2 corrected formatting

For completeness, I should post a follow-up to Chris's answer above. Chris's argument shows that the map dp: TcC → Tp(c)B is an injection, where p is (the restriction of) the projection EB. But the conditions I originally proposed do not force this to be a surjection except in the finite-dimensional case (when a (co)dimension count shows that dim C is at least dim B), as the following example shows.

Let N be a compact connected manifold, B = N x N, E = Smooth Maps ( [0,1] → N ), and the projection is π(e) = (e(0),e(1)). Pick a function L : T_N_ TNR, and define f : ER to by f(_e_) = ∫[0,1] L(_e_'(t),e(t)) dt. Then the set C consists of solutions to the Euler-Lagrange equations dt[\partialv L] = \partialq L. Generically, this is a non-degenerate second-order differential equation, and so dim C = 2 dim N = dim B. This is in fact the application I am interested in.

On the other hand, let's pick a one-form b and a function c on N, and let's suppose that the exterior derivative d_b_ is nondegenerate, so that d_b_ is a symplectic structure on N. Let L(v,q) = _b_v + c. Then one can check (it's straightforward) that the Euler-Lagrange equations are a non-degenerate first order differential equation on N, equivalent to the Hamilton equation for the sympectic manifold (N,d_b_) with Hamiltonian c (or perhaps -c depending on your conventions). Thus a solution is determined by its initial location, and so dim C = dim N.

On the other hand, the Hessian H defined in Chris's answer is now a nondegenerate first-order linear differential equation, and the condition says that the only solution φ to this equation with φ(0) = 0 = φ(1) is the trivial solution φ = 0. For a general Lagrangian, the Hessian is a second-order operator, and this condition is nontrivial, but when the Lagrangian is first-order, the Hessian necessarily has no kernel — a solution is determined by a single value.

Thus the condition that Chris thought I was imposing — that C be a manifold with dimension the same as B — cannot be dropped.

A final remark is that in finite dimensions, C is cut out by dim F = dim E - dim B equations, and so dim C is at least dim B. The point is that this dimension count fails when dim F = ∞.

1

For completeness, I should post a follow-up to Chris's answer above. Chris's argument shows that the map dp: TcC → Tp(c)B is an injection, where p is (the restriction of) the projection EB. But the conditions I originally proposed do not force this to be a surjection except in the finite-dimensional case (when a (co)dimension count shows that dim C is at least dim B), as the following example shows.

Let N be a compact connected manifold, B = N x N, E = Smooth Maps ( [0,1] → N ), and the projection is π(e) = (e(0),e(1)). Pick a function L : T_N_ → R, and define f : ER to by f(_e_) = ∫[0,1] L(_e_'(t),e(t)) dt. Then the set C consists of solutions to the Euler-Lagrange equations dt[\partialv L] = \partialq L. Generically, this is a non-degenerate second-order differential equation, and so dim C = 2 dim N = dim B. This is in fact the application I am interested in.

On the other hand, let's pick a one-form b and a function c on N, and let's suppose that the exterior derivative d_b_ is nondegenerate, so that d_b_ is a symplectic structure on N. Let L(v,q) = _b_v + c. Then one can check (it's straightforward) that the Euler-Lagrange equations are a non-degenerate first order differential equation on N, equivalent to the Hamilton equation for the sympectic manifold (N,d_b_) with Hamiltonian c (or perhaps -c depending on your conventions). Thus a solution is determined by its initial location, and so dim C = dim N.

On the other hand, the Hessian H defined in Chris's answer is now a nondegenerate first-order linear differential equation, and the condition says that the only solution φ to this equation with φ(0) = 0 = φ(1) is the trivial solution φ = 0. For a general Lagrangian, the Hessian is a second-order operator, and this condition is nontrivial, but when the Lagrangian is first-order, the Hessian necessarily has no kernel — a solution is determined by a single value.

Thus the condition that Chris thought I was imposing — that C be a manifold with dimension the same as B — cannot be dropped.

A final remark is that in finite dimensions, C is cut out by dim F = dim E - dim B equations, and so dim C is at least dim B. The point is that this dimension count fails when dim F = ∞.