Ytukyg

Let $(M^n,g)$ be a Riemannian manifold, and $T$ a symmetric $(1,1)$-tensor field, i.e., $langle T(X),Yrangle = langle X,T(Y)rangle $. For convenience, denote $$Delta_Tu=sum_ilangle nabla_{e_i}nabla u, Te_irangle $$
and
$$mathrm{Ric}_T(X,Y)=sum_ilangle R(X,e_i)(Te_i), Yrangle , $$
where $u$ is a smooth function on $M$ and ${e_i}$ is a local ON frame field.

Now assume that $T$ is a Codazzi operator, i.e., for any $X,Yin Gamma(TM)$, $(nabla_XT)Y=(nabla_YT)X$. We choose ${e_i}_{i=1}^n$ be a local orthonormal frame field of $M$ such that $nabla_{star }e_i=0$ at the considered point. For the distance function r(x) from a fixed point $x_0$, by the definition, we have ($nabla_XT$ is symmetric since $T$ is symmetric)
begin{equation*}
begin{split}
Delta_{nabla_{partial_r}T}r=&sum_{i=1}^nlangle nabla_{e_i}partial_r,(nabla_{partial_r}T)e_irangle=sum_{i=1}^nlangle nabla_{e_i}partial_r,(nabla_{e_i}T)partial_rrangle \
=&sum_{i=1}^ne_ilangle partial_r,(nabla_{e_i}T)partial_rrangle -sum_{i=1}^nlangle partial_r,(nabla_{e_i}nabla_{e_i}T)partial_rrangle -sum_{i=1}^nlangle partial_r,(nabla_{e_i}T)(nabla_{e_i}partial_r)rangle .
end{split}
end{equation*}
However,
begin{equation*}
begin{split}
sum_{i=1}^nlangle partial_r,(nabla_{e_i}T)(nabla_{e_i}partial_r)rangle =&sum_{i=1}^nlangle (nabla_{e_i}T)partial_r,nabla_{e_i}partial_rrangle \
=&sum_{i=1}^nlangle (nabla_{partial_r}T)e_i,nabla_{e_i}partial_rrangle =Delta_{nabla_{partial_r}T}r.
end{split}
end{equation*}
Hence, we obtain
begin{equation}
begin{split}
Delta_{nabla_{partial_r}T}r=frac{1}{2}sum_{i=1}^ne_ilangle partial_r,(nabla_{e_i}T)partial_rrangle -frac{1}{2}sum_{i=1}^nlangle partial_r,(nabla_{e_i}nabla_{e_i}T)partial_rrangle
end{split}
end{equation}
We now compute the two terms of the R.H.S. of the above equality. Firstly, notice that $nabla_{partial_r}partial_r=0$, we have
begin{equation}
begin{split}
sum_{i=1}^ne_ilangle partial_r,(nabla_{e_i}T)partial_rrangle =&sum_{i=1}^ne_ilangle partial_r,(nabla_{partial_r}T)e_irangle =sum_{i=1}^ne_ilangle (nabla_{partial_r}T)partial_r,e_irangle \
=&sum_{i=1}^ne_i (partial_rlangle Tpartial_r, e_irangle )-sum_{i=1}^ne_ilangle Tpartial_r,nabla_{partial_r}e_irangle \
=&sum_{i=1}^npartial_r(e_ilangle Tpartial_r,e_irangle )-sum_{i=1}^nlangle Tpartial_r, nabla_{e_i}nabla_{partial_r}e_irangle \
=&sum_{i=1}^npartial_rlangle (nabla_{e_i}T)partial_r,e_irangle +sum_{i=1}^npartial_rlangle Tnabla_{e_i}partial_r, e_irangle \
&+sum_{i=1}^npartial_rlangle Tpartial_r,nabla_{e_i}e_irangle -sum_{i=1}^nlangle Tpartial_r, nabla_{e_i}nabla_{partial_r}e_irangle \
=&sum_{i=1}^nlangle (nabla_{partial_r}nabla_{e_i}T)partial_r,e_irangle +partial_r(Delta_Tr)\
&+sum_{i=1}^nlangle Tpartial_r,nabla_{partial_r}nabla_{e_i}e_irangle -sum_{i=1}^nlangle Tpartial_r, nabla_{e_i}nabla_{partial_r}e_irangle \
=&sum_{i=1}^nlangle (nabla_{partial_r}nabla_{e_i}T)partial_r,e_irangle +partial_r(Delta_Tr)+mathrm{Ric}(partial_r, Tpartial_r).
end{split}
end{equation}
Secondly,
begin{equation}
begin{split}
sum_{i=1}^nlangle partial_r,(nabla_{e_i}nabla_{e_i}T)partial_rrangle =&sum_{i=1}^nlangle partial_r,nabla_{e_i}((nabla_{e_i}T)partial_r)rangle -sum_{i=1}^nlangle partial_r,(nabla_{e_i}T)nabla_{e_i}partial_rrangle \
=& sum_{i=1}^nlangle partial_r,nabla_{e_i}((nabla_{partial_r}T)e_i)rangle -sum_{i=1}^nlangle partial_r,(nabla_{nabla_{e_i}partial_r}T)e_irangle \
=&sum_{i=1}^nlangle (nabla_{e_i}nabla_{partial_r}T)e_i-(nabla_{nabla_{e_i}partial_r}T)e_i,partial_rrangle \
=&sum_{i=1}^nlangle (nabla_{partial_r}nabla_{e_i}T)e_i,partial_rrangle -sum_{i=1}^nlangle (R(partial_r,e_i)T)e_i,partial_rrangle \
=&sum_{i=1}^nlangle (nabla_{partial_r}nabla_{e_i}T)e_i,partial_rrangle +mathrm{Ric}(partial_r,Tpartial_r)-mathrm{Ric}_T(partial_r,partial_r).
end{split}
end{equation}
From the above three equalities we obtain
begin{equation*}
begin{split}
Delta_{nabla_{partial_r}T}r
=frac{1}{2}partial_r(Delta_Tr)
+frac{1}{2}mathrm{Ric}_T(partial_r,partial_r).
end{split}
end{equation*}

Now, my question is that when $T=mathrm{Id}_{TM}$ the above equation becomes
begin{equation*}
begin{split}
partial_r(Delta_r)+mathrm{Ric}(partial_r,partial_r)=0.
end{split}
end{equation*}
But it is well known that the Bochner formula for the distance function
begin{equation*}
begin{split}
|mathrm{Hess}r|^2+partial_r(Delta_r)+mathrm{Ric}(partial_r,partial_r)=0.
end{split}
end{equation*}
This obtain a contradiction.

What is wrong with the above derivation? Thanks in advence.