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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 $T$ 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{equation} \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} \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_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} \end{equation} Secondly, \begin{equation} \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} \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.

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  • $\begingroup$ I don't see how you get the following equality. Can you explain a bit more? \begin{equation} \begin{split} &\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 \end{split}\end{equation} $\endgroup$ Commented Dec 18, 2018 at 15:58
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    $\begingroup$ I am not very sure if this question is suitable for MO $\endgroup$ Commented Dec 18, 2018 at 17:47
  • $\begingroup$ $\langle T\partial_r,e_i\rangle $ is a function and $\{e_i\}$ is normal frame.@Willie Wong $\endgroup$
    – G Z
    Commented Dec 18, 2018 at 18:33
  • $\begingroup$ Sorry, my question may be a little superficial, I will take a warning from it later. @Praphulla Koushik $\endgroup$
    – G Z
    Commented Dec 18, 2018 at 20:21

1 Answer 1

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Let me re-do the computation using index notation.

Your goal is to compute $\partial_r (\Delta_T r)$ which we can rewrite as

$$ \nabla^c( T^{ab} \nabla^2_{ab} r) \nabla_c r = \nabla^c T^{ab} \nabla^2_{ab} r \nabla_c r + T^{ab} (\nabla^c \nabla_a \nabla_b r) \nabla_c r$$

The first term on the right is $\Delta_{\nabla_{\partial_r} T} r$. So we focus our attention on the second term.

$$ \nabla^c\nabla_a \nabla_b r = \nabla_a\nabla^c \nabla_b r + [\nabla^c, \nabla_a ] \nabla_b r $$

So (the $\pm$ is just me forgetting which sign is the right one)

$$ T^{ab} (\nabla^c\nabla_a \nabla_b r) \nabla_c r = T^{ab}(\nabla_a \nabla^c \nabla_b r) \nabla_c r \pm \mathrm{Ric}_T(\partial_r, \partial_r) $$

The term in the middle is

$$ \nabla_a \nabla^c \nabla_b r \nabla_c r = \nabla_a (\underbrace{\nabla^c \nabla_b r \cdot \nabla_c r}_{ = 0}) - \nabla^c \nabla_b r \nabla_a \nabla_c r.$$

The term that $=0$ is so due to the fact that $\nabla r$ is geodesic. So your identity should be

$$ \partial_r (\Delta_T r) + \mathrm{Ric}_T(\partial_r, \partial_r) = \Delta_{\nabla_{\partial_r} T} r - T^{ab} g^{cd} \nabla^2_{bd} r \nabla^2_{ac} r $$

which now reduces correctly to Bochner's identity when $T = g$.

Remark Notice that nowhere in this derivation is either of the property you started with, namely that $T$ is symmetric and $T$ is Codazzi, used.

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  • $\begingroup$ Prof. Wong, thanks a lot for your answer. But, in fact, I already know the equality that you derived. It can also be derived directly from the formula concerning $\Delta_T$. See the formula (2.1) in \url{ac.els-cdn.com/S0362546X18301573/…}. My goal is to know what $\Delta_Tr$ is going to be when $T$ is Codazzi. That's why I asked this question to find out where my computation was wrong. Anyway, thank you very much. $\endgroup$
    – G Z
    Commented Dec 18, 2018 at 20:01

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