I come across an interesting problem below which confuses me a lot.

Let $L$ be a submanifold of $\mathbb{R}^n$ with codimension > 1 . Prove that: (1) if $x$ $\notin L$, for almost every line $l$ passing through $x$, we have $l$ $\cap$ $L$ = $\emptyset$. (2) if $x$ $\in L$, for almost every line $l$ passing through $x$, we have $l$ $\cap$ $L$ = {$x$}.

The definition of "almost every" is not given clearly. I guess it means that when we equate lines passing through $x$ with $\mathbb{R}^n$\{$x$} the points can't satisfy the results above are measured zero.

Edit: It is more proper to consider measure zero in $\mathbb{RP}^{n-1}$

Considering conclusion containing "almost every", I attempted to use parametric transversality theorem to solve it. If we can prove that almost every line intersects transversally with $L$ while it is impossible for $l$ $\cap$ $L$ $\neq$ $\emptyset$ due to the restriction on the codimension. However, I found this is ridiculous because when $n=3$ , $L$ also a line whose codimension is 2 and $x \in L$ , every line passing through $x$ does not intersect transversally with $L$ . So we can't solve both situations at the same time as preconceived.

Is my idea correct? If it is, how should I construct the function satisfying the conditon for parametric transversality theorem? Appreciate for any help!

Parametric Transversality Theorem: Suppose that $F:X\times S\to Y$ is a smooth map of manifolds and $Z$ is a submanifold of $Y$, all manifolds without boundary. If $F$ is transverse to $Z$ then for almost every $s\in S$ the map $f_s : x\mapsto F(x,s)$ is transverse to $Z$.


1 Answer 1


Let $x\in \mathbb R^n$. Define $F: (L\setminus \{x\}) \times \mathbb R\to \mathbb R^n$, $$ F(y, t) = tx + (1-t)y.$$

Clearly this map is smooth. Since $(L\setminus \{x\} \times \mathbb R)$ has dimension strictly less then $n= \dim \mathbb R^n$, the image of $F$ is of measure zero.

That means (i) when $x\notin L$, the union of the sets of lines in $\mathbb R^n$ passing through $x$ and $L$ is of measure zero in $\mathbb R^n$, and when (ii) $x\in L$, the union of sets of lines joining $x$ to some $y\in L$ in $\mathbb R^n$ is of measure zero.

In general, there is a more natural way to talk about measure zero in this setting: for each $x\in \mathbb R^n$, let $\mathbb{RP}^{n-1}_x$ be the set of lines in $\mathbb R^n$ passing through $x$. Then $\mathbb {RP}^{n-1}_x$ is a smooth manifold of dimension $n-1$ and "sets of measure zero" is well defined. Define $$ G : L\setminus\{x\} \to \mathbb{RP}^{n-1}_x, $$ where for each $y\in L$, $G(y)$ is the line in $\mathbb R^n$ passing though $x, y$. Again one can check that $G$ is smooth. Since $L$ has codimension $>1$, $\dim L < n-1=\dim \mathbb{RP}^{n-1}_x$. Thus the image is of measure zero.

Note that we use just a weak form of Sard theorem (about smooth maps $f : M \to N$ with $\dim M < \dim N$).


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