# Ngô Quốc Anh

## January 20, 2010

### RG: Divergence of an (p,s)-tensor

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:24

Followed by this topic where we define the divergence of a vector field $X$ as follows

$\displaystyle {\rm div} X = dx^i \left( \nabla_{\frac{\partial}{\partial x^i}} X\right)$

that is, in local coordinates,

$\displaystyle {\rm div} X = \left\langle {\frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}},d{x^i}} \right\rangle = \frac{{\partial {X^i}}}{{\partial {x^i}}} + {X^j}\Gamma _{ij}^i$

it’s time to take opportunity to define the divergence of an $(p,s)$-tensor where $p\geq 1$. We assume $\alpha$ is an $(p,0)$-tensor, that is,

$\displaystyle \alpha= {\alpha _{{i_1}{i_2} \cdots {i_p}}}d{x^{{i_1}}} \otimes d{x^{{i_2}}} \otimes\cdots\otimes d{x^{{i_p}}}$.

We then defined the divergence of $\alpha$ to be an $(p-1,0)$-tensor given as the following

$\displaystyle {\rm div}{(\alpha)_{{i_1}{i_2} \cdots {i_{p - 1}}}} = {g^{\alpha \beta }}{\nabla _\alpha }{\alpha _{\beta {i_1}{i_2} \cdots {i_{p - 1}}}} = {\nabla _\alpha }{\alpha _{\alpha {i_1}{i_2} \cdots {i_{p - 1}}}}$.

In particular, for a $1$-form $X$, that is an $(1,0)$-tensor with the form $X=X_idx^i$ in local coordinates, one has

$\displaystyle {\rm div}(X) = {g^{\alpha \beta }}{\nabla _\alpha }{X_\beta }$.

In general, if $\alpha$ is an $(p,s)$-tensor with $p\geq 1$, then the divergence of $\alpha$ to be an $(p-1,s)$-tensor given as the following

$\displaystyle {\rm div}{(\alpha)_{{i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s} = {g^{\alpha \beta }}{\nabla _\alpha }{\alpha _{\beta {i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s} = {\nabla _\alpha }{\alpha _{\alpha {i_1}{i_2} \cdots {i_{p - 1}}}}^{j_1j_2\cdots j_s}$.

Note that, for a vector field $X=X^i\frac{\partial}{\partial x^i}$ which is an $(0,1)$-tensor, we need to change to an $(1,0)$-tensor, an $1$-form. To this purpose, we need to change $\frac{\partial}{\partial x^i}$ to $g_{ij}dx^j$. Therefore,

$\displaystyle X = {X^i}\frac{\partial }{{\partial {x^i}}} = {X^i}{g_{ij}}d{x^j}$.

Since

$\displaystyle {\nabla _\alpha }({X^i}{g_{i\beta }}d{x^\beta }) = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }}d{x^\beta } + {X^i}{g_{i\beta }}{\nabla _\alpha }d{x^\beta } = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }}d{x^\beta } - {X^i}{g_{i\beta }}\Gamma _{\alpha i}^\beta d{x^i}$

then

$\displaystyle {\nabla _\alpha }{X_\beta } = \frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }} - {X^i}{g_{ij}}\Gamma _{\alpha \beta }^j$.

Thus

$\displaystyle {\rm div}(X) = {g^{\alpha \beta }}{\nabla _\alpha }{X_\beta } = {g^{\alpha \beta }}\left( {\frac{{\partial {X^i}}}{{\partial {x^\alpha }}}{g_{i\beta }} - {X^i}{g_{ij}}\Gamma _{\alpha \beta }^j} \right) = \frac{{\partial {X^\alpha }}}{{\partial {x^\alpha }}} - {X^\alpha }\Gamma _{\alpha \beta }^\beta$.