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The top answer to What does the integral of position with respect to time mean? says that “Acceleration is an element of (if we use a connection to identify the horizontal subbundle of () with ”. What exactly does this mean? What is the horizontal subbundle of $T^2 M$ and how can it be determined by the connection? What would acceleration look like if the connection isn’t used?

I suppose it’d be useful to first know what $T^2 M$ is in this context. I know that it’s the tangent bundle of the tangent bundle and how it relates to the tangent bundle, but I can’t wrap my head around how it connects to the original manifold. I’ve seen that it’s like the set of all accelerations like how the tangent bundle is the set of all velocities, but $dim(T^2 M)=4dim(M)$ so it can’t be a simple set of vectors. I’ve also seen it as somehow related to second order derivative operators like how the tangent bundle is related to the first order ones, but I haven’t seen that fleshed out either.

So what is the second order tangent bundle, how can the connection determine a subbundle, and how does this all connect to acceleration?

Lave Cave
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    Briefly: if $\xi\in TM$, then space $T_{\xi}(TM)$ has a particular subspace $V_{\xi}$ called the vertical subspace: it is $\ker d\pi_{\pi(\xi)}$, where $\pi\colon TM\to M$ is the natural projection. It is called "vertical" since moving in its direction does not change the base point on $M$. A connection on $TM$ is equivalent to the data of a supplement $H_{\xi}$ (isomorphic to $T_{\pi(\xi)}M$) for all $\xi\in TM$, which is called "horizontal": the acceleration is defined as the projection of $\gamma''(t)\in T_{\xi}TM= V_{\xi}\oplus H_{\xi}$ on this horizontal subspace (with $\xi=\gamma'(t)$). – Didier Sep 24 '22 at 20:41

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