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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?

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  • $\begingroup$ I do not have an answer to your precise question, but let me point out that there is a distinguished map between the tangent and cotangent bundles given by the symplectic form, which is generically defined in classical systems. $\endgroup$ Dec 16, 2020 at 6:11

3 Answers 3

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

  2. 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 $$\mathbb{g}~=~g_{ij}~\mathrm{d}x^i\odot \mathrm{d}x^j.$$

  3. 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 $\mathbb{g}$.

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

  5. 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.

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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.

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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 $Q$).

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