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One way to define the tangent space of a manifold at a point $p\in M$ is the following: We define an equivalent relation on the space of curves passing $p$ as follows: Two curves $\alpha, \beta$ are equivalents iff they have tangencity of order at least one that is $\parallel \alpha(t)-\beta(t)\parallel=o(|t|)$, in a local smooth coordinate.

Then the tangent bundle is the disjoint union of these tangent spaces.

One can generalize this by increasing the order of tangencity, that is $\parallel \alpha(t) -\beta (t)\parallel=o(|t|^{k})$.

Does the later gives us the $k-th$ tangent bundle? In this way, what would we get when $k$ is not an integer. For non integer $k$, do we get a well known structure(as a fiber bundle or some thing like this)?

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    $\begingroup$ (+1) Thank you for asking this. I had a similar question in mind long time ago but I never found its answer! $\endgroup$
    – user82740
    Commented Dec 4, 2015 at 13:39
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    $\begingroup$ I am pretty sure this gives you a "jet bundle" not a higher order tangent bundle. $\endgroup$ Commented Dec 4, 2015 at 14:28
  • $\begingroup$ @StevenGubkin Thanks for your comment. We have a jet bundle for integer k, yes? but what about noninteger? $\endgroup$ Commented Dec 4, 2015 at 15:06
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    $\begingroup$ And another quibble: in the title you probably mean "noninteger order", not "dimension". For smooth curves the order of contact is always an integer. So you will be looking at curves which are actually not smooth. IMHO it maybe more worthwhile to think dually: instead of thinking about Holder continuous curves, perhaps a better approach is to see what happens if we try to define a locally ringed space based on $\mathbb{R}$ (for example) and its sheaf of $C^{0,\alpha}$ functions. $\endgroup$ Commented Dec 4, 2015 at 15:08
  • $\begingroup$ @WillieWong Thank you for your comment. Can I ask you to expand your comments in the form of an answer with more details? $\endgroup$ Commented Dec 6, 2015 at 16:32

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