For unit speed curves $\gamma'(s)\cdot\sigma'(s) \geq \cos(2Ks)$ with bounded curvature

Question

Suppose that we have two unit speed curves $\gamma,\sigma:[0,l] \to \mathbb{R}^n$ or $\mathbb{R}^2$ if necessary and that $\gamma'(0) = \sigma'(0)$. If we assume that the curvature for both curves is bounded from above by $K > 0$, then I would like to say that for $s\in [0,\frac{\pi}{2K}]$ we have an estimate of the form $$ \gamma'(s)\cdot\sigma'(s) \geq \cos(2 K s) $$

This inequality would be equivalent to showing that for $s\in [0,\frac{\pi}{2K}]$ we have that $$ \gamma'(s)\cdot\sigma'(s) \geq \tilde{\gamma}'(s)\cdot\tilde{\sigma}'(s) $$ where $\tilde{\gamma},\tilde{\sigma}$ are unit speed curves on two circles of radius $\frac{1}{K}$ which touch at the point $\gamma(0) = \sigma(0)$ which are traveling in oppposite directions but are constantly curving with curvature $K$. The inner product inequality is also equivalent to showing that $$ ||\gamma'(s)-\sigma'(s)|| \leq || \tilde{\gamma}'(s)- \tilde{\sigma}'(s) || $$

This can be seen by squaring both sides and organizing terms since $$||\gamma'(s)||^2 = ||\sigma'(s)||^2 = ||\tilde{\gamma}'(s)||^2 = ||\tilde{\sigma}'(s)|| = 1$$

For example, in 2D this could be $\tilde{\gamma}(t) = (\frac{1}{K}\cos(Kt),\frac{1}{K}\sin(Kt))$ and $\tilde{\sigma}(t) = (\frac{2-\cos(Kt)}{K},\frac{1}{K}\sin(Kt))$. Then $\tilde{\gamma}'(t) = (-\sin(Kt),\cos(Kt))$, $\tilde{\sigma}'(t) = (\sin(Kt),\cos(Kt))$ and so $\tilde{\gamma}'(0) = \tilde{\sigma}'(0) = (1,0)$ and by double angle identities we have that $\tilde{\gamma}'(t)\cdot \tilde{\sigma}'(t) = \cos(2Kt)$.

So far I have been able to show that if $\kappa_\gamma,\kappa_\sigma$ represent the curvature at time $s$ and $N_\gamma,N_\sigma$ are the unit normal curves for $\gamma$ and $\sigma$ respectively, then we have that $$ \gamma'(s)\cdot\sigma'(s) \geq 1 - 2Ks. $$ This inequality follows from the Fundamental Theorem of Calculus \begin{align*} \gamma'(s)\cdot\sigma'(s) &= \gamma'(0)\cdot\sigma'(0) + \int_0^s\Big(\gamma'(t)\cdot\sigma'(t)\Big)'dt\\ &= 1 + \int_0^s \gamma''(t)\cdot\sigma'(t) + \gamma'(t)\cdot\sigma''(t)dt\\ &= 1 + \int_0^s \kappa_\gamma(t)N_\gamma(t)\cdot\sigma'(t) + \kappa_\sigma(t)\gamma'(t)\cdot N_\sigma(t)dt\\ &\geq 1 - \int_0^s|\kappa_\gamma(t)|+|\kappa_\sigma(t)|dt\\ &\geq 1 - \int_0^s2Kdt\\ &= 1 - 2Ks \end{align*}

I appreciate any recommendations.

Answer 1

I am going to answer this question now that I have tried a very different approach based on using that the curve has unit-speed and some basic results from Riemannian geometry.

First note that since $\gamma$ and $\sigma$ are unit speed curves, then $\gamma'(s),\sigma'(s) \in\mathbb{S}^n$ for all $s\in\mathbb{R}$. This then allows us to define smooth curves $\alpha,\beta:\mathbb{R}\to\mathbb{S}^n$ by $\alpha(s) = \gamma'(s)$ and $\beta(s) = \sigma'(s)$. In particular, since $\gamma'(0) = \sigma'(0)$, then we have that $\alpha(0) = \beta(0)$.

Next we introduce the standard round metric on $\mathbb{S}^n$ which is defined for $p,q\in \mathbb{S}^n\subset\mathbb{R}^{n+1}$ to be the quantity $$ d_{\mathbb{S}^n}(p,q) := \arccos(p\cdot q) $$ where $p\cdot q$ is the classical dot product on $\mathbb{R}^{n+1}$. Since $x\to \arccos(x)$ is strictly decreasing on $[-1,1]$, then our goal of minimizing $\gamma'(s)\cdot \sigma'(s)$ is equivalent to maximizing $d_{\mathbb{S}^n}(\alpha(s),\beta(s))$ subject to the constraint that the curvature of $\gamma$ and $\sigma$ are both bounded above by $K$. Now the curvature for a unit speed curve at time $s$ is the quantity $||\gamma''(s)||$. Now the velocity vector of $\alpha$ is $\alpha'(s) = \gamma''(s) \in T_{\gamma'(s)}\mathbb{S}^n$ and with respect to the round metric on $\mathbb{S}^n$ we see that the curvature bound of $K$ implies that $|\alpha'(s)| = ||\gamma''(s)||\leq K$. That is, the speed of $\alpha$ and $\beta$ are both bounded above by $K$.

Now $\overline{B_{\mathbb{S}^n}(\alpha(0),sK)}$ is the set of all points which can be reached by time $s$ when traveling a long a curve of speed at most $K$. Thus $\alpha(s),\beta(s) \in \overline{B_{\mathbb{S}^n}(\alpha(0),sK)}$ for all $s > 0$. Using facts from Riemannian geometry, as long as $sK\leq \frac{\pi}{2}$, then $\text{diam}(\overline{B_{\mathbb{S}^n}(\alpha(0),sK)}) = 2sK$. Thus provided $s\in [0,\frac{\pi}{2K}]$, we will have that $$ d_{\mathbb{S}^n}(\alpha(s),\beta(s)) \leq \text{diam}(\overline{B_{\mathbb{S}^n}(\alpha(0),sK)}) = 2sK $$ Then by the definition of the round metric as well as that $\alpha(s) = \gamma'(s)$ and $\beta(s) = \sigma'(s)$, we find that for $s\in [0,\frac{\pi}{2K}]$ that $$ \gamma'(s)\cdot\sigma'(s) \geq \cos(2Ks) $$ We can similarly show that for the same assumptions on $\sigma$ that if $s\in [0,\frac{\pi}{K}]$ then $$ \sigma'(0)\cdot\sigma'(s)\geq \cos(Ks). $$ I will also mention that the lower bound is achieved when traveling along geodesics in opposite directions for the main result and for the statement above, the result holds when traveling along a geodesic as well.