# transverse intersections

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

Question 1: Let $M$ be a compact $n$-dimensional smooth manifold in $\mathbb{R}^{n+1}$, and take a point $p \notin M$. Prove that there is always a line $l_p$ passing through $p$ such that $l_p \cap M \neq \emptyset$, and $l_p$ intersects transversally with $M$.

Question 2: Let $M$ be a compact $n$-dimensional smooth manifold in $\mathbb{R}^{n+m}$, and take a point $p \notin M$. Prove that for all $1\leq k\leq m$, there is always a hyperplane $P_p$ with $\text{dim}(P_p)=k$ passing through $p$ such that $P_p \cap M \neq \emptyset$, and $P_p$ intersects transversally with $M$.

Note: “Transverse” means “the tangent spaces intersect only at 0”.

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

So it seems reasonable to develop the dimensional technique to attach question 1. In 1 dimension, it will relate to investigating the ordinary differential equation:

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

Where $p=(a,b)$, $M$ has a parameterization $M={x,f(x)}$. If there is a counterexample for question 1, then there is another solution that satisfies the ODE in the sense: for every line $l$, there is an intersection point $a_l\in l\cap M$, and $f$ satisfies the ODE at $a_l$.

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

Proof 1 (provided by fedja):
Area trick. (Weakness: it seems we could not prove the transversally intersection point has 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 vector $n_x$ 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$intersects$M$has nonempty interior. For each point$q \notin M$, the projection$\Pi:M \to S_{q,\epsilon_q}^{n-1}$onto the sphere centered at$q$, of radius$\epsilon_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 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$(all these points have the same degree$d_q$), the line joining$p$and$q’$intersects$M$. Projection of$B$on$S_p^{n-1}$is an open set which we sought. For the general case (partial solution), a similar reasoning should work, however, notice that for$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$for which the intersection is nonempty has a nonempty interior. Last time we proved that there is a zero-dimensional sphere containing$p$, namely${p,q}$, which has a nonzero linking number with$M, and by deforming it 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 an $m-1$-dimensional sphere with a 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 neighborhood of $P$; in particular, we get an open set of parameters $s$ for which $f_s$ intersects $M. Well, we don’t actually need a round sphere, but we do need a smooth sphere that lies in an$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”.