Why is it a surface?

Question

Consider $\vec{X} \in \mathbb{R}^{d}$, a vector having $d$ dimensions. Why is it that the constraint equation $g(\vec{X}) = 0$ represents a $d-1$ dimensional surface in $d$ dimensional hyperspace? I know it is obvious to visualize in lower dimensions(i.e. 2,3) but can we argue it rigorously?

Answer 1

The first thing to understand is the case when $g$ is linear. If $T:\Bbb{R}^d\to\Bbb{R}$ is linear then, the rank-nullity theorem gives you the answer: $\dim \ker T=\dim(\Bbb{R}^d)-\text{rank}(T)=d-\text{rank}(T)$. Now, if $T$ is non-zero, then the image is all of $\Bbb{R}$, so by definition the rank is $1$, so the kernel (which is the zero level set) is a $(d-1)$-dimensional subspace.

In the general non-linear case, this is a consequence of the implicit function theorem, which in particular gives the regular value theorem/level-set theorem and this in particular says if $g:\Bbb{R}^d\to\Bbb{R}$ is a smooth function such that it's derivative is nowhere vanishing (you forgot to include this condition as part of the hypotheses), then the level sets $g^{-1}(\{c\})$ are $(d-1)$-dimensional submanifolds of $\Bbb{R}^d$, i.e they are "hypersurfaces".

Of course, things can be generalized to more dimensions. Anyway, the intuition comes from linear algebra, and particularly, the rank-nullity theorem.