Representation for a Smooth Isometric Mapping from a Connected Planar Domain to a Surface

Abstract

A representation theorem for a smooth isometric mapping of a flat, connected domain $\mathcal {D}$ in two-dimensional Euclidean point space $\mathcal {E}^{2}$ into a surface $\mathcal {S}$ in three-dimensional Euclidean point space $\mathcal {E}^{3}$ is presented. The form of the mapping is shown to be necessary and sufficient to describe any such smooth isometry. Importantly, this work is not based upon the hypothesis that the mapped surface is ruled. In general, a mapping from a flat planar domain into a ruled surface is far from being isometric, and the property of being ruled is a partial consequence of our representation theorem.

Appendix: Lemmas Invoked in Establishing the Necessity of the Representation

Consider a smooth mapping r from $\mathcal {D}$ to $\mathcal {E}^{3}$, the gradient Q of which obeys (10) for all a in $\mathcal {V}^{2}$. Let {ı,ȷ} be a fixed orthonormal basis for $\mathcal {V}^{2}$. Then, by (10),

$$ |\boldsymbol {Q}\boldsymbol {\imath }|=1, \qquad |\boldsymbol {Q}\!\boldsymbol {\jmath }|=1, $$

(A.1)

and

$$ 2\boldsymbol {Q}\boldsymbol {\imath }\cdot \boldsymbol {Q}\boldsymbol {\jmath }=|\boldsymbol {Q}(\boldsymbol {\imath }+\boldsymbol {\jmath })|^2-|\boldsymbol {Q}\boldsymbol {\imath }|^2-|\boldsymbol {Q}\!\boldsymbol {\jmath }|^2=|\boldsymbol {\imath }+\boldsymbol {\jmath }|^2-2=0. $$

(A.2)

Let n be a mapping from $\mathcal {D}$ to $\mathcal {U}^{3}$ defined in accord with (20)₂, so that

$$ \boldsymbol {n}=\boldsymbol {Q}\boldsymbol {\imath }\times\!\boldsymbol {Q}\!\boldsymbol {\jmath }. $$

(A.3)

Importantly, (A.3) is but one of many equivalent ways by which n may be defined. For example, suppose that b and ${ \boldsymbol {b}_{\scriptscriptstyle \perp }}$ are elements of $\mathcal {U}^{2}$ defined such that

$$ \boldsymbol {b}:=\cos\alpha \boldsymbol {\imath }+\sin\alpha \boldsymbol {\jmath }\quad\text{and}\quad { \boldsymbol {b}_{\scriptscriptstyle \perp }}:=-\sin\alpha \boldsymbol {\imath }+\cos\alpha \boldsymbol {\jmath }$$

(A.4)

for some α in [0,2π]. Then, $\boldsymbol {b}\cdot { \boldsymbol {b}_{\scriptscriptstyle \perp }}=0$ and it is easy to see that

$$ \boldsymbol {Q}\boldsymbol {b}\times\!\boldsymbol {Q}{ \boldsymbol {b}_{\scriptscriptstyle \perp }}=\boldsymbol {n}. $$

(A.5)

In particular, choosing b=e and ${ \boldsymbol {b}_{\scriptscriptstyle \perp }}={ \boldsymbol {e}_{\scriptscriptstyle \perp }}$ in (A.5) yields (36).

We next establish some lemmas useful for the proof of the necessity of our representation contained in Sect. 3.

Lemma 1

At each point in $\mathcal {D}$, there exists a symmetric linear transformation of $\mathcal {V}^{2}$ to itself, namely a member of Sym₂, such that

$$ \nabla\!\boldsymbol {Q}=\boldsymbol {n}\otimes \boldsymbol {S}. $$

(A.6)

Proof

In view of (A.1)–(A.3), define mappings m _i , i=1,2,3, from $\mathcal {D}$ to $\mathcal {U}^{3}$ by

$$ \boldsymbol {m}_1=\boldsymbol {Q}\boldsymbol {\imath }, \qquad \boldsymbol {m}_2=\boldsymbol {Q}\!\boldsymbol {\jmath }, \quad \text{and}\quad \boldsymbol {m}_3=\boldsymbol {m}_1\times \boldsymbol {m}_2= \boldsymbol {n}. $$

(A.7)

Since the value of ∇Q=∇∇r at any point in $\mathcal {D}$ can be viewed as a linear transformation from Sym₂ to $\mathcal {V}^{3}$, it admits a representation of the form

$$ \nabla\!\boldsymbol {Q}=\sum _{i=1}^3\boldsymbol {m}_i\otimes \boldsymbol {S}_i, $$

