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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 $P\cap M$.

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  • $\begingroup$ 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
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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.

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