Editing Affine connection (section)

==Surface theory revisited==
If {{mvar|M}} is a surface in {{math|'''R'''<sup>3</sup>}}, it is easy to see that {{mvar|M}} has a natural affine connection. From the linear connection point of view, the covariant derivative of a vector field is defined by differentiating the vector field, viewed as a map from {{mvar|M}} to {{math|'''R'''<sup>3</sup>}}, and then projecting the result orthogonally back onto the tangent spaces of {{mvar|M}}. It is easy to see that this affine connection is torsion-free. Furthermore, it is a metric connection with respect to the Riemannian metric on {{mvar|M}} induced by the inner product on {{math|'''R'''<sup>3</sup>}}, hence it is the Levi-Civita connection of this metric.

===Example: the unit sphere in Euclidean space===
Let {{math|⟨ , ⟩}} be the usual [[scalar product]] on {{math|'''R'''<sup>3</sup>}}, and let {{math|'''S'''<sup>2</sup>}} be the unit sphere. The tangent space to {{math|'''S'''<sup>2</sup>}} at a point {{mvar|x}} is naturally identified with the vector subspace of {{math|'''R'''<sup>3</sup>}} consisting of all vectors orthogonal to {{mvar|x}}. It follows that a vector field {{mvar|Y}} on {{math|'''S'''<sup>2</sup>}} can be seen as a map {{math|''Y'' : '''S'''<sup>2</sup> → '''R'''<sup>3</sup>}} which satisfies

: <math>\langle Y_x, x\rangle = 0\,, \quad \forall x\in \mathbf{S}^2.</math>

Denote as {{math|d''Y''}} the differential (Jacobian matrix) of such a map. Then we have:

:'''Lemma'''. The formula
::<math>(\nabla_Z Y)_x = \mathrm{d}Y_x(Z_x) + \langle Z_x,Y_x\rangle x</math>
:defines an affine connection on {{math|'''S'''<sup>2</sup>}} with vanishing torsion.

:::'''Proof'''. It is straightforward to prove that {{math|∇}} satisfies the Leibniz identity and is {{math|''C''<sup>∞</sup>('''S'''<sup>2</sup>)}} linear in the first variable. So all that needs to be proved here is that the map above does indeed define a tangent vector field. That is, we need to prove that for all {{mvar|x}} in {{math|'''S'''<sup>2</sup>}}
::::<math>\bigl\langle(\nabla_Z Y)_x,x\bigr\rangle = 0\,.\qquad \text{(Eq.1)}</math>

:::Consider the map
::::<math>\begin{align} f: \mathbf{S}^2&\to \mathbf{R}\\ x &\mapsto \langle Y_x, x\rangle\,.\end{align}</math>

:::The map ''f'' is constant, hence its differential vanishes. In particular
::::<math>\mathrm{d}f_x(Z_x) = \bigl\langle (\mathrm{d} Y)_x(Z_x),x(\gamma'(t))\bigr\rangle + \langle Y_x, Z_x\rangle = 0\,.</math>

:::Equation 1 above follows. [[Q.E.D.]]