# Metric tensors and linear (differential) operators defined on a manifold and its osculating sphere

Given a 1D Riemannian manifold $\Gamma$ embedded in 2D Euclidean space (e.g. a parametric curve on a plane $\mathbb{R}^{2}$ ), and point $x_{0}\in \Gamma$, we denote $S^{1}(x_{0})$ the circle osculating $\Gamma$ at $x_0$.

Denote also $g = \lbrace g_{ij}\rbrace$ and $h = \lbrace h_{ij}\rbrace$ the induced metric tensors of $\Gamma$ and $S^{1}(x_{0})$ respectively, with $g^{ij}$ and $h^{ij}$ their inverse matrices and $G$, $H$ their determinants.

### My questions are:

1. Are the values of induced metric tensors $\lbrace g_{ij} \rbrace$ and $\lbrace h_{ij} \rbrace$ equal at the point $x_0\in \Gamma$?

• If yes, does it imply that for any linear (differential) operator defined on a manifold via metric tensor we will have the equality again? In particular, the Laplace-Beltrami operator: $> \forall f \in \mathcal{C}^{\infty}(\mathbb{R}^{2})$

$$\Delta_{\Gamma} \,f := \dfrac{1}{\sqrt{G}} \partial_{i} \Big( \sqrt{G} g^{ij} \partial_{i} \,f \Big) \bigg|_{x=x_{0}} \color{red}{\stackrel{(?)}{=}} \dfrac{1}{\sqrt{H}} \partial_{i} \Big( \sqrt{H} h^{ij} \partial_{i} \, f \Big) \bigg|_{x=x_{0}} =: \Delta_{S^{1}(x_0)}\, f.$$

1. If the answer for the previous question is yes, would it be possible to estimate the discrepancy of the values of these operators within the $\varepsilon$-neighborhood of $x_{0}$?

2. Is it possible to have similar conclusions for the manifolds of higher dimensions?

### Attempted solutions:

1. Yes. Since the both $g$ and $h$ are defined as restriction Euclidean metric from $\mathbb{R}^{2}$ to the tangent spaces of $\Gamma$ and $S^{1}(x_{0})$, we conclude that $g\big|_{x=x_0} = h\big|_{x=x_0}$ if and only if the tangent spaces of $\Gamma$ and $S^{1}(x_0)$ coincide at $x_0$. It follows from the definition of an osculating circle that the unit normal vectors of $\Gamma$ and $S^{1}(x_{0})$ at $x_0$ are equal, so as the tangent vectors. Does the equity of tangent vectors imply the equity of tangent planes though?

• Yes, by definition? Or do we need to establish the proximity of operators not only at the point $x_0$, but also in some neighborhood?
2. Consider a point $x_1 \in \Gamma$ such that $\big|x_0 - x_1\big|<\varepsilon$ for some $\varepsilon>0$. Given a linear (differential) operator $\mathcal{F}_{\Omega}$ defined on a manifold $\Omega$ in terms of its metric tensor, we want to estimate the norm of difference of operators $\mathcal{F}_{\Gamma}$ and $\mathcal{F}_{S^{1}(x_{0})}$ at the point $x_1$:

$$\bigg\| \mathcal{F}_{\Gamma}\big|_{x_1} - \mathcal{F}_{S^{1}(x_{0})}\big|_{x_1} \bigg\| = \sup_{ 0 \not\equiv f \in \mathcal{C}^{\infty}(\mathbb{R}^{2})} \Bigg\lbrace \dfrac{\Big\| \big[\mathcal{F}_{\Gamma} - \mathcal{F}_{S^{1}(x_{0})} \big]\big|_{x_1} f \Big\|}{\| f\|} \Bigg\rbrace ,$$ where $\big[\mathcal{F}_{\Gamma} - \mathcal{F}_{S^{1}(x_{0})} \big] \big|_{x_1} f = \big[\mathcal{F}_{\Gamma} \big] \big|_{x_1} f - \big[ \mathcal{F}_{S^{1}(x_{0})} \big] \big|_{x_1} f$.