(A.8)

where S _i is a mapping of $\mathcal {D}$ to Sym₂ for each i=1,2,3. Choose and fix arbitrary elements a and b of $\mathcal {V}^{2}$. Then, since

$$\begin{aligned} \nabla(\boldsymbol {Q}\boldsymbol {a}\cdot \boldsymbol {Q}\boldsymbol {b}) &= \bigl(\nabla(\boldsymbol {Q}\boldsymbol {a}) \bigr)^{{\scriptscriptstyle \top }}\!\boldsymbol {Q}\boldsymbol {b}+ \bigl(\nabla(\boldsymbol {Q}\boldsymbol {b}) \bigr)^{{\scriptscriptstyle \top }}\!\boldsymbol {Q}\boldsymbol {a} \\ &= \bigl((\nabla\!\boldsymbol {Q})\boldsymbol {a}\bigr)^{{\scriptscriptstyle \top }}\!\boldsymbol {Q}\boldsymbol {b}+ \bigl((\nabla\!\boldsymbol {Q})\boldsymbol {b}\bigr)^{{\scriptscriptstyle \top }}\!\boldsymbol {Q}\boldsymbol {a}, \end{aligned}$$

(A.9)

(A.8) and the recognition that Qa and Qb are orthogonal to m ₃ for all a and b in $\mathcal {V}^{2}$ yield

$$\begin{aligned} \nabla(\boldsymbol {Q}\boldsymbol {a}\cdot \boldsymbol {Q}\boldsymbol {b})&= \sum _{i=1}^3 \bigl(( \boldsymbol {m}_i\otimes \boldsymbol {S}_i\boldsymbol {a})^{{\scriptscriptstyle \top }}\!\boldsymbol {Q}\boldsymbol {b}+( \boldsymbol {m}_i\otimes \boldsymbol {S}_i\boldsymbol {b})^{{\scriptscriptstyle \top }}\!\boldsymbol {Q}\boldsymbol {a}\bigr) \\ &=\sum _{i=1}^3 \bigl((\boldsymbol {S}_i\boldsymbol {a}\otimes \boldsymbol {m}_i)\boldsymbol {Q}\boldsymbol {b}+(\boldsymbol {S}_i\boldsymbol {b}\otimes \boldsymbol {m}_i)\boldsymbol {Q}\boldsymbol {a}\bigr) \\ &=\sum _{i=1}^2 \bigl((\boldsymbol {Q}\boldsymbol {b}\cdot \boldsymbol {m}_i)\boldsymbol {S}_i\boldsymbol {a}+(\boldsymbol {Q}\boldsymbol {a}\cdot \boldsymbol {m}_i) \boldsymbol {S}_i\boldsymbol {b}\bigr) \\ &=\sum_{i=1}^2\boldsymbol {S}_i(\boldsymbol {a}\otimes \boldsymbol {b}+\boldsymbol {b}\otimes \boldsymbol {a})\boldsymbol {Q}^{{\scriptscriptstyle \top }}\!\boldsymbol {m}_i, \end{aligned}$$

(A.10)

which, with (11) and (A.7)_1,2, implies that S ₁ and S ₂ must satisfy

$$ \boldsymbol {S}_1(\boldsymbol {a}\otimes \boldsymbol {b}+\boldsymbol {b}\otimes \boldsymbol {a})\boldsymbol {\imath }+\boldsymbol {S}_2( \boldsymbol {a}\otimes \boldsymbol {b}+\boldsymbol {b}\otimes \boldsymbol {a})\boldsymbol {\jmath }=\bf0 $$

(A.11)

for all a and b in $\mathcal {V}^{2}$. With the particular choices a=b=ı, a=b=ȷ, and a=ı and b=ȷ in sequence, (A.11) results in requirements,

$$ \boldsymbol {S}_1\boldsymbol {\imath }=\mathbf{0},\qquad \boldsymbol {S}_2\boldsymbol {\jmath }=\bf0,\quad \mbox{and}\quad \boldsymbol {S}_{\mathrm{1}}\boldsymbol {\jmath }+\boldsymbol {S}_{\mathrm{2}}\boldsymbol {\imath }=\bf0 , $$

(A.12)

which, since S ₁ and S ₂ are mappings of $\mathcal {D}$ to Sym₂, can be met if and only if

$$ \boldsymbol {S}_1=\mathbf{0} \quad\text{and}\quad \boldsymbol {S}_2=\bf0. $$

(A.13)

Finally, using (A.7)₃ and (A.13) in (A.8) and setting S ₃=S, we obtain

$$ \nabla\!\boldsymbol {Q}=\boldsymbol {n}\otimes \boldsymbol {S}, $$

(A.14)

which completes the proof. ■

Lemma 2

The gradient ∇n of n defined by (A.3) admits a representation of the form

$$ \nabla \boldsymbol {n}=-\boldsymbol {Q}\boldsymbol {S}, $$

(A.15)

where Q denotes the gradient of r and S is the mapping of $\mathcal {D}$ to Sym₂ encountered in Lemma  1.

