Levi-Civita connection from idempotents Let $(M,g)$ be a closed Riemannian manifold. Let $V$ be a smooth complex vector bundle over $M$. We can write $V$ as the range of an idempotent $E$ in a matrix algebra $M_n(C^\infty(M))$ acting on a trivial bundle $M\times\mathbb{C}^n$. Here $C^\infty(M)$ denotes the complex-valued functions on $M$.
Let $D$ be the trivial connection on $M\times\mathbb{C}^n$ defined by applying the de Rham differential to each entry separately. Then the expression $EDE$ defines a connection on $V$.
Some books call the connection $EDE$ the "Levi-Civita connection" on $V$, "by analogy to the classical situation", and I am trying to understand how this analogy works. In particular, when $V$ is the tangent bundle $TM$ (ignoring the fact that this is a real vector bundle), the Levi-Civita connection depends on the metric $g$, while the expression $EDE$ does not seem to depend on $g$.
Question: Is there a precise sense in which $EDE$ produces the "classical" Levi-Civita connection on $TM$?
 A: $\newcommand{\R}{\mathbb{R}}$
Yes, it uniquely defines a connection in the case where $V$ is the tangent bundle(the connection won't depend on the choice of orthogonal idempotent) and its the Levi Civita connection. But in general different idempotents from the same metric will define different connections(I think, I'm not sure on this one though).
Let $ V^l \to M^k $ be the vector bundle of rank $l$ over a $k$ manifold and take an isometric embedding of the vector bundle into $\R^{2n}$ with the standard metric.   Let $s$ be a section of the vector bundle.  Let $e: M \to Mat_{n\times n}$ be the orthogonal  projection onto the tangent bundle of $M$.
If we differentiate $s$ in a given direction using the ambient $\R^{2n}$, (we can just take all the directions into account by taking the exterior derivative $ds$), then the result will not be in the vector bundle any more.  But $eds$ will be.  Hence we define the derivative of $s$ in the $v$ direction to be $\nabla_v s =e\frac{ds}{dv}$
Let us consider the case where $E$ is the tangent bundle of the manifold and $M^k \subset \R^n$.  It is possible to write down the expression $e\frac{ds}{dv}$ just using the metric on the manifold.
To show this we show that $eds$ satisfies the axioms of the Levi Civita connection:

If $X,Y,Z$ are sections of our vector bundle, then
\[ X \langle Y,Z \rangle = \langle \nabla_X Y, Z \rangle + \langle
Y, \nabla_X Z \rangle \]

Proof:
\[ \nabla_X Y-\nabla_Y X=[X,Y]  \]
If we expand the rhs,   \[ \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle = \langle e \frac{dY}{dX},  Z \rangle + \langle Y , e \frac{dZ}{dX} \rangle = \]
\[\langle  \frac{dY}{dX}, e^T Z \rangle + \langle e^TY ,  \frac{dZ}{dX} \rangle \]
Now $e^T=e$ because $e$ is an orthogonal projection.  So this is
\[=\langle  \frac{dY}{dX}, eZ \rangle + \langle eY ,  \frac{dZ}{dX} \rangle=\langle  \frac{dY}{dX}, Z \rangle + \langle Y ,  \frac{dZ}{dX} \rangle=X \langle Y,Z \rangle \]
\[ \nabla_X Y-\nabla_Y X= e \frac{dY}{dX} -e \frac{dX}{dY}=e[X,Y]=[X,Y] \]
Exercise:
Using the above identities, and cyclically permuting $X,Y,Z$ write down an expression for $\nabla_X Y$ in terms of the inner product.
Using the above expression, write down a matrix $A$ of 1-forms in terms of the inner product such that
\[\nabla s=ds-As \]
If you do this exercise, the matrix $A$ you get will be the matrix of christoffel symbols.
Another good exercise is to write out what $A$ is in terms of the idempotent.  (You should get $(1-e)de$)
Note that you can't cyclically permute the indices $X,Y,Z$ for an arbitrary vector bundle.  I don't think that you get a canonical connection on a vector bundle just given the metric(I think this was one of your questions), and at the very least, the trick above doesn't work.
Aside:  I came up with this answer after reading through the comment above about it being in Dirac's General relativity book.  I've been trying for the past few months and I can't quite tell what he is doing.  I did see though that his argument factors through proving that the axioms of the Levi Civita connection are satisfied without saying that that is what he's doing.
