Weak curvatures of irregular curves in high-dimensional Euclidean spaces

Abstract

We deal with a robust notion of weak normals for a wide class of irregular curves defined in Euclidean spaces of high dimension. Concerning polygonal curves, the discrete normals are built up through a Gram–Schmidt procedure applied to consecutive oriented segments, and they naturally live in the projective space associated with the Gauss hyper-sphere. By using sequences of inscribed polygonals with infinitesimal modulus, a relaxed notion of total variation of the jth normal to a generic curve is then introduced. For smooth curves satisfying the Jordan system, in fact, our relaxed notion agrees with the length of the smooth jth normal. Correspondingly, a good notion of weak jth normal of irregular curves with finite relaxed energy is introduced, and it turns out to be the strong limit of any sequence of approximating polygonals. The length of our weak normal agrees with the corresponding relaxed energy, for which a related integral-geometric formula is also obtained. We then discuss a wider class of smooth curves for which the weak normal is strictly related to the classical one, outside the inflection points. Finally, starting from the first variation of the length of the weak jth normal, a natural notion of curvature measure is also analyzed.

Appendix: Proof of Proposition 3

Assuming $N=3$, according to the notation from (4), the fifth-order expansions of $\mathbf{c}$ at s give:

$$\begin{aligned} \displaystyle \mathbf{v}_0(h)= & {} {{\dot{\mathbf{c}}}}+\frac{\mathbf{c}^{(3)}}{6}\,h^2+\mathbf{a}\,h^4+\mathbf{o}(h^4) \\ \displaystyle \mathbf{v}_1(h)= & {} -{{\dot{\mathbf{c}}}}+2\mathbf{c}^{(2)}\,h-\frac{13}{6}\,{\mathbf{c}^{(3)}}\,h^2+\frac{5}{3}\,\mathbf{c}^{(4)}\,h^3-\mathbf{b}\,h^4+\mathbf{o}(h^4) \\ \displaystyle \mathbf{v}_2(h)= & {} {{\dot{\mathbf{c}}}}+2\mathbf{c}^{(2)}\,h+\frac{13}{6}\,{\mathbf{c}^{(3)}}\,h^2+ \frac{5}{3}\,\mathbf{c}^{(4)}\,h^3+\mathbf{b}\,h^4+\mathbf{o}(h^4) \\ \displaystyle \mathbf{v}_3(h)= & {} -{{\dot{\mathbf{c}}}}+4\mathbf{c}^{(2)}\,h-\frac{49}{6}\,{\mathbf{c}^{(3)}}\,h^2+\frac{34}{3}\,\mathbf{c}^{(4)}\,h^3+\mathbf{o}(h^3) \end{aligned}$$

where $\mathbf{a}$ and $\mathbf{b}$ depend on $\mathbf{c}^{(5)}(s)$. We thus get:

$$\begin{aligned} \Vert \mathbf{v}_0(h)\Vert ^2=1-\frac{\Vert \mathbf{c}^{(2)}\Vert ^2}{3}\,h^2+\Bigl ( 2 \mathbf{a}\bullet {{\dot{\mathbf{c}}}}+\frac{1}{36}\,\Vert \mathbf{c}^{(3)}\Vert ^2\Bigr )\,h^4+o(h^4) \end{aligned}$$

and

$$\begin{aligned} \Vert \mathbf{v}_0(h)\Vert ^{-2}=1+\frac{\Vert \mathbf{c}^{(2)}\Vert ^2}{3}\,h^2+\Bigl ( \frac{1}{9}\,\Vert \mathbf{c}^{(2)}\Vert ^4-\frac{1}{36}\,\Vert \mathbf{c}^{(3)}\Vert ^2-2 \mathbf{a}\bullet {{\dot{\mathbf{c}}}}\Bigr )\,h^4+o(h^4) \end{aligned}$$

whence (5) holds. We also have

$$\begin{aligned} \mathbf{v}_1(h)\bullet \mathbf{v}_0(h)=-1+\frac{7}{3}\,\Vert \mathbf{c}^{(2)}\Vert ^2h^2+\frac{1}{3}\,\Bigl ( \mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)} + 5\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}}\Bigr )\,h^3 -a\,h^4 +o(h^4) \end{aligned}$$

where

$$\begin{aligned} a:= \frac{13}{36}\,\Vert \mathbf{c}^{(3)}\Vert ^2+(\mathbf{a}+\mathbf{b})\bullet {{\dot{\mathbf{c}}}} \end{aligned}$$

