This problem may be an embarrassing one, but I could not prove it even 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 generalise it to:

Question 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 pointing out, assume "transverse" means "the tangent spaces intersect only at 0".

We focus on question 1 for simplicity.

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 prove 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 has some special linear structure, I do not know if this point of view will be helpful.

I will appreciate for any useful answers and comments.

  • $\begingroup$ I assume "transitive" means "transversally". $\endgroup$ – Igor Rivin Nov 28 '17 at 18:06
  • $\begingroup$ It depends a bit on what exactly you mean by a "smooth manifold". Clearly a straight open interval on the plane and a point on the extension of that interval isn't something you are interested in. So, what exactly do you assume to exclude such stupid configurations? $\endgroup$ – fedja Nov 28 '17 at 18:16
  • $\begingroup$ @fedja, sorry I forget to assume $M$ is compact, thanks! This is not mean I am not interested in whether this kind of type is true for not compact case, but I could not describe very exactly what condition will make this type of result to be true (it seem it is necessary to assume $M$ is not a "cone" central at $p$ to make this result to be true), anyway. Thanks! $\endgroup$ – Hu xiyu Nov 28 '17 at 18:24
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    $\begingroup$ One obvious approach would be this: if a line $\ell$ through $p$ is not transversal to $M$, then the area of $M$ in the cone of some small aperture $\varepsilon>0$ vastly exceeds $\varepsilon^n$. So if the projection of $M$ to the sphere centered at $p$ has positive area, you run into a contradiction immediately (well, with some minimally intelligent use of either covering lemmas or uniform bounds on smoothness, whichever you like more). $\endgroup$ – fedja Nov 28 '17 at 18:28
  • $\begingroup$ @Huxiyu P.S. Nothing to be embarrased about, by the way. I can tell you much more shameful examples of my own mathematical blindness and I suspect the other people can do it too if you find a way to open them up :-) $\endgroup$ – fedja Nov 28 '17 at 18:49

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 $0$".

Update (thanks to Fan Zheng): This follows from the $k=m$ case. Take any $m$-plane $P$ that intersects $M$ and does it transversally. Then, any $k$-plane inside $P$ intersects $M$ "transversally" in the above sense. We choose any $k$-plane $\tilde P \subset P$ which contains both $p$ and any point $x \in P \cap M$, and we're done.

  • $\begingroup$ Appreciate! this is a excellent answer! Sorry, for $k<m$ case the transverse us just mean "the tangent spaces intersect only at 0", I have corrected it. Would it be true that $k<m$ case is the directly result of $k=m$ case? By the way, I thought the area trick which is mentioned by fedja still effective, we can use the area trick to proof question 2 which lead to another proof. Thanks a lot! $\endgroup$ – Hu xiyu Nov 29 '17 at 3:37
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    $\begingroup$ @Huxiyu If you define transversality this way, then the $k<m$ case is indeed a consequence of the $k=m$ case. $\endgroup$ – Fan Zheng Nov 29 '17 at 4:57
  • $\begingroup$ @FanZheng Indeed, I'm updating the answer. $\endgroup$ – Piotr Nov 30 '17 at 5:26

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