Cotangent lift of the commutator of 2 vector fields Given a smooth manifold $X$ and a vector field $u$ on $X$, we can perform a cotangent lift to get a vector field $\tilde u$ on $T^*X$. Explicitly, this can be done by thinking of $u$ as defining an infinitesimal diffeomorphism on $X$, which then induces a diffeomorphism on $T^*X$, which in turn gives us a vector field on $T^*X$.
My question then is as follows. Given vector fields $u$ and $v$ on $X$ and their respective cotangent lifts $\tilde u$ and $\tilde v$, is the cotangent lift of the vector field $[u,v]$ equal to $[\tilde u,\tilde v]$?
This seems to be true, but I don't see how to justify it. Giving the canonical symplectic structure to $T^*X$, one can check that $\tilde u$ and $\tilde v$ are symplectomorphisms, and an easy calculation shows that so is $[\tilde u, \tilde v]$. So, this seems to suggest that the above claim might be true.
 A: The answer is yes, but one needs to describe more explicitly what is the 'cotangent lift'.
Denote by $\pi : T^*X \to X$ the canonical projection, by $\lambda$ be the tautological 1-form on $T^*X$ and by $\omega = d\lambda$ the tautological symplectic form. Given a smooth function $f$ on $T^*X$, define its associated Hamiltonian vector field $X_f$ implicit via $-df = X_f \lrcorner \, \omega$.
Given a vector field $u$ on $X$, consider any lift $u'$ of $u$ on $T^*X$ i.e. $\pi_{\ast}u' = u$ (such a lift exists, as one can see by consider any connection on the cotangent bundle) and define the smooth function $$f_u : T^*X \to \mathbb{R} : \xi \mapsto \lambda_{\xi}(u'_{\xi}) = \xi(u_{\pi(\xi)}) \, .$$
We shall define the 'cotangent lift' $\tilde{u}$ of $u$ as the Hamiltonian vector field associated to $f_u$ i.e. $\tilde{u} := X_{f_u}$. Easy calculations in cotangent coordinates show that $\tilde{u}$ is indeed a lift of $u$, and that it is indeed the infinitesimal version of the 'cotangent lift' of a diffeomorphism on $X$.
Slightly more generally than before: given any vector field $v$ on $T^*X$, one can define a smooth map $s_v =  \lambda(v) : T^*X \to \mathbb{R}$. In the case when $v = X_f$ for some function $f$, application of Cartan's formula yields $$d(s_{X_f}) =  \mathcal{L}_{X_f}\lambda + df .$$
Taking in particular $f = f_u$ implies that $\mathcal{L}_{\tilde{u}}\lambda = 0$, showing how 'natural/tautological' the cotangent lift is.
By the naturality of the Lie bracket for related vector fields, we get that $[\tilde{u}, \tilde{v}]$ is a lift of $[u,v]$ for vector fields $u,v$ on $X$; we aim to show that the former is the cotangent lift of the latter. Since $\tilde{u}, \tilde{v}$ are Hamiltonian vector fields, it is well-known that $[\tilde{u}, \tilde{v}] = X_{\{f_u, f_v \}}$ where $\{-, - \}$ is the Poisson bracket. It therefore suffices to prove that $\{f_u, f_v\} = f_{[u,v]}$, as we would then have $[\tilde{u}, \tilde{v}] = \widetilde{[u,v]}$. We compute
$$ \begin{align}
f_{[u,v]} &= \lambda([\tilde{u}, \tilde{v}]) = \lambda(\mathcal{L}_{\tilde{u}}\tilde{v}) = \mathcal{L}_{\tilde{u}}(\lambda(\tilde{v})) - (\mathcal{L}_{\tilde{u}}\lambda)(\tilde{v}) \\
&= \mathcal{L}_{\tilde{u}}(\lambda(\tilde{v})) = \tilde{u} \lrcorner \, d(f_v) = - \omega(\tilde{v}, \tilde{u}) = \{ f_u, f_v \} \, . 
\end{align} $$