and hence

$$\begin{aligned} \frac{\mathbf{v}_1(h)\bullet \mathbf{v}_0(h)}{\Vert \mathbf{v}_0(h)\Vert ^2}=-1+2\Vert \mathbf{c}^{(2)}\Vert ^2h^2+\frac{1}{3}\,\Bigl ( \mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)} + 5\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}}\Bigr )\,h^3 -b\,h^4+ o(h^4) \end{aligned}$$

where

$$\begin{aligned} b:= \frac{1}{3}\,\Vert \mathbf{c}^{(3)}\Vert ^2+(\mathbf{b}-\mathbf{a})\bullet {{\dot{\mathbf{c}}}}-\frac{2}{3}\,\Vert \mathbf{c}^{(2)}\Vert ^4 \end{aligned}$$

that gives

$$\begin{aligned} \mathbf{N}_1(h)= & {} \displaystyle 2\mathbf{c}^{(2)}h-2\bigl (\Vert \mathbf{c}^{(2)}\Vert ^2{{\dot{\mathbf{c}}}}+\mathbf{c}^{(3)}\bigr )\,h^2+ \frac{1}{3}\,\Bigl ( 5 \mathbf{c}^{(4)} - 5(\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}})\,{{\dot{\mathbf{c}}}}- (\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}){{\dot{\mathbf{c}}}}\Bigr )\,h^3 \\&\displaystyle +\Bigl ( b\,{{\dot{\mathbf{c}}}} -\frac{1}{3}\,\Vert \mathbf{c}^{(2)}\Vert ^2\mathbf{c}^{(3)}+\mathbf{a}-\mathbf{b}\Bigr )\,h^4+\mathbf{o}(h^4). \end{aligned}$$

As a consequence, we get

$$\begin{aligned} \displaystyle \Vert \mathbf{N}_1(h)\Vert ^2=4\Vert \mathbf{c}^{(2)}\Vert ^2h^2- 8 {\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}}\,h^3+ 4\Bigl (\Vert \mathbf{c}^{(3)}\Vert ^2-\Vert \mathbf{c}^{(2)}\Vert ^4+\frac{5}{3}\, \mathbf{c}^{(4)}\bullet \mathbf{c}^{(2)} \Bigr )\,h^4+ o(h^4) \end{aligned}$$

whence

$$\begin{aligned} \displaystyle \Vert \mathbf{N}_1(h)\Vert ^{-2}= & {} \frac{1}{4\Vert \mathbf{c}^{(2)}\Vert ^2h^2}\Bigl [ 1+ 2\frac{\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}}{\Vert \mathbf{c}^{(2)}\Vert ^2}\,h+ \Bigl ( 4\frac{(\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)})^2}{\Vert \mathbf{c}^{(2)}\Vert ^4}\nonumber \\&-\frac{5}{3}\,\frac{ \mathbf{c}^{(4)}\bullet \mathbf{c}^{(2)} }{\Vert \mathbf{c}^{(2)}\Vert ^2}-\frac{\Vert \mathbf{c}^{(3)}\Vert ^2}{\Vert \mathbf{c}^{(2)}\Vert ^2}+ {\Vert \mathbf{c}^{(2)}\Vert ^2} \Bigr )\,h^2+ o(h^2)\Bigr ] \end{aligned}$$

and definitely (6) holds, where

$$\begin{aligned} \mathbf{d}:= & {} -\frac{1}{6}\,\frac{\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}}{\Vert \mathbf{c}^{(2)}\Vert }\,\mathbf{t}+\varOmega \,\mathbf{n}_1+\Bigl (\frac{5}{6}\,\frac{\mathbf{c}^{(4)}\bullet \mathbf{c}^{(3)\perp }}{\Vert \mathbf{c}^{(2)}\Vert \,\Vert \mathbf{c}^{(3)\perp }\Vert }\nonumber \\&- \frac{\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}}{\Vert \mathbf{c}^{(2)}\Vert ^3}\,\Vert \mathbf{c}^{(3)\perp } \Vert \Bigr )\,\mathbf{n}_2+ \frac{5}{6}\,\frac{\Vert \mathbf{c}^{(4)\perp }\Vert }{\Vert \mathbf{c}^{(2)}\Vert }\,\mathbf{n}_3 \end{aligned}$$

