# Riemannian geometry a beginners guide - Morgan F.

ISBN 0-86720-242-4

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Let Gi denote the Gauss curvature of the projection 5, of S into

Ri x R2 x {0} x • • • x R,- x • • • x {0} = R?.

Then the Gauss curvature G of S at p is

n

G = 2 Gi

/=3

simply because the dot product of two vectors is just the sum of the products of the components.

4.1. Theorem. Let S be a C2 surface in R". The first variation of the area of S with respect to a compactly supported C2 vectorfield V on S is given by integrating V against the mean curvature vector:

61(5) =

5

Proof. Since the formula is linear in V, we may consider variations in the Xi, x2,. . . directions separately. For the xlf x2 directions, which correspond to sliding the surface along itself, 51(5) = 0, as the formula says. Let V be a small variation in the x3 direction, and consider an infinitesimal square area dx 1 dx2 at p, where we may assume that the x3 component of II is diagonal:

Ki 0

-0 k2_

To first order, it is displaced to an infinitesimal area

(1 + I'V|kj) dx 1 (1 + |'V| k2) dx2 ~ (1 - V • H)dxi dx2.

The formula follows.

The concepts of local coordinates and the first fundamental form

SURFACES IN R" 27

extend without change from R2 to R". Likewise Proposition 3.5 generalizes as follows.

4.2. Proposition. For any local coordinates U\, u2 about a point p in a C2 surface S in R”, the second fundamental tensor II at p is similar to

Remark. TPS± and hence P change from point to point. If TPS± is just the jc3, . . . ,jc„-plane, then

P(fll, 02, 03, 04, • • •) = (0? 0? 03,04, • • •)•

If Xi, x2 give an orthogonal basis for TPS, then

If x1? x2 are not orthogonal, compute P by replacing x2 by

where P denotes projection onto TPS± and

d2\

(1)

p X2Xn - 2(xi • X2)Xi2 + x2x22

x?xi - (Xi • X2)2

G = det (g~1P(D2x))

(2)

(Pxu) • (Px22) - (Px12)2 X?xi - (xx • x2)2

(3)

Xi • Xi x2 x2

X2-Xi

x2 -

Xi • X!

Example. Consider the surface

{(w, z) G C2: w = ez}.

28 CHAPTER 4

We will use x = Re z and y = Imz as coordinates. Then x = (x, y, e? cos y, ex sin y)

Xi = (1,0, ex cos y, ex sin y) x2 = (0,1, -e*sin y, e*cosy)

Xu = (0,0, e? cos y, ex sin y)

X12 = (0, 0, —ex siny, ex cosy)

X22 = (0,0, -ex cos y, —e* sin y)

Note that x2 = x| = 1 + e2*, Xi • x2 = 0. Hence,

(1 + e2*)2

This is a minimal surface. (As a matter of fact, every complex analytic variety is a minimal surface.)

To compute G, first compute P(xl7). Since Xi, x2 are orthogonal,

n/ \ xn • xt Xu • x2

P(xn) = Xu---------------X!-----------X2

X! -X! X2 X2

(-e2*, 0, ex cos y, e? sin y)

~ 1 + e2* '

Similarly,

_ (0, — e2*, —e? sin y, ex cos y) 1 + e2

_ (e , 0, —ex cos y, —ex sin y) 1 + e2

P(X22)= . .

Hence,

= (~e4x - e^) - (e4x + e2*) ^ le2*

(1 + e2")2[(l + e^f - 0] (1 + e2*)3'

4.3. Gauss’s Theorema Egregium. Finally, Gauss’s Theorema Egregium, with the same proof as in Section 3.6, says that G is intrinsic, given in local coordinates uu u2 in which G = / to first order by the formula

SURFACES IN R" 29

EXERCISES

1. Compute the mean curvature vector and the Gauss curvature at each of the following:

a. At the origin for the graph

(z, w) = f(x, y) = (x2 + 2y2, 66x2 — 24xy + 59y2).

[Then compare with Exercise 3.1(b, c).]

b. At a general point of {(z, w>) G C2: w = z2}.

2. Show that for the graph of a complex analytic function /,

{w=/(z)}CC2,

H = 0,

and

G=-2|/"(z)|2(1 + |/'(z)|2)-3.

In particular, the graph of a complex analytic function is a minimal surface. (Compare with the example after Proposition

4.2.)

3. Minimal surface equation. Show that the graph of a function /: R2^R"-2 is a minimal surface if and only if

(1 + l/,|2)/» - 2(fx ■ /„)/,, + (1 + \fA2)fyy = #.

[Compare with formula 3.5(3) for the special case n = 3.]

5

This chapter extends the theory to C2 w-dimensional surfaces S in R". As before, choose orthonormal^coordinates on R" with the origin at p and S tangent to the jci,jc2,...,jcm-plane at p. Locally S is the graph of a function

A unit vector v tangent to S at p, together with the vectors normal to 5 at p, spans a plane, which intersects S in a curve. The curvature vector k of the curve, which we call the curvature in the direction v, is just the second derivative

k = (D2f)p(v, v).

The bilinear form (D2f)p on TPS with values in is called the second fundamental tensor II of S at p, given in coordinates as a symmetric m x m matrix with entries in TPS±:

11 =

tl

dxf

d2f

d2f

_ dXx dXr

d*i dXr

d2f

dx.

32 CHAPTER 5

The trace of II is called the mean curvature vector H. [Some treatments define H as (trace II)/«.]

Hypersurfaces. For hypersurfaces (n ~ m + 1), II is just the unit normal n times a scalar matrix, called the second fundamental form and also denoted by II. H = Hn, where H is the {scalar) mean curvature. If we choose coordinates to make the second fundamental form diagonal,

11 =

Kl

0

0

then //=*! + ••• + Km. If the unit normal n = («i,. .., nn) is extended locally as a unit vectorfield, then dnn/dxn = 0, while for 1 < i < n - 1, dnildXi = —Kt [compare to equation 2.0(2)]. Hence

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