Witten’s Problems, Set Two — N^{o}. 4 (solution written by P. Deligne)

(a) Let P be the subbundle of the tangent bundle T of W generated by D, and P_{2} be generated by D and D^{2} := \frac{1}{2}[D, D]. We have

B_{−} ⊕ P_{2} ⟶^{∼} T

and by the Frobenius pairing, P_{2}/P ≃ P ⊗ P. The filtration T ⊃ P_{2} ⊃ P of T, with successive quotients B_{−}, P_{2}/P and P, gives a canonical isomorphism

Ber(T) ≃ Ber(B_{−}) ⊗ Ber(P_{2}/P) ⊗ Ber(P)
≃ B_{−} ⊗ P_{2}/P ⊗ P^{∨}.

As P_{2}/P ≃ P^{⊗2}, this gives

Ber(T)^{∨} = B−^{∨}

and the lagrangian density

⟨dX _{B^{−}}, dX _{P^{∨}}⟩

is canonically a section of the bundle of volume elements.

(b) Not all W are locally isomorphic to the flat model given. Indeed, P_{2} and B_{−} are integrable distributions, giving locally a decomposition of W as

W^{1,1} × W^{1}.

The odd distribution P is in the tangent space of the fibers of the projection pr_{2} to W^{1}. All fibers are identified with W^{1,1} by the projection to W^{1,1} and the given flat model corresponds to the case where the induced distributions on the fibers pr_{2}^{−1}(t) ⟶^{∼} W^{1,1} are all equal.

Let us compute the Euler-Lagrange equations for the extremals for the “flat” W given. Because D^{2} = ∂_{+}, the relevant trivialization of Ber(T) is the one corresponding to the basis (∂_{−}, ∂_{+}, D) or, what amounts to the same, corresponding to the basis (∂_{u}, ∂_{v}, ∂_{θ}) (triangular change of basis). The lagrangian density is

⟨∂_{−}X, DX⟩ . (volume form [u, v, θ] given by the coordinates).

If we let X depend on an additional even parameter a, the variation \frac{d}{da} of the lagrangian density is given as follows at a = 0: let X be what X is at a = 0, and δX be δ_{a}X: a section of X^{∗}T_{M}. Defining derivatives of vectors using the Levi-Civita connection, using that this connection preserves ⟨ , ⟩ and that it is torsion free (cf. exercise 1 of set 1), we get at a = 0 that

\frac{d}{da} ℒ = (⟨∂_{a}∂_{−}X, DX⟩ + ⟨∂_{−}X, ∂_{a}DX⟩) [u, v, θ]
= (⟨∂_{−}∂_{a}X, DX⟩ + ⟨∂_{−}X, D∂_{a}X⟩) [u, v, θ]
= (⟨∂_{−}δX, DX⟩ + ⟨∂_{−}X, DδX⟩)  [u, v, θ]
= (∂_{−} ⟨δX, DX⟩ + D ⟨∂_{−}X, δX⟩)  [u, v, θ]
− (⟨δX, ∂_{−}DX⟩ − ⟨D∂_{−}X, δX⟩) [u, v, θ]

where the first term is

d (⟨δX, DX⟩ ∂_{−}⊥[u, v, θ] + ⟨∂_{−}X, δX⟩ D⊥[u, v, θ]) ,

because the Lie derivative ℒ_{D}([u, v, θ]) vanishes, and the second is

δℒ = −2 ⟨δX, ∂_{−}DX⟩ [u, v, θ].

This gives the equation of motion

(1)  ∂_{−}DX = 0.

If one works over a purely even parameter space B, a map X: W → M factors as

W → W_{red} ⟶^{x} M,

the first map being (u, v, θ) → (u, v). Indeed, the even functions on W × B are just the functions on W_{red} × B, as W × B is of dimension (∗, 1). The equation (1) amounts to

(1)_{red}  ∂_{−}∂_{+}x = 0

for x: the reduced space Y_{red} of the space Y of extremals is the space of “harmonic” maps from W_{red} (viewed as a quotient of W) to M. Beware that “harmonicity” is of an hyperbolic character, as we are in a Minkowsky space setting.

