If two hypersurfaces in $\mathbb{R}^n$ intersect in one point, then its normals at this point are the same except, possibly, by the sign

Question

I read that if two hypersurfaces in $\mathbb{R}^n$ intersect in one point, then its normals at this point are the same except, possibly, by the sign. Although this is very intuitive, I am curious to know how to prove this formally.

I thought the following: I think that two regular surfaces in $\mathbb{R}^3$ intersect in one point if they have contact of order $\geq 1$ at this point, which means

according this exercise in Do Carmo's book. The exercise $c$ imply that $\frac{x_u \times x_v}{|x_u \times x_v|} = \pm \frac{\overline{x}_u \times \overline{x}_v}{|\overline{x}_u \times \overline{x}_v|}$

This led me to ask me the following questions:

1. What formal definition for two hypersurfaces in $\mathbb{R}^n$ intersect?

2. What is the definition for normal vector to $M \subset \mathbb{R}^n$ in $p$? The natural attempt to define the normal is define the cross product in $\mathbb{R}^n$, but this is not possible for each $n$ according these notes.

Only to be clear, my question is how to prove formally the affirmation in the title of this topic, but I think if I can answer the two questions above, then I can do an argument inspired by the case in $\mathbb{R}^3$.

Thanks in advance!

Answer 1

1. The intersection of hypersurfaces is meant just in the set theory sense. But you need here that they intersect in a single point, so $M_1\cap M_2=\{p\}$.

2. A hypersurface $M\subset\mathbb R^n$ has a tangent space $T_xM$ in each point $x\in M$, which is a linear subspace in $\mathbb R^n$ of dimension $n-1$. So there are exactly two unit vectors that are normal to $T_xM$, so in each point there is a natural unit normal up to sign.

For me, the more natural formulation of the statement would actually be that if $M_1\cap M_2=\{p\}$ then $M_1$ and $M_2$ have the same tangent space at $p$. This indeed exactly says that they have contact of order $\geq 1$ in $p$. Just as a hint: I believe that the simplest way to see this is by realizing the hypersurfaces as zero sets of regular functions in a neighborhood of $p$.

Edit: Continuation of hint: Realize $M_1$ lcoally around $p$ as $f^{-1}(\{0\})$ and $M_2$ as $g^{-1}(\{0\})$ with $df(p),dg(p)\neq 0$. Then $T_pM_1=ker(df(p))$ and $T_pM_2=ker(dg(p))$. Now you can consier $(f,g)$ as a function to $\mathbb R^2$ and of course its zero locus coincides with $M_1\cap M_2$ locally around $p$. By construction, the kernel of $d(f,g)(p)$ is the intersection $\ker(df(p))\cap\ker(dg(p))=T_pM_1\cap T_pM_2$. If the two tangent spaces are differnt, then this intersection has dimension less than $n-1$. In that case, $d(f,g)$ is surjective (and the dimension of the kernel is $n-2$). Hence $(f,g)$ is regular in a neighborhood of $p$ and $M_1\cap M_2$ is a submanifold of dimesnion $n-2$. If $n>2$, this contradicts the fact that $M_1\cap M_2=\{p\}$.