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} Rright)=Delta nabla^{k} R-rleft(nabla^{k} Rright)+sum_{j=0}^{lfloor k / 2rfloor}left(nabla^{j} Rright) otimes_{g}left(nabla^{k-j} Rright)

$$$$

begin{array}{l}

frac{partial}{partial t}(|nabla^k R |^2) =Deltaleft|nabla^{k} Rright|^{2}-2left|nabla^{k+1} Rright|^{2}-(k+2) rleft|nabla^{k} Rright|^{2}

+left(nabla^{k} Rright) otimes_{g}left(sum_{j=0}^{lfloor k / 2rfloor}left(nabla^{j} Rright) otimes_{g}left(nabla^{k-j} Rright)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 / 2rfloor}left(nabla^{j} Rright) otimes_{g(t)}left(nabla^{n-j} Rright)

$$

$$

begin{array}{c}

nabla^{n} R^{2}=displaystyle{sum_{j=0}^{lfloor n / 2rfloor}left(nabla^{j} Rright) otimes_{g(t)}left(nabla^{n-j} Rright) }\

left(frac{partial}{partial t} Gammaright) otimes_{g(t)}left(nabla^{j} Rright)=(nabla R) otimes_{g(t)}left(nabla^{j} Rright)

end{array}

$$

$$

frac{partial}{partial t}left(nabla_{k_{1}} nabla_{k_{2}} ldots nabla_{k_{n}} Rright)=nabla_{k_{1}}left{partial_t nabla_{k_{2}} ldots nabla_{k_{n}} Rright}-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 $displaystyle{sum_{j=0}^{lfloor k / 2rfloor}left(nabla^{j} Rright) otimes_{g}left(nabla^{k-j} Rright)}$ to avoid taking up too much space. The remaining terms are of the form:

$$ begin{aligned}

&left(partial_{t} Gammaright) otimes nabla^{k-1} R=nabla R otimes nabla^{k-1} R=Sigma\

&nabla^r(partial_t Gamma) otimes nabla^{k-r-1} R = nabla^{r+1}R otimes nabla^{k-r-1}R = Sigma end{aligned}

$$

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} Rright|^{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} Rright) \

&=(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.