# Ngô Quốc Anh

## June 19, 2013

### The Bochner formula for vector fields

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 2:50

The Bochner formula for a gradient vector field is a important tool in geometric analysis which basically says that

$\displaystyle\frac{1}{2}\Delta (|\nabla u{|^2}) = |{\nabla ^2}u{|^2} + \langle \nabla \Delta u,\nabla u\rangle + {\rm Ric}(\nabla u,\nabla u)$

for any function $f$. Whenever $|\nabla u| \ne 0$, we have a  formula for $|\nabla u|$ as follows

$\displaystyle\frac{1}{2}\Delta (|\nabla u|) = \frac{1}{|\nabla u|} \left( \nabla u \cdot \nabla u ( \Delta u) + {\rm Ric}(\nabla u,\nabla u) + |\nabla \nabla u|^2 - \bigg| \Big\langle \nabla \nabla u, \frac{\nabla u}{|\nabla u|}\Big\rangle\bigg|\right).$

Today, we discuss a variant of it called the Bochner formula for vector fields. I found this identity in a recent preprint of Li Ma [here].

Theorem. Let $(M, g)$ be a Riemannian manifold of dimension $n$. Let $X$ be a smooth vector field on $M$. Then we have

$\displaystyle\frac{1}{2}\Delta (|X|^2) = |\nabla X|^2 + {\rm div}({\mathbb L_X}g)(X) - {\nabla _X}{\rm div}X - {\rm Ric}(X,X)$

where ${\mathbb L_X}g$ is the Lie derivative of the vector field $X$ with respect to underlying metric $g$.

We note that the term ${\rm div}({\mathbb{L}_X}g)(X)$ is understood as following: Since $g$ is a tensor of type $(0,2)$, its Lie derivative with respect to $X$, $\mathbb L_X g$ is also a tensor of type $(0,2)$. As such, before taking the divergence, we take $({\mathbb{L}_X}g)(X)$ and this is an $(0,1)$-tensor which is able to take divergence.

Proof. Fix $x \in M$. Choose local orthonormal frame $\left\{ \frac{\partial}{\partial x_i} \right\}$, that means

$g_{ij} = \delta_{ij}$

and

$\displaystyle\nabla_\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}=0$

at $x$. Then we calculate everything at $x$. Write

$\displaystyle X=X^i\frac{\partial}{\partial x_i}$.

For an $(0,2)$-tensor $g=g_{ij}dx^i \otimes dx^j$, its Lie derivative with respect to $X$ is given by

$\begin{array}{lcl} {\mathbb{L}_X}g &=& \displaystyle {g_{ij,k}}{X^k}d{x^i} \otimes d{x^j} + {g_{ij}}\frac{{\partial {X^i}}}{{\partial {x^k}}}d{x^k} \otimes d{x^j} + {g_{ij}}\frac{{\partial {X^j}}}{{\partial {x^k}}}d{x^i} \otimes d{x^k} \hfill \\&=& \displaystyle\left( {{g_{ij,k}}{X^k} + {g_{kj}}\frac{{\partial {X^k}}}{{\partial {x^i}}} + {g_{ik}}\frac{{\partial {X^k}}}{{\partial {x^j}}}} \right)d{x^i} \otimes d{x^j} \hfill \\\end{array}$

i.e., Lie derivatives preserve the type of tensors. Using this, we can calculate ${\rm div}({\mathbb{L}_X}g)(X)$ as follows

$\begin{array}{lcl} {\rm div}({\mathbb{L}_X}g)(X) &=& \displaystyle\sum_i\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}({\mathbb{L}_X}g)} \right)\left( {\frac{\partial }{{\partial {x^i}}},X} \right) \hfill \\ &=& \displaystyle \sum_i {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {{\mathbb{L}_X}g\left( {\frac{\partial }{{\partial {x^i}}},X} \right)} \right) - \sum_i {\mathbb{L}_X}g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right). \end{array}$

By definition, the fact that $\mathbb L_XX=0$, and making use of normal coordinates, we have

$\begin{array}{lcl} \displaystyle {\mathbb{L}_X}g\left( {\frac{\partial }{{\partial {x^i}}},X} \right) &=& \displaystyle {\mathbb{L}_X}\left[ {g\left( {\frac{\partial }{{\partial {x^i}}},X} \right)} \right] - g\left( {{\mathbb{L}_X}\frac{\partial }{{\partial {x^i}}},X} \right) \hfill \\&=&\displaystyle \nabla _X\left[ {g\left( {\frac{\partial }{{\partial {x^i}}},X} \right)} \right] + g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,X} \right)\\&=&\displaystyle {g\left( {\frac{\partial }{{\partial {x^i}}},\nabla _XX} \right)} + g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,X} \right). \hfill \\ \end{array}$

Keep in mind that in the previous calculation we have used $\mathbb L_X f=\nabla_X f$ for any smooth function $f$ and

$\displaystyle\mathbb L_X \frac{\partial}{\partial x^i} = \Big[X, \frac{\partial}{\partial x^i}\Big] =\nabla_X \frac{\partial}{\partial x^i}-\nabla_\frac{\partial}{\partial x^i} X = - \nabla_{\frac{\partial}{\partial x^i}} X.$

