I was reading a paper about minimal hypersurfaces. The author said that the Codazzi equation for a hypersurface $M$ in a manifold with constant sectional curvature is given by

$(\nabla A)(X,Y)=(\nabla A)(Y,X)$

where $A$ is the shape operator of $M$ and $(\nabla A)(X,Y)=\nabla_X(AY)-A(\nabla_XY)$. Then the author said that if $M$ is a $minimal$ hypersurface then in a local orthonormal frame $\{e_1,e_2,...,e_n\}$, we have

\begin{equation} (\nabla A)(e_i,e_i)=0 \end{equation}

I couldn't figure out why this should be true. If $M$ is minimal then all I know is that in the orthonormal frame $\sum_{i=1}^n g(Ae_i,e_i)=0$. I tried to differentiate it and get the expression but I was unable to do so. Can you please help me with this ?


  • $\begingroup$ $\sum_i\ \nabla_XA (e_i,e_i)=0$. Since $M$ is space form, so all eigenvalue of $\nabla_XA$ are equal. $\endgroup$ – HK Lee May 10 '18 at 2:40
  • $\begingroup$ @HKLee Why is $M$ a space form ? I guess the ambient manifold is of constant sectional curvature. Why can we say the same thing about $M$ ? $\endgroup$ – seeker May 10 '18 at 2:43

I find it's easier to think about the second fundamental form $\def\II{\mathrm{I\!I}}\II,$ i.e. the $(0,2)$-tensor corresponding to $A$ via the metric: $$\II(X,Y) = g(AX,Y).$$ Since $A$ is self-adjoint, $\II$ is symmetric, and thus the $(0,3)$-tensor $\nabla \II$ is symmetric in its last two slots. The constant-curvature Codazzi equation can be written in terms of $\II$ as $$\nabla \II(X,Y,Z) = \nabla \II(Y,X,Z),$$ which tells you that $\nabla \II$ is symmetric in its first two slots; so we conclude that $\nabla \II$ is totally symmetric.

Now that we know this, the desired result is easy: minimality tells us that the metric trace of $\II$ is zero, so the trace of $\nabla \II$ (in any two slots, thanks to symmetry) is zero because the Levi-Civita connection commutes with metric traces. In particular, we have $$\sum_i \nabla \II(e_i,e_i,Y)=g\left(\sum_i\nabla A(e_i,e_i),Y\right)=0$$ for arbitrary $Y$, and thus the trace of $\nabla A$ vanishes.

  • $\begingroup$ thanks a lot for a very clear explanation. I should've thought a little harder. $\endgroup$ – seeker May 10 '18 at 12:15

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