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I have a question regarding a comment in Lee's book "Riemannian Geometry - an Introduction to Curvature".

On page 52, Lee introduces the Euclidean Connection as the map $\overline{\triangledown} : \mathcal{T}(M) \times \mathcal{T}(M) \to \mathcal{T}(M)$ that acts locally on a vector field $Y$ by $$ \overline{\triangledown}(X,Y) = \overline{\triangledown}_X (Y_i \partial_i):= (XY^i) \partial_i $$

But then, on page 67 he makes the following remark:

The Euclidean connection on $\mathbb{R}^n$ has one very nice property with repsect to the Euclidean metric: it satisfies the product rule $$ \overline{\triangledown}_X \langle Y,Z \rangle = \langle \overline{\triangledown}_X Y,Z \rangle + \langle Y, \overline{\triangledown}_X Z \rangle $$

The thing that confuses me a little is I don't know what the left hand means, as I understood the connection is defined on $\mathcal{T}(M)$, but $\langle Y,Z \rangle \notin \mathcal{T}(M)$. How can I understand the left hand side of the above equation?

One guess that I have is that I should understand it in terms of the connection on the tensor bundle $T^2(M)$, as defined in Lemma 4.6 (page 53) , but then I have

Thanks for your help!

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Harlekin Hello! I'm finding it a little confusing his references to the above equations. Why do not the lists? And you do not mean the right side of the equation? – user27456 Apr 4 '12 at 21:33
Think of $<Y,Z>$ as being a function defined on $\mathbb R\times \mathbb R$, and think of $\overline \nabla_X$ as being the directional derivative w.r.t. $X$. – treble Apr 4 '12 at 23:53
See Lee's book, p. 53, Lemma 4.6 (b). The LHS means the usual action of vector fields on smooth functions ("directional derivative") – Yuri Vyatkin Apr 4 '12 at 23:54
Yuri is right. That is, one can talk about $\overline{\nabla}_Xf := Xf$ where $f$ is a smooth function, and in this case $f = \langle Y, Z \rangle$. – Jesse Madnick Apr 5 '12 at 16:35

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