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be $M$ a surface and $P$ a plane which intersects $M$ at a single point $p$. Show that $P$ is the tangent plane in order in p.

Tangent plane in order is $p+T_{p}M$

I try of suppose that $q\neq p$ so $q-p\in T_{p}M$ an make a contradiction but don't result

thx

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1 Answer 1

I assume you are talking about a 2 dimensional surface $M$ embedded into $\mathbb R^3$. Here is one argument, but there might be simpler ones.

Without loss of generality you may assume $p = 0$ and $P$ is the $x,y$-plane. Since $M$ intersects $P$ only in $0$ it follows that all points in $M$ either have nonnegative $z$-coordinate or all points in $M$ have nonpositive $z$-coordinate (otherwise $M$ could not be a manifold, since $M\setminus \{0\}$ would be disconnected). One may assume the latter, i.e. the $z$-coordinate of all points in $M$ is $\leq 0$.

There are different methods to define the tangent plane. I assume you know, that the tangent plane is the plane spanned by derivatives of curves lying inside $M$. Take two curves $\alpha, \beta \subset M$ with $\alpha(0) = \beta(0) = 0$ such that $T_pM$ is spanned by $\dot \alpha(0)$ and $\dot \beta(0)$. It follows that the $z$-coordinate of $\alpha$ and $\beta$ attains its maximum in $\alpha(0)$ resp. $\beta (0)$. Hence $\dot \alpha (0)$ and $\dot \beta (0)$ lie inside the $x,y$-plane so $T_pM \subseteq P$. For dimensional reasons equality must hold.

In higher dimensions analogeous arguments work.

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