$R_{ijkl}$ is curvature ,$\nabla $ is Riemannian connection ,$f$ is a function on Riemannian manifold. $g^{ij}$ is inverse of $g_{ij}$.
If $\nabla_i\nabla_jv_k-\nabla_j\nabla_iv_k=R_{ijkl}g^{lm}v_m$, How to get that $\nabla_i\nabla_j\nabla_kf-\nabla_j\nabla_i\nabla_kf=R_{ijkl}\nabla_lf$ ?
In fact , I don't know whether exactly I have asked what I want to know .The question is from the paper.In the first picture , what is the mean of $\nabla_i v^j$, I think it should be $\nabla_i (v^j\frac{\partial}{\partial x^j})$.And the same is the $\nabla_iv_j$.Besides, how to get the 1.1.3 ? And the second picture is about my first quesion.
Using $$\nabla_i\nabla_j v_k - \nabla_j\nabla_i v_k = R_{ijkl}g^{lm} v_m,$$ and put $v_k = \nabla_k f$ gives $$\nabla_i\nabla_j \nabla_k f - \nabla_j\nabla_i \nabla_k f= R_{ijkl}g^{lm} \nabla_m f = R_{ijkl} \nabla_l f.$$ ($g^{lm} = \delta_{lm}$ used). Now put $$R_{ij} + \nabla_i\nabla_jf = 0\Rightarrow \nabla_i\nabla_jf = -R_{ij}$$ into the previous equation gives $$\nabla_i(-R_{jk}) - \nabla_j(-R_{ik}) = R_{ijkl} \nabla_l f.$$ And that is the same as the second last equation from your linked page. Now take trace on $(j,k)$ gives (as $R_{ii} = R$, $R_{il} = R_{ijjl}$) $$-\nabla_iR + \nabla_jR_{ij} = R_{ij} \nabla_j f.$$ Then using the Bianchi identity you'll get $$(*)\ \ \ \ \ \nabla_i R - 2R_{ij} \nabla_j f = 0.$$ Now $$\begin{split} \nabla_i |\nabla f|^2 &= \nabla_i \left( \nabla_j f\right)^2 \\ &= 2 \nabla_j f \nabla_j \nabla_i f \\ &=-2 \nabla_j f (R_{ij}) \end{split}$$ (Using the definition of gradient Ricci soliton)
As a result, $$\begin{split} \nabla_i (|\nabla f|^2 + R) &= \nabla_i |\nabla f|^2 + \nabla_i R \\ &= -2 R_{ij} \nabla_j f + \nabla_i R \\ &=0 \end{split}$$ by $(*)$. Thus the function $|\nabla f|^2 + R$ is a constant function, hence (1.1.15).