Orthonormal basis given by the integral of a vector field along a curve orthogonal to the tangent

Question

A friend and I are trying to understand the construction on section 3.2 of this paper by Bernatzki and Ye. Suppose $\gamma$ is a smooth simple closed curve in $\mathbb{R}^3$. Then this paper claims there exist vector fields $v_i: \gamma \to \mathbb{R}^3$, $i = 1,2,3$, such that $v_i \perp \dot{\gamma}$ everywhere, and the three vectors $V_i = \int_{\gamma} v_i(s) ds$, $s$ being the arc-length parameter, form an orthonormal basis. Furthermore, we can make the support of the $v_i$ arbitrarily small. Why is this true?

We know that each $v_i$ can be written as a linear combination $f_iN + g_i B$, with $(T,N,B)$ being the Frenet frame, $f_i$ and $g_i$ being functions, but the integrals $\int f_iN + g_i B$ don't seem to have particularly nice forms, even after applying the Frenet-Serret equations and integration by parts. I think almost any choice of $v_i$ should yield $V_i$ forming a basis of $\mathbb{R}^3$, but there seems to be difficulty in proving this rigorously.

Answer 1

This is an unusual result, and I'm surprised the authors dismissed it as if it were standard and well-known. Here's the main idea. You can sort out the final details. Having the supports be very small is required and is the main trick, from my point of view. (Think delta functions, or smooth approximations thereto.)

For a closed space curve, it is well-known that the tangent indicatrix is "balanced" with respect to every direction. Indeed, assuming $\gamma$ is arclength parametrized (with length $L$), we have $\int_0^L T(s)\,ds = \int_0^L \gamma'(s)\,ds = 0$, and so $\int_0^L (T\cdot A)\,ds = 0$ for any vector $A$. The same principle applies to the vector field $\kappa N$ (but not to $N$ itself), since $T'=\kappa N$. In particular, if you fix $s_1$ and take $A=N(s_1)$, then since $(\kappa N)(s_1)\cdot A>0$ and the integral is $0$, by the intermediate value theorem there must be $s_2$ so that $(\kappa N)(s_2) \cdot A = 0$.

Let $\rho_1$ be a bump function with integral $1$ supported in an arbitrarily small neighborhood of $s_1$, and let $\rho_2$ similarly be supported in an arbitrarily small neighborhood of $s_2$. Let $v_1 = \rho_1 N$ and let $v_2 = \rho_2 N$. You can rescale to make $V_1$ and $V_2$ unit vectors; they won't quite be orthogonal, but I believe you can alter $v_2$ by adding a small multiple of $B$ to fix it.

Now, can we find $s_3$ so that $\alpha N(s_3)+\beta B(s_3)$ is orthogonal to both $N(s_1)$ and $N(s_2)$ for some scalars $\alpha,\beta$? If not, the normal plane to $\gamma$ at every $s$ never contains the vector $A=N(s_1)\times N(s_2)$. But this means that $T(s)\cdot A$ is always nonzero, contradicting our original observation that the tangent indicatrix is always balanced. So there is such an $s_3$, and once again we use our bump function trick to create $v_3$ and correct as necessary.

If you find some fatal flaws with this, let me know. And certainly let us know if you figure out a trivial solution, but I certainly didn't see one.

Answer 2

I am sure Ted Shifrin's answer is correct, and here are the missing details on orthogonality.

From his answer, we can see that we can find $s_1,s_2,$ and $s_3$ such that some linear combination of $N(s_i)$ and $B(s_i)$ forms an orthonormal basis of $\mathbb{R}^3$. The tricky part is that after taking an approximation to the identity supported around small neighborhoods of the $s_i$ and integrating, we might no longer have that the integrated vectors are orthogonal.

Let us first assume that we found $s_1,s_2$ such that $N(s_1)$ and $N(s_2)$ are orthogonal. Assume $\rho_1$ a smooth function of compact support around $s_1$, and we can modify $\rho_1$'s support to be arbitrarily small and such that the mollified vector $V_1 = \int_0^L \rho_1(s)N(s)ds$ is arbitrarily close to $N(s_1)$. Now take another smooth approximation to the identity $\rho_2$ with compact support on an arbitrarily small neighborhood of $s_2$. We claim that we can adjust $v_2$ by writing $v_2 = \rho_2(s)N(s) + f(s)N(s) + g(s)B(s)$, where $f$ and $g$ are smooth with support is contained within that of $\rho_2$, such that $V_2 = \int_0^L \left[ \rho_2(s)N(s) + f(s)N(s) + g(s)B(s) \right]ds$ is orthogonal to $V_1$.