Proof

Differentiating (A.3) and using (A.6), we find that

$$\begin{aligned} \nabla \boldsymbol {n}&= \bigl((\boldsymbol {n}\times\!\boldsymbol {Q}\boldsymbol {\jmath })\otimes \boldsymbol {\imath }- (\boldsymbol {n}\times\!\boldsymbol {Q}\boldsymbol {\imath })\otimes\!\boldsymbol {\jmath }\bigr)\boldsymbol {S} \\&=-(\boldsymbol {Q}\boldsymbol {\imath }\otimes \boldsymbol {\imath }+\boldsymbol {Q}\!\boldsymbol {\jmath }\otimes\!\boldsymbol {\jmath })\boldsymbol {S} \\&=-\boldsymbol {Q}(\boldsymbol {\imath }\otimes \boldsymbol {\imath }+\boldsymbol {\jmath }\otimes \boldsymbol {\jmath })\boldsymbol {S} \\&=-\boldsymbol {Q}\boldsymbol {S}, \end{aligned}$$

(A.16)

which completes the proof. ■

Lemma 3

Let Q and n be given by Q=∇r and (A.3), with r being three-times continuously differentiable. Then, there exists mappings k from $\mathcal {D}$ to $\mathbb {R}$ and e from $\mathcal {D}$ to $\mathcal {U}^{2}$ such that the mapping S of $\mathcal {D}$ to Sym₂ encountered in Lemma 1 can be expressed as

$$ \boldsymbol {S}=k \boldsymbol {e}\otimes \boldsymbol {e}. $$

(A.17)

Proof

Let W=ȷ⊗ı−ı⊗ȷ. Granted that r is three-times continuously differentiable, we differentiate (A.6) and use (A.15) to find that

$$\begin{aligned} \nabla\nabla\!\boldsymbol {Q}[\boldsymbol {W}] &=-(\nabla \boldsymbol {n})\boldsymbol {W}\!\boldsymbol {S}+\boldsymbol {n}\otimes\nabla \boldsymbol {S}[\boldsymbol {W}] \\&=\boldsymbol {Q}\boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}+\boldsymbol {n}\otimes\nabla \boldsymbol {S}[\boldsymbol {W}]. \end{aligned}$$

(A.18)

Since S is in Sym₂ and W is skew, $(\boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S})^{{\scriptscriptstyle \top }}=-\boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}$ and since {W} provides a basis for the collection of skew linear transformations of $\mathcal {V}^{2}$ to $\mathcal {V}^{2}$, there exists a scalar λ such that

$$ \boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}=\lambda \boldsymbol {W}. $$

(A.19)

Applying the Cayley–Hamilton theorem to SW while bearing in mind that tr(SW)=0 and that detW=1, gives

$$ \boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}\boldsymbol {W}+(\det\bf \boldsymbol {S})\boldsymbol {I}=\bf0. $$

(A.20)

Next, taking the trace of (A.20) while invoking (A.19) yields

$$\begin{aligned} \det \boldsymbol {S}&=-\frac{1}{2}\text {tr}(\boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}\boldsymbol {W}) \\\det \boldsymbol {S}&=\frac{1}{2}\text {tr}(\boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}\boldsymbol {W}^{{\scriptscriptstyle \top }} ) \\&=\frac{1}{2}\boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}\cdot \boldsymbol {W} \\&=\frac{1}{2}\lambda \boldsymbol {W}\cdot \boldsymbol {W} \\&=\lambda, \end{aligned}$$

(A.21)

with which (A.19) becomes

$$ \boldsymbol {S}\boldsymbol {W}\!\boldsymbol {S}=(\det \boldsymbol {S})\boldsymbol {W}. $$

(A.22)

Thus,

$$ \nabla\nabla\!\boldsymbol {Q}[\boldsymbol {W}]=(\det \boldsymbol {S})\boldsymbol {Q}\boldsymbol {W}+\boldsymbol {n}\otimes\nabla \boldsymbol {S}[\boldsymbol {W}] $$

(A.23)

and, because $\nabla\nabla\!\boldsymbol {Q}[\boldsymbol {W}]=\bf0$, we see that

$$ \det \boldsymbol {S}=0 $$

(A.24)

and that $\nabla \boldsymbol {S}[\boldsymbol {W}]=\bf0$. Finally, we recognize that (A.24) implies that there exist mappings k from $\mathcal {D}$ to $\mathbb {R}$ and e from $\mathcal {D}$ to $\mathcal {U}^{2}$ such that the spectral representation of S has the form

$$ \boldsymbol {S}=k \boldsymbol {e}\otimes \boldsymbol {e}, $$

(A.25)

which completes the proof. ■

Remark 6

In the foregoing proof, ∇∇Q is a linear transformation which can be viewed as a linear transformation of linear transformations of $\mathcal {V}^{2}$ to itself to linear transformations of $\mathcal {V}^{2}$ to $\mathcal {V}^{3}$.