Rank condition in the Implicit Mapping Theorm

[cmiller5277]

Hi there. I've recently come across the Implicit Mapping theorem in my studies and noticed that there is a condition that the rank of the image must be the maximum possible. I'm not directly seeing why this condition is needed, so I was wondering if anyone could provide me with an example of why this condition cannot be dropped. Thanks

 

[Timbuqtu]

The simplest case of the theorem applies to a differentiable realvalued function on R^2, say F(y,x). Then if \partial F/\partial y \neq 0 at some point (y_0,x_0), there is a neighbourhood I of x_0 in R such that there exist a differentiable f:I -> R^2 with f(x_0)=y_0 and F(f(x),x)=0 for x in I.

Now consider for instance F(y,x)=x. Then F(y,x)=0 corresponds to the vertical x=0 line in the x,y-plane. Obviously this is not (locally) the graph of some function f(x). The reason for this is that the tangentlines are vertical, which is equivalent to \partial F/\partial y = 0. So the condition is necessary.

 

[cmiller5277]

That does make sense, thank you.

 

[mathwonk]

that is a bad example for the question asked. i.e. he asked about the implicit function theorem, and its relation to maximality of rank of the map.

the map here is F(x,y) and its rank is the rank of the matrix of partials with entries dF/dx and dF/dy.

the theorem says that if either of these partials is non zero at p, then the level set F= F(p) is locally a smooth manifold, i.e. the graph of a function, either y(x) or x(y), in your example x(y).

A real example where the theorem fails would be a function like F(x,y) = xy, where the gradient at the origin is (0,0), hence has rank zero, not 1.

the pair of lines xy=0 is not locally the graph of any function either of x or of y near (0,0).the correct hypothesis for the rank theorem however is not that the rank be maximal but that it be locally constant. note that since rank is upper semi continuous, rank maximal at a point, implies rank locally maximal hence rank locally constant.