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}.

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