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Let $N, M \subseteq R^n$be transversal submonifolds of $R^n$ We say that $N$ and $M$ are transversal if $T_pN + T_pM =T_pR^n$ for all $p \in N\cap M$. Why is $N \cap M$ a manifold?

C. Falcon
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Janusz Tracz
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1 Answers1

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Let $p$ be the codimension of $M$ and let $q$ be the codimension of $N$. Let us fixed $x\in M\cap N$, there exists $f=(f_1,\cdots,f_p)\colon\mathbb{R}^n\rightarrow\mathbb{R}^p$ and $g=(g_1,\cdots,g_q)\colon\mathbb{R}^n\rightarrow\mathbb{R}^q$ such that $\mathrm{rg}(\mathrm{d}_xf)=p$ and $\textrm{rg}(\mathrm{d}_xg)=q$. Moreover, $f=0$ is an equation for $M$ in a neighborhood of $x$ and $g=0$ is an equation for $N$ in a neighborhood of $x$. Let define $\varphi:=(f_1,\cdots,f_p,g_1,\cdots,g_q)\colon\mathbb{R}^n\rightarrow\mathbb{R}^{p+q}$, then $\varphi=0$ is an equation for $M\cap N$ in a neighborhood of $x$. Besides, the hypothesis $T_xM+T_xN=\mathbb{R}^n$ implies that: $$\dim(\ker(\mathrm{d}_xf)\cap\ker(\mathrm{d}_xg))=n-\dim\ker(\mathrm{d}_xf)-\dim\ker(\mathrm{d}_xg).$$ Which can be written as: $$\dim(\ker(\mathrm{d}_x\varphi))=n-p-q.$$ Therefore, $\varphi$ has rank $p+q$. Whence the result.

C. Falcon
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    Intuitively (the content of this answer is that), transversality means that the two submanifolds locally look like two orthogonal hyperplanes in a vector space. – tomasz Nov 09 '16 at 00:28
  • You forgot some instances of $\operatorname{dim}$ in the penultimate equation. – Danu Nov 09 '16 at 02:05
  • Can you explain the implication involving the dimension of kernel intersection @C.Falcon – Devendra Singh Rana Sep 11 '19 at 13:57
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    @Devengra Singh Rana I just used the following formula $\dim(E+F)=\dim(E)+\dim(F)-\dim(E\cap F)$ and here the transversely ensures that the dimension of $E+F$ is $n$. – C. Falcon Sep 12 '19 at 14:09
  • @C.Falcon Okay, Clearly the result follows if $M$ and $N$ are manifolds without boudary, but can your arguments be applied when $M$ or $N$ is a manifold with boudary or with corners? Indeed a subset $S$ of $\Bbb R^n$ is a $k$-manifold without boudary if and only if there exist a function $h$ from an open neighborhood of $S$ into $\Bbb R^{n-k}$ and whose derivative has maximum rank and whose value are zero onto $S$ but unfortunately I do not know any analogous result for manifolds with boundary: so what can you say about? Excuse me for the bother. – Antonio Maria Di Mauro Jul 24 '21 at 18:12
  • @C.Falcon you are missing a minus sign in the equation involving the intersection of the kernel. – ImHackingXD Nov 05 '23 at 21:47