I have a question regarding the definition of the covariant derivative of tensor fields, as given by John Lee in the book Riemannian Manifolds: An Introduction to Curvature
On page 53 he states the following lemma:
Let $\triangledown$ be a linear connection on $M$. There is a unique connection in each tensor bundle $T^k_l(M)$, also denoted $\triangledown$, such that the following conditions are satisfied:
(a) On TM, $\triangledown$ agrees with the given connection
(b) on $T^0M$, $\triangledown$ is given by ordinary differentiation on functions:
$$
\triangledown_X f = Xf
$$
This connection satisfies the following additional property:
For any $F \in \mathcal{T}^k_l(M)$, vector fields $Y_i$ and 1-forms $\omega^j$,
$$
\begin{eqnarray}
&(\triangledown_X F) (\omega^1, \dots \omega^l,Y_1, \dots Y_k) =
X( F (\omega^1, \dots \omega^l,Y_1, \dots Y_k)) \\
&- \sum_{i = 1}^l F (\omega^1, \dots , \triangledown_X \omega^i, \dots \omega^l,Y_1, \dots Y_k) \\
&- \sum_{i = 1}^k F (\omega^1, \dots , \omega^l,Y_1, \dots, \triangledown_X Y_i, \dots Y_k)
\end{eqnarray}
$$
Now, if I apply this property, say, to the Euclidean metric and the Euclidean connection, I obtain the following.
Let $Y = Y^i \partial_i$ and $Z = Z^i \partial_i$ be vector fields in local coordinates. Let the Euclidean Connection $\overline{\triangledown}$ be given by
$$
\overline{\triangledown}_X Y = (X Y^i) \partial_i
$$
Let $g$ denote the Euclidean metric, so that
$$
g(Y,Z) = \sum_i Y^iZ^i
$$
Then, the above property gives
$$
\overline{\triangledown}_X g(Y,Z) = X(\sum_i Y^iZ^i) – \sum_i (XY^i)Z^i – \sum_i Y^i(XZ^i) = 0
$$
.. this does look strange – what am I doing wrong ?
Best Answer
Your computation is correct. The fact that the euclidean connection and metric satisfy $\overline{\nabla}_Xg = 0$ for all $X$ can be phrased as saying that the euclidean connection is compatible with the euclidean metric.
See Page 67 of your book for further details (especially Lemma 5.2).