Similarly, we also have

$\begin{array}{lcl} \displaystyle {\mathbb{L}_X}g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right) &=& \displaystyle {\mathbb{L}_X}\left( {g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right)} \right) - g\left( {{\mathbb{L}_X}\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right) - g\left( {\frac{\partial }{{\partial {x^i}}},{\mathbb{L}_X}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right)\\&=& \displaystyle {\nabla_X}\left( {g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right)} \right) +g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X} \right)+ g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X}}X} \right)-g\left( \frac{\partial}{\partial x^i}, \nabla_X\nabla_\frac{\partial}{\partial x^i}X\right)\\&=& \displaystyle {\big| {\nabla_\frac{\partial}{\partial x^i} X} \big|^2} + g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X}}X} \right). \end{array}$

Therefore,

$\displaystyle {\rm div}({\mathbb{L}_X}g)(X) = \sum_i {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {g\left( {\frac{\partial }{{\partial {x^i}}},\nabla_XX} \right)} + g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,X} \right)\right) - {\left| {\nabla X} \right|^2} - \sum_i g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X}}X} \right)$.

On the other hand

$\displaystyle\frac{1}{2}\Delta {\left| X \right|^2} =\frac{1}{2}\Delta g(X,X)=\frac{1}{2}\sum_i\nabla_\frac{\partial}{\partial x^i}\nabla_\frac{\partial}{\partial x^i} g(X,X)=\sum_i {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left[ {g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,X} \right)} \right]$

and

$\displaystyle\begin{array}{lcl} \displaystyle \sum_i \left[{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {g\left( {\frac{\partial }{{\partial {x^i}}},\nabla_XX} \right)} \right) - g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X}}X} \right)\right] &=& \displaystyle \sum_i \left[g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{\frac{\partial }{{\partial {x^i}}}}}{\nabla _X}X} \right) - g\left( {\frac{\partial }{{\partial {x^i}}},{\nabla _{{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X}}X} \right)\right] \hfill \\&=& \displaystyle \sum_i g\left( {\frac{\partial }{{\partial {x^i}}},\nabla _{\frac{\partial }{{\partial {x^i}}},X}^2X} \right) \hfill \\&=& \displaystyle {\rm Ric}(X,X) + \sum_i g\left( {\frac{\partial }{{\partial {x^i}}},\nabla _{X,\frac{\partial }{{\partial {x^i}}}}^2X} \right) \hfill \\&=& \displaystyle {\rm Ric}(X,X) + {\nabla _X}({\rm div}X). \hfill \\ \end{array}$

Combining all above quantities we have

$\displaystyle {\rm div}({\mathbb{L}_X}g)(X) = \frac{1}{2}\Delta {\left| X \right|^2} - {\left| {\nabla X} \right|^2} + {\rm Ric}(X,X) + {\nabla _X}({\rm div}X)$

which is the same as our formula.

As can be seen, the normal coordinates have been used several times to avoid massive calculation, for example, we have

$\displaystyle g\left( {{\nabla _X}\frac{\partial }{{\partial {x^i}}},X} \right) = g\left( {{\nabla _{{X^j}\frac{\partial }{{\partial {x^j}}}}}\frac{\partial }{{\partial {x^i}}},X} \right) = g \bigg({X^j}\underbrace {{\nabla _{\frac{\partial }{{\partial {x^j}}}}}\frac{\partial }{{\partial {x^i}}}}_0,X\bigg) = 0.$

The above identity has the following useful consequence: If $(M,g)$ is closed then by integrating both sides, we obtain

$\displaystyle -\int_M {\rm div}({\mathbb L_X}g)(X) dv_g = \int_M |\nabla X|^2 dv_g - \int_M {\nabla _X}{\rm div}X dv_g - \int_M {\rm Ric}(X,X)dv_g .$

Notice that

$\displaystyle\int_M {\nabla _X}{\rm div}X dv_g=-\int_M |{\rm div}X|^2 dv_g.$

To see this, we write ${\nabla _X}{\rm div}X = \langle X, \nabla {\rm div} X \rangle$ thanks to the fact that ${\rm div} X$ is a function, then we use the divergence theorem to get

$\displaystyle\int_M \langle X, \nabla {\rm div} X \rangle dv_g=-\int_M \langle {\rm div} X,{\rm div} X\rangle dv_g.$

Hence

$\displaystyle -\int_M {\rm div}({\mathbb L_X}g)(X) dv_g = \int_M |\nabla X|^2 dv_g +\int_M |{\rm div}X|^2 dv_g - \int_M {\rm Ric}(X,X)dv_g .$

Further calculation also shows that

$\begin{array}{lcl}\displaystyle -\int_M {\rm div}({\mathbb L_X}g)(X) dv_g &=&\displaystyle -\int_M X_j{\rm div}({\mathbb L_X}g)(\frac{\partial}{\partial x_j}) dv_g \\ &=& \displaystyle -\int_M X_j \nabla_{\frac{\partial}{\partial x_i}}({\mathbb L_X}g)(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) dv_g\\ &=& \displaystyle \int_M (\nabla_{\frac{\partial}{\partial x_i}} X_j) ({\mathbb L_X}g)(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) dv_g\\ &=& \displaystyle\frac{1}{2}\int_M (\nabla_{\frac{\partial}{\partial x_i}} X_j + \nabla_{\frac{\partial}{\partial x_j}} X_i) \mathbb L_X g_{ij}\\ &=& \displaystyle\frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g\end{array},$

i.e.

$\displaystyle \frac{1}{2}\int_M |\mathbb L_X g|^2 dv_g= \int_M |\nabla X|^2 dv_g +\int_M |{\rm div}X|^2 dv_g - \int_M {\rm Ric}(X,X)dv_g .$