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I am reading Eschenburg and Heintze's proof of the Cheeger-Gromoll splitting theorem. Lemma 1 states:

Let $f\in C^\infty(M)$ with $||grad(f)||=1$. If c is an integral curve of the gradient, then it is a minimising geodesic and $$\begin{align} -Ricc(c', c') & =(\Delta f\circ c)'+||Hess_f\circ c||^2 \\ & \le (\Delta f\circ c)' +\frac{1}{n-1}(\Delta f\circ c)^2 \end{align}$$

The first part, about the minimising geodesics, has already been answered here

They begin the proof by choosing some $t_0\in \mathbb{R}$ and they take a neighbourhood $U$ of $c(t_0)$, where they choose and orthonormal frame $\{ E_1,E_2,...,E_n\}$ such that $E_n=gradf$ and $E_i$ are parallel along $gradf$. Then the calculation give $$\begin{align} Ricc(E_n,E_n) &=\sum_{i=1}^{n} \left< R(E_i,E_n)E_n,E_i\right> \\ &=\sum_i\left( -\left< \nabla_{E_n}\nabla _{E_i}E_n,E_i \right> -\left< \nabla_{\nabla_{E_i}E_n}E_n,E_i\right> \right) \\ &=-E_n\left(\sum_i \left< \nabla_{E_i}E_n,E_i\right>\right) -\sum_i\left< \nabla_{E_i}E_n,\nabla_{E_i}E_n\right> \\ &=-E_n(\Delta f)-||Hess_f||^2 \end{align}$$

Tis way we prooved th equality of the Lemma. I understand intuitively the existence of such a frame $\{ E_1,E_2,...,E_n\}$ but I would like to see a rigorous proof. Mainly I don't understand how one gets from this to the inequality. All the authors say is that from the Schwarz inequality we have $$\begin{align} Ricc(E_n,E_n) & \leq -E_n(\Delta f)-\sum_{i=1}^{n-1}\left<Hess_f(E_i),E_i\right>\\ & \leq -E_n(\Delta f) -\frac{1}{n-1}(\Delta f)^2 \end{align}$$ But I can't see how it is done

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For the second question, note that $\operatorname{Hess}f$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis $\{E_1, \cdots, E_n\}$. We can also choose $E_n = c' = \nabla f$ since $c'$ is an eigenvector with eigenvalue $0$: $$ \operatorname{Hess} f(c') = c' c' f - \nabla_{\nabla_{c'}c'} f = 0$$

Thus we write

$$ \operatorname{Hess} f(E_i, E_j) = \delta_{ij} \lambda_i.$$

with $\lambda_n = 0$. In terms of $\lambda_i$'s we have $$ \Delta f = \sum_{i=1}^{n-1} \lambda_i, \ \ \ \| \operatorname{Hess} (f)\|^2 = \sum_{i=1}^{n-1} \lambda_i^2.$$

Thus we have (By Cauchy Schwarz inequality)

\begin{align} \Delta f &= \sum_{i=1}^{n-1} \lambda_i\\ &= (\lambda_1, \cdots, \lambda_{n-1}) \cdot (1, \cdots, 1) \\ &\le |(\lambda_1, \cdots, \lambda_{n-1})| | (1, \cdots, 1)| \\ &= \sqrt{\lambda_1^2 + \cdots + \lambda_{n-1}^2} \sqrt{n-1} \\ \Rightarrow (\Delta f)^2 &\le (n-1) \|\operatorname{Hess} f\|^2 \end{align}

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