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Let the zero sets $F=\{x \in \mathbb{R}^n: f(x) = 0\}$, $G = \{x \in \mathbb{R}^n : g(x) = 0\}$, where $f$ and $g$ are $m$-dimensional real, analytic, continuous, and nonlinear vector functions. Under some assumptions, these sets define hypersurfaces of zero Lebesgue measure in $\mathbb{R}^n$. I was wondering:

  1. Are $F$ and $G$ always submanifolds embedded in $\mathbb{R}^n$ or are there exceptions - in the latter case, are there conditions that guarantee that they are submanifolds ?
  2. What is the dimension of $F\cap G$ ? As pointed out here Measure of the intersection of two manifolds, if $F$ and $G$ are $(n-1)$-dimensional manifolds and their intersection is transversal, then $\text{dim}(F\cap G) = n-2$. However, is there anything that can be said if the intersection is not transversal ? In general, I am interested in some sort of inequality $\text{dim}(F\cap G) \leq n-2$, assuming that $F \subseteq G$, that $G \subseteq F$ do not hold.
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  • $\begingroup$ I think you should edit your question. What are your assumptions about $f$ and $g$? And by measure: Do you mean Hausdorff measure? Lebesgue measure? $\endgroup$
    – Dieter Kadelka
    Commented Oct 14, 2020 at 10:24
  • $\begingroup$ Thanks! I meant Lebesgue measure. Regarding the functions, besides continuity, they are real and analytic. What other assumptions would make answering the question easier ? $\endgroup$
    – Chris
    Commented Oct 14, 2020 at 10:50
  • $\begingroup$ You'd need $f$ and $g$ of maximal rank (eg submersions). Note that $f$ and $g$ define a function $(f,g)\colon \mathbb{R}^n \to \mathbb{R}^{2m}$ and $F\cap G$ is the zero-set of this function. If $F$ and $G$ are $(n-1)$-dimensional, then $m=1$. So I would imagine that away from singularities, $F\cap G$ is $(n-2)$-dimensional. $\endgroup$
    – David Roberts ♦
    Commented Oct 14, 2020 at 11:40
  • $\begingroup$ Thanks! I didn't really get why $m=1$ but then I guess $F$ and $G$ are equivalent to the sets $\{x:\|f(x)\| = 0\}$, $\{x:\|g(x)\| = 0\}$, which are zero-sets of real 1D functions. If $f$, $g$ are close to singularities, what would that imply for $F\cap G$ ? In general, is there any case when the dimension of $F\cap G$ would be $n-1$ ? $\endgroup$
    – Chris
    Commented Oct 14, 2020 at 12:20

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