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Chapter 2: Problem 17

Two regular surfaces \(S_{1}\) and \(S_{2}\) intersect transversally if whenever \(p \in S_{1} \cap S_{2}\) then \(T_{p}\left(S_{1}\right) \neq T_{p}\left(S_{2}\right)\). Prove that if \(S_{1}\) intersects \(S_{2}\) transversally, then \(S_{1} \cap S_{2}\) is a regular curve.

Short Answer

Expert verified
\( S_{1} \cap S_{2} \) is a regular curve because their tangent spaces are distinct at each point of intersection.

Step by step solution

01

Define the surfaces

Let the two regular surfaces be denoted by \(S_{1}\) and \(S_{2}\). Assume that \(S_{1}\) and \(S_{2}\) intersect transversally at a point \(p\). This means that at the point \(p, T_{p}(S_{1}) eq T_{p}(S_{2})\).
02

Understand transversal intersection

Transversal intersection means the tangent planes at the intersection point are not the same. Mathematically, this means \(T_{p}(S_{1}) eq T_{p}(S_{2})\), where \(T_{p}(S_{1})\) and \(T_{p}(S_{2})\) are the tangent planes of \(S_{1}\) and \(S_{2}\) at \(p\) respectively.
03

Intersection forms a curve

Since \(S_{1}\) and \(S_{2}\) intersect transversally, their intersection \(S_{1} \cap S_{2}\) forms a set of points where both surfaces meet. At these intersection points, the dimension of \( T_{p}(S_{1} \cap S_{2}) \) must be one less than the dimension of \(S_{1}\), since we are considering two surfaces intersecting (both surfaces having two dimensions).
04

Show it is a regular curve

Given that both \(S_{1}\) and \(S_{2}\) are regular surfaces, the intersection \( S_{1} \cap S_{2} \) must have a well-defined tangent vector at each point of intersection. Hence, each point in the intersection has a tangent vector that varies smoothly along the intersection. This implies the intersection \( S_{1} \cap S_{2} \) forms a smooth curve, i.e., a regular curve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Regular Surfaces
A regular surface refers to a two-dimensional manifold that is smooth and differentiable. In simpler terms, a regular surface is like a smooth, infinitely stretchable sheet. Every small patch of this surface can be described by a smooth function.

For instance, think of the surface of a sphere or a torus. These are examples of regular surfaces because they don't have any sharp edges or sudden changes in direction.

Mathematically, a surface is regular if, around every point, there exists a neighborhood that can be parametrized by two variables. This means you can smoothly map points from a flat plane to the surface.
Tangent Planes
A tangent plane is a flat surface that 'just touches' a point on a regular surface. At this touching point, the tangent plane has the same slope as the surface.

Imagine touching a basketball at one point with a flat sheet of paper. The paper represents the tangent plane to the basketball's surface at that specific point.

The tangent plane is crucial because it helps in understanding how surfaces interact and intersect. Mathematically, for a surface defined by a function, the tangent plane at a point can be found using the partial derivatives of the function.
Regular Curve
A regular curve is a smooth, one-dimensional curve that can be described mathematically by a continuous and differentiable function.

Consider a piece of string stretched out without any kinks. That's akin to a regular curve in space. It's smooth, and if you move along it, there's no sudden change in direction.

In mathematics, a curve is regular if its tangent vector is non-zero at every point. This ensures that as you move along the curve, you can always determine its direction.
Transversal Intersection
Transversal intersection refers to the way two regular surfaces intersect such that their tangent planes are distinct at the points of intersection.

Imagine two sheets of paper crossing each other at a point, but not lying flat against each other at that point. Instead, they form an 'X' like shape, this ensures that there is a clear intersection line.

Mathematically, if two surfaces, denoted as \(S_1\) and \(S_2\), intersect transversally at a point \(p\), then the tangent planes at \(p\) for each surface, \(T_p(S_1)\) and \(T_p(S_2)\), are different. This distinctness is essential as it implies that the intersection forms a regular curve, which is smooth and has a well-defined tangent.

