Peter Teichner. Back to the variational complex. The case E = M × N → M (trivial bundle). “Sigma-models” Φ = Γ(M, E) = C^{∞}(M, N). The easiest Lagrangian has kinetic term only: M and N are Riemannian or Lorentz manifolds. L: J^{1}E = Hom(T M, T N) → R = (f ↦ (x ↦ −∥f(x)∥^{2}/2)). V_{g} is the volume form on M. We have λ = LV_{g} ∈ Ω^{d,0}JE. Hence d_{V} λ = d_{H}(α) + E(λ) for some

α ∈ Ω^{d−1,1}JE.

Case 1: M = N = R. We have x: M → R, y: M × N → R, ẏ: J^{1}E → R: ẏ[x, ϕ] = ∂_{x}ϕ(x) = ϕ̇(x). Hence J^{1}E = R^{3} has coordinate functions x, y, ẏ. L = −ẏ^{2}/2, λ = −ẏ^{2}dx/2. θ = dy − ẏdx ∈ Ω^{0,1} and

θ̇ = dẏ − ÿdx. δλ = −ẏθ̇dx = −ẏdẏdx. Set α = −ẏθ. Hence d_{H}α = −(ÿdy + ẏdẏ)dx. δλ = d_{H}α + ÿdy dx so that E(λ) = ÿdy dx. E(λ) contains no terms dẏ, dÿ, etc.

From E(λ) we get E(Σ^{1}) ∈ Ω^{1}(Φ) by integration E(Σ)_{ϕ}(ψ) = ∫ E_{ϕ}(ψ). E(Σ) vanishes at those fields whose second derivative is 0. Hence the classical solutions are linear maps R → R.

Case 2: M is arbitrary, N = R. E(Σ)_{ϕ}(ψ) = ∫_{Σ}(Δ_{g}ϕ(x))ψ(x)V_{g}. Case 3: M = R, N is arbitrary. E(Σ)_{ϕ}(ψ) = ∫_{Σ}(⟨∇_{ϕ̇(x)}ϕ̇(x), ψ(x)⟩V_{g}. Case 4: E(Σ)_{ϕ}(ψ) = ∫_{Σ}⟨tr_{g} ∇(T ϕ), ψ⟩V_{g}. Here T ϕ ∈ Γ(M, T^{∗}M ⊗ ϕ^{∗}T N), ∇T ϕ ∈ Γ(M, T^{∗}M ⊗ T^{∗}M ⊗

ϕ^{∗}T N).

Now we want to find α. α = ẏθ gives α(0) ∈ Ω(Φ) by evaluating near 0. Lemma: θ_{[x,ϕ]}(ψ) = ψ(x) is independent of ϕ. θ̈_{(x,ϕ)}(ψ) = ψ̈(x). Hence α(0) ∈ Ω^{1}(Φ) comes from the bilinear pairing Φ ⊗ Φ → R. It factors through R × R ⊂ Φ.

(ϕ, ψ) ↦ ϕ̇(0)ψ(0). dα ∈ Ω^{2}(Φ) is constant. dϕ_{ϕ}(ψ̇^{0}, ψ̇^{1}) = ⟨ψ^{0}, ψ^{1}⟩ − ⟨ψ^{1}, ψ^{0}⟩.

Generalization: For M = R we have the same formula. For N = R we have α(Σ)_{ϕ}(ψ) = ∫_{Σ}(∗dϕ)_{Σ} ∧ ψ.