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Parameterized Surfaces 📂Geometry

Parameterized Surfaces

Definition1

  • For an open set UR2U \subset \mathbb{R}^{2}, let’s say the connected set AR2A\subset \mathbb{R}^{2} satisfies the following:

    UAUandA is piecewise differentiable U \subset A \subset \overline{U} \quad \text{and} \quad \partial A \text{ is piecewise differentiable}

    s:AMs : A \to M is called a parameterized surface within the manifold MM.

  • A vector field VV along ss, maps each qAq \in A to V(q)Ts(q)MV(q) \in T_{s(q)}M and is a differentiable function in the following sense:

    If ff is a differentiable function over MM, then the function qAV(q)fRq \in A \mapsto V(q)f \in \mathbb{R} is also differentiable.

Description

Let’s call (u,v)(u, v) the Cartesian coordinates over R2\mathbb{R}^{2}. For a fixed v0v_{0}, the function us(u,v0)u \mapsto s(u,v_{0}) is a curve over MM. Considering the standard basis of the tangent space over R2\mathbb{R}^{2}, it leads to {u,v}\left\{ \frac{\partial }{\partial u}, \frac{\partial }{\partial v} \right\}. Simplify the function value ds(u)ds(\frac{\partial }{\partial u}) of derivative dsds of ss as

su \dfrac{\partial s}{\partial u}

Then su\dfrac{\partial s}{\partial u} is a vector field following curve us(u,v0)u \mapsto s(u,v_{0}). Similarly, sv\dfrac{\partial s}{\partial v} is defined as a vector field along ss. In differential geometry, similar concepts are considered for coordinate patch maps.

Consider a vector field VV along ss. Now, we will define the covariant derivatives DVu\dfrac{D V}{\partial u}, DVv\dfrac{D V}{\partial v} of this VV. Think about the reduction mapping of VV over the curve us(u,v0)u \mapsto s(u, v_{0}). Then DVdu(u,v0)\dfrac{DV}{du}(u,v_{0}) can be defined as the covariant derivative of this reduction mapping. As this can be considered for all v0v_{0}, DVu(u,v)\dfrac{DV}{\partial u}(u, v) is defined for all (u,v)A(u,v) \in A. DVv\dfrac{D V}{\partial v} is defined in the same manner.

Symmetry

If MM is a manifold with a symmetric connection, and s:AMs : A \to M is a parameterized surface, then the following holds:

Dvsu=Dusv \dfrac{D}{\partial v}\dfrac{\partial s}{\partial u} = \dfrac{D}{\partial u}\dfrac{\partial s}{\partial v}

Proof

Let x:URnM\mathbf{x} : U \subset \mathbb{R}^{n} \to M be a coordinate system including a neighborhood of point s(A)s(A). In other words, the coordinate system is chosen such that s(A)x(U)s(A) \subset \mathbf{x}(U). Then x1s(u,v)\mathbf{x}^{-1} \circ s(u,v) is a point of Rn\mathbb{R}^{n}, and it’s represented as follows:

x1s(u,v)=(s1(u,v),,sn(u,v)) \mathbf{x}^{-1} \circ s (u,v) = \left( s^{1}(u,v), \dots, s^{n}(u,v) \right)

Also, s:AMs : A \to M and the derivative of ss is ds:T(u,v)ATs(u,v)Mds : T_{(u,v)}A \to T_{s(u,v)}M. The coordinates of AA are (u,v)(u, v), and the coordinates of MM are x1=(s1,,sn)\mathbf{x}^{-1} = \left( s^{1}, \dots, s^{n} \right), therefore,

ds=[s1us1vsnusnv] ds = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} & \dfrac{\partial s^{1}}{\partial v} \\ \vdots \\ \dfrac{\partial s^{n}}{\partial u} & \dfrac{\partial s^{n}}{\partial v} \end{bmatrix}

The basis of T(u,v)AT_{(u,v)}A is, the coordinate vector of {u,v}\left\{ \dfrac{\partial }{\partial u}, \dfrac{\partial }{\partial v} \right\} is [10]\begin{bmatrix} 1 \\ 0\end{bmatrix}. Thus,

su=ds(u)=[s1us1vsnusnv][10]=[s1usnu]=isiuxi \dfrac{\partial s}{\partial u} = ds(\dfrac{\partial }{\partial u}) = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} & \dfrac{\partial s^{1}}{\partial v} \\ \vdots & \vdots \\ \dfrac{\partial s^{n}}{\partial u} & \dfrac{\partial s^{n}}{\partial v} \end{bmatrix} \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} \dfrac{\partial s^{1}}{\partial u} \\ \vdots \\ \dfrac{\partial s^{n}}{\partial u} \end{bmatrix} = \sum_{i} \dfrac{\partial s^{i}}{\partial u}\dfrac{\partial }{\partial x_{i}}

Now, calculating Dv(su)\dfrac{D }{\partial v}\left( \dfrac{\partial s}{\partial u} \right), by the properties of covariant derivatives,

Dv(su)=Dv(jsjuxj)=j2sjvuxj+jsjuDvxj=j2sjvuxj+i,jsjus/uxj=j2sjvuxj+i,jsjusiuxixj=j2sjvuxj+i,jsjusiuxixj=k2skvuxk+i,j,ksjusiuΓijkxk \begin{align*} \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) &= \dfrac{D }{\partial v} \left( \sum_{j} \dfrac{\partial s^{j}}{\partial u}\dfrac{\partial }{\partial x_{j}} \right) \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{j} \dfrac{\partial s^{j}}{\partial u}\dfrac{D }{\partial v}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \nabla_{{\partial s}/{\partial u}}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \nabla_{\frac{\partial s^{i}}{\partial u}\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{j} \dfrac{\partial^{2} s^{j}}{\partial v \partial u}\dfrac{\partial }{\partial x_{j}} + \sum_{i,j} \dfrac{\partial s^{j}}{\partial u} \frac{\partial s^{i}}{\partial u}\nabla_{\frac{\partial }{\partial x_{i}}}\dfrac{\partial }{\partial x_{j}} \\ &= \sum_{k} \dfrac{\partial^{2} s^{k}}{\partial v \partial u}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} \dfrac{\partial s^{j}}{\partial u} \frac{\partial s^{i}}{\partial u}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}} \\ \end{align*}

Similarly, Du(sv)\dfrac{D }{\partial u}\left( \dfrac{\partial s}{\partial v} \right) is

Dv(su)=k2skuvxk+i,j,ksjvsivΓijkxk \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) = \sum_{k} \dfrac{\partial^{2} s^{k}}{\partial u \partial v}\dfrac{\partial }{\partial x_{k}} + \sum_{i,j,k} \dfrac{\partial s^{j}}{\partial v} \frac{\partial s^{i}}{\partial v}\Gamma_{ij}^{k} \dfrac{\partial }{\partial x_{k}}

At this point, as these are differentiable functions where sk:R2Rs^{k} : \mathbb{R}^{2} \to \mathbb{R}, 2uv=2vu\dfrac{\partial ^{2}}{\partial u \partial v} = \dfrac{\partial ^{2}}{\partial v \partial u} holds. Therefore,

Dv(su)=Dv(su) \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right) = \dfrac{D }{\partial v} \left( \dfrac{\partial s}{\partial u} \right)


  1. Manfredo P. Do Carmo, Riemannian Geometry (Eng Edition, 1992), p67-68 ↩︎