Given $S\subset\mathbb{R}^n$ a differentiable manifold and a tangent vector $v$ at a given point $x\in S$. Question: Is there a curve $\eta:[0,\epsilon)\to S$ starting at $x$ with tangent $v$ such that $\eta$ is twice differentiable at $0$?
$\begingroup$
$\endgroup$
9
-
$\begingroup$ It depends on what amount of differentiability you assume for the manifold $S$. If $S$ is a $C^k$-differentiable manifold then there is such a $C^k$-curve $\eta$, as you can easily see in local coordinates at $x$. $\endgroup$– Lee MosherCommented Mar 21, 2013 at 14:55
-
$\begingroup$ The manifold $S$ is given by the set of zeros of a smooth function $H:\mathbb{R}^n\to\mathbb{R}^n$ with locally constant rank jacobian. $\endgroup$– Shake BabyCommented Mar 21, 2013 at 15:10
-
1$\begingroup$ @Shake: What do you mean by a smooth function (how many derivatives)? If you mean infinitely differentiable, then the answer is trivial. If you mean a function that is merely $C^1$ then it becomes interesting. $\endgroup$– MishaCommented Mar 21, 2013 at 15:32
-
$\begingroup$ @Lee: $S$ in in $\mathbb R^n$ so we can talk about differentiability of the curve of any order regardless of what $S$ is. $\endgroup$– Ryan BudneyCommented Mar 21, 2013 at 17:18
-
$\begingroup$ @Ryan: Yes, I was just pointing out a special case, given that the OP did not specify the level of differentiability of $S$. In some contexts, to say that $S \subset \mathbb{R}^n$ is differentiable means that it is $C^\infty$ differentiable, and I was unsure whether the OP knew that. $\endgroup$– Lee MosherCommented Mar 21, 2013 at 17:36
|
Show 4 more comments
1 Answer
$\begingroup$
$\endgroup$
2
From the comment of the OP, $H:\mathbb R^m\to \mathbb R^m$ is $C^\infty$ and has constant rank Jacobian. Thus $S=H^{-1}(0)$ is a $C^\infty$ manifold of codimension the rank of the Jacobian; see 1.13 of (here). Thus such a curve exists (even a $C^\infty$-curve).
answered Mar 21, 2013 at 19:55
-
1$\begingroup$ I think the OP suggested $H$ is just $C^2$. So the pre-image will only be $C^2$. $\endgroup$ Commented Mar 21, 2013 at 20:02
-
$\begingroup$ Note that both implicit function theorem and, hence, constant rank theorem, work without reduction in smoothness. Therefore, in OP's case, $S$ is locally a graph of a $C^2$-function and, hence, obviously contains $C^2$-curves tangent to every tangent vector. This (implicit function theorem) is undergraduate level material or 1-st year graduate material depending on the country. All in all, the question is not suitable for MO, math.stackexchange would be the right place. Voting to close. $\endgroup$– MishaCommented Mar 21, 2013 at 21:57