I've been thinking of an obvious way to define higher order tangent space with smooth curves, but it seems that this definition does not coincide with the repeated tangent bundle construction.
So, let $M$ be a smooth manifold, and let $\gamma: [0,1] \rightarrow M$ be a smooth curve with $\gamma(0) = p$. We can always define an equivalence relation on such paths as $\gamma_1 \equiv \gamma_2$ iff for all smooth functions $f: M \rightarrow \mathbb{R}$, there exists an open neighborhood of $p$ on which $(f \circ \gamma_1)'(0) = (f \circ \gamma_2)'(0)$ and $(f \circ \gamma_1)''(0) = (f \circ \gamma_2)''(0)$.
This would define a second order tangent space at $p$ for $M$ this way, and I guess that intuitively it would be a vector space of dimension $2 dim M$.
Now, this means that the above construction would produce a second order tangent bundle of dimension $3 dim M$ (if it exists), while repeating the usual (first order) tangent bundle construction on $TM$ would produce a tangent bundle of dimension $4 dim M$.
Am I right here or did I miss something? Why is there such difference? (I think i understand the difference geometrically and why the dimensions disagree. It seems to me that the good notion of acceleration is given by the paths construction, that could naturally be embedded into $TTM$. Is that true?)
Best Answer
No, in the case that $M\subset\Bbb R^n$ (or $\Bbb C^n$ or $\Bbb P^n$) is (locally) the image of a parametrization $\phi\colon U\to\Bbb R^n$, $U\subset R^k$ open, you need to look at the span of the first and all second partial derivatives of $\phi$. So, the second-order tangent bundle will generically be a bundle of rank $k+\frac12 k(k+1)$. (Of course, this "embedded" notion fails to be a bundle when the submanifold has unexpected shrinking of the osculating space.) The appropriate abstract construction is that of a jet bundle (in this case the $2$-jet). See, for example, Wiki, Hirsch's Differential Topology, and/or Guillemin/Golubitsky's Stable mappings and their singularities.