"Rolling without slipping" is a powerful idea, but the phrase doesn't necessarily lead one to the intended mental model. In particular,
torsion is something that is at issue only for manifolds of dimension 3 or higher. Perhaps you can imagine taking a 3-manifold, and rolling it along a hyperplane in 4-space --- but
the metaphor becomes strained, partly because most Riemannian 3-manifolds cannot be
smoothly isometrically embedded in $\mathbb E^4$.
Another way to think of it is this: suppose you have a smooth parametrized
curve say in a Riemannian
3-manifold, $\alpha: [0,T] \rightarrow M^3$. Then the claim is that there exists
exists a matching curve $\beta: [0,T] \rightarrow \mathbb E^3$ together with a map
$\phi$ from a neighborhood of the image of $\beta$ to a neighborhood of the image of $\alpha$
that takes the Euclidean metric to the metric of $M^3$ up to first order along the curve.
Furthermore, $\beta$ is uniquely determined up to an isometry of $\mathbb E^3$.
Basically, $\beta$ is what you get if you "roll" $M^3$ along $\mathbb E^3$ along $\gamma$
in a way
that best maintains contact between the two spaces: that is, "without slipping".
To see that the curve $\beta$ (assuming it exists) is uniquely determined by $\alpha$,
we can imagine trying to send a neighborhood of $\beta$ to a neighborhood of another
curve $\gamma: [0,T] \rightarrow \mathbb E^3$ in a way that maintains first order
contact of the metric. If you look at curves parallel to $\beta$ in $\mathbb E^3$, the first
derivative of their arc length
is negative in the direction that $\beta$ is curving. The logarithmic derivative of
arc length, for curves displaced along a normal vector field that remains as parallel to
itself as possible, is the magnitude of the curvature.
If you try to twist the normal coordinate system, this corresponds to the concept
of torsion. It's easiest to visualize along a straight line: if you twist a neighborhood
of a line in space,
you distort the metric on each concentric cylinder, by changing the angles between
cross-section circles and generating lines. I.e., threads that wind around a hose
at angles $\pm \pi/4$ are effective at preventing twisting (= torsion). The same
principle holds for any curve in space: the first order behavior of the metric in a
neighborhood of the curve locks in the Frenet characterization of the curve (well,
the curvature and torsion as a function of arc length, but these
are different from but related to the curvature and torsion of a connection).
Why does the matching curve exist? You can check derviatives etc. but
better to just imagine it. Basically, you could reparametrize $\alpha$ by arc length,
then project a neighborhood of $\alpha$ back to $\alpha$ by sending each point to the closest
point of $\alpha$, and parametrize the lines of projection by their arc length.
On each concentric tube, there's a unique unit vector field orthogonal to the preimages
of projection to $\alpha$. Scale this vector field so that it commutes with
projection, to get a full set of
cylindrical coordinates for a neighborhood of $\alpha$. The only first-order
invariant for the metric that is free is the first derivative of scaling function.
Using that, you can match the first derivative
by using the curvature and torsion of a curve in space.
This process defines the affine connection on the tangent bundle. The Levi-Civita
connection is the linear part of the affine connection, which is automatically by
definition torsion free. The non-torsion-free connections are ones that
impart twists on little neighborhoods of curves. This is usually expressed
by translating it into a formula about covariant derivatives of two vector fields
not being as commutative as it should be.
This really calls for pictures. Any volunteers?
As far as I know, this sort of structure was first invoked by Dirac in order to take a square root of the Laplacian, and this he was doing in order to write down Lorentz invariant Klein-Gordon equations. It is a useful exercise to try to solve the equation $D^2 = \Delta$ on a Euclidean space $V$ for a first order operator $D$; you will find that the coefficients have to satisfy certain relations that cannot be satisfied by ordinary real or complex numbers. The algebraic structure required to obtain these relations is provided by an algebra $A$ with $V$ as a linear subspace such that $v^2 = -||v||^2 1$ in the algebra. In other words, you need to take a "square root" of your quadratic form.
