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A well known result in geometric measure theory asserts that if $(M^{n+1}, g)$ is a closed Riemannian manifold and $\alpha \in H_n(M)$ is a nonzero homology class, then there exists a closed embedded minimal hypersurface $\Sigma$ (smooth outside a singular set of dimension less than or equal to $n-7$) minimizing area in $[\Sigma]=\alpha$.

My question is if this result can be (or has been) generalized for manifolds with nonempty boundary. Precisely, I wonder if the following is true:

Let $(M^{n+1}, g)$ be a compact Riemannian manifold with nonempty boundary. Given $\alpha \in H_n(M, \partial M)$ a nonzero relative homology class, there exists an embedded and free boundary minimal hypersurface $\Sigma$ (smooth outside a singular set of dimension less than or equal to $n-7$) minimizing area in $[\Sigma]=\alpha$.

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    $\begingroup$ Yes, it works: double your manifold across the boundary and observe that the classical argument works equivariantly, with respect to the natural involution of the double. $\endgroup$
    – Moishe Kohan
    Commented Feb 17, 2020 at 14:02
  • $\begingroup$ Will the result surface be free boundary? $\endgroup$
    – Eduardo Longa
    Commented Feb 17, 2020 at 16:12
  • $\begingroup$ Doubling a manifold with boundary results in a manifold without boundary. Thus, the double of $\Sigma$ has no boundary. $\endgroup$
    – Moishe Kohan
    Commented Feb 17, 2020 at 16:54
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    $\begingroup$ Yes: Compare it with a competitor $\Sigma_1\subset M$. Then $Area(\Sigma_1)=Area(D\Sigma_1)/2\ge Area(D\Sigma_0)/2=Area(\Sigma_0)$, where $D\sigma_i$ is the double of $\Sigma_i$ inside $DM$, the double of $M$. $\endgroup$
    – Moishe Kohan
    Commented Feb 17, 2020 at 19:17
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    $\begingroup$ Well, since it is an area-minimizer in its homology class.... $\endgroup$
    – Moishe Kohan
    Commented Feb 18, 2020 at 2:29

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