Here $\big[\mathcal{F}_{\Gamma} \big] \big|_{x_1}$ and $\big[ \mathcal{F}_{S^{1}(x_{0})} \big] f \big|_{x_1}$ denote the operator $\mathcal{F}$ defined on $\Gamma$ and $S^{1}(x_{0})$ respectively, estimated at the point $x_1$ applied to a function $f$. $$\bigg\| \mathcal{F}_{\Gamma}\big|_{x_1} - \mathcal{F}_{S^{1}(x_{0})}\big|_{x_1} \bigg\| \leq \Big\| \mathcal{F}_{\Gamma}\big|_{x_1}\Big\| + \Big\| \mathcal{F}_{S^{1}(x_{0})}\big|_{x_1} \Big\|$$

If we assume that the operator $\mathcal{F}$ is defined as a linear combination of the entries of metric tensor matrix, i.e. $\mathcal{F} : = a^{ij}g _{ij}$, then the norm

$$\Big\| \mathcal{F}_{\Gamma}\big|_{x_1} \Big\| \leq |a^{ij}| \cdot \big\|g _{ij}({x_1})\big\| = \big| a^{ij}g_{ij}(x_1) \big|, \quad i,j = \overline{1,2}.$$

Since we are talking about Riemannian manifolds, we know that the metric tensor has to vary smoothly from point to point. If $\Gamma$ can be parametrized in terms of its arclength $s$ with $s_0, s_1$ corresponding to $x_0,x_1$ respectively, then we can apply Taylor expansion to every entry of the metric tensor:

$$g_{ij}(s_1) = g_{ij}(s_0) + (s_1 - s_0) \cdot \partial_{s} g_{ij}\big|_{s=s_1} + \frac{1}{2} (s_1 - s_0)^{2} \cdot \partial_{s}^{2} g_{ij}\big|_{s=s_1} + \dots$$ If there exists a constant $C>0$ such that $| \partial_{s} g_{ij} | \leq C$ $\forall i,j = 1,2$, we get

$$\big\| g_{ij}(s_1) - g_{ij}(s_0)\big\| = C \varepsilon + \frac{1}{2} C^{2} \varepsilon^{2} + \dots \approx \mathcal{O}(C\varepsilon).$$

If, in addition, we assume the existance of a constant $A$ s.t. $a^{ij}\leq A$, then

$$\Big\| \mathcal{F}_{\Gamma}\big|_{x_1} \Big\| \leq 4 A C\varepsilon, \quad \Big\| \mathcal{F}_{S^{1}(x_{0})}\big|_{x_1} \Big\| \leq 4 A C\varepsilon,$$ so that $$\boxed{ \Big\| \mathcal{F}_{\Gamma}\big|_{x_1} - \mathcal{F}_{S^{1}(x_{0})}\big|_{x_1} \Big\| \leq 8 A C \varepsilon }$$

Is there a way to obtain similar estimates for a wider class of linear operators $\mathcal{F}$, and/or with less assumptions on the metric tensor?

3. As for the manifolds of higher dimensions, I assume that we would have to reformulate the problem itself. For example, in order to generalize the problem for case of 2D surface in 3D Euclidean space we will have to use two principle curvatures and construct something like osculating ellipsoid (perhabs, an oblate spheroid) with its major and minor axis inverse proportional to the principle curvatures of the manifold. Similarly we can go over higher number of dimensions by osculating the manifold with an $n$-dimensional ball stretched along the principle axis of the manifold.

• Can you write down the definition of $S^1(x_0)$? Don't quite understand what you mean. – user99914 Apr 13 '15 at 6:16
• @John, let $\Gamma$ be a curve in $\mathbb{R}^{2}$. $\forall p\in\Gamma$ denote $\vec{\mathbf{n}}(p)$ - unit normal, $\kappa(p)$ - curvature, and $R=R(p):=\frac{1}{\kappa(p)}$. Then the center point (of our prospective osculating circle) is $o=p+R\,\vec{\mathbf{n}}(p)$, and the osculating circle is $$S^{1}(p):=\big\lbrace q\in\mathbb{R}^{2}\ \big|\quad\|p-q\|=R\big\rbrace.$$ In other words, $S^{1}(p)$ is a circle touching $\Gamma$ at the point $p$, whose unit normal at $p$ is equal to the one of $\Gamma$, and whose radius is the inverse curvature of $\Gamma$ at $p$. – Vlad Apr 13 '15 at 9:38
• Correction, in the comment above I meant to write $$S^{1}(p):=\big\lbrace q\in\mathbb{R}^{2}\ \big|\quad\|o-q\|=R\big\rbrace.$$ – Vlad Apr 13 '15 at 10:47