(Definitions from Marsden & Ratiu, ``Introduction to Mechanics and Symmetry''): A Lagrangian is regular if the Hessian $\partial^2 L/(\partial \dot{q}^i \partial \dot{q}^j)$ is weakly non degenerate (i.e. - I think? - has zero determinant somewhere); this condition is equivalent to the condition that the Legendre transformation $\dot{q} \mapsto p := \partial L/\partial \dot{q}$ be locally invertible.

A Lagrangian is hyperregular if the Legendre transformation is a diffeomorphism; similarly, a Hamiltonian is hyperregular if the inverse Legendre transformation is a diffeomorphism.

I have three (related) questions:

  1. Is there a more "intrinsic" characterisation of the hyperregularity condition for Lagrangians (i.e. one that does not explicitly relate it to the Legendre transformation)? In particular, is hyperregularity equivalent to the Hessian being strongly non-degenerate (i.e. - again, I think? - its having zero determinant everywhere)?
  2. Is there a natural counterpart to the regularity condition for Hamiltonians?
  3. Again, is there an intrinsic characterisation of the hyperregularity condition for Hamiltonians?
  • $\begingroup$ Comment to the question (v1): M&R also consider infinitely many $q^{i}{}^{\prime}$s. Are you only interested in the finite-dimensional case? $\endgroup$
    – Qmechanic
    Commented Jul 30, 2014 at 18:42
  • $\begingroup$ @Qmechanic: yes, at least to begin with. (I mean, I'd also be interested in results pertaining to the infinite-dimensional case, but I'm primarily thinking about the finite-dimensional case at the moment.) $\endgroup$ Commented Jul 31, 2014 at 8:55

1 Answer 1


Summary: Properties the Legendre transformation are directly connected to properties of the Lagrangian because the former is just the (velocity-)gradient of the latter. Hyperregularity corresponds to global properties of the Lagrangian function.

At least in finite dimensions, weak non-degeneracy of the Hessian (considered as a quadratic form) means the Hessian matrix has non-vanishing determinant. If this is true in an open region, this implies that the Legendre transformation${}^{1}$ is locally invertible (as you say) and indeed that it is a local diffeomorphism.

To see how this is related intrinsically to the Lagrangian, first consider the case of a single velocity coordinate. Then the Hessian is just the second derivative${}^{2}$, and if it doesn't vanish on a local region this means the Lagrangian is convex (or concave) there. Thus, regularity of the Lagrangian is equivalent to local convexity, and this guarantees that the first derivative (the Legendre transformation) is invertible and differentiable. Inversely, for smooth Lagrangians the first derivative cannot fail to be invertible and differentiable unless the second derivative vanishes at some point, i.e., degeneracy of the Hessian.

This equivalence extends to $n$ velocity coordinates: the Hessian of a Lagrangian is non-degenerate in a local region if and only if the gradient of the Lagrangian can be interpreted as a local diffeomorphism (the Legendre transformation) from one subets of $\mathbb{R}^n$ to another.${}^{3}$ A special case of this is when the Lagrangian is globally convex [e.g., $L(\vec{x},\vec{v}) = m \vec{v}^2/2 - V(\vec{x})$], in which case the Hessian is positive definite.

The stronger property of hyperregularity for the Lagrangian is defined to be when the Legendre transform is a diffeomorphism. The inverse Legendre transform is then also a diffeomorphism, and this defines a hyperregular Hamiltonian. Merely regular Lagrangians are not guaranteed to have Hamiltonian formulations. Furthermore, this is symmetric: regular Hamiltonians can be defined to be those with a non-degenerate Hessian (in momentum), and these need not have Lagrangian formulations unless the Hamiltonian is hyperregular.${}^4$

The condition of hyperregularity rules out Legendre transformations that are local diffeomorphism but are not bijective. What does it mean for a regular Lagrangian to induce a non-bijective Legendre transform? An annoying example is $L(x,v) = e^v$. In this case, $v=\ln p$ and this breaks simply because $p<0$ is ill defined. However, this illustrates how the fact that $\partial_v L$ is bounded from below implies the corresponding Legendre transformation $p= e^v$ is a locally diffeomorphism but is not a diffeomorphism on all of $\mathbb{R}$.

There exist more illuminating examples of regular but not hyperregular Lagrangians which might help illustrate this global property.${}^5$

${}^{1}$ Note, I am following anygivenpoint's usage of the geometric meaning of "Legendre transformation", i.e., the map between the tangent bundle of configuration-velocity pairs $\{(x,v)\}$ and the cotangent bundle of configuration-momentum pairs $\{(x,p)\}$. This is distinct from the (closely related) functional meaning of "Legendre transformation", i.e., the function transformation that takes a convex function $f$ to convex function $g$ defined up to additive constant by $f' = (g')^{-1}$. The geometric Legendre transformation induces the functional Legendre transformation. The functional meaning is usually what physicists (except perhaps those working on geometry) mean when they say "Legendre transformation", and is the transformation that maps between the Hamiltonian and Lagrangian.

${}^2$ Here derivatives always made with respect to velocity or momentum coordinates, not configuration coordinates. That is, we are taking the fiber derivative.

${}^3$ I think this uses the inverse function theorem.

${}^4$ See p. 209-222 of "Foundations of Mechanics" by Abraham & Marsden, especially parts 3.5.2, 3.5.9, 3.6.1, 3.6.5 - 3.6.8.

${}^5$ Footnote 16 of Gotay & Nester, "Presymplectic lagrangian systems. I : the constraint algorithm and the equivalence theorem" Annales de l'I.H.P. Physique théorique, Volume 30 (1979) no. 2 , p. 129-142 might be useful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.