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}) $
> 
> 
> $$ \Delta\_{\Gamma} \,f := \dfrac{1}{\sqrt{G}} \partial\_{i} \Big(
>  \sqrt{G} g^{ij} \partial\_{i} \,f \Big) \bigg|\_{x=x\_{0}}
>  \color{red}{\stackrel{(?)}{=}} \dfrac{1}{\sqrt{H}} \partial\_{i} \Big(
>  \sqrt{H} h^{ij} \partial\_{i} \, f \Big) \bigg|\_{x=x\_{0}}
>  =: \Delta\_{S^{1}(x\_0)}\, f. $$
> 
> 
> 2. 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} $?
> 3. 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.




---


I have already [asked this question](https://math.stackexchange.com/questions/1232483/metric-tensors-and-linear-differential-operators-defined-on-a-manifold-and-its) a while ago on [math.stackexchange.com](http://math.stackexchange.com), but did not get an answer.



Differential Operators On A Curve And On Osculating Circle
The one-dimensional case is actually quite confusing so I will jump right to the general case. 


Consider hypersurfaces $M\_1,M\_2\subset \mathbb{R}^{n+1}$ which make "order-two" contact at $0 \in \mathbb{R}^{n+1}$. Let me define this in the following way:



> 
> 
> > 
> > Order-two contact at $0$ means that (after rotating) we can find functions $u\_1,u\_2:B\_\epsilon(0)\subset \{x^{n+1}=0\}\to \mathbb{R}$ so that locally, $M\_i$ is the graph $\{(x,u\_i(x)):x\in B\_\epsilon(0)\}$. Moreover, $u\_1$ and $u\_2$ agree up to second order: $u\_1(0) = u\_2(0) = 0$, $Du\_1(0) = Du\_2(0) = 0$, and $D^2u\_1(0)=D^2u\_2(0)$.
> > 
> > 
> > 
> 
> 
> 


It is a good exercise to check that this means that (1) $M\_1$ and $M\_2$ are tangent at $0$, with $T\_0M\_i=\{x^{n+1}=0\}$ and (2) the second fundamental form $A:Sym^2(T\_0 M\_i)\to\mathbb{R}$ are the same, after making the identification of the tangent planes as in (1). The second statement can be verified by deriving the formula for the second fundamental form of a graph.


Now, as you say, (1) implies that the metric tensors are the same at $0$ *after we identify the tangent planes with $\{x^{n+1}=0\}$.* 


Now, as for the Laplacian, a very convenient formula to relate the hypersurface Laplacian to the intrinsic Laplacian is:
$$
\tag{\*}\Delta\_{\mathbb{R}^{n+1}} f = \Delta\_{M\_i}f+D^2\_{\mathbb{R}^{n+1}}f(\nu\_{M\_i},\nu\_{M\_i})+H\_{M\_i}df(\nu\_{M\_i})
$$
Here, $H\_{M\_i}$ is the mean curvature (i.e., the trace of the second fundamental form). 




---


Here is a derivation of (\*). Write $D$ for the Euclidean connection and $\nabla$ for the induced connection on $M$ (I will drop the $i$ here). Then, recall that the *definition* of the second fundamental form is
$$
D\_XY = \nabla\_X Y - A(X,Y)\nu
$$
Thus, the Euclidean Hessian and the $M$ Hessian are related as
$$
D^2f(X,Y) = (Ddf)(X,Y) = X(df(Y)) - df(D\_XY) = X(df(Y)) - df(\nabla\_XY) + A(X,Y)df(\nu) = \nabla^2f(X,Y) +A(X,Y)df(\nu).
$$
Now, the Laplacian is the trace of the Hessian. Note that the traces are over different spaces. Choose an orthonormal basis of $T\_pM$ $e\_1,\dots,e\_n$ and complete it to an orthonormal basis of $T\_p\mathbb{R}^{n+1}$ by adding $\nu$. 


Thus,
$$
\Delta\_{\mathbb{R}^{n+1}} f = \sum\_{i=1}^n D^2 f (e\_i,e\_i) +D^2f(\nu,\nu) = \sum\_{i=1}^n \nabla ^2 f (e\_i,e\_i) +D^2f(\nu,\nu) + \sum\_{i=1}^n A(e\_i,e\_i)df(\nu) = \Delta\_M f +D^2f(\nu,\nu) + H\_Mdf(\nu) 
$$
as claimed.




---


Putting (\*) together with our previous claims (i.e., the normal vector and mean curvature must agree at $0$), we see that the Laplacians of $f$ restricted to $M\_1$ and $M\_2$ agree at $0$!




---


This answers everything but (3). I claim that $$|\Delta\_{M\_1}f(x)-\Delta\_{M\_2}f(y)|\leq C(|x|+|y|)$$ for $x,y$ sufficiently close to $0$, where $C$ depends on $M\_i$ and on $f$. To see this, you may simply appeal to (\*) and note that $\Delta\_{\mathbb{R}^{n+1}}f$, $D^2\_{\mathbb{R}^{n+1}}f$, $\nu\_{M\_i}$, $H\_{M\_i}$, and $df$, are all Lipschitz continuous, so their value at $x$ (rel. $y$) is close to their value at $0$, up to a factor of $|x|$ (rel. $y$). 


