I'm trying to prove the following identities (under the normalized Ricci flow on surfaces, on which $\partial_t g = (r-R)g$ holds true, where $r$ denotes the average scalar curvature and has the same sign as the Euler characteristic):
$$ \frac{\partial}{\partial t}\left(\nabla^{k} R\right)=\Delta \nabla^{k} R-r\left(\nabla^{k} R\right)+\sum_{j=0}^{\lfloor k / 2\rfloor}\left(\nabla^{j} R\right) \otimes_{g}\left(\nabla^{k-j} R\right) $$
$$ \begin{array}{l} \frac{\partial}{\partial t}(|\nabla^k R |^2) =\Delta\left|\nabla^{k} R\right|^{2}-2\left|\nabla^{k+1} R\right|^{2}-(k+2) r\left|\nabla^{k} R\right|^{2} +\left(\nabla^{k} R\right) \otimes_{g}\left[\sum_{j=0}^{\lfloor k / 2\rfloor}\left(\nabla^{j} R\right) \otimes_{g}\left(\nabla^{k-j} R\right)\right] \end{array} $$
where by $A \otimes_g B$ we refer to any tensor field which is a finite linear combination of contractions and metric contractions of the tensor product $A \otimes B$. Now, I've already proven that the following hold:
$$\partial_{t} \nabla R=\Delta \nabla R+\frac{3}{2} R \nabla R-r \nabla R$$ $$ \nabla^{n} \Delta R-\Delta \nabla^{n} R=\sum_{j=0}^{\lfloor n / 2\rfloor}\left(\nabla^{j} R\right) \otimes_{g(t)}\left(\nabla^{n-j} R\right) $$ $$ \begin{array}{c} \nabla^{n} R^{2}=\displaystyle{\sum_{j=0}^{\lfloor n / 2\rfloor}\left(\nabla^{j} R\right) \otimes_{g(t)}\left(\nabla^{n-j} R\right) }\\ \left(\frac{\partial}{\partial t} \Gamma\right) \otimes_{g(t)}\left(\nabla^{j} R\right)=(\nabla R) \otimes_{g(t)}\left(\nabla^{j} R\right) \end{array} $$ $$ \frac{\partial}{\partial t}\left(\nabla_{k_{1}} \nabla_{k_{2}} \ldots \nabla_{k_{n}} R\right)=\nabla_{k_{1}}\left\{\partial_t \nabla_{k_{2}} \ldots \nabla_{k_{n}} R\right\}-\sum_{l=2}^{n}\left(\partial_{t} \Gamma_{k_{1} k_{l}}^{m}\right) \nabla_{k_{2}} \ldots \nabla_{k_{l-1}} \nabla_{m} \ldots \nabla_{k_{n}} R $$
So, to prove the first formula for the evolution of $\nabla^k R$, I used recursive applications of this last identity just above, but I'd like someone to check my work. I noticed there would be terms of the form:
\begin{align*} \nabla_{k_1} \cdots \nabla_{k_{n-1}}(\partial_t \nabla_{k_n} R) &= \nabla_{k_1} \cdots \nabla_{k_{n-1}}(\Delta \nabla_{k_n}R + \frac{3}{2} R \nabla_{k_n} R - r \nabla_{k_n R}) \\ &= \nabla_{k_1} \cdots \nabla_{k_{n-1}} ( \nabla_{k_n} \Delta R + \Sigma + \frac{3}{2} R \nabla_{k_n} R - r \nabla_{k_n R} )\\ &=\nabla^k \Delta R + \Sigma - r(\nabla^{k} R) \\ &=\Delta \nabla^k R + \Sigma - r(\nabla^k R) \end{align*}
where by $\Sigma$ I'm denoting $\displaystyle{\sum_{j=0}^{\lfloor k / 2\rfloor}\left(\nabla^{j} R\right) \otimes_{g}\left(\nabla^{k-j} R\right)}$ to avoid taking up too much space. The remaining terms are of the form:
$$ \left(\partial_{t} \Gamma\right) \otimes \nabla^{k-1} R=\nabla R \otimes \nabla^{k-1} R=\Sigma $$
and so we have proved the first identity. But I didn't manage to prove the second one. We have:
$$\begin{aligned} \frac{\partial}{\partial t}\left|\nabla^{k} R\right|^{2} &= \frac{\partial}{\partial t}\left(g^{i_1 p_1} \cdots g^{i_k p_k} \nabla_{i_1} \cdots \nabla_{i_k} R \nabla_{p_1} \cdots \nabla_{p_k} R\right) \\ &=(R-r)k \|\nabla^k R \|^2 + 2 \langle \nabla^k R, \partial_t(\nabla^k R) \rangle \end{aligned} $$
and since
\begin{align*} 2 \langle \nabla^k R, \partial_t(\nabla^k R) \rangle &= 2 \langle \nabla^k R, \Delta \nabla^k R + \Sigma - r \nabla^k R \rangle \\ &=2 \langle \nabla^k R, \Delta \nabla^k R \rangle + 2 \langle \nabla^k R, \Sigma \rangle - 2r \|\nabla^k R\|^2 \end{align*}
We're then left to prove that:
$$2 \langle \nabla^k R, \Delta \nabla^k R \rangle = -kR \|\nabla^k R\|^2 + \Delta \|\nabla^k R\|^2 - 2 \|\nabla^{k+1} R\|^2$$
but I've been stuck on this one for a while. I'd really appreciate some help on this! Thanks in advance.