Manifold defined by Euclidean distance (norm) between function output and a value

Question

Suppose $f:\mathbb{R}^n \to \mathbb{R}^m$ is a smooth function. Then the level set is $(n-m)$-dimensional manifold $$ \mathcal{M} = \left\{x\in\mathbb{R}^n \, :\, f(x) = y\right\} $$ Let $\parallel \cdot \parallel_2$ be the $\mathrm{L}_2$ norm. Then clearly the following set contains the same points of $\mathcal{M}$, so they are the same set $$ \mathcal{N} = \left\{x\in \mathbb{R}^n\,:\, \parallel f(x) - y \parallel_2 = 0\right\} $$

Computationally or Theorertically, is there any reason to work with $\mathcal{M}$ rather than $\mathcal{N}$ (or vice versa)?

The key differences are these:

• $\mathcal{M}$ has dimension $n-m$, instead $\mathcal{N}$ has dimension $n-1$
• The gradient at a point on $\mathcal{M}$ is $J_f\in\mathbb{R}^{m\times n}$ is a matrix. Instead $\mathcal{N}$'s gradient at a point on $\mathcal{N}$ is $J_{g}\in\mathbb{R}^{1\times n}$ is a *vector, where $g = ||\cdot||_2 \circ f$

Basically this is what I am thinking about: if the function defining the manifold is between Euclidean spaces, is there any reason to consider the manifold $\mathcal{M}$ with Jacobian matrix, rather than the manifold $\mathcal{N}$ with gradient vector?

Answer 1

The flaw in your reasoning, as many commenters noted, is that $\mathcal N$ is not $n-1$ dimensional. I do not for sure how you arrived at this conclusion, but I am guessing you are using a statement like the following:

Let $g:\mathbb R^a\to\mathbb R^b$. If $z\in \text{Im}(g)$, then the set $\{x\in\mathbb R^a:g(x)=z\}$ is an $n-m$ dimensional manifold.

You conclude that since $g:x\to \Vert f(x)-y\Vert_2$ maps from $\mathbb R^n$ to $\mathbb R^1$, that $\mathcal N$ has dimension $n-1$. However, the above statement is not always true; it is an oversimplification of the following true theorem:

Let $g:\mathbb R^a\to\mathbb R^b$ be smooth. If $z\in \text{Im}(g)$ is a regular value of $g$, then the set $\{x\in\mathbb R^a:g(x)=z\}$ is an $n-m$ dimensional manifold. (source)

Note the two differences between the false statement and true theorem:

1. $g$ must be smooth, and
2. $z$ must be a regular value.

$z$ being a regular point means that whenever $g(x)=z$, then the derivative $dg(x)$ is a surjective map. This means that, as a $b\times a$ matrix, we must have $a\geq b$ and the rows of $dg(x)$ are linearly independent. These two points are very relevant to your question because

1. the function $g:x\mapsto\Vert f(x)-y\Vert_2$ you use to define $\mathcal N$ is not smooth. In particular, it is not differentiable at points $x$ such that $f(x)-y=0$ which are precisely the points you care about.
2. It is common to make the norm function smooth by squaring it. But $0$ is not a regular value of $h:x\mapsto\Vert f(x)-y\Vert_2^2$. We can see this by calculating: $dh(x)=(f(x)-y)^T df(x)$. Here, $(f(x)-y)^T$ is an $1\times m$ vector and $df(x)$ is an $m\times n$ matrix. When $h(x)=0$, $f(x)-y=0$, so $dh(x)=0$, which is not surjective.

So while your assertion that $\mathcal M=\mathcal N$ is correct, your observations under "key differences" are not. In fact, unless we know that $y$ is a regular value of $f$, we cannot even say that $\mathcal M$ is an $n-m$ dimensional manifold.

Now to answer your actual question, is there a reason to work with $\mathcal M$ rather than $\mathcal N$? The answer is no. While it is sometimes the case that one mathematical object can have two equivalent definitions, each of which is useful in a different context, in this case, the two definitions are so similar that it is trivial to derive one from the other. Therefore, there is no advantage to using one or the other. That said, the definition given for $\mathcal M$ is likely to be considered more "natural" in most contexts simply because it is less complicated.