Transverse intersections：从 Sard 定理到 generic position

旧博客原文

原题：transverse intersections

https://en.wikipedia.org/wiki/Transversality_(mathematics)

This problem may be a embarrassed one, but I even could not prove it for the 1 dimensional case.

Here is the problem:

>**Question 1** $M$ is a compact $n$ -dimensional smooth manifold in $R^{n+1}$ , take a point $p\notin M$. prove there is always a line $l_p$ pass $p$ and $l_p\cap M\neq \emptyset$ , and $l_p$ intersect transversally with $M$ .

You can naturally generated it to:
>**Qusetion 2** $M$ is a compact $n$-dimensional smooth manifold in $R^{n+m}$ , take a point $p\notin M$ . Prove $\forall 1\leq k\leq m$, there is always a hyperplane $P_p, dim(P_p)=k$ pass $p$ and $P_p\cap M\neq \emptyset$ , and $P_p$ intersect transversally with $M$ .

Thanking for Piotr pointed out, assuming “transverse” means “the tangent spaces intersect only at 0”.

We focus on question 1 for simplified.

Even in 1 dimension it is not easy at least for me, **warning**: a line $l$ pass $p$ may be intersect $M$ at several points combine a set $A_l$ , $A_l$ could be finite, countable or even it is not countable (consider $M$ is induced by a smooth function for which the zeros set is Cantor set.)… And if there is one point $a\in A_l$ , $l$ is tangent with the tangent line of $M$ at $a$ , then $l$ is not intersect transversally with $M$ .

**My attempt**:
I could use a dimensional argument and Sard’s theorem to establish a similar result but instead of a fix point $p$ , we proof for generic point in $R^{n+1}$ which is not in $M$ we can choose such a line.

So it seems reasonable to develop the dimensional technique to attach the question 1, in 1 dimensional, it will relate to investigate the ordinary differential equation:

$\frac{f(x)-b}{x-a}=f'(x)$

Where $p=(a,b)$ , $M$ have a parameterization $M=\{x,f(x)\}$ . If there is a counterexample for the question 1, then there is another solution which satisfied the ODE in the sense:

at least for every line $l$ there is a intersection point $a_l\in l\cap M$ , $f$ satisfied ODE at $a_l$ .

This is just like the uniqueness of the solution of such a ODE is destroyed at some subspace of a line which have some special linear structure, I do not know if this point of view with be helpful.

I will appreciate for any useful answers and comments.

Proof 1(provided by fedja)

Area trick.(weakness:it seems we could not proof the transtivasally intersection point have positive measure by this way).

Proof 2(provided by Piotr)

#For the codimension 1 case.#

###Using Thom transversality theorem.###
Consider the maps $f_s:\mathbb{R} \to \mathbb{R}^n$ parametrized by $s \in S^{n-1}$ and given by $f_s(t) = p + t \cdot s$ . The map $F(s,t) = f_s(t)$ , $F:S^{n-1} \times \mathbb{R} \to \mathbb{R}^n$ is clearly transverse to $M$ , thus Thom’s transversality says that $f_s$ is transverse to $M$ for almost all $s$ . Now it suffices to prove that for an open set in $S^{n-1}$ , the line given by $f_s$ intersects $M$ . Proven below.

###Using Sard’s theorem directly.###
Thom’s transversality is usually proven using Sard’s theorem. Here is the idea.

Consider the projection $\Pi:\mathbb{R}^n \setminus \{p\}\to S^{n-1}_p$ onto a sphere centered at $p$ . A line $l_p$ through $p$ intersects $M$ transversally if the two points $l_p \cap S^{n-1}_p$ are regular values of $\Pi$ (indeed, the critical points of $\Pi$ are exactly the points $x \in M$ at which the normal $\vec n_x$ is perpendicular to the radial direction (with respect to $p$)). By Sard’s theorem, the set of regular values is dense in $S^{n-1}_p$ .

We need to choose any point $s$ on the sphere for which both $s$ and $-s$ are regular values, and the line $f_s$ through $p$ and $s$ actually intersects $M$ . It suffices to prove that the set of points $s$ for which this line intersects $M$ contains an open set. We could now use the Jordan-Brouwer Separation Theorem and we would be done, but we can do it more directly (and in a way that seems to generalize).

###The set of points $s \in S$ for which $f_s$ intersects $s$ has nonempty interior.###
For each point $q \notin M$ the projection $\Pi:M \to S_{q,\varepsilon_q}^{n-1}$ onto the sphere centered at $q$ , of radius $\varepsilon_q$ small enough so that the sphere does not intersect $M$ , has some (topological) degree $d_q$ . It is easy to check that if one takes any point $x \in M$ and considers the points $x \pm \delta \vec n_x$ for small $\delta$ , the degrees of the corresponding maps differ by $1$ . It follows that we can find a point $q$ for which $d_q \neq d_p$ , which guarantees that for every point $q'$ in a small open ball $B$ around $q$ (all these points have same degree $d_q$ ), the line joining $p$ and $q'$ intersects $M$ . Projection of $B$ on $S_p^{n-1}$ is an open set which we sought.

