If there is a natural choice of Riemannian metric on configuration manifolds, what is the relation between Legendre transform and induced isomorphism? It is well known that Lagrangian mechanics is formulated on the tangent bundle of the configuration space $\rm TQ$, while Hamiltonian mechanics is formulated on the cotangent bundle $\rm T^*Q$. The lack of a metric (in general) means that there are no musical isomorphisms $\flat:{\rm T} M \to {\rm T}^* M$ and $\sharp:{\rm T}^* M \to {\rm T} M$, so many of the tools of Riemannian geometry are inaccessible - but we can still effect a Legendre transform.
My question is whether there exists a class of configuration manifolds which have a natural choice of Riemannian metric, and if so, what is the relation between the Legendre transform and induced isomorphism?
 A: Having a linear musical isomorphism is not required by the Lagrangian theory, but in mechanics a large class of systems have such an isomorphism.
Let $Q$ be the configuration manifold and let $m\in\Gamma(S^2T^\ast Q)$ be a Riemannian metric on $Q$ (i.e. positive definit), called mass.
If a mass tensor is given, then define the kinetic energy as $$ T:TQ\rightarrow\mathbb R,\quad T(v)=\frac{1}{2}m(v,v). $$
Let $U\in C^\infty(Q)$ be a smooth function on configuration space. The Lagrangian $L:TQ\rightarrow\mathbb R$ is said to be natural if it is of the form $$ L=T-\pi^\ast U, $$ where $\pi:TQ\rightarrow Q$ is the canonical projection.
Since the pullback $\pi^\ast:\Omega(Q)\rightarrow \Omega(TQ)$ is an injection of the exterior algebra of $Q$ into the exterior algebra of $TQ$, we may employ the useful abuse of notation and write $\pi^\ast U=U$.
The Legendre map is defined as follows. Let $L_q:=L|_{T_qQ}$ be the restriction of the Lagrangian to the tangent space $T_qQ$, then $L_q:T_qQ\rightarrow \mathbb R$ and we may consider the differential $$ dL_q|_{\dot q}:T_qQ\rightarrow\mathbb R,  $$which is a linear functional on $T_qQ$ given by $$ dL_q|_\dot q(v)=\frac{d}{dt}L_q(\dot q+tv)|_{t=0}. $$
The corresponding map $\mathbb FL(\dot q)=dL_{\pi(\dot q)}|_\dot q$, which is a $\mathbb FL:TQ\rightarrow T^\ast Q$ is the Legendre map and we often write $$ p(\dot q)=\mathbb F L(\dot q),$$i.e. this is the canonical momentum. The Legendre map is a fibre bundle morphism (preserves fibres) but is not a vector bundle morphism in general because it need not be fibrewise linear.
Now if the Lagrangian $L=T-U$ is the above form, then $$ p(\dot q)(v)=\frac{d}{dt}L(\dot q+tv)|_{t=0}=\frac{d}{dt}\left(\frac{1}{2}m(\dot q+tv,\dot q+tv)\right)|_{t=0}=\frac{1}{2}\frac{d}{dt}(m(\dot q,\dot q)+2tm(\dot q,v)+t^2 m(v,v))|_{t=0}=m(\dot q,v), $$ thus we obtain $$ p(\dot q)=m(\dot q,\cdot), $$ i.e. the canonical momentum associated to the velocity $\dot q$ is just the ordinary "lowering" of $\dot q$ via the Riemannian metric $m$.
Here we have used that $U$ is constant along the fibres, so even if we slightly generalize this system by defining the Lagrangian to be $$ L(\dot q)=T(\dot q)-U(\dot q), $$ where $U$ is a velocity-dependent potential, the Legendre map and the metric-induced raising/lowering would no longer agree.
A: *

*An abstract generic configuration space/manifold $M$ does not have a natural/canonical choice of metric structure.


*If the configuration space $M$ is paracompact, we can use the partition of unity to prove that there exists a globally defined positive definite metric tensor $$\mathbb{g}~=~g_{ij}~\mathrm{d}x^i\odot \mathrm{d}x^j.$$


*In order to write down a kinetic term $T$ for the Lagrangian, we typically need a metric tensor, say $$T~=~\frac{1}{2} g_{ij}v^iv^j.$$
So in physics we often assume that the configuration space $M$ is equipped with a metric tensor $\mathbb{g}$.


*As OP already mentions the Legendre transformation $TM\to T^{\ast}M$ does not rely on the existence of a metric tensor.


*When the canonical/conjugate momentum $p_i=\frac{\partial L}{\partial v^i}$ differs from the mechanical/kinetic momentum $g_{ij}v^j$, the Legendre transformation differs from the musical isomorphism. This happens quite often, e.g. for a point charge in an EM background.
A: It is not necessary to have a canonical linear isomorphism induced by a metric  whend dealing with the Legendre diffeomorphism.
Suppose that $$L : TQ \to \mathbb{R}$$ is given such that its Hessian matrix $$\frac{\partial^2 L}{\partial \dot{q}^i\partial \dot{q}^j} \qquad \mbox{is not singular on an atlas of $Q$}\tag{1}$$ (and thus it is non singular in all natural charts on $Q$).
Consider the map (Legendre transformation)  $$G : TM \to T^*M$$ that, in coordinates, reads
$$G : (q, \dot{q}) \mapsto (q, p_k) := \left.\left(q,\frac{\partial L}{\partial \dot{q}^k}\right|_{(q, \dot{q})}\right)$$
One sees by direct inspection  that this map is well defined when changing local coordinates on $TM$ and on $T^*M$ correspondingly.
The crucial fact is that, as each $T_qQ$ is open and convex,  (1) implies that $G$ is a diffeomorphism from the whole $TQ$ onto an open set  $G(T) \subset T^*Q$.
The phase space is this open set and not the whole $T^*Q$ in general as it instead happens if $L$ is quadratic in $\dot{q}$ (in that case there is a metric and the Legendre transform is the musical linear isomorphism at each point of $Q$).