Before defining $f$, observe

\begin{align*} \langle V_1, V_2 \rangle &= \langle V_1, \int_0^L \left[ \rho_2(s)N(s) + f(s)N(s) + g(s)B(s) \right]ds \rangle\\ &= \int_0^L \langle V_1, \rho_2(s)N(s) \rangle + \langle V_1, f(s)N(s) + g(s)B(s) \rangle ds. \end{align*}

So if we can find a smoothly varying solution $(f(s),g(s))$ to the equation $$-\langle V_1, \rho_2(s)N(s) \rangle = \langle V_1, f(s)N(s) + g(s)B(s) \rangle,$$ we will have proved orthogonality. However, we want to avoid the situation where the only solution we can choose is $f = -\rho_2, g = 0$, since this would force $v_2$ and hence $V_2$ to be zero.

To prove we can avoid this scenario, we need to prove an elementary lemma of linear algebra.

Lemma: If $(f,g) = (-\rho,0)$ is the only solution to $\langle V, (f+\rho)N + gB \rangle = 0$, then $V$ is a multiple of $N$.

Proof: We prove the contrapositive. Write $V = aT + bN + cB$ and suppose $V$ is not a multiple of $N$, so either $a \neq 0$ or $c \neq 0$. Our equation is equivalent to $b(f+\rho) + cg = 0$. If $c = 0$ (and thus $a \neq 0$), then $f = -\rho, g = 1$ is a solution. If $c \neq 0$, then $f = 0, g = \frac{-b\rho}{c}$ is a solution. $\square$

(The lemma is actually vacuously true; it will never be the case that $(-\rho,0)$ is the only solution, but now I don't need to prove that). Now we can assume that $\rho_1$ is chosen so that $V_1$ is close enough to $N(s_1)$ such that it is not a scalar multiple of any $N(s)$ for $s$ in the support of $\rho_2$.

For $s$ outside the support of $\rho_2$, we can take $f(s) = g(s) = 0$, which will mean the supports of $f$ and $g$ will lie inside that of $\rho_2$. If $\rho_2(s) > 0$, then $V_1$ is not a scalar multiple of $N(s)$, so by the lemma we can choose $f(s)$ and $g(s)$ to be something other than the trivial solution $f(s) = -\rho_2(s), g(s) = 0$.

To see these solutions are smoothly varying in $s$, we can observe that the choice of $f(s)$ and $g(s)$ corresponds to picking a point in the translated subspace $V_1^\perp \cap \text{span}(N(s),B(s)) -\rho_2(s)N(s)$ which avoids $-\rho_2(s)N$ on a set of positive $s$-measure. For fixed $s$, $V_1^\perp \cap \text{span}(N(s),B(s)) -\rho_2(s)N(s)$ is an affine subspace of dimension at least $1$ which smoothly varies. (Imagine a line in the $(N,B)$ plane which smoothly moves around, translated by $-\rho_2(s)$ in the $N$ direction). The only issue which could occur is when this line becomes the entire $(N,B)$ plane, corresponding to the situation when $V_1$ is parallel to $T$. If we are avoiding the point $-\rho_2(s)N$, then the line could become a plane for an isolated $s_0$, and then discontinuously jump to another line. If this occurs, we would be forced to select $(f(s_0),g(s_0)) = (-\rho_2(s_0),0)$ for this singular $s$ in order to maintain continuity of $f$ and $g$, but as this only occurs at an isolated $s_0$, we could still maintain that $(f(s),g(s)) \neq (-\rho_2(s),0)$ for $s$ in a punctured neighborhood of this $s_0$. And of course, should $V_1$ be parallel to $T(s)$ for a set of positive $s$-measure, meaning the intersection becomes a plane for more than an instant, then we could still choose points such that we avoid having $(f,g) = (-\rho_2,0)$, ensuring our selection is on the appropriate line once $T(s)$ ceases to be parallel to $V_1$.

