Tangent space of tangent vector

Question

Let $M$ be a smooth manifold. There's a (split) short exact sequence $$0\to T_aM\to T_v(TM)\stackrel {D_v}{\to} T_aM \to 0,$$ where $v\in TM$ and $a=\pi(v)\in M$. I'm trying to understand what this exact sequence means, but I don't even have an intuition about $T_v(TM)$. Can someone explain an example of tangent space of tangent vector (e.g. tangent vector of $S^1$)? How to take a tangent vector of a tangent vector geometrically?

Answer 1

Every vector bundle over a smooth manifold is itself a manifold. So $TM$, the tangent bundle of a manifold $M$, is also a smooth manifold and therefore has its own tangent bundle, $T(TM)$.

Think of a tangent space at an individual vector $v_a\in T_aM$ as the $2\dim(M)$ space of possible deformations of $v_a$. Either $v_a$ can move within $T_aM$, it can change its base point, or both.$^*$

In a local trivialization, $v_a$ can be written as the pair $(a,v)$ where $a\in M$ and $v\in T_aM$. If we take a smooth path through $v_a$ in $TM$, say $(a(t),v(t))$, the tangent vector to this path is merely $(\dot{a},\dot{v})$.

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$^*$ If I understand what you mean by "$D_v$," this sentence is exactly what your split short exact sequence means.