For an open set U⊂R2, let’s say the connected setA⊂R2 satisfies the following:
U⊂A⊂Uand∂A is piecewise differentiable
s:A→M is called a parameterized surface within the manifoldM.
A vector field V along s, maps each q∈A to V(q)∈Ts(q)M and is a differentiable function in the following sense:
If f is a differentiable function over M, then the function q∈A↦V(q)f∈R is also differentiable.
Description
Let’s call (u,v) the Cartesian coordinates over R2. For a fixed v0, the function u↦s(u,v0) is a curve over M. Considering the standard basis of the tangent space over R2, it leads to {∂u∂,∂v∂}. Simplify the function value ds(∂u∂) of derivative ds of s as
∂u∂s
Then ∂u∂s is a vector field following curve u↦s(u,v0). Similarly, ∂v∂s is defined as a vector field along s. In differential geometry, similar concepts are considered for coordinate patch maps.
Consider a vector field V along s. Now, we will define the covariant derivatives ∂uDV, ∂vDV of this V. Think about the reduction mapping of V over the curve u↦s(u,v0). Then duDV(u,v0) can be defined as the covariant derivative of this reduction mapping. As this can be considered for all v0, ∂uDV(u,v) is defined for all (u,v)∈A. ∂vDV is defined in the same manner.
Symmetry
If M is a manifold with a symmetric connection, and s:A→M is a parameterized surface, then the following holds:
∂vD∂u∂s=∂uD∂v∂s
Proof
Let x:U⊂Rn→M be a coordinate system including a neighborhood of point s(A). In other words, the coordinate system is chosen such that s(A)⊂x(U). Then x−1∘s(u,v) is a point of Rn, and it’s represented as follows:
x−1∘s(u,v)=(s1(u,v),…,sn(u,v))
Also, s:A→M and the derivative of s is ds:T(u,v)A→Ts(u,v)M. The coordinates of A are (u,v), and the coordinates of M are x−1=(s1,…,sn), therefore,
ds=∂u∂s1⋮∂u∂sn∂v∂s1∂v∂sn
The basis of T(u,v)A is, the coordinate vector of {∂u∂,∂v∂} is [10]. Thus,