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Let $\tau_Q \colon TQ \rightarrow Q$ be tangent bundle, $T\tau_Q : TTQ \rightarrow TQ$ be its derivative. $VQ$, defined as $\mathrm{\mathop{Ker}}\; T\tau_Q$, is a subbundle of the second tangent bundle $\tau_{TQ}\colon TTQ \rightarrow TQ$, and $VQ$ is tangent to the fibers of $\tau_Q$. How does one prove it without use of coordinates?

Feels like it should be pretty clear from the definition but I still don't get what should be written down to make a consistent proof. I could only check that $a$ in $a^i\frac{\partial}{\partial q^i} + b^i\frac{\partial}{\partial \dot{q}^i} \in VQ$ must vanish, and hence $q=\mathrm{const}$ are the equations of $VQ$'s integral manifolds in $TQ$ which are exactly the fibers of $\tau_Q$.

It is also extremely interesting to me how would the ‘purely algebraic’ proof look like—where one treats $\tau_Q$ as a module over $C^\infty(Q)$ and $\tau_{TQ}$ as a module over $C^\infty(TQ)$. Unfortunately I still don't even know how are the algebras $C^\infty(Q)$ and $C^\infty(TQ)$ related.

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Isn't the first part of your question just the submersion variant of the implicit function theorem? – t.b. Apr 14 '12 at 14:32

If $\gamma : [0,1] \to T_qQ$ is a path in a fiber of $TQ$ then $$T_{\tau_Q}(\gamma'(0)) = \frac{d}{dt}\vert_{t=0} \tau_Q(\gamma(t)) = \frac{d}{dt}\vert_{t=0} q = 0$$ so $\gamma'(0) \in VQ$. Conversely, a dimensionality count will show that every element in $VQ$ is the derivative of a path in a fiber of $TQ$.

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