De Donder–Weyl theory

Remarks

The map $z \mapsto p$ is a fiberwise Legendre transformation. For regular $\mathcal{L}$, a smooth $\varphi$ solves the Euler–Lagrange equations of $\mathcal{L}$ if and only if $\varphi$ together with $p^\mu_a = \partial \mathcal{L} / \partial z^a_\mu (x, \varphi, \partial \varphi)$ solves the De Donder–Weyl equations.

Example: for $n = 1$, the variable $x^1 = t$ is time, there is one momentum $p_a$ per coordinate, $H$ is the usual Hamiltonian, and the equations are Hamilton's equations $\dot\varphi^a = \partial H / \partial p_a$, $\dot p_a = -\partial H / \partial y^a$.

Example: for $n = 4$, $k = 1$, $\eta = \mathrm{diag}(1, -1, -1, -1)$ and $\mathcal{L} = \frac{1}{2} \eta^{\mu\nu} z_\mu z_\nu - \frac{1}{2} m^2 y^2$, one has $p^\mu = \eta^{\mu\nu} z_\nu$, which is invertible. Then $H = \frac{1}{2} \eta_{\mu\nu} p^\mu p^\nu + \frac{1}{2} m^2 y^2$, and the equations $\partial_\mu \varphi = \eta_{\mu\nu} p^\nu$, $\partial_\mu p^\mu = -m^2 \varphi$ give the Klein–Gordon equation $\eta^{\mu\nu} \partial_\mu \partial_\nu \varphi + m^2 \varphi = 0$.

Non-example: for $n = 2$, $k = 1$ and $\mathcal{L} = z_1$, the polymomenta are $p^1 = 1$, $p^2 = 0$ for all $z$, so the map $z \mapsto p$ is not injective and clause 2 fails.