In brief, a spinor bundle on a Riemannian manifold is a setting for taking a square root of the Riemannian metric. To be precise, it is a bundle $S$ on which tangent vectors act as bundle morphisms in such a way that $v^2 s = -||v||^2 s$. In Dirac's equation, the coefficients of $D$ were given by certain matrices (the "Pauli spin matrices"), and thus he was thinking of $D$ as taking values in a vector space which carries a representation of the algebra $A$. Thus the spinor bundle is a global version of that vector space.
That tells you what properties the spinor bundle is supposed to have, but it doesn't tell you what the bundle actually is. If you look it up in a book, you will find that the spinor bundle is an associated bundle to a principal $Spin(n)$ (or $Spin^c(n)$) bundle via the spin representation, but to me that is only a little more helpful than defining a Riemannian metric to be a reduction of structure group from the principal $GL(n)$-bundle of frames to a $O(n)$-bundle.
Here is what I would consider to be a more concrete and well-motivated description. Let us return to the algebra $A$ associated to a Euclidean space $V = \mathbb{R}^n$ as above. The universal example of such an algebra is the Clifford algebra $Cl(V)$, equipped with a natural left action of $V$. Choosing an orthonormal basis for $V$, one can describe $\mathbb{R}_n := Cl(V)$ as the universal algebra over $\mathbb{R}$ generated by symbols $e_1, \ldots, e_n$ subject to the relations $e_j^2 = -1$ and $e_j e_k = -e_k e_j$ for $i \neq j$. It is not hard to see that $Cl(V)$ is isomorphic as a vector space (but not as an algebra) to the exterior algebra of $V$, and thus $Cl(V)$ inherits a natural $\mathbb{Z}/2\mathbb{Z}$ grading, given by products of even / odd numbers of generators. Notice that right multiplication by the $j$th generator is an odd anti-involution, so a choice of orthonormal basis for $V$ gives $Cl(V)$ the structure of a $n$-multigraded super algebra.
We can define a (real) spinor bundle of a $n$-manifold to be a bundle which is locally isomorphic to the trivial bundle whose fibers are given by $\mathbb{R}_n$ equipped with a left action of the tangent bundle and a $n$-multigrading structure coming from a choice of local orthonormal frame. There is an obvious notion of complex spinor bundle as well: just use the complex Clifford algebra $\mathbb{C}_n$. Note that the fiber dimension of this bundle will be twice that of the bundle obtained via the spin representation, but the multigrading operators can be used to "reduce" my version of the spinor bundle down to the usual version. There are lots of reasons why I believe it is more convenient to think of a spinor bundle as a bundle of Clifford algebras with extra supersymmetry data, but I will briefly focus on a topological reason that I think cuts to the heart of the matter.
The existence of a real spinor bundle on a manifold $M$ (a "Spin structure") is a rather severe condition. The complexification of a real spinor bundle is a complex spinor bundle, but not all complex spinor bundles ("Spin$^c$ structures") arise in this way. For example, any complex manifold has a spin$^c$ structure, but even $\mathbb{C}P^2$ fails to have a spin structure. An orientation on $M$ can be recovered from a choice of spin$^c$ structure, and indeed "spin$^c$-able" is only a little bit stronger than orientable - most orientable manifolds that you can name are probably spin$^c$-able. My point in bringing this up is to relate spinor bundles to K-homology, the generalized homology theory dual to topological K-theory. In ordinary homology theory, a choice of orientation on an $n$-manifold $M$ is the same thing as a choice of fundamental class in $H_n(M)$. Similarly, a choice of real / complex spinor bundle on a $n$-manifold $M$ is the same thing as a choice of fundamental class in the $n$th degree real / complex K-homology of $M$ (the multigrading data are crucial here). This observation is the starting point for some of the more conceptual proofs of the Atiyah-Singer index theorem, but this answer has gone on long enough. I hope it helps!
Best Answer
The easiest way to see that the norm of the curvature corresponds to the energy is to consider the special case of an abelian U(1)-Yang-Mills theory (i.e. electrodynamics). If you write out the norm squared of the curvature in terms of the $E$ and $B$ fields you get the expression $E^2 + B^2$. This is exactly the familiar energy density of an electrodynamic field, see for example wikipedia.