A mistake I can't find about the Bochner formula












3












$begingroup$


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.






differential-geometry riemannian-geometry







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    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.






    differential-geometry riemannian-geometry







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      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.






      differential-geometry riemannian-geometry







      share|cite|improve this question











      $endgroup$




      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.






      differential-geometry riemannian-geometry




      differential-geometry riemannian-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 7:10







      G. Zhao

















      asked Dec 18 '18 at 6:34









      G. ZhaoG. Zhao

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