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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 ?

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up vote 3 down vote accepted

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).

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