# Convex function on Riemannian manifold

When I read the 9.5 of Topping's Lectures on the Ricci flow, I have some problem.

Assume $$W$$ is a vector bundle over manifold $$(M,g)$$, and $$A$$ is connection on $$W$$. $$\{e_1,...,e_l\}$$ is a frame of $$W_p$$, and extend it to a local frame for $$W$$ by radial parallel translation using the connection $$A$$, and $$E\in \Gamma(W)$$.

$$\Psi :W\rightarrow \mathbb R$$ is parallel function, namely if $$\omega_1\in W$$ can be parallel translated (using the connection A) into $$\omega_2\in W$$, then $$\Psi(\omega_1)=\Psi(\omega_2)$$. Denote the restriction of $$\Psi$$ to the fibre $$W_p$$ as $$\Psi_p$$, then there is $$\nabla d (\Psi\circ E)(p)= Hess(\Psi_p)(E(p)) (AE(p), AE(p)) \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + d\Psi_p(E(p))(A^2E(p)+\frac{1}{2}R_A(\cdot,\cdot)E(p)) \tag{9.5.5}$$ If assume $$\Psi_p$$ weakly convex, after taking the trace of $$(9.5.5)$$, how to get $$\Delta_M(\Psi\circ E)(p)\ge d\Psi_p(E(p))(\Delta_A E(p)) ~~~? \tag{9.5.6}$$

What I try:

First, the weak convex means $$\text{Hess}(\Psi_p)(E(p)) (AE(p), AE(p)) \ge 0 \tag{1}$$ but I don't know how to deal $$\text{tr } d\Psi_p(E(p))R_A(\cdot,\cdot)E(p)) \ge 0 \tag{2}$$ I guess, the convex on manifold is not same with linear space. Maybe the weak convex on manifold mean $$(1)$$ and $$(2)$$. But I can't find the definition of convex on Riemannian manifold.

PS(2022-6-25): After some thinking, I think it is irrelevant with convex on manifold, since $$W_p$$ is vector space. Besides, by maltreat the Weitzenbock formula, I give a rough process :

Weitzenbock formula: assuming $$\omega=\omega_i dx^i$$ and $$\overline\omega$$ is the dual vector of $$\omega$$, then $$\Delta \omega (Y) = tr \nabla^2 \omega(Y) - R(\overline\omega,Y)$$

Roughly, (or similarly), I have $$(\Delta_A E)(\omega) = (tr A^2 E)\omega - g^{ij} R(e_i,\omega)E(e_j)$$ By Bianchi, I have $$R(e_i, \omega) E(e_j) + R(\omega, E)e_i(e_j) + R(E,e_i)\omega(e_j)=0$$ Therefore, (treat $$R(e_i, \omega) E(e_j)$$ as $$\langle R(e_i, \omega) E, e_j \rangle$$ ) $$g^{ij}R(e_i,\omega) E(e_j) = -\frac{1}{2}g^{ij} R(e_i, e_j)E(\omega)$$ At last, I get $$\Delta_A E = tr A^2 E+ \frac{1}{2} tr R(\cdot, \cdot)E$$ So, the $$(2)$$ is redundant. And since it is ordinary convex, $$(1)$$ must be right. Therefore, we can get $$(9.5.6)$$. But the process is very rough. Who can rigorous it? Thanks.