\def\R{{\bf R}}

Daniel Berwick Evans.

Supersymmetries and supercharges.

$N$ is a spacetime (a supermanifold) (a Riemannian or Lorentzian manifold).
$F$ is a superspace of fields.
Lagrangian $L\in\Omega^{0,n}(F\times N)$.

We have classical solutions $M\subset F$.

Variational 1-forms.  $\gamma\in\Omega^{1,n-1}_{\rm loc}(F\times N)$.

Pick $\gamma$ such that $E(L)=\delta L+d\gamma\in\Omega^{1,n}_{\rm loc}(F\times M)$
does not contain variations of derivatives of fields, just variations of the fields themselves.

Remarks: (1)~If $L$ only depends on the 1-jet of fields, $\gamma$ is uniquely determined.
(2)~By Taken's theorem $\gamma$ is generally only determined up to a $d$-closed ($d$-exact) form.
(3)~For the usual mechanics the restriction of $\gamma$ to~$M$ is $\sum_i p_idq^i$.

Examples: (1)~Classical particles on a Riemannian manifold~$M$.
$N=\R$, $F=\mathop{\rm Map}(\R,M)$, $L=(1/2)|\dot x|^2dt$.

$\delta L=\langle\delta_\nabla\dot x,\dot x\rangle dt
=\langle\delta_\nabla dx,\dot x\rangle=-\langle d\delta_\nabla x,\dot x\rangle
=-d\langle\delta x,\dot x\rangle+\langle\delta x,d_\nabla\dot x\rangle
=\gamma+E(L)$, $E(L)=\langle\delta x,\nabla_{\dot x}\dot x$.
Here $\dot x dt=dx$.

Superparticle.  $N=\R$, $F=\{x\colon\R\to M,\psi\in\Gamma(x^*\pi TM)\}$,
$L=(1/2)(|\dot x|^2+\langle\psi,\nabla_{\dot x}\psi\rangle)dt$.
$\delta L=-d\langle\psi,\delta_\nabla\psi\rangle+\langle\delta_\nabla\psi,d_\nabla\psi\rangle
+\langle\delta x,R(\psi,\psi)dx\rangle-d\langle\delta x,\dot x\rangle
+\langle\delta x,\nabla\dot x,\dot x\rangle$.

Equations of motion: $\nabla_{\dot x}\dot x=(1/2)R(\psi,\psi)\dot x$, $\nabla_{\dot x}\psi=0$.
$\gamma_{\rm sp}=\langle\dot x,\delta x\rangle+(1/2)\langle\psi,\delta_\nabla\psi\rangle$.

Symplectic structures on the space of classical solutions~$M$.
$\omega=\delta\gamma\in\Omega^{2,n-1}_{\rm loc}(F\times N)$.
$\omega=(d+\delta){\cal L}$ on $M\times N$, ${\cal L}=L+\gamma$.
Hence $(d+\delta)\omega=0$ on $M\times N$.

Classical mechanics: $d\omega=0$ on~$M$, hence $\omega(t)=\omega$.
In general, $\Omega(t):=\int_{\{t\}\times S}\omega$.  By Stokes this does not depend on~$t$.

\bye