(33)

with the coefficient $\varOmega$ of $\mathbf{n}_1$ equal to

$$\begin{aligned} \varOmega := \frac{\bigl (\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}\bigr )^2}{\Vert \mathbf{c}^{(2)}\Vert ^4}\,\Bigl (\frac{3}{2}\,\Vert \mathbf{c}^{(2)}\Vert ^2-1\Bigr )+\frac{1}{2}\,\Vert \mathbf{c}^{(2)}\Vert ^2-\frac{1}{2}\,\frac{\Vert \mathbf{c}^{(3)} \Vert ^2}{\Vert \mathbf{c}^{(2)}\Vert ^2}. \end{aligned}$$

(34)

Moreover, in order to compute $\mathbf{N}_2(h)$, we check:

$$\begin{aligned} \mathbf{v}_2(h)\bullet \mathbf{v}_0(h)= & {} 1-\frac{7}{3}\,\Vert \mathbf{c}^{(2)}\Vert ^2h^2+\frac{1}{3}\bigl ( \mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}+5\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}}\bigr )+o(h^3) \\ \frac{\mathbf{v}_2(h)\bullet \mathbf{v}_0(h)}{\Vert \mathbf{v}_0(h)\Vert ^2}= & {} 1-2\Vert \mathbf{c}^{(2)}\Vert ^2h^2+\frac{1}{3}\bigl ( \mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}+5\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}}\bigr )+o(h^3) \end{aligned}$$

and hence

$$\begin{aligned}&-\frac{\mathbf{v}_2(h)\bullet \mathbf{v}_0(h)}{\Vert \mathbf{v}_0(h)\Vert ^2}\,\mathbf{v}_0(h)= -{{\dot{\mathbf{c}}}}+\Bigl (2\Vert \mathbf{c}^{(2)}\Vert ^2\,{{\dot{\mathbf{c}}}}-\frac{1}{6}\,\mathbf{c}^{(3)}\Bigr )\,h^2 \nonumber \\&\quad -\frac{1}{3}\,\bigl ( \mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}+5\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}} \bigr )\,{{\dot{\mathbf{c}}}}\,h^3 +\mathbf{o}(h^3). \end{aligned}$$

Furthermore,

$$\begin{aligned} \mathbf{v}_2(h)\bullet \mathbf{N}_1(h)=4\Vert \mathbf{c}^{(2)}\Vert ^2h^2 +\Bigl (\frac{5}{3}\,\mathbf{c}^{(4)}\bullet \mathbf{c}^{(2)} + \Vert \mathbf{c}^{(2)}\Vert ^4 - \Vert \mathbf{c}^{(3)}\Vert ^2 \Bigr )\,h^4+o(h^4) \end{aligned}$$

so that

$$\begin{aligned} \bigl ( \mathbf{v}_2(h)\bullet \mathbf{N}_1(h)\bigr )\,\mathbf{N}_1(h)=4\Vert \mathbf{c}^{(2)}\Vert \,h^2\Bigl \{ 2\mathbf{c}^{(2)}h-2\bigl (\Vert \mathbf{c}^{(2)}\Vert ^2{{\dot{\mathbf{c}}}}+\mathbf{c}^{(3)}\bigr ) \,h^2+\mathbf{A}\,h^3+o(h^3)\Bigr \} \end{aligned}$$

where

$$\begin{aligned} \mathbf{A}:= & {} \frac{1}{3}\,\Bigl ( 5 \mathbf{c}^{(4)} - 5(\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}})\,{{\dot{\mathbf{c}}}}- (\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}){{\dot{\mathbf{c}}}}\Bigr )\\&+2\,\Bigl (\Vert \mathbf{c}^{(2)}\Vert ^2-\frac{\Vert \mathbf{c}^{(3)}\Vert ^2}{\Vert \mathbf{c}^{(2)}\Vert ^2} \Bigr )\,\mathbf{c}^{(2)}+\frac{10}{3}\,\frac{\mathbf{c}^{(4)}\bullet \mathbf{c}^{(2)}}{\Vert \mathbf{c}^{(2)}\Vert ^2}\,\mathbf{c}^{(2)}. \end{aligned}$$

We thus obtain:

$$\begin{aligned} \frac{ \mathbf{v}_2(h)\bullet \mathbf{N}_1(h)}{\Vert \mathbf{N}_1(h)\Vert ^2}\,\mathbf{N}_1(h) = 2\mathbf{c}^{(2)}\,h+\Bigl (4\frac{\mathbf{c}^{(2)}\bullet \mathbf{c}^{(3)}}{\Vert \mathbf{c}^{(2)}\Vert ^2}\,\mathbf{c}^{(2)} -2\bigl (\Vert \mathbf{c}^{(2)}\Vert ^2{{\dot{\mathbf{c}}}}+\mathbf{c}^{(3)}\bigr )\Bigr )\,h^2+\mathbf{B}\,h^3+\mathbf{o}(h^3), \end{aligned}$$