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Most popular questions from this chapter

One way to define a system of coordinates for the sphere \(S^{2}\), given by \(x^{2}+y^{2}+(z-1)^{2}=1\), is to consider the so-called stereographic projection \(\pi: S^{2} \sim\\{N\\} \rightarrow R^{2}\) which carries a point \(p=(x, y, z)\) of the sphere \(S^{2}\) minus the north pole \(N=(0,0,2)\) onto the intersection of the \(x y\) plane with the straight line which connects \(N\) to \(p\) (Fig. 2-12). Let \((u, v)=\pi(x, y, z)\), where \((x, y, z) \in S^{2} \sim\\{N\\}\) and \((u, v) \in x y\) plane. a. Show that \(\pi^{-1}: R^{2} \rightarrow S^{2}\) is given by $$ \pi^{-1}\left\\{\begin{array}{l} x=\frac{4 u}{u^{2}+v^{2}+4} \\ y=\frac{4 v}{u^{2}+v^{2}+4} \\ z=\frac{2\left(u^{2}+v^{2}\right)}{u^{2}+v^{2}+4} \end{array}\right. $$ b. Show that it is possible, using stereographic projection, to cover the sphere with two coordinate neighborhoods.

Let \(Q\) be the union of the three coordinate planes \(x=0, y=0, z=0\). Let \(p=(x, y, z) \in R^{3}-Q .\) a. Show that the equation in \(t\), $$ \frac{x^{2}}{a-t}+\frac{y^{2}}{b-t}+\frac{z^{2}}{c-t} \equiv f(t)=1, \quad a>b>c>0 $$ has three distinct real roots: \(t_{1}, t_{2}, t_{3}\). b. Show that for each \(p \in R^{3}-Q\), the sets given by \(f\left(t_{1}\right)-1=0\), \(f\left(t_{2}\right)-1=0, f\left(t_{3}\right)-1=0\) are regular surfaces passing through \(p\) which are pairwise orthogonal.

(Gradient on Surfaces.) The gradient of a differentiable function \(f: S \rightarrow R\) is a differentiable map grad \(f: S \rightarrow R^{3}\) which assigns to each point \(p \in S\) a vector grad \(f(p) \in T_{p}(S) \subset R^{3}\) such that \(\langle\operatorname{grad} f(p), v\rangle_{p}=d f_{p}(v) \quad\) for all \(v \in T_{p}(S)\) Show that a. If \(E, F, G\) are the coefficients of the first fundamental form in a parametrization \(\mathbf{x}: U \subset R^{2} \rightarrow S\), then grad \(f\) on \(\mathbf{x}(U)\) is given by $$ \operatorname{grad} f=\frac{f_{u} G-f_{v} F}{E G-F^{2}} \mathbf{x}_{u}+\frac{f_{v} E-f_{u} F}{E G-F^{2}} \mathbf{x}_{v} $$ In particular, if \(S=R^{2}\) with coordinates \(x, y\), $$ \operatorname{grad} f=f_{x} e_{1}+f_{y} e_{2} $$ where \(\left\\{e_{1}, e_{2}\right\\}\) is the canonical basis of \(R_{2}\) (thus, the definition agrees with the usual definition of gradient in the plane). b. If you let \(p \in S\) be fixed and \(v\) vary in the unit circle \(|v|=1\) in \(T_{p}(s)\), then \(d f_{p}(v)\) is maximum if and only if \(v=\operatorname{grad} f /|\operatorname{grad} f|(\) thus, grad \(f(p)\) gives the direction of maximum variation of \(f\) at \(p)\). c. If grad \(f \neq 0\) at all points of the level curve \(C=\\{q \in S ; f(q)=\) const. \(\\},\) then \(C\) is a regular curve on \(S\) and grad \(f\) is normal to \(C\) at all points of \(C\).

Show that the area \(A\) of a bounded region \(R\) of the surface \(z=f(x, y)\) is $$ A=\iint_{Q} \sqrt{1+f_{x}^{2}+f_{y}^{2}} d x d y $$ where \(Q\) is the normal projection of \(R\) onto the \(x y\) plane.

Show that the tangent planes of a surface given by \(z=x f(y / x), x \neq 0\), where \(f\) is a differentiable function, all pass through the origin \((0,0,0)\).
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