# A mistake I can't find about the Bochner formula

Let $$(M^n,g)$$ be a Riemannian manifold, and $$T$$ a symmetric $$(1,1)$$-tensor field, i.e., $$\langle T(X),Y\rangle = \langle X,T(Y)\rangle$$. For convenience, denote $$\Delta_Tu=\sum_i\langle \nabla_{e_i}\nabla u, Te_i\rangle$$ and $$\mathrm{Ric}_T(X,Y)=\sum_i\langle R(X,e_i)(Te_i), Y\rangle ,$$ 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 $$X,Y\in \Gamma(TM)$$, $$(\nabla_XT)Y=(\nabla_YT)X$$. 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) $$\begin{equation*} \begin{split} \Delta_{\nabla_{\partial_r}T}r=&\sum_{i=1}^n\langle \nabla_{e_i}\partial_r,(\nabla_{\partial_r}T)e_i\rangle=\sum_{i=1}^n\langle \nabla_{e_i}\partial_r,(\nabla_{e_i}T)\partial_r\rangle \\ =&\sum_{i=1}^ne_i\langle \partial_r,(\nabla_{e_i}T)\partial_r\rangle -\sum_{i=1}^n\langle \partial_r,(\nabla_{e_i}\nabla_{e_i}T)\partial_r\rangle -\sum_{i=1}^n\langle \partial_r,(\nabla_{e_i}T)(\nabla_{e_i}\partial_r)\rangle . \end{split} \end{equation*}$$ However, $$\begin{equation*} \begin{split} \sum_{i=1}^n\langle \partial_r,(\nabla_{e_i}T)(\nabla_{e_i}\partial_r)\rangle =&\sum_{i=1}^n\langle (\nabla_{e_i}T)\partial_r,\nabla_{e_i}\partial_r\rangle \\ =&\sum_{i=1}^n\langle (\nabla_{\partial_r}T)e_i,\nabla_{e_i}\partial_r\rangle =\Delta_{\nabla_{\partial_r}T}r. \end{split} \end{equation*}$$ Hence, we obtain $$$$\begin{split} \Delta_{\nabla_{\partial_r}T}r=\frac{1}{2}\sum_{i=1}^ne_i\langle \partial_r,(\nabla_{e_i}T)\partial_r\rangle -\frac{1}{2}\sum_{i=1}^n\langle \partial_r,(\nabla_{e_i}\nabla_{e_i}T)\partial_r\rangle \end{split}$$$$ We now compute the two terms of the R.H.S. of the above equality. Firstly, notice that $$\nabla_{\partial_r}\partial_r=0$$, we have $$$$\begin{split} \sum_{i=1}^ne_i\langle \partial_r,(\nabla_{e_i}T)\partial_r\rangle =&\sum_{i=1}^ne_i\langle \partial_r,(\nabla_{\partial_r}T)e_i\rangle =\sum_{i=1}^ne_i\langle (\nabla_{\partial_r}T)\partial_r,e_i\rangle \\ =&\sum_{i=1}^ne_i (\partial_r\langle T\partial_r, e_i\rangle )-\sum_{i=1}^ne_i\langle T\partial_r,\nabla_{\partial_r}e_i\rangle \\ =&\sum_{i=1}^n\partial_r(e_i\langle T\partial_r,e_i\rangle )-\sum_{i=1}^n\langle T\partial_r, \nabla_{e_i}\nabla_{\partial_r}e_i\rangle \\ =&\sum_{i=1}^n\partial_r\langle (\nabla_{e_i}T)\partial_r,e_i\rangle +\sum_{i=1}^n\partial_r\langle T\nabla_{e_i}\partial_r, e_i\rangle \\ &+\sum_{i=1}^n\partial_r\langle T\partial_r,\nabla_{e_i}e_i\rangle -\sum_{i=1}^n\langle T\partial_r, \nabla_{e_i}\nabla_{\partial_r}e_i\rangle \\ =&\sum_{i=1}^n\langle (\nabla_{\partial_r}\nabla_{e_i}T)\partial_r,e_i\rangle +\partial_r(\Delta_Tr)\\ &+\sum_{i=1}^n\langle T\partial_r,\nabla_{\partial_r}\nabla_{e_i}e_i\rangle -\sum_{i=1}^n\langle T\partial_r, \nabla_{e_i}\nabla_{\partial_r}e_i\rangle \\ =&\sum_{i=1}^n\langle (\nabla_{\partial_r}\nabla_{e_i}T)\partial_r,e_i\rangle +\partial_r(\Delta_Tr)+\mathrm{Ric}(\partial_r, T\partial_r). \end{split}$$$$ Secondly, $$$$\begin{split} \sum_{i=1}^n\langle \partial_r,(\nabla_{e_i}\nabla_{e_i}T)\partial_r\rangle =&\sum_{i=1}^n\langle \partial_r,\nabla_{e_i}((\nabla_{e_i}T)\partial_r)\rangle -\sum_{i=1}^n\langle \partial_r,(\nabla_{e_i}T)\nabla_{e_i}\partial_r\rangle \\ =& \sum_{i=1}^n\langle \partial_r,\nabla_{e_i}((\nabla_{\partial_r}T)e_i)\rangle -\sum_{i=1}^n\langle \partial_r,(\nabla_{\nabla_{e_i}\partial_r}T)e_i\rangle \\ =&\sum_{i=1}^n\langle (\nabla_{e_i}\nabla_{\partial_r}T)e_i-(\nabla_{\nabla_{e_i}\partial_r}T)e_i,\partial_r\rangle \\ =&\sum_{i=1}^n\langle (\nabla_{\partial_r}\nabla_{e_i}T)e_i,\partial_r\rangle -\sum_{i=1}^n\langle (R(\partial_r,e_i)T)e_i,\partial_r\rangle \\ =&\sum_{i=1}^n\langle (\nabla_{\partial_r}\nabla_{e_i}T)e_i,\partial_r\rangle +\mathrm{Ric}(\partial_r,T\partial_r)-\mathrm{Ric}_T(\partial_r,\partial_r). \end{split}$$$$ From the above three equalities we obtain $$\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.