Over a general parameter space B, a map X: W → M (i.e. a map from W × B to M) can be described by

(a) its restriction x to the subvariety θ = 0 of W (i.e. its restriction to W_{red} × B (not: (W × B)_{red});

(b) its derivative ψ = ∂_{θ}X, restricted to θ = 0. It is an odd section of x^{∗}T_{M}. If we use local coordinates on M, this means writing

X^{i} = x^{i} + θψ^{i},

with x^{i} and ψ^{i} depending only on u and v.

Let \overline{ℒ} be obtained from the lagrangian density by integrating along the fibers of (u, v, θ) → (u, v). It is

\overline{ℒ} = (− ⟨∂_{−}x, ∂_{+}x⟩ + ⟨∇_{−}ψ, ψ⟩) du dv .

We can view \overline{ℒ} as a lagrangian density for the space of fields (x, ψ) on the (u, v)-space W_{red}: x a map to M, ψ an odd section of x^{∗}T_{M}. For general reasons, the lagrangian density \overline{ℒ} give rise to the same space of extremals as ℒ does, and to the same 2-form on the space of extremals.

As in Set 1, N^{o} 2, the Euler Lagrange equations for (x, ψ) are of the form

(1) ∇_{−}∂_{+}x = quadratic in ψ (involving the curvature),

(2) ∇_{−}ψ = 0.

If the parameter space B is purely even, the odd section ψ of x^{∗}T_{M} (on W_{red} × B) is zero, and one finds again that Y_{red} is a space of “harmonic” maps.

Over a general B, if I is the ideal defining B_{red} and if one works modulo I^{2}, (1) again reduces to ∇_{−}∂_{+}x = 0. This gives the normal bundle of Y_{red} in Y: at x in Y_{red}, it is ΠV, for V the space of sections s of x^{∗}T M such that ∇_{−}s = 0.

Because of the quadratic terms in (1), I don’t know any natural way to map Y to Y_{red}, to view Y as a bundle over Y_{red}. Taking (x, ψ) ↦ x = (x, 0) will not do.

However, the quadratic terms in (1) don’t spoil the kind of Cauchy data for which (1)

(2) is a well posed system of equations. With the notations of (d): u = t + x, v = t − x, a Cauchy data consist in x, ẋ and ψ at t = 0. In terms of X: W → M, this is X and its derivatives Ẋ and ∂_{θ}X at t = θ = 0. The projection (x, ẋ, ψ) ↦ (x, ẋ): Cauchy data for Y_{red} → Cauchy data for Y, gives a projection Y → Y_{red}. It is, however, not “natural” in the sense that it is not invariant by time evolution: it depends on having chosen the line t = 0.

Suppose the parameter space B is of dimension (∗, b). Let I be the ideal defining B_{red} in B. To solve (1) (2) for a given Cauchy data, one can proceed as follows:

(1) Solve (1) (2) over B_{red}. After pull back to B_{red}, we have ψ = 0 by parity reason and one has to solve for an “harmonic” map: ∇_{−}∂_{+}x_{0} = 0.

(2) One can view x_{0} as being over B/I^{2}: modulo I^{2}, the even part of C^{∞}(W × B) is C^{∞}(W × B_{red}). One solves (1) (2) modulo I^{2}, using x_{0} given by (1) and solving (2). This amounts to solving an equation ∇_{−}ψ = 0 on W_{red} × B_{red}, for ψ a section of x_{0}^{∗}T_{M} ⊗ pr_{2}^{∗}(I/I^{2}). We now have (x_{1}, ψ_{1}) modulo I^{2}.

(3) If we already have (x_{r}, ψ_{r}) which is a solution mod I^{r+1}, i.e. a r-jet of solution along B_{red} ⊂ B, we extend it in any way to (x^{0}, ψ^{0}) modulo I^{r+2}. Then correct (x^{0}, ψ^{0}) into a solution (x_{r+1}, ψ_{r+1}), by a correction trivial modulo I^{r+1}. At each step, this amounts to solving on W_{red} × B_{red} an equation

∇_{−}∂_{+}x = second member,  or
∇_{−}ψ = second member,

with x or ψ in x_{0}^{∗}T_{M} ⊗ pr_{2}^{∗}I^{r+1}/I^{r+2}.

This process terminates when r = b.

Assuming M complete, and some control at spatial infinity, we can now identify the space Y of extremals with the space of Cauchy data.