#For the general case (partial solution).

I think a similar reasoning should work, however, notice that for $k < m$ we cannot make $P_p$ intersect transversally with $M$ because of dimensional reasons: the dimensions of $M$ and $P_p$ don’t add up to at least $n+m$ . Recall that transversality implies Thus, either (1) you want to consider $k \geq m$ , or (2) define “transversal intersection” for such manifolds saying that the tangent spaces have to intersect at an empty set.

Also, for $k>n$ we can just take any plane $P_p$ which works for $k=m$ and just extend it to a $k$ -dimensional plane.

###Assuming $k = m$ .###
A similar reasoning should work for $f_s:\mathbb{R}^m \to \mathbb{R}^{n+m}$ with $s = (s_1, \ldots, s_m)$ going over all families of pairwise perpendicular unit vectors, and $f_s(t_1,\ldots,t_m) = p+\sum_{j=1}^m t_i \cdot s_i$ . Thom’s transversality says that for almost all choices of $s$ , the plane $f_s$ is transverse to $M$ .

### The nonempty interior issue. ###
The only thing left is to prove that the set of $s$ for which the intersection is nonempty has nonempty interior. Last time we proved that there is a zero-dimensional sphere containing $p$ , namely $\{p,q\}$ , which has nonzero linking number with $M$ , and by deforming if to spheres $\{p,q'\}$ and taking lines through pairs $p,q'$ , we got an open set of parameters for which the line intersects $M$ .

Here should be able to do a similar trick by finding a $m-1$ -dimensional sphere with nonzero linking number with $M$ . The ball that bounds that sphere has to intersect $M$ , thus the plane $P$ containing the sphere has to intersect $M$ . By perturbing the sphere we get spheres with the same linking numbers, and get all the planes that lie in a neighbourhood of $P$ ; in particular, we get an open set of parameters $s$ for which $f_s$ intersects $M$ .

Well, we don’t actually need a *round* sphere, but we do need a *smooth* sphere that lies in a $m$ -dimensional plane. There’s some trickery needed to do this, but I am sure something like this can be done.

Maybe somebody else can do it better?

### For $k<m$ ###

I don’t really know how to attack this case, assuming “transverse” means “the tangent spaces intersect only at $0$ “.

补充说明

以下是新整理的中文说明；上方旧博客原文保持不变。

Transversality 是微分拓扑中最基础也最有用的原则：如果两个几何对象不是被特殊关系强迫相切，那么经过任意小扰动后，它们通常会横截相交。许多“存在一个好方向”的问题，本质上都是 transversality 和 Sard 定理的影子。

Transverse intersections：从 Sard 定理到 generic position — Transversality 把几何相交问题转化为参数空间中的 regular value 问题。

1. 定义

若 $A,B\subset X$ 是光滑子流形，在交点 $x\in A\cap B$ 处称它们 transverse，如果

$$T_xA+T_xB=T_xX.$$

对一条曲线和一条直线来说，这表示交点处切线方向不同；对高维子流形来说，它表示两者的切空间张成了整个环境空间。

2. 固定点与过点直线的问题

一个自然问题是：给定 compact smooth submanifold $M\subset\mathbb R^m$ 和一点 $p\notin M$，是否存在过 $p$ 的直线与 $M$ 横截相交？

困难在于，一条直线可能与 $M$ 有多个交点，甚至交点集合不一定有限。如果某个交点处直线方向落在 $T_xM$ 中，就会失去横截性。因此不能只检查一个交点，而要同时控制所有交点。

3. Sard 定理的用法

把方向空间看成 projective space。对每个 $x\in M$，从 $p$ 指向 $x$ 的方向给出映射

$$\Phi:M\to \mathbb RP^{m-1},\qquad \Phi(x)=[x-p].$$

若选择的方向是 $\Phi$ 的 regular value，那么对应射线与 $M$ 的交点满足横截条件。Sard 定理告诉我们，critical values 的测度为零，所以 generic direction 是好的。

4. 为什么“generic”比“显式构造”容易

显式找一条好直线可能很难，因为坏方向集合由所有相切条件组成。Sard 定理的强处在于，它不需要列出坏方向，只需证明坏方向是某个光滑映射的 critical values。

这就是微分拓扑常见的思想：把几何条件变成参数空间中的 regular value 问题。

5. 高维推广

对过 $p$ 的 $k$-平面，也可以考虑相应的 Grassmannian 参数空间。横截性条件仍然可以写成 evaluation map 的 regular value 条件。Thom transversality theorem 进一步说明，在函数空间中 transverse maps 构成 residual set。

所以这类问题的正确答案通常不是找一条神奇的线，而是证明坏参数集合很小，因而几乎所有选择都好。

Transverse intersections：从 Sard 定理到 generic position

旧博客原文

补充说明

1. 定义

2. 固定点与过点直线的问题

3. Sard 定理的用法

4. 为什么“generic”比“显式构造”容易

5. 高维推广

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