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Chapter 1

The principal curvatures.

1.1 Volume of a thickened hypersurface

We want to consider the following problem: Let YRn be an oriented hyper- surface, so there is a well defined unit normal vector, lν(y) , at each point of Y. Let Yh denote the set of all points of the form

y+tlν(y) , 0th.

We wish to compute Vn(Yh) where Vn denotes the n-dimensional volume. We will do this computation for small h, see the discussion after the examples.

Examples in three dimensional space.

1. Suppose that Y is a bounded region in a plane, of area A. Clearly


in this case.

2. Suppose that Y is a right circular cylinder of radius r and height ' with outwardly pointing normal. Then Yh is the region between the right circular cylinders of height ' and radii r and r+h so

V3(Yh) = π[(r+h)2-r2]

= 2πrh+πh2

= hA+h212rA

= A(h+12kh2) ,

where A=2πr is the area of the cylinder and where k=1/r is the curvature of the generating circle of the cylinder. For small h, this formula is correct, in fact,




whether we choose the normal vector to point out of the cylinder or into the cylinder. Of course, in the inward pointing case, the curvature has the opposite sign, k=-1/r.

For inward pointing normals, the formula breaks down when h>r, since we get multiple coverage of points in space by points of the form y+tlν(y) .

3. Y is a sphere of radius R with outward normal, so Yh is a spherical shell, and

V3(Yh) = 43π[(R+h)3-R3]

= h4πR2+h24πR+h343π

= hA+h21RA+h313R2A

= 13A[3h+31Rh2+1R2h3],

where A=4πR2 is the area of the sphere.

Once again, for inward pointing normals we must change the sign of the coefficient of h2 and the formula thus obtained is only correct for h1R.

So in general, we wish to make the assumption that h is such that the map

Y×[O, h]Rn, (y, t)y+tlν(y)

is injective. For Y compact, there always exists an h0>0 such that this condition holds for all h<h0. This can be seen to be a consequence of the implicit function theorem. But so not to interrupt the discussion, we will take the injectivity of the map as an hypothesis, for the moment.

In a moment we will define the notion of the various averaged curvatures, H1, ..., Hn-1, of a hypersurface, and find for the case of the sphere with outward pointing normal, that

H1=1R, H2=1R2,

while for the case of the cylinder with outward pointing normal that

H1=12r, H2=0,

and for the case of the planar region that


We can thus write all three of the above the above formulas as




1.2 The Gauss map and the Weingarten map.

In order to state the general formula, we make the following definitions: Let Y be an (immersed) oriented hypersurface. At each xY there is a unique (positive) unit normal vector, and hence a well defined Gauss map


assigning to each point xY its unit normal vector, lν(x) . Here Sn-1 denotes the unit sphere, the set of all unit vectors in Rn.

The normal vector, lν(x) is orthogonal to the tangent space to Y at x. We will denote this tangent space by TYx. For our present purposes, we can regard TYx as a subspace of Rn: If tγ(t) is a differentiable curve lying on the hypersurface Y, (this means that γ(t)Y for all t) and if γ(0)=x, then γ(0) belongs to the tangent space TYx. Conversely, given any vector vTYx, we can always find a differentiable curve γ with γ(0)=x, γ(0)=v. So a good way to think of a tangent vector to Y at x is as an ""infinitesimal curve“ on Y passing through x.


1. Suppose that Y is a portion of an (n-1) dimensional linear or affine sub- space space sitting in Rn. For example suppose that Y=Rn-1 consisting of those points in Rn whose last coordinate vanishes. Then the tangent space to Y at every point isjust this same subspace, and hence the normal vector is a constant. The Gauss map is thus a constant, mapping all of Y onto a single point in Sn-1.

2. Suppose that Y is the sphere of radius R (say centered at the origin). The Gauss map carries every point of Y into the corresponding (parallel) point of Sn-1. In other words, it is multiplication by 1/R:


3. Suppose that Y is a right circular cylinder in R3 whose base is the circle of radius r in the x1, x2 plane. Then the Gauss map sends Y onto the equator of the unit sphere, S2, sending a point x into (1/r)π(x) where π : R3R2 is projection onto the x1, x2 plane.

