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Having a manifold $M$ and some vector bundle $E$ over $M$ I am familiar with the definition of a connection given by a function $\nabla:\chi(M)\times\Gamma(E)\rightarrow \Gamma(E)$ that satisfies the leibniz rule on the second entry and is linear on the first entry.

But I have been having problem with an alternative definition that I give as appears in my book.

Definition: A connection is a map $d_A:\Gamma(E)\rightarrow\Gamma(T^*M\otimes E)$ such that $d_A(f\sigma+\tau)=(df)\otimes \sigma + fd_A(\sigma)+d_A(\tau)$.

It seems clear enough that both definition are equivalent under the identification $d_A(\sigma)(X)=\nabla_X(\sigma)$

But when I am going to compute something I get in trouble. For instance I couldn't solve the following question:

Let $M$ a manifold embedded in $\mathbb R^N$ and let E=TM the tangent bundle. We thus have $$ M\times \mathbb R^N = TM\oplus NM $$ where $NM$ is the normal bundle. then if we have a section $\sigma\in \Gamma(E)\subset \Gamma(M\times\mathbb R^N)$ we can compute its exterior derivative $d\sigma \in \Gamma(T^*M\otimes (TM\oplus NM))$ (so far so good) but then it says $d_A(\sigma)=(d\sigma)^T$ (Where T denotes the projection is a connection an E) is a connection on E, in fact de Levi Civita connection.

I have two questions: First about the projection, I am guessing is just projecting just the second factor of the tensor product onto $TM$ along $NM$ but I am not quite sure. The second is about the statement of the Levi- Civita connection, I have been trying to prove it but I am struggling even with the notation and haven't been able to prove it.

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    $\begingroup$ The definition using $d_A$ is covered in detail in R.W.R. Darling, Differential Forms and Connections, Chapter 9. The fact that the projection becomes a Levi-Cevita connection on a Riemannian manifold is covered in Chapter 8 of Volume II of Spivak, including Addenda. Hope this helps, I don't feel confident enough to explain it myself, but Darling develops the bundle-valued exterior derivative (that would be $d_A$) because it makes invariants into forms, and the way to put the connections together (Spivak) is to consider the tangents of the horizontal lifts of all curves through a point. $\endgroup$ – hkr Jul 27 '16 at 4:35

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