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The curvature of a minimal disc $S^2 \subset \mathbf{R}^3$ can be bounded in terms of the curvature of its boundary via the Gauss–Bonnet formula: \begin{equation} \frac{1}{2}\int_S \lvert A \rvert^2 \leq -2\pi + \int_{\partial S} k_g, \end{equation} where $k_g$ is the geodesic curvature of $\partial S$. (In fact this is an identity rather than an inequality.)

Is there an analog of this valid for minimal hypersurfaces $S^n \subset \mathbf{R}^{n+1}$?

For example, is there an inequality like $\int_S \lvert A \rvert^2 \leq C + \int_{\partial S} \lvert A_{\partial S} \rvert^2$, with $C$ depending on the topology of $S$? (This might be missing some normalization, but I'm just trying to give an idea of the 'spirit' of the inequality that I am interested in.)

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  • $\begingroup$ The geodesic curvature $k_g$ is bounded above by the curvature of $\partial S$ in $\mathbb R^3$. This way you can get an inequality. $\endgroup$
    – Sebastian Goette
    Commented Feb 6, 2023 at 8:09
  • $\begingroup$ @SebastianGoette Sure, but what I am asking about is a higher-dimensional analog of this. $\endgroup$
    – Leo Moos
    Commented Feb 6, 2023 at 8:41
  • $\begingroup$ I do not have an answer but one should use here the Chern-Gauss-Bonnet theorem for (even-dimensional) manifolds with boundary. $\endgroup$
    – Ivan Izmestiev
    Commented Feb 6, 2023 at 18:26
  • $\begingroup$ I don't know a higher-dimensional version. I just wanted to give a version where the boundary term does not depend on the solution. $\endgroup$
    – Sebastian Goette
    Commented Feb 7, 2023 at 9:22
  • $\begingroup$ @IvanIzmestiev The Chern-Gauss-Bonnet theorem combined with the theorema egregium should give a polynomial of degree $2k$ if the minimiser has dimension $2k$. I am not sure if that is helpful. $\endgroup$
    – Sebastian Goette
    Commented Feb 7, 2023 at 9:24

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