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分类： Differential topology

  • 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。