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RBega2
RBega2
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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 $0$.

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 (and is the lowest as it doesn't change sign) with eigenvalue $0$.

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 $0$.

Source Link
RBega2
RBega2
  • 2.5k
  • 1
  • 9
  • 18

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 (and is the lowest as it doesn't change sign) with eigenvalue $0$.