# About the implicit function theorem

1. Jun 17, 2008

### quasar987

In the implicit function theorem, we take a C^1 function F:R^n x R^m --> R^m and given a condition on the partial derivatives at a point (x_0,y_0) such that F(x_0,y_0)=0, we conclude that the relation F(x,y)=0 implicitely defines a function f(x)=y in a nbh of x_0. I.e. the last m variables are determined by the first n in a nbh of x_0.

What if I want to express the last k variable (where k differs from m) in terms of the others?

Is this impossible?

2. Jun 20, 2008

### olliemath

Interesting.. I don't know of any results and it's not immediately obvious..
I managed to sketch out the following partial results, but for more perhaps analyse the proof of the original theorem?

If k<m then you get a continuous function \pi\circ F:R^n+m-->R^k where \pi is the canonical projection function R^m-->R^k. The corresponding Jacobian is the same as the original but with some rows removed, so its inverse is the same as the original inverse with some columns removed. It follows that there exists a continuous function f:R^n+m-k-->R^k defined on some appropriate neighbourhood by the implicit function theorem.

Suppose k>m and that f is defined on the simply connected neighbourhood U of x_0 in R^n. Take the projection of U onto the first j:=n+m-k coordinates - call it V_1 - and the projection of U onto the last i=k-m coordinates - call it V_2.
If there exists any continuous function g:V_1-->V_2 (and this is not a trivial requirement when j>i, may be of use in constructing a counterexample) then (p,g(p)) is an element of U for all p in V_1 and y:=f(p,g(p)) is an element of R^m. We have that F( (p,g(p)),y )=0 for all p in V_1, so the function h:R^j-->R^k defined by h(p)=(g(p),y) is then the one required.

As an illustrative example take the unit sphere in R^3; given a point in the x-y plane near (0,0) we get a point on the 'northern hemisphere', this induces the continuous map f:R^2-->R. Take any continous function g:[-\delta,\delta]-->[-\epsilon,\epsilon] from the x axis to the y axis. Then we get a curve in the plane defined by v(p)=(p,g(p)) near zero and a corresponding curve through (0,0,1) defined by (v,f(v)) on the sphere. The map h:R-->R^2 defined by h(p)=(g(p),f(y)) is then as required. Note well that the choices of g give parametrizations of curves on the sphere through (0,0,1) - of which there are many, so the uniqueness property of the original implicit function theorem fails..

sorry for the long post!

3. Jun 20, 2008

### mathwonk

yes this is a standard version of the theorem, the theorem says if the derivative has locally constant rank k near a, then the function looks locally, smoothly, like projection on some k variables.

look in a good calculus book, like dieudonne, foundations of modern analysis, under the "rank theorem".

the usual theorem uses the fact that a smooth function whose derivative has maximal rank at a, also has maximal rank near a, so the locally constant rank hypothesis is automatic in that case.