where

$$\begin{aligned} \displaystyle \mathbf{B}:= & {} \frac{1}{3}\,\Bigl ( 5 \mathbf{c}^{(4)} - 5(\mathbf{c}^{(4)}\bullet {{\dot{\mathbf{c}}}})\,{{\dot{\mathbf{c}}}}- 13(\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}){{\dot{\mathbf{c}}}}\Bigr ) \\&\displaystyle +4\,\Bigl (\Vert \mathbf{c}^{(2)}\Vert ^2-\frac{\Vert \mathbf{c}^{(3)}\Vert ^2}{\Vert \mathbf{c}^{(2)}\Vert ^2} +2\,\frac{\bigl (\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}\bigr )^2}{\Vert \mathbf{c}^{(2)}\Vert ^4} \Bigr )\,\mathbf{c}^{(2)} -4\,\frac{\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}}{\Vert \mathbf{c}^{(2)}\Vert ^2}\,\mathbf{c}^{(3)}. \end{aligned}$$

Putting the terms together, we get:

$$\begin{aligned} \mathbf{N}_2(h)=4\mathbf{c}^{(3)\perp } h^2+4\,\mathbf{D}\,h^3+\mathbf{o}(h^3) \end{aligned}$$

where

$$\begin{aligned} \mathbf{D}:=\bigl (\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}\bigr ){{\dot{\mathbf{c}}}} +\Bigl (\frac{\Vert \mathbf{c}^{(3)}\Vert ^2}{\Vert \mathbf{c}^{(2)}\Vert ^2}-\Vert \mathbf{c}^{(2)}\Vert ^2 \Bigr )\,\mathbf{c}^{(2)} +\frac{\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}}{\Vert \mathbf{c}^{(2)}\Vert ^2} \,\mathbf{c}^{(3)} -2\,\frac{\bigl (\mathbf{c}^{(3)}\bullet \mathbf{c}^{(2)}\bigr )^2}{\Vert \mathbf{c}^{(2)}\Vert ^4}\,\mathbf{c}^{(2)} \end{aligned}$$

and definitely (7) holds, where in terms of the orthonormal basis $(\mathbf{t},\mathbf{n}_1,\mathbf{n}_2,\mathbf{n}_3)$ we obtain formula (8) for $\mathbf{D}$.

Finally, formula (9) follows by arguing as in the proof of Proposition 2. In fact, the Gram–Schmidt procedure yields that $(\mathbf{t}(h),\mathbf{n}_1(h),\mathbf{n}_2(h),\mathbf{n}_3(h))$ is an orthonormal basis of ${{\mathbb {R}}}^4$, whence $\mathbf{n}_3(h)=\mathbf{n}_3+\mathbf{o}(1)$.

More precisely, we have $\mathbf{n}_3(h)=\pm *(\mathbf{t}(h)\wedge \mathbf{n}_1(h)\wedge \mathbf{n}_2(h))$, where $*$ is the Hodge operator in ${{\mathbb {R}}}^4$, whereas $*(\mathbf{t}\wedge \mathbf{n}_1\wedge \mathbf{n}_2)=\pm \mathbf{n}_3$, with the same sign ± in the previous two formulas, by our choice in (4). Using that

$$\begin{aligned} \mathbf{t}(h)=\mathbf{t}+\mathbf{o}(h),\quad \mathbf{n}_1(h)=\mathbf{n}_1+\alpha \,\mathbf{n}_2\,h+\mathbf{o}(h), \quad \mathbf{n}_2(h)=\mathbf{n}_2+(\beta \,\mathbf{t}+\gamma \,\mathbf{n}_1)\,h+\mathbf{o}(h) \end{aligned}$$

for some real numbers $\alpha ,\beta ,\gamma \in {{\mathbb {R}}}$, we get $\mathbf{t}(h)\wedge \mathbf{n}_1(h)=\mathbf{t}\wedge \mathbf{n}_1+\alpha \,\mathbf{t}\wedge \mathbf{n}_2\,h+\mathbf{o}(1)\wedge \mathbf{o}(1)\,h$ and hence $\mathbf{t}(h)\wedge \mathbf{n}_1(h)\wedge \mathbf{n}_2(h)=\mathbf{t}\wedge \mathbf{n}_1\wedge \mathbf{n}_2+\mathbf{o}(1)\wedge \mathbf{o}(1)\wedge \mathbf{o}(1)\,h$, whence actually (9) holds true, as required.