\def\R{{\bf R}}
\def\Hom{\mathop{\rm Hom}}
\def\tr{\mathop{\rm tr}}

Peter Teichner.

Back to the variational complex.

The case $E=M\times N\to M$ (trivial bundle).
``Sigma-models'' $\Phi=\Gamma(M,E)=C^\infty(M,N)$.

The easiest Lagrangian has kinetic term only: $M$~and~$N$ are Riemannian or Lorentz manifolds.

$L\colon J^1E=\Hom(TM,TN)\to\R=(f\mapsto(x\mapsto-\|f(x)\|^2/2))$.

$V_g$ is the volume form on~$M$.  We have $\lambda=LV_g\in\Omega^{d,0}JE$.
Hence $d_V\lambda=d_H(\alpha)+E(\lambda)$ for some $\alpha\in\Omega^{d-1,1}JE$.

Case~1: $M=N=\R$.  We have $x\colon M\to\R$, $y\colon M\times N\to\R$,
$\dot y\colon J^1E\to\R$: $\dot y[x,\phi]=\partial_x\phi(x)=\dot\phi(x)$.

Hence $J^1E=\R^3$ has coordinate functions $x$,~$y$,~$\dot y$.
$L=-\dot y^2/2$, $\lambda=-\dot y^2dx/2$.
$\theta=dy-\dot y dx\in\Omega^{0,1}$ and $\dot\theta=d\dot y-\ddot y dx$.
$\delta\lambda=-\dot y\dot\theta dx=-\dot y d\dot y dx$.
Set $\alpha=-\dot y\theta$.
Hence $d_H\alpha=-(\ddot y dy+\dot y d\dot y)dx$.
$\delta\lambda=d_H\alpha+\ddot y dy\,dx$ so that $E(\lambda)=\ddot y dy\,dx$.
$E(\lambda)$ contains no terms $d\dot y$, $d\ddot y$, etc.

From $E(\lambda)$ we get $E(\Sigma^1)\in\Omega^1(\Phi)$
by integration $E(\Sigma)_\phi(\psi)=\int E_\phi(\psi)$.
$E(\Sigma)$ vanishes at those fields whose second derivative is~0.
Hence the classical solutions are linear maps $\R\to\R$.

Case~2: $M$ is arbitrary, $N=\R$.  $E(\Sigma)_\phi(\psi)=\int_\Sigma(\Delta_g\phi(x))\psi(x)V_g$.

Case~3: $M=\R$, $N$ is arbitrary.  $E(\Sigma)_\phi(\psi)=\int_\Sigma(\langle\nabla_{\dot\phi(x)}\dot\phi(x),\psi(x)\rangle V_g$.

Case~4: $E(\Sigma)_\phi(\psi)=\int_\Sigma\langle\tr_g\nabla(T\phi),\psi\rangle V_g$.
Here $T\phi\in\Gamma(M,T^*M\otimes\phi^*TN)$, $\nabla T\phi\in\Gamma(M,T^*M\otimes T^*M\otimes\phi^*TN)$.

Now we want to find~$\alpha$.
$\alpha=\dot y\theta$ gives $\alpha(0)\in\Omega(\Phi)$ by evaluating near~0.

Lemma: $\theta_{[x,\phi]}(\psi)=\psi(x)$ is independent of~$\phi$.
$\ddot\theta_{(x,\phi)}(\psi)=\ddot\psi(x)$.

Hence $\alpha(0)\in\Omega^1(\Phi)$ comes from the bilinear pairing $\Phi\otimes\Phi\to\R$.
It factors through $\R\times\R\subset\Phi$.  $(\phi,\psi)\mapsto\dot\phi(0)\psi(0)$.
$d\alpha\in\Omega^2(\Phi)$ is constant.
$d\phi_\phi(\dot\psi^0,\dot\psi^1)=\langle\psi^0,\psi^1\rangle-\langle\psi^1,\psi^0\rangle$.

Generalization: For $M=\R$ we have the same formula.
For $N=\R$ we have $\alpha(\Sigma)_\phi(\psi)=\int_\Sigma(*d\phi)_\Sigma\wedge\psi$.

\bye
