Projection of a gradient and the gradient of a projection Hi,
I am trying to figure out if there are any functions, and then for which, where one can say that the gradient of the projection is the same as the projection of the gradient.
In this case a projection of the function f(x,y,z) is an integral $p(x,y) = \int f(x,y,z) dz$.
It seems to me that it might not be true in the general case. But are there any such functions?
Another way of looking at it is that I would like to know if there are an analogy to the projection-slice theorem, but for gradients.
I hope the question is correctly formulated, and my apologies if I missed out something very obvious. It does seem to me like this question should have popped up before and there should be some know result. I might have missed it in the textbook.
Helps and pointers appreciated.
 A: I think you need to reformulate your problem. 
The standard definition of a vector field being `projectible' [eg Warner.
See $\pi$-projectible ] requires, in the case of the projection $(x,y,z) \to (x,y)$
it to have the form $F_1 (x,y){{\partial} \over {\partial x}}  + F_2 (x,y){{\partial} \over {\partial y}} + F_3(x,y,z) {{\partial} \over {\partial z }}$. 
For your ``projection'' to be finite, you need your $f(x,y,z)$ to depend on $z$
in such a way that the integral is finite.  As a consequence, its gradient will
have $F_1$ and $F_2$ either zero, or depending on $z$. 
Combining your def. of `projection of a function' with the standard def. of
projection of a vector field you get the result that the only projected vector field
you can get is  the zero vector fields. 
A: If you look in the index under either projectible'' or perhaps $\pi$-projectible' in
F. Warner's bookFoundations of Differentiable Manifolds and Lie Groups'' or many other differential geometry or topology books, I believe you will find a reference.
Here goes an answer though .   Say you have a vector field $X$ 
on manifold $M$  and a smooth onto map $F:M \to N$. How would you push $X$ forward to get a
vector field on $N$?  There is  only one real choice and that  is given  by the formula:
$$(F_* X)(y) = dF_x (X (x)),  y \in N$$
where  we take an $x$ with $F(x) = y$ and where  $dF_x: T_x M \to T_{F(x)} N$ is the differential of $F$ (Jacobian matrix, in coordinates) at that $x$.  But if $F$ is not  1-1, then there are many choices of $x$, and different choices give different results for $dF_x (X (x))$.  If the result is the same, independent of $x \in F^{-1}(y)$, for all $y \in N$ , then the vector field $X$ is called $F$-projectible''.
The simplest examples of $F$-projectible vector fields 
arise when $F$ is the quotient map for some free Lie group action,
and the vector field $X$ is invariant under that group action.  In your case,
the group action is that of the real line by  translation in the $z$-direction. 
