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Let $M$ be a smooth manifold, $\gamma:[0,a] \to M$ a smooth path. Assume we are given another smooth path $\phi:(-\epsilon,\epsilon) \to M$ that starts from an endpoint of $\gamma$, i.e $\phi(0)=\gamma(a)$.

Does there exists a smooth variation $f:(-\epsilon,\epsilon)\times [0,a]\rightarrow M$ of $\gamma$ such that $f(s,0)=\gamma(0)$ and $f(s,a)=\phi(s)$? (By variation I mean $f(0,t)=\gamma(t)$).

I am not insisting that the variation wille be defined on all $(-\epsilon,\epsilon)$, every variation which is defined for some "time" will be fine.

I think a problem might arise when a variation will have to pass through a "non-existent" region and so will not be smooth (Imagine a disk removed from the plane which lies inside an area where the variation is supposed to pass through. At some point there will be a "jump" - since each curve in the variation has a "discrete" choice: to pass above or below the forbidden region).

I think that allowing the variation to be defined on arbitrarily small intevals is supposed to overcome this, but I am unsure about how to do this.


Motivation:

In some scenarios, you want the endpoint of every curve in your variation to stay in a certain submanifold which intersects $\gamma(a)$. (See the comments here and here).


Remark:

The standard way of building smooth variations is different from what isrequired here;

For any given smooth variational field $V(t)∈T_{γ(t)}M$, we define $f(s,t)=exp_{γ(t)}(sV(t))$ where $exp$ is the exponential map w.r.t some Riemannian metric on $M$).

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  • $\begingroup$ Incompleteness ("non-existent regions") will prevent the obvious exponential map argument from working, but a different route should work - the union of the curves must have some open neighbourhood, and we can build the variation inside this neighbourhood. One idea that might work: concatenate $\gamma$ with $\phi|_{[0,s]}$ and scale the parameter to keep the domain as $[0,a]$. Doing this for each $s$ gives a piecewise-smooth variation satisfying the required conditions, so if you can work out how to smooth it out while preserving $f(0,t) = \gamma(t)$ then you're done. $\endgroup$ Jun 26, 2016 at 2:36
  • $\begingroup$ Your idea is interesting (this is not how I usually imagine variations). I think it might work; for each $s \neq 0$, we should go through $\gamma|_{[0,a-s]}$, and then smooth out $\phi_{[0,s] }* \gamma|_{[a-s,a]} $ (i.e create a path which is glued smoothly at $\gamma(a-s)$ with $\gamma$ and which ends at $\phi(s)$). Thus, when $s$ gets smaller (in absolute value) the smoothings are done on smaller and smaller intervals. I think such a construction might be non-smooth at some points, but existence of a finite number of points of non-smoothness does not really matter for my applications. $\endgroup$ Jun 26, 2016 at 7:27

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I don't know exactly, how general you want to situation to be. Under weak assumptions, which basically say that the image of $\phi$ stays away from $\gamma(0)$, I think this is easy to get as follows: The derivative $\phi'$ defines a vector field along the curve $\phi$. Choose open subsets $U\supset V\supset \phi((-\epsilon,\epsilon))$ such that $\gamma(0)\notin U$, $\bar U$ is compact $V\subset U$. (The assumption that $U$ can be chosen such that $\gamma(0)\notin U$ is exactly the restriction I was referring to in the beginning.)

Extend $\phi'$ arbitrarily to a local vector field defined on $U$ and then multiply by a bump function with support contained in $U$ which is identically $1$ on $V$. The result can be extended by zero to a globally defined vector field $\xi$ on $M$. By construction, this vector field has compact support and hence it defines a global flow, call that $\psi_t$. Now define the variation by $f:(-\epsilon,\epsilon)\times [0,a]\to M$ by $f(s,t):=\psi_s(\gamma(t))$. This is clearly smooth by the usual properties of flows and $f(0,t)=\gamma(t)$ for all $t\in [0,a]$. Also by construction $\xi(\gamma(0))=0$, so $f(s,0)=\gamma(0)$ for all $s$. Finally, along the image of $\phi$, $\xi$ coincides with $\phi'$ which means that $\phi$ is an integral curve of $\xi$. But this shows that $f(s,a)=\psi_s(\gamma(a))=\phi(s)$.

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