TITLE: Critical points on a fiber bundle
QUESTION [7 upvotes]: Consider a (smooth) bundle E→_B_, and a (smooth) function f: E → R on the total space.  Then it makes sense to talk about the derivatives of f along the fibers.  Let C be the subspace of E consisting of all points for which all fiber-wise derivatives of f vanish, so that upon intersecting with any fiber C consists of the critical points of the restriction of f to the fiber.  If the fiber is n-dimensional, then C is carved out by n equations, and so generically has codimension n in E.
Let's say that c is a point in C so that the second derivative of f in the fiber is nondenegerate (i.e. has non-degenerate Hessian; i.e. f restricts to a Morse function on the fiber through c).  Does it follow that the projection C→_B_ is a local diffeomorphism near c?
The answer is yes when everything is finite-dimensional (and I believe the statement is iff).  I am interested in the case when B is finite-dimensional but the fibers of E are infinite-dimensional.
Edit:  This is a response to Andrew's question below (since answering in the comments proves difficult).
I'm using the word "Morse" loosely, largely because I don't actually know Morse theory.  I would suggest that the definition I give is better than what's traditionally used.  What I actually mean is this:
Let M be a smooth manifold and f:M→R a smooth map.  What type of object is the second derivative f(2)?  In general, you should not talk about it by itself, because it does not transform as a tensor, although the pair (f(1),f(2)) is a vector in the 2-jet bundle over M.  But if c is a critical point of f, then f(2)(c) is naturally a symmetric bilinear form TcM x TcM → R.  Thus it is a map TcM→T*cM.  All I ask is that this map have zero kernel.
But if this condition is too weak, whereas a reasonable stronger condition works, I'd love to hear it.

REPLY [2 votes]: 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 E → B.  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 : TN → R, and define f : E → R 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 = ∞.