For $V_3$, we have that a linear combination $U = \alpha N(s_3) + \beta B(s_3)$ is orthogonal to $N(s_1)$ and $N(s_2)$. Assume $U$ is of unit length, and take smooth functions $\alpha(s), \beta(s)$ such that $U(s) = \alpha(s)N(s) + \beta(s)B(s)$ is a unit frame and that $U(s_3) = U$. Likewise, set $W = -\beta N + \alpha B$ so that $U(s) \perp W(s)$ for all $s$. Now consider $\rho_3(s)U(s)$ with $\rho_3$ being a smooth approximation of the identity with arbitrarily small support around $s_3$. Again, we need to choose correction terms $f(s)$ and $g(s)$ with $v_3(s) = \rho_3(s)U(s) + f(s)U(s) + g(s)W(s)$ so $V_3 \perp V_1,V_2$. The difficulty here is that since we need to satisfy two simultaneous orthogonality equations, we have that our selection of points $f(s),g(s)$ must be on the mutual intersection of the planes $V_1^\perp - \rho_3(s)U(s)$, $V_2^\perp - \rho_3(s)U(s)$, and $\text{span}(N,B) = \text{span}(U,W)$, which is in general the single point $-\rho_3(s)U(s)$, and we could be forced into the situation that $V_3 = 0$, so we cannot hope to define $f$ and $g$ so that $v_3$ is pointwise orthogonal to both $V_1$ and $V_2$.

However, there already exists a unique (up to sign) unit $V_3$ orthogonal to $V_1$ and $V_2$. Choose the option that is closer to $U(s_3)$, and note that the distance to $U(s_3)$ can be made arbitrarily small by decreasing the supports of the $\rho_i$. Then we need to find another $(f,g)$ with support around a small $\varepsilon$-neighborhood of $s_3$ satisfying $V_3 = \int_{s_3 - \varepsilon}^{s_3 + \varepsilon} \left[ \rho_3(s)U(s) + f(s)U(s) + g(s)W(s) \right]ds$.

Furthermore, assume the supports of $\rho_1$ and $\rho_2$ are small enough such that $V_3$ is close enough to $U(s_3)$ to ensure the quantity $\langle V_3, U(s) \rangle$ is positive and bounded away from zero for $s \in (s_3-\varepsilon, s_3+\varepsilon)$.

To measure the deviance of $\int v_3 ds$ from $V_3$, we can take its inner product with $V_3$ and see how close it is to $1$. Observe

\begin{align*} \langle V_3, \int_0^L v_3(s)ds \rangle &= \int_{s_3-\varepsilon}^{s_3+\varepsilon} \langle V_3, (\rho_3(s) + f(s))U(s) + g(s)W(s) \rangle ds. \end{align*} We need to solve the equation $\langle V_3, (\rho_3(s) + f(s))U(s) + g(s)W(s) \rangle = \rho_3(s)$, as this would ensure the integrand is $\rho_3(s)$, and thus have total integral $1$, proving $\langle V_3, \int v_3 \rangle = 1$.

Notice that for all $s$, $V_3 = \langle V_3,T(s)\rangle T(s) + \langle V_3, U(s) \rangle U(s) + \langle V_3, W(s) \rangle W(s)$. Thus, our equation becomes $\langle V_3, U(s) \rangle(\rho_3(s) + f(s)) + \langle V_3, W(s) \rangle g(s) = \rho_3(s)$. As $\langle V_3, U(s) \rangle \neq 0$, we have that $f(s) = \frac{\rho_3(s)}{\langle V_3, U(s) \rangle} - \frac{\langle V_3, W(s) \rangle}{ \langle V_3, U(s) \rangle}g(s) - \rho_3(s)$. Picking $g(s)$ to be any smooth function with compact support on $(s_3-\varepsilon,s_3+\varepsilon)$, we have smoothly defined $f(s)$. We can also see that $f(s) \to 0$ as $s \to s_3 \pm \varepsilon$, so $f$ also has compact support. After picking an appropriate $g$ to ensure $|\int v_3| = 1$, we are done.