2 2 2 Fix mint: 3:: I]. Consider the ellipsoid E = {(5.31}, z} E R3 | :2 + i, + E2 = 1}_ to} Prove that E is a smooth surface. (Hint: Adapt the surface patches for the sphere which were given in exercise 4.1.2.} {o} Compute one of the {nontrivial} transition maps between allowable patches in 1your atlas. b {c} Find the tangent plane TPE for p = [I], \"5 , i). Definition 4.1.1 A subset S of R" is a surface if, for every point p E S, there is an open set U in R2 and an open set W in R$ containing p such that SOW is homeomorphic to U. A subset of a surface S of the form Sn W, where W is an open subset of R3, is called an open subset of S. A homeomorphism o : U - SnW as in this definition is called a surface patch or parametrization of the open subset SOW of S. A collection of such surface patches whose images cover the whole of S is called an atlas of S. Example 4.1.2 Every plane in R" is a surface with an atlas consisting of a single surface patch. In fact, let a be a point on the plane, and let p and q be two unit vectors that 4.1 What is a surface? 69 are parallel to the plane and perpendicular to each other. If v is any point of the plane, v - a is parallel to the plane, and so v - a = up + vq for some scalars u and v. Thus, the desired surface patch is o(u, v) = a + up + vq, and its inverse map is o (v) = ((v -a) .p, (v - a) - q). These formulas make it clear that o and o-1 are continuous, and hence that is a homeomorphism. (We shall not verify this in detail.) The following example shows why we have to consider surfaces, and not just surface patches. Example 4.1.3 A circular cylinder is the set of points of R' that are at a fixed distance (the radius of the cylinder) from a fixed straight line (its aris). For example, the circular cylinder of radius 1 and axis the z-axis, which we shall call the unit cylinder, is S = {(r,y, 2) ( R3 [a ty' = 1}. The simplest parametrization of S is o(u, v) = (cos u, sin u, v). Clearly, o(u, v) E S for all (u, v) E R3, and every point of S is of this form. Moreover, o is continuous. However, o is not injective, and so is not a home- omorphism, because o (u, v) = o(u + 2x, v) for all (u, v). To get an injective map we can restrict u to lie in an interval of length