Question

EXERCISE 3.35. Let S be a te a plane such that pE P and S lies on one regular surface and peS. Suppose P C R3 (closed) side of P. Prove that ToS=P.

Answer 1

SCR
is a regular surface and
pES
. Without loss of generality we can assume that is the origin.

Now suppose
TpS+P
. Since both
TpS
and are planes through the origin in $\mathbb{R}^3$, their intersection is exactly one dimensional. Let
1E TS
and $v_2 \in P$ be perpendicular to $T_p S \cap P$. Let be the span of
{U1, U2
.

By definition of we have that $Q + T_p S = \mathbb{R}^3$ . Hence the plane and the surface intersect transversally and their intersection is a 1 dimensional manifold. In other words this manifold is actually a curve call it .

But now there are two tangents to at the point , namely the line produced by extending in both directions (since the span of $v_2$ is in and does not intersect and $v_1$ is a tangent vector) . This cannot happen as a curve can have only one tangent at a point.

Hence we must have