# Cheeger-Gromoll splitting proof

Background information: $E_i$ is the parallel orthonormal frame along $c$ and $E_n=\nabla f\circ c$.

Lemma: Let $M$ be a Riemannian manifold and $f\in C^\infty(M)$ with $||\text{grad} f||=1$. If $c$ is an integral curve of $\text{grad}f$, then $c$ is a geodesic realizing the distance between any two of its points and

\begin{align} -\text{Ric}(\dot{c},\dot{c})&=(\Delta f\circ c)'+||\text{Hess}f\circ c||^2\\ &\geq (\Delta f\circ c)'+\frac{1}{n-1}(\Delta f\circ c)^2\\ \end{align}

where $\Delta f=\text{tr} \text{ Hess} f$.

Proof

This is a proof from Eschenburg and Heintze's paper, An Elementary Proof of the Cheeger-Gromoll Splitting theorem. Below are my questions and concerns. I would really appreciate it if someone could help me understand the steps. Thanks.

Question 1: Using Petersons' text it seems that $$||\text{Hess}f||^2=\sum_{i,j=1}^n\langle \nabla_{E_i}E_n,E_j\rangle^2$$ so I don't see how the author gets his computation from.

Question 2 I don't see why $-||\text{Hess}f||^2\leq- \sum_{i=1}^n\langle\text{Hess}f(E_i),E_i\rangle^2$

Question 3 I don't see how they get the inequality $-\sum_{i=1}^n\langle\text{Hess}f(E_i),E_i\rangle^2\leq -\frac{1}{n-1}(\Delta f)^2$.

Question 1 Given a function $f : M \to \mathbb R$ and $g$ a metric, the Hessian of $f$, $H = \text{Hess}_f$ is a $(1,1)$-tensor $TM \to TM$ defined as $$\text{Hess}_f (X) =\nabla_X \nabla f.$$ For any $(1,1)$ tensor, one can define the norm $\|H\|$ to be $$\| H\|^2 = \sum_{i,j=1}^n \langle H(E_i), E_j\rangle^2,$$ where $\{E_1, E_2, \cdots, E_n\}$ is any orthonormal basis. Thus in our situation $$\|\text{Hess}_f\|^2 = \sum_{i,j=1}^n \langle \text{Hess}_f(E_i), E_j\rangle^2 = \sum_{i,j=1}^n \langle \nabla_{E_i} \nabla f , E_j\rangle^2 = \sum_{i,j=1}^n \langle \nabla_{E_i} E_n , E_j\rangle^2$$ as in our case $\nabla f = E_n$ along $c$.

Question 2 follows from the defintion: $$\begin{split} \|\text{Hess}_f\|^2 &= \sum_{i,j=1}^n \langle \text{Hess}_f(E_i) , E_j\rangle^2 \\ &\ge \sum_{i=1}^n \langle \text{Hess}_f (E_i), E_i\rangle^2. \end{split}$$

Question 3 Using $\Delta f = \text{tr}\text{Hess}_f$, $$\begin{split} (\Delta f)^2 &= (\text{tr} \text{Hess}_f)^2 \\ &= \left(\sum_{i=1}^n \langle\text{Hess}_f(E_i), E_i\rangle\right)^2 \\ &= \left(\sum_{i=1}^{n-1} \langle\text{Hess}_f(E_i), E_i\rangle\right)^2 \end{split}$$

since the last term is zero:
$$\langle \text{Hess}_f (E_n), E_n \rangle= \langle \nabla_{E_n} E_n,E_n\rangle =\frac 12 E_n \langle E_n, E_n\rangle = 0.$$

Now by Cauchy Schwarz inequality, $$\begin{split} (\Delta f)^2 &= \left(\sum_{i=1}^{n-1} \langle\text{Hess}_f(E_i), E_i\rangle\right)^2 \\ &\le \left(\sum_{i=1}^{n-1} 1^2 \right)\left(\sum_{i=1}^{n-1} \langle \text{Hess}_f (E_i), E_i\rangle^2 \right)\\ &= (n-1) \left(\sum_{i=1}^{n-1} \langle \text{Hess}_f (E_i), E_i\rangle^2 \right). \end{split}$$

• Hi John, so your answer helps with everything except the jump the author makes from line 3 to line 4. His definition of the norm of the Hessian is different from what you and I defined. – Enigma Dec 1 '15 at 7:15
• Can you state the definition that the authors used? @Enigma – user99914 Dec 1 '15 at 7:17
• So the author basically says that $||\text{Hess}f||^2=\sum_{i,j=1}^n\langle \nabla_{E_i}E_n,E_j\rangle\langle\nabla_{E_j}E_n,E_i\rangle$, which doesn't seem to coincide with our defintion. This can be seen in the equality from line 3 to line 4 in the proof given by the author. – Enigma Dec 1 '15 at 7:30
• That's because $\langle \text{Hess}_f (X), Y\rangle = \langle \text{Hess}_f (Y), X\rangle$ as $\langle \nabla_X \nabla f, Y\rangle = X(Yf ) - \langle \nabla f, \nabla _X Y\rangle$ (see the calculation here. @Enigma – user99914 Dec 1 '15 at 7:39