# Ngô Quốc Anh

## January 13, 2010

### R-G: A few words about writing covariant derivative of tensors in a short form

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, Riemannian geometry — Ngô Quốc Anh @ 14:21

Before going further, I would like to mention a convention in writing covariant derivative of tensors in a short form. What I mean is how to understand the following notation $\nabla_i g_{\alpha\beta}$ where $g$ is a Riemannian metric.

We all know that we can apply covariant derivative to a scalar function, for example, $\nabla_i f$ is nothing but the partial derivative with respect to $x^i$. However, the notation $\nabla_i g_{\alpha\beta}$ is a little bit different. For $g$, a Riemannian metric, which is also a $(0,2)$ tensor, in the full form, we can write

$\displaystyle g=g_{\alpha\beta}dx^\alpha \otimes dx^\beta$.

The sub indexes $\alpha\beta$ in the notation $\nabla_i g_{\alpha\beta}$ tell us that we are talking about the $\alpha\beta$-component of the covariant derivative of $(0,2)$ tensor $g$ with respect to the vector field $\frac{\partial}{\partial x^i}$.

If you have $\nabla_i h^{\alpha\beta}$, then you are working on some $(2,0)$-tensor $h$ of the form

$\displaystyle h=h^{\alpha\beta} \frac{\partial}{\partial x^\alpha} \otimes \frac{\partial}{\partial x^\beta}$.

We now go back to the case $\nabla_i g_{\alpha\beta}$. Precisely, one should write

$\displaystyle {\nabla _i}{g^{\alpha \beta }} = ({\nabla _{\frac{\partial }{{\partial {x^i}}}}}g)\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right)$.

Since $\{dx^\alpha\}$ is a basis for the dual space of a space spanned by the basis $\{\frac{\partial}{\partial x^\beta}\}$, it is clear to see that the right hand side of the above convention is just the $\alpha\beta$-component, that is, the coefficient of the term

$\displaystyle {\nabla _i}{g^{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }$.

Now we show that $\nabla_i g^{\alpha\beta}=0$ whenever $g$ is a Riemannian metric. This is equivalent to show that every coefficients of $\nabla_i g$ equal to zero, in other word, $\nabla_i g=0$. Since $g$ is a $(0,2)$-tensor, then we can compute covariant derivative of $g$ as follows

$\displaystyle\begin{gathered}{\nabla _i}g = {\nabla _i}\left( {{g_{\alpha \beta }}d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\\qquad = \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha } \otimes d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } + {g_{\alpha \beta }}{\nabla _i}\left( {d{x^\alpha }} \right) \otimes d{x^\beta } + {g_{\alpha \beta }}d{x^\alpha } \otimes {\nabla _i}\left( {d{x^\beta }} \right) \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }}} \right)d{x^\alpha } \otimes d{x^\beta } - \left( {{g_{\alpha \beta }}\Gamma _{ik}^\alpha+ {g_{\alpha \beta }}\Gamma _{ik}^\beta } \right)d{x^k} \otimes d{x^\beta } \hfill \\ \qquad= \left( {\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} - \Gamma _{i\alpha }^k{g_{k\beta }} - \Gamma _{i\beta }^k{g_{k\alpha }}} \right)d{x^\alpha } \otimes d{x^\beta }. \hfill \\ \end{gathered}$

Since $g$ is a metric connection, i.e.

$\displaystyle Xg\left( {Y,Z} \right) = g\left( {{\nabla _X}Y,Z} \right) + g\left( {Y,{\nabla _X}Z} \right)$

one has

$\displaystyle \frac{\partial }{{\partial {x^i}}}g\left( {\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) = g\left( {{\nabla _i}\frac{\partial }{{\partial {x^\alpha }}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},{\nabla _i}\frac{\partial }{{\partial {x^\beta }}}} \right)$.

Thus

$\displaystyle\frac{\partial }{{\partial {x^i}}}{g_{\alpha \beta }} = g\left( {\Gamma _{i\alpha }^k\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^\beta }}}} \right) + g\left( {\frac{\partial }{{\partial {x^\alpha }}},\Gamma _{i\beta }^k\frac{\partial }{{\partial {x^k}}}} \right) = \Gamma _{i\alpha }^k{g_{k\beta }} + \Gamma _{i\beta }^k{g_{k\alpha }}$.

In other word the $\alpha\beta$-component of $\nabla_i g$ equals to zero. At the last word, in order to calculate covariant derivative of some $(p,q)$-tensors, you need to use the following three properties

1. ${\nabla _X}\left( {fT} \right) = X(f)T + f{\nabla _X}T$.
2. ${\nabla _X}\left( {T \otimes Q} \right) = {\nabla _X}T \otimes Q + T \otimes {\nabla _X}Q$.
3. For $(1,0)$– or $(0,1)$-tensors:

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^j}}} = \Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}$

and

$\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}d{x^j}=-\Gamma _{ik}^jd{x^k}$.

## 1 Comment »

1. […] I've been reading some thing and made some conclusions, help me to understand this please. https://anhngq.wordpress.com/2010/0…ariant-derivative-of-tensors-in-a-short-form/. First of all, by ∇ρδgμν we refer to the scalar that results from the evaluation of […]

Pingback by Issues with the variation of Christoffel symbols | Physics Forums — April 25, 2018 @ 23:15

This site uses Akismet to reduce spam. Learn how your comment data is processed.