$. (I believe these are calibrated because of the Hardt-Simon foliations.)


However I cannot think of any examples of area-minimising surfaces which are not calibrated.


**Question**:
What are they? I am especially interested in the codimension one case, where $\Sigma^n \subset M^{n+1}$. In which settings, or under which hypotheses, is an area-minimising surface not be calibrated?


*Remark*: I can formulate a more technically precise question, at the price of using some terms from geometric measure theory. Let $B \subset \mathbf{R}^{n+k}$ be the unit ball, and $T \in \mathbf{I}\_n(B)$ be an integral current with $\partial T = 0$ in $B$. Suppose that $T$ is *area-minimising* in the sense that for some $\epsilon > 0$ and all currents $S \in \mathbf{I}\_{n+1}(B)$ with $\mathrm{spt} \, S \subset \subset B$ and $\mathrm{dist}(\mathrm{spt} \, S,\mathrm{spt} \, T) \leq \epsilon $, $\mathrm{Area} \, (T + \partial S) \geq \mathrm{Area} \, T$. Is there a neighbourhood of $T$ on which it admits a calibration? Here again I would be most interested in the case $k = 1$.



Non-calibrated area-minimising surface
Actually, a better example along the lines Otis suggests would be the geodesic $\mathbb{RP}^1\subset\mathbb{RP}^2$. Of course, $\mathbb{RP}^1$ is orientable and it is homologically mass-minimizing, but it can't be calibrated on any open set $U\subset\mathbb{RP}^2$ containing $\mathbb{RP}^1$ because twice it is not even stable.


Of course, this also works for any $\mathbb{RP}^{2n-1}\subset\mathbb{RP}^{2n}$, and there are higher codimension examples of closed geodesics in (orientable) lens spaces that are homologically mass-minimizing but that cannot be calibrated on any open neighborhood of the geodesic. One can even foliate $\mathbb{RP}^3$ by homologically mass-minimizing geodesics that cannot be calibrated on any open neighborhood.


What one probably needs to assume, at least, is that every multiple of $\Sigma$ is homologically area-minimizing in some neighborhood before one could hope to construct a 'neighborhood' calibration.


**Remark (10/12/20):** I just recalled one example of possible interest for this question, since the OP is interested in what can happen in Euclidean space. A student of mine, Timothy Murdoch, in his PhD thesis "Twisted calibrations and the cone on the Veronese surface" (Rice University, 1988), showed that the $3$-dimensional cone in $\mathbb{R}^5$ on the Veronese surface in $S^4$ is area-minimizing, but, of course, it's not orientable. However, its 'double cover' is a cone on the $2$-sphere and so is orientable. I don't know whether this double cover is area-minimizing in $\mathbb{R}^5$ or not. It obviously cannot be calibrated, even if it is area-minimizing.


Explicitly, here is the example: Think of $\mathbb{R}^5$ as $S^2\_0(\mathbb{R}^3)$, the traceless $3$-by-$3$ matrices with real entries endowed with the quadratic form $\langle a,b\rangle = \mathrm{tr}(ab)$, which is invariant under $\mathrm{SO}(3)$ with the irreducible action $A\cdot a = AaA^{-1}$ for $A\in\mathrm{SO}(3)$ and $a\in S^2\_0(\mathbb{R}^3)$. Then the Veronese cone $C\subset S^2\_0(\mathbb{R}^3)$ is the set of matrices $a$ with eigenvalues $t^2,t^2, -2t^2$ for some $t\ge0$. It is a cone on an $\mathrm{SO}(3)$-homogeneous minimal surface $\mathbb{RP}^2\subset S^4$ known as the Veronese surface. (Note that $C$ and $-C$ intersect only at the origin.) $C$ is smooth except at the origin, and, if you define the 'double cover' by counting each smooth point as two points with different orientations, then the double cover is homeomorphic to $\mathbb{R}^3$, parametrized by the quadratic map $s:\mathbb{R}^3\to S^2\_0(\mathbb{R}^3)$ defined by
$$
s(x) = |x|^2\, I\_3 - 3\,x\,x^T\quad\text{for}\ x\in\mathbb{R}^3.
$$
Tim showed that, if you take the (literal) Riemannian double cover of $S^2\_0(\mathbb{R}^3)\setminus (-C)$, then the double cover of $C\setminus\{0\}\simeq \mathbb{R}^3\setminus\{0\}$ can be calibrated in the ambient double cover as a Riemannian manifold.


