2
$\begingroup$

Following Anton Petrunin’s suggestion, I revise the question to make it less vague.

Let $M^{m}$ be an $m$-dimensional Riemannian manifold, and let $\gamma$ be a unit-speed curve $I \to M^{m}$. We say that $\gamma$ is planar if there are a unit vector field $N$ along $\gamma$ and a function $\kappa$ such that $$\begin{aligned} D_{t}\gamma' &= \kappa N,\\ D_{t} N &= -\kappa \gamma', \end{aligned}$$ where $D_{t}$ is the covariant derivative along $\gamma$.

I know that, if $M^{m}$ has constant curvature, then studying planar curves in $M^{m}$ is really the same as studying planar curves in $M^{2}$; indeed, any planar curve in $M^{m}$ is contained in a two-dimensional totally geodesic submanifold of $M^{m}$.

I am interested in the case where the curvature is not necessarily constant. Is there any good reason to study planar curves in $M^{m}$ instead of just $M^{2}$?

More precisely, suppose that $\gamma$ is a planar curve in $M^{m}$. Does there exists a two-dimensional Riemannian submanifold $S$ of $M^{m}$ such that $\gamma(I) \subset S$ and such that $$R_{S}(X,Y,Z,W) = R_{M}(X,Y,Z,W)$$ for all $X,Y,Z,W \in \mathfrak{X}(S)$? Here $R_{S}$ and $R_{M}$ denote the Riemann curvature tensors of $S$ and $M^{m}$, respectively.

$\endgroup$
2
  • 1
    $\begingroup$ It is too vague, try to write more. $\endgroup$ Jul 17, 2022 at 13:26
  • $\begingroup$ @AntonPetrunin Thanks for the feedback, I have edited the question. $\endgroup$
    – MK7
    Jul 21, 2022 at 20:51

1 Answer 1

2
$\begingroup$

Here are some comments about the OP's question that don't give a definitive answer to the final question (although the answer may well be 'no', see below), but do provide more information, at least in the simplest possible case, in which $M$ has dimension $3$.

First, by the Gauss equation in dimension $3$, a surface $S\subset M$ has the property $R_S(W,X,Y,Z) = R_M(W,X,Y,Z)$ for all tangent vector fields $W,X,Y,Z$ on $S$ if and only if the second fundamental form of $S$ in $M$ has rank at most $1$. Elsewhere, the OP has called such submanifolds 'essentially flat' (to be abbreviated EF below). To simplify the discussion to follow, I want a name for surfaces $S\subset M$ whose second fundamental form has rank equal to $1$ everywhere on $S$. I propose 'nondegenerate essentially flat', to be abbreviated NDEF.

Now, in the real-analytic setting (i.e., all given data are real-analytic, including the metric on $M$), the Cartan-Kähler theorem implies the following result:

Proposition: Let $\gamma:I\to M$ be an embedded unit speed curve defined on some interval $I\subset\mathbb{R}$ and suppose that $\nu:I\to TM$ is a unit vector field along $\gamma$ that satisfies $\nu(t)\cdot\gamma'(t) = 0$ while $\nu(t)\cdot \nabla_{\gamma'(t)}\gamma'(t)\not=0$ for all $t\in I$. Then there exists an embedded NDEF surface $S\subset M$ that contains $\gamma(I)$ and has $\nu(t)\perp T_{\gamma(t)}S$ for all $t\in I$. Moreover, any other NDEF surface $\tilde S\subset M$ with these properties agrees with $S$ on some neighborhood of $\gamma(I)$ in $M$.

Note that this answers the OP's final question in the affirmative in the real-analytic setting as long as the function $\kappa$ is non-vanishing. However, I suspect that the OP actually wants to ask an even more specific question in the case of a planar curve: Does there exist a EF surface $S\subset M$ containing $\gamma(I)$ such that both $\gamma'$ and $N$ are tangent to $S$ along $\gamma(I)$? Even in the real-analytic setting, the Cartan-Kähler theorem cannot be applied to construct such a surface because choosing $\nu = \gamma'\times N$ would have $\nu(t)\cdot \nabla_{\gamma'(t)}\gamma'(t) = 0$, so that the above Proposition cannot be applied.

Now, the Proposition almost certainly will not hold in the smooth setting without further hypotheses. The reason is that the PDE that the Cartan-Kähler theorem is solving is an initial value problem for a second order PDE that is of degenerate type, i.e., neither elliptic nor hyperbolic and the characteristics of the PDE are of multiplicity $2$. (I hesitate to call this type of system 'parabolic' because, while everyone agrees that $u_{xx}-u_t=0$ is a parabolic equation, not everyone agrees that $u_{xx} = 0$ is a parabolic equation.) While I have not carefully written out the details, I'm pretty sure that I could construct a (real-analytic) Riemannian manifold $M^3$ and a (real-analytic) unit speed curve $\gamma:I\to M$ and a (smooth but not real-analytic) normal vector field $\nu:I\to TM$ along $\gamma$ that satisfy the conditions of the Proposition but for which the desired NDEF surface $S$ does not exist.

Now, in the smooth setting, given an NDEF surface $S\subset M^3$, $S$ will be foliated by the null curves of its second fundamental form, and all of these null curves will be 'planar' by the OP's definition. In fact, these null curves are exactly the characteristics of the PDE that defines NDEF surfaces to begin with. Thus, the more `refined' question mentioned above is actually the 'characteristic initial value problem' that is typical of parabolic PDE. (For example, the usual IVP for the heat equation $u_t = u_{xx}$ is to specify $u$ along $t=0$, which is clearly a characteristic IVP. It is instructive to note that, even in this very simple setting, one generally gets a solution to the initial value problem only for $t\ge 0$, i.e., there will be no solution matching the initial conditions for $t<0$, even when the specified initial value is real-analytic.)

What I suspect is that the best one can do (in either the real-analytic or smooth setting) for the 'generic' Riemannian $3$-manifold and planar curve $\gamma:I\to M$ would be to prove that, under certain conditions of non-degeneracy that would be tedious to spell out here, there will exist a (real-analytic or smooth) NDEF surface $S\subset M$ with boundary $\gamma(I)$ such that $N$ is tangent to $S$ along $\gamma(I)$, and, 'generically', it will not be possible to extend $S$ as a (real-analytic or smooth) NDEF surface that contains $\gamma(I)$ in its interior.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.