I have the following situation: There is an embedded submanifold (in my particular case it is of codimension one but I do not think it matters here) $M\subseteq R^n$, and a vector subspace $P\subsetneq R^n$.

There are also a point $p\in P\cap M $ and a tangent vector point $v\in P\cap T_pM $. (Since $M$ is a submanifold of $R^n$ I am using the standard identification and considering $T_pM\subsetneq R^n$ so $P\cap T_pM $ makes sense).

It is also given that the intersection $P\cap M $ is a submanifold of $M$. My question is: does $v\in T_p(P\cap M) $?

My ideas so far are: $v \in P∩T_pM$ implies we can build two curves: $\alpha :I \mapsto P, \beta:I \mapsto M $ such that $\alpha (0)=\beta(0)=p, \alpha' (0)=\beta'(0)=v$. But I am not sure it is possible to find a single curve which passes throug $p$ with velocity $v$ that lies inside $P\cap M$.

  • $\begingroup$ Of course you are right, the (local) uniqueness of solutions ensures that my two curves coincide on a small enough interval. So we have a curve $ \alpha :I \mapsto P\cap M $ which is smooth when considered as a curve in $M$, and as a curve in $P$ separately. but how do we know it is smooth as a curve into the intersection manifold $P\cap M$? $\endgroup$ – Asaf Shachar Apr 1 '15 at 10:09
  • $\begingroup$ I don't think anything like uniqueness of solutions applies here; you have a tangent vector, not a tangent vector field. It's certainly not the case that your curves need to coincide on a small enough interval. Even two curves just in $M$ through a given tangent vector $v$ need not coincide on a small interval. $\endgroup$ – mollyerin Apr 1 '15 at 11:41

No, this isn't enough to guarantee that $v \in T_p(P \cap M)$. As a (degenerate-looking) counterexample, take $M$ to be the graph of $y = x^2$ in $\mathbb{R}^2$, $p$ the origin, and $P$ the subspace corresponding to the $x$-axis. Then $(1,0)$ is in both $P$ and $T_pM$, but it's not in $T_p(P \cap M)$ because the latter is too small.

On the other hand, I believe the following general fact is true: If $M$ and $N$ are embedded submanifolds of $\mathbb{R}^n$ which intersect transversely at $p$ (which means $T_p M + T_p N = \mathbb{R}^n$), then $M \cap N$ is a submanifold in a neighborhood of $p$ and $T_p(M \cap N) = T_pM \cap T_pN$. Your result follows by letting $N = P$.

Sketch of proof of this fact: write $M = f^{-1}(0)$ and $N = g^{-1}(0)$, on some neighborhood $U$ of $p$, where $f: U \to \mathbb{R}^{n - k}$ and $g : U \to \mathbb{R}^{n-l}$, where $k$ and $l$ are the dimensions of $M$ and $N$, respectively, and $0$ is a regular value of both $f$ and $g$. The point then is that the transversality assumption guarantees that $0$ is also a regular value of $(f, g) : U \to \mathbb{R}^{2n-k-l}$, which guarantees that $M \cap N \cap U$ is a submanifold and that the desired equality on tangent spaces holds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.