Where do extra dimensions in gradient come from?

Question

The gradient of a scalar function $f\colon \mathbb{R}^n \to \mathbb{R}$ is a vector-valued function $\nabla f\colon \mathbb{R}^n \to \mathbb{R}^n$. Since applying a function can't increase information ($\nabla f$ can't contain information not in $f$), the $n$ dimensions in $\nabla f$ must not be independent -- they must be a relatively "diffuse" (or "redundant") representation of (a subset of) the information in $f$'s single dimension. Is this an accurate understanding?

If so, what is the pattern of dependency among the dimensions in $\nabla f$? That is, what constraints exist among them?

Several answers suggest that $\nabla$ "mixes in" information from the initial space $\mathbb{R}^n$, but I don't see precisely how that information (just a flat Euclidean topology, right?) is represented in $\nabla f$.

One answer points out that $\nabla f$ localizes information that is nonlocal in $f$ -- I get that, but it seems more like a rearrangement of information within the dimensions of $\nabla f$ than a constraint across them -- after all, when $n=1$, no additional dimensions are required for the representation of $\nabla f$.

Moderators: if these edits aren't clear enough to remove the "hold" status, please comment as to what is unclear -- 3 answers have so far accurately interpreted what I was asking, I just haven't yet fully understood them and through comments/edits am trying to prompt refinements to be more complete and/or easier to understand.

Answer 1

Perhaps the following fact is what you are after:

Not every map $\mathbb R^n \rightarrow \mathbb R^n$ is the gradient of a function $\mathbb R^n\rightarrow\mathbb R$.

That is, even though the gradient "creates dimensions", as you put it, it is also constrained in its form, which sort of cancels out the new dimensions. For instance, in two dimensions, suppose we have a function $f(x,y)$ whose gradient is $(a(x,y),b(x,y))$. It can quickly be seen that $a_y(x,y)-b_x(x,y)=0$ given that $f_{xy}=f_{yx}$ (where the subscripts indicate partial derivatives). Thus, for instance, $(y,0)$ is not the gradient of any function since, if it were, the mixed partial derivatives wouldn't match. This can be generalized by considering exact differential forms and the exterior derivative, but I'll just leave those keywords if you wish to investigate further.

From this, we notice that, even though the gradient has two components, knowing one component already tells us almost everything we need to know about the other - if we know $a$, then we know that: $$b(x,y)=\left(\int_{0}^xa_y(x,y)\,dx\right)+c(y)$$ for some function $c$ - which is already reducing the amount of "missing" information from a function $b:\mathbb R^2\rightarrow \mathbb R$ to one merely $c:\mathbb R\rightarrow \mathbb R$. So, even though the gradient has a more complicated representation than the original function, not everything in the representation is independent, so there's no new information added.

Answer 2

The dimensions are already there. It's $\mathbb{R}^n$ after all. The fact that your function is mapping it onto $\mathbb{R}$ doesn't change the fact that there's plenty of information from every one of the $n$ directions. In fact, that's exactly what the gradient is revealing to you. It's not creating any dimensions nor is it creating any more information; it's just revealing the structure of how your function changes along the $n$ axes. So, in a sense, it's the other way around. It's not that the gradient is creating dimensions or information; rather, it's that your scalar function is compressing or hiding data, in a manner of speaking.

Answer 3

The gradient, aka, Jacobian of a map from $f: \mathbb R^n \rightarrow \mathbb R^m : f:(x_1, x_2,..,x_n) \rightarrow (f_1(x_1,..,x_n), f_2(x_1,x_2,..,x_n),...,f_m(x_1,x_2,..,x_n))$ is the $m \times n$-matrix $Jf: (Jf(i,j):= \frac {\partial f_i}{\partial x_j} )$

In that sense, it contains information from both the initial space and the target space.