# Canonical isomorphisms vs Legendre transform

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.

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.