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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.

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