Intuition behind functional dependence

Question

What is the intuition behind functional independence ?

(This is defined in the following way: Let $k\leq n$. The $C^1$ functions $F_1,\ldots,F_k:\mathbb{R}^n\rightarrow \mathbb{R}$ are functionally independent if the matrix whose columns are the gradients $\nabla F_1,\ldots,\nabla F_k$ has full rank, i.e. rank $k$, on the whole domain of definition.

From what I gather from this answer, this is the same as saying that $F:=(F_1,\ldots,F_k):\mathbb{R}^n\rightarrow \mathbb{R}^k$ is submersion, but that doesn't help me much either, because I also don't have any intuition concerning submersions.)

So what does it really mean if the functions are functional indepedent - or conversely, dependent ? Is there, in the latter case, then also a relationship like $g(\nabla F_1,\ldots,\nabla F_k)=0$ -- or maybe like $g(F_1(x),\ldots F_k (x))=0$ for some $x$ -- for $g$ ranging in some specific set, similar to the case of linear independence ( in which $g$ would be from the set $\{g:\mathbb{R}^k\rightarrow \mathbb{R}:g(x_1,\ldots,x_k)=\sum \lambda_i x_i \text{ for some nonzero } \lambda_i \in \mathbb{R}\}$).

Answer 1

Functional dependence of $k$ given functions $F_i:\>\Omega\to{\mathbb R}$ $\>(1\leq i\leq k)$ with common domain $\Omega\subset{\mathbb R}^n$ means, intuitively, that there is a nontrivial function $$g:\quad{\mathbb R}^k\to{\mathbb R},\qquad y\mapsto g(y)$$ such that $$g\bigl(F_1(x),F_2(x),\ldots, F_k(x)\bigr)\equiv 0\qquad \forall x\in\Omega\ .\tag{1}$$ "Nontrivial" for $g$ means that $\nabla g(y)\ne0$ for all $y\in{\mathbb R}^k$. Taking the derivative of $(1)$ we see that $$dg\bigl(F(x)\bigr)\cdot dF(x)\equiv 0\in{\cal L}({\mathbb R}^n,{\mathbb R}^k)\qquad\forall x\in\Omega\ .$$ In terms of matrices this says that the rows of the matrix $\bigl[dF(x)\bigr]$ are linearly dependent with coefficients given by $\nabla g\bigl(F(x)\bigr)\ne0$, for each $x\in\Omega$.

Now the rows of the matrix $\bigl[dF(x)\bigr]$ are nothing else but the gradients $\nabla F_i(x)$. Therefore functional dependence of the $F_i$ in the above sense implies that the gradients $\nabla F_i(x)$ are linearly dependent, at each point $x\in\Omega$.

In your definition of "functional independence" you don't want even a hint of such a thing. Therefore you insist that at all points $x\in\Omega$ the $k$ gradients $\nabla F_i(x)$ should be linearly independent.

Answer 2

The derivative of $F$ is $DF=[\nabla F_1,\cdots,\nabla F_k]^T$ ; let $a\in\mathbb{R}^n$, $b=F(a)$ and $V=F^{-1}(b)$. $V$ is the intersection of the $k$ hypersurfaces $F_i(x)=b_i$. The normal vector in $a$ to such a hypersurface is $\nabla F_i(a)\in \mathbb{R}^n$. Here $rank(DF_a)=k$, that is $W=span(\nabla F_1(a),\cdots,\nabla F_k(a))$ has dimension $k$. According to the implicit function theorem, in a neighborhhood of $a$, $V$ is a variety of dimension $n-k$, that is $V$ is $C^1$-isomorphic to an open subset of $\mathbb{R}^{n-k}$. Moreover the tangent space of $V$ in $a$ is the orthogonal of $W$.

For instance, let $n=3,k=2$. $V$ is the intersection of $2$ surfaces in the standard space. The normal vectors $u_1,u_2$ in $a$ are not parallel ; then $V$ is locally a line and the cross product $u_1\times u_2$ is tangent to this line.

EDIT 1. In other words, an approximation of the equation of $V$ is $DF_a(x-a)=0$, that is, for every $i\leq k$, $<\nabla F_i(a),x-a>=0$.

@ user36772 , I just read your last four lines ! You speak about the case when (in my instance) the $2$ previous normals are parallel. Then the surfaces are tangent in $a$ and we know nothing about the intersection. In other words, when the hypothesis of a theorem are not satisfied, then (is it surprising ?) the theorem does not work.

EDIT 2. (answer to user36772). A level set is a subvariety of codimension $1$, that is an hypersurface.

The tangent space to an hypersurface is the hyperplane that is orthogonal to a normal vector ; then the tangent space to $V$ is the intersection of these hyperplanes. From the geometrical point of view, to say that the $C^1$ functions are independent in a neighborhood of $a$ is equivalent to identify each hypersurface with its tangent hyperplane in $a$ and to say that the linear equations associated to these hyperplanes are linearly independent. This property is stable in the following sense: if we move slightly our hypersurfaces, then the intersection remains similar to the original one.

If these equations are not linearly independent, then the instability comes at a gallop. For instance, consider, in $\mathbb{R}^3$, $a=0, F_1=y-x^2,F_2=y-x^4$. Then, locally, the intersection of the surfaces is the line $Oz$. Yet, if you move one surface, then the intersection may be locally void.

Think also to the GPS ; we need $5$ satellites. Geometrically, the intersection of $3$ spheres suffice. Yet a fourth measure allows synchronization of clocks. Why the fifth ? Because if two among the satellites are "close", then the associated spheres are nearly tangent and the intersection is unstable.

Answer 3

Functional independence between two functions means that the level set of each of the functions intersects transversely the level set of the other function. In the plane, this means that a function $f$ functionally independent of another function $g$ cannot be written as $f=F(g)$ where $F$ is another function, because in this case, the level sets would be the same curves, for both $f$ and $g$. Similarly, in $k$ dimensions, functional independence means that the intersection between the $k$ $(k-1)$-dimensional level sets of the $k$ functions is a point, not a line or a plane, i. e., the functions $F_1$, $F_2$,...,$F_k$ contains $k$ independent informations of your ambient space, and can be taken locally as coordinates.