Another good way to think of the tangent space is in terms of a local

parameterization which means that we are given a map X:MRn where M is some open subset of Rn-1 and such that X(M) is some neighborhood of x in Y. Let y1, ..., yn-1 be the standard coordinates on Rn-1. Part of the requirement that goes into the definition of parameterization is that the map X be regular, in the sense that its Jacobian matrix

dX:=(Xy1, ... Xyn-1)



whose columns are the partial derivatives of the map X has rank n-1 every- where. The matrix dX has n rows and n-1 columns. The regularity condition amounts to the assertion that for each zM the vectors,

Xy1(z) , ... Xyn-1(z)

span a subspace of dimension n-1. If x=X(y) then the tangent space TYx is precisely the space spanned by

Xy1(y) , ... Xyn-1(y) .

Suppose that F is a differentiable map from Y to R. We can then define its differential, dFx : TYxRm. It is a linear map assigning to each vTYx a value dFx(v)Rm: In terms of the ""infinitesimal curve“ description, if v=γ(0) then

dFx(v)=dFγdt(0) .

(You must check that this does not depend on the choice of representing curve, γ.)

Alternatively, to give a linear map, it is enough to give its value at the elements of a basis. In terms of the basis coming from a parameterization, we have

dFx(Xyi(y))=FXyi(y) .

Here FX: MRm is the composition of the map F with the map X. You must check that the map dFx so determined does not depend on the choice of parameterization. Both of these verifications proceed by the chain rule.

One immediate consequence of either characterization is the following im- portant property. Suppose that F takes values in a submanifold Z R. Then


Let us apply all this to the Gauss map, lν, which maps Y to the unit sphere, Sn-1. Then


But the tangent space to the unit sphere at lν(x) consists of all vectors perpendicular to lν(x) and so can be identified with TYx. We define the Wein- garten map to be the differential of the Gauss map, regarded as a map from TYx to itself:

Wx:=dlνx, Wx : TYxTYx.

The second fundamental form is defined to be the bilinear form on TYx given by

IIx(v, w):=(Wxv, w) .



In the next section we will show, using local coordinates, that this form is symmetric, i.e. that

(Wxu, v)=(u, Wxv) .

This implies, from linear algebra, that Wx is diagonizable with real eigenvalues. These eigenvalues, k1=k1(x), ..., kn-1=kn-1(x) , of the Weingarten map are

called the principal curvatures of Y at the point x.


1. For a portion of (n-1) space sitting in Rn the Gauss map is constant so its differential is zero. Hence the Weingarten map and thus all the principal curvatures are zero.

2. For the sphere of radius R the Gauss map consists of multiplication by 1/R which is alinear transformation. The differential of alinear transformation is that same transformation (regarded as acting on the tangent spaces). Hence the Weingarten map is 1/R×id and so all the principal curvatures are equal and are equal to 1/R.

3. For the cylinder, again the Gauss map is linear, and so the principal curvatures are 0 and 1/r.

We let Hj denote the jth normalized elementary symmetric functions of the principal curvatures. So

H0 = 1

H1 = 1n-1(k1+...+kn-1)

Hn-1 = k1k2...kn-1

and, in general,

Hj=(n-1j)1i1<...<ijn-1ki1...kij. (1.1)

H1 is called the mean curvature and Hn-1 is called the Gaussian curvature.

All the principal curvatures are functions of the point xY. For notational simplicity, we will frequently suppress the dependence on x. Then the formula for the volume of the thickened hypersurface (we will call this the ""volume formula“ for short) is:

Vn(Yh)=1ni=1X(ni)hiYHi-1dn-1A (1.2)

where dn-1A denotes the ( n-1 dimensional) volume (area) measure on Y.

A immediate check shows that this gives the answers that we got above for the the plane, the cylinder, and the sphere.



1.3 Proof of the volume formula.

We recall that the Gauss map, lν assigns to each point xY its unit normal vector, and so is a map from Y to the unit sphere, Sn-1. The Weingarten map, Wx, is the differential of the Gauss map, Wx=dlνx, regarded as a map of the tangent space, TYx to itself. We now describe these maps in terms of a local parameterization of Y. So let X : MRn be a parameterization of class C2 of a neighborhood of Y near x, where M is an open subset of Rn-1. So x=X(y), yM, say. Let


so that N: MSn-1 is a map of class C1. The map


gives a frame of TYx. The word ""frame“ means an isomorphism of our ""stan- dard“ (n1)-dimensional space, Rn-1 with our given (n1)-dimensional space, TYx. Here we have identified T(Rn-1)y with Rn-1, so the frame dXy gives us a particular isomorphism of Rn-1 with TYx.

Giving a frame of a vector space is the same as giving a basis of that vector space. We will use these two different ways of using the word"" frame“ inter- changeably. Let e1, ..., en-1 denote the standard basis of Rn-1, and for X and N, let the subscript i denote the partial derivative with respect to the ith Cartesian coordinate. Thus


for example, and so X1(y), ..., Xn-1(y) ""is“ the frame determined by dXy (when we regard TYx as a subspace of R). For the sake of notational sim- plicity we will drop the argument y. Thus we have

dX(ei) = Xi,

dN(ei) = Ni,

and so

WxXi = Ni.

Recall the definition, IIx(v, w)=(Wxv, w) , of the second fundamental form. Let (Lij) denote the matrix of the second fundamental form with respect to the basis X1, ...Xn-1 of TYx. So

Lij = IIx(Xi, Xj)

= (WxXi, Xj)

= (Ni, Xj)


Lij=-(N, 2Xyiyj) , (1.3)



the last equality coming from differentiating the identity

(N, Xj)0

in the ith direction. In particular, it follows from (1.3) and the equality of cross derivatives that

(WxXi, Xj)=(Xi, WxXj)

and hence, by linearity that

(Wxu, v)=(u, Wxv)u, vTYx.

We have proved that the second fundamental form is symmetric, and hence the Weingarten map is diagonizable with real eigenvalues.

Recall that the principal curvatures are, by definition, the eigenvalues of the Weingarten map. We will let


denote the matrix of the Weingarten map with respect to the basis X1, ..., Xn-1. Explicitly,


If we write N1, ..., Nn-1, X1, ..., Xn-1 as column vectors of length n, we can write the preceding equation as the matrix equation

(N1, ..., Nn-1)=(X1, ..., Xn-1)W. (1.4)

The matrix multiplication on the right is that of an n×(n-1) matrix with an (n-1)×(n-1) matrix. To understand this abbreviated notation, let us write it out in the case n=3, so that X1, X2, N1, N2 are vectors in R3:

X1=(X11X12X13), X2=(X21X22X23), N1=(N11N12N13), N2=(N21N22N23)

Then (1.4) is the matrix equation


Matrix multiplication shows that this gives

N1=W11X1+W21X2, N2=W12X1+W22X2,

and more generally that (1.4) gives Ni=jWjiXj in all dimensions.

Now consider the region Yh, the thickened hypersurface, introduced in the preceding section except that we replace the full hypersurface Y by the portion X(M) . Thus the region in space that we are considering is

{X(y)+λN(y), yM, 0<λh}.



It is the image of the region M×(0, h] Rn-1×R under the map

(y, λ)X(y)+λN(y) .

We are assuming that this map is injective. By (1.4), it has Jacobian matrix (differential)

J=(X1+λN1, ..., Xn-1+λNn-1, N)=

(X1, ..., Xn-1, N)((In-1+λW)001). (1.5)

The right hand side of (1.5) is now the product of two n by n matrices.

The change of variables formula in several variables says that

Vn(h)=M0h|detJ|dhdy1...dyn-1. (1.6)

Let us take the determinant of the right hand side of (1.5). The determinant of the matrix (X1, ..., Xn-1, N) is just the (oriented) n dimensional volume of the parallelepiped spanned by X1, ..., Xn-1, N. Since N is of unit length and is perpendicular to the Xs, this is the same as the (oriented) n-1 dimensional volume of the parallelepiped spanned by X1, ..., Xn-1. Thus, ""by definition“,

|det(X1, ..., Xn-1, N)|dy1...dyn-1=dn-1 A. (1.7)

(We will come back shortly to discuss why this is the right definition.) The second factor on the right hand side of (1.5) contributes

det(1+λW)=(1+λk1)...(1+λkn-1) .

For sufficiently small λ, this expression is positive, so we need not worry about the absolute value sign if h small enough. Integrating with respect to λ from 0 to h gives (1.2).

We proved (1.2) if we define dn-1A to be given by (1.7). But then it follows from (1.2) that

ddhVn(Yh)|h=0=Ydn-1A. (1.8)

A moment's thought shows that the left hand side of (1.8) is exactly what we want to mean by ""area“: it is the ""volume of an infinitesimally thickened region“. This justifies taking (1.7) as a definition. Furthermore, although the definition (1.7) is only valid in a coordinate neighborhood, and seems to depend on the choice of local coordinates, equation (1.8) shows that it is independent of the local description by coordinates, and hence is a well defined object on Y. The functions Hj have been defined independent of any choice of local coordinates. Hence (1.2) works globally: To compute the right hand side of (1.2) we may have to break Y up into patches, and do the integration in each patch, summing the pieces. But we know in advance that the final answer is independent of how we break Y up or which local coordinates we use.



1.4 Gauss's theorema egregium.

Suppose we consider the two sided region about the surface, that is


corresponding to the two different choices of normals. When we replace lν(x) by -lν(x) at each point, the Gauss map lν is replaced by -lν, and hence the Wein- garten maps Wx are also replaced by their negatives. The principal curvatures change sign. Hence, in the above sum the coefficients of the even powers of h cancel, since they are given in terms of products of the principal curvatures with an odd number of factors. For n=3 we are left with a sum of two terms, the coefficient of h which is the area, and the coefficient of h3 which is the integral of the Gaussian curvature. It was the remarkable discovery of Gauss that this curvature depends only on the intrinsic geometry of the surface, and not on how the surface is embedded into three space. Thus, for both the cylinder and the plane the cubic terms vanish, because (locally) the cylinder is isometric to the plane. We can wrap the plane around the cylinder without stretching or tearing.

It was this fundamental observation of Gauss that led Riemann to investigate the intrinsic metric geometry of higher dimensional space, eventually leading to Einstein's general relativity which derives the gravitational force from the curvature of space time. A first objective will be to understand this major theorem of Gauss.

An important generalization of Gauss's result was proved by Hermann Weyl in 1939. He showed: if Y is any k dimensional submanifold of n dimensional space (so for k=1, n=3Y is a curve in three space) , let Y(h) denote the ""tube“ around Y of radius h, the set of all points at distance h from Y. Then, for small h, Vn(Y(h)) is a polynomial in h whose coefficients are integrals over Y of intrinsic expressions, depending only on the notion of distance within Y. Let us multiply both sides of (1.4) on the left by the matrix (X1, ..., Xn-1)T to obtain


where Lij=(Xi, Nj) as before, and

Q=(Qij):=(Xi, Xj)

is called the matrix of the first fundamental form relative to our choice of local coordinates. All three matrices in this equality are of size (n-1)×(n-1) . If we take the determinant of the equation L=QW we obtain

detW=detLdetQ, (1.9)

an expression for the determinant of the Weingarten map (a geometrical prop- erty of the embedded surface) as the quotient of two local expressions. For the case n-1=2, we thus obtain a local expression for the Gaussian curvature, K=detW.



The first fundamental form encodes the intrinsic geometry of the hypersur- face in terms of local coordinates: it gives the Euclidean geometry of the tangent space in terms of the basis X1, ..., Xn-1. If we describe a curve tγ(t) on the surface in terms of the coordinates y1, ..., yn-1 by giving the functions tyj(t), J=1, ..., n-1 then the chain rule says that



y(t)=(y1(t), ..., yn-1(t)) .

Therefore the (Euclidean) square length of the tangent vector γ(t) is

γ(t)2=i,j=1x-1Qij(y(t))dyidt(t)dyjdt(t) .

Thus the length of the curve γ given by


can be computed in terms of y(t) as

i,j--1n-1Qij(y(t))dyidt(t)dyjdt(t) dt

(so long as the curve lies within the coordinate system).

So two hypersurfaces have the same local intrinsic geometry if they have the same Q in any local coordinate system.

In order to conform with a (somewhat variable) classical literature, we shall make some slight changes in our notation for the case of surfaces in three di- mensional space. We will denote our local coordinates by u, v instead of y1, y2 and so Xu will replace X1 and Xv will replace X2, and we will denote the scalar product of two vectors in three dimensional space by a. instead of (, ) . We write

Q =


(EFFG) (1.10)

E := F := G :=

XuXu (1.11