I am a bit confused about the statement of Theorem 1.1 in this paper by Brian White. For convenience, I will restate it here.

Theorem: Let $\Omega$ be an open subset of a Riemannian $3$-manifold. Let $g_i$ be a sequence of smooth Riemannian metrics on $\Omega$ converging smoothly to a Riemannian metric $g$. Let $M_i \subseteq \Omega$ be a sequence of properly embedded surfaces such that $M_i$ is minimal with respect to $g_i$. Suppose also that the area and the genus of $M_i$ are uniformly bounded on compact subsets of $\Omega$. Then (after passing to a subsequence) the $M_i$ converge to a smooth, properly embedded $g$-minimal surface $M$. For each connected component $\Sigma$ of $M$, either

  1. the convergence to $\Sigma$ is smooth with multiplicity one, or
  2. the convergence is smooth (with some multiplicity $> 1$) away from a discrete set $S$.

In the second case, if $\Sigma$ is two-sided, then it must be stable.

My question concerns the last sentence of the statement above. It seems very surprising to me the conclusion about the stability. What if $g$ is a metric with strictly positive Ricci curvature?

I was thinking about the following 2-dimensional situation. Imagine to have a sequence of metrics $g_i$ on the 2-sphere such that the limit metric $g$ has positive curvature. Imagine that each $g_i$ carries two closed geodesics which are converging to an unstable geodesic with multiplicity 2. See this poor quality picture: enter image description here

The compactness theorem above says that one cannot construct such an example on the $3$ sphere and using minimal surfaces instead of geodesics. Can anyone give me an intuition of what is going on here?

Thank you!


What's going on with the picture you drew is that because the $g_i$ have regions of negative curvature (in the "valleys" where the $M_i$ sit) that become arbitrarily close to the "hilltop" where $M$ sits), one must have that the curvature of $g$ along $M$ is identically zero. This means that $M$ is actually weakly stable as the constant function $1$ is an eigenfunction of the stability operator (and is the lowest as it doesn't change sign) with eigenvalue $0$.

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  • $\begingroup$ Thank you for your answer! So essentially you are saying that the limit metric $g$ in my example cannot actually be the standard metric of $\mathbb{S}^2$? $\endgroup$ – Onil90 May 1 '19 at 14:31
  • 2
    $\begingroup$ @Onil90 Yes. Think of $f_\epsilon(x)=x^4-2 \epsilon^2x^2$. This has strict local minima at $x=\pm \epsilon$ and a strict local maxima at $x=0$ when $\epsilon>0$. However, $f_0=x^4$ has only a degenerate local minima at $x=0$. $\endgroup$ – RBega2 May 1 '19 at 17:15
  • $\begingroup$ Oh, right! Thanks a lot for the clarification!! $\endgroup$ – Onil90 May 1 '19 at 19:53

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