How to show $g^{ij}g^{kl}h_{ik}h_{jl}=\frac{(g^{ij}h_{ij})^2}{n} \Rightarrow \text{it is a round sphere}$?

Question

Consider a compact n-dim Riemannian manifold $(M,g)$, and immersion to $R^{n+1}$, $h_{ij}$ is the second fundamental form. if $$ g^{ij}g^{kl}h_{ik}h_{jl}=\frac{(g^{ij}h_{ij})^2}{n} $$ how to show $M$ is a round sphere ? In fact, I want to show $M$ has same sectional curvature, in view of $M$ is compact, then $M$ is sphere. Choice a orthonormal coordinate, by Gauss equation, I have $$ R_{ijkl}=h_{il}h_{jk}-h_{ik}h_{jl} $$ Then , $$ K=\frac{R_{ijij}}{n(n-1)}=\frac{h_{ij}^2-h_{ii}h_{jj}}{n(n-1)}=\frac{h^2_{ij}(1-n)}{n(n-1)}=-\frac{h^2_{ij}}{n} \ne constant $$ How should I do it ?

Answer 1

We assume $n\ge 2$ as in Huisken. Indeed your claim is not true for $n=1$. The equality

$$g^{ij}g^{kl}h_{ik}h_{jl}=\frac{(g^{ij}h_{ij})^2}{n}$$

is the equality case of the Cauchy Schwarz inequality: Take an orthonormal basis so that $g_{ij}= \delta_{ij}$ and $h_{ij}$ is also diagonal with eigenvalues $\lambda_1, \cdots, \lambda_n$. The above is

$$ \sum_i \lambda_i^2 = \frac{1}{n} (\lambda_1+\cdots+\lambda_n)^2. $$

This implies $\lambda_1 = \cdots =\lambda_n$ and thus $h_{ij} = \frac 1nHg_{ij}$, where $H = g^{ij} h_{ij}$. That is, $M$ is a umbilical hypersurfaces. In $n=2$, this implies that $M$ is a part of plane or sphere (I am not sure if that is true in general dimensions). Since $M$ is compact, $M$ has to be the round sphere.

For $n\ge 3$, we use your calculation

$$R_{ijkl} = \frac{H^2}{n^2} (g_{il}g_{jk} - g_{ik} g_{jl})$$

In particular, we have

$$ R_{ijji } = \frac{H^2}{n^2}$$

for all $i\neq j$. When $n\ge 3$, your assertion follows from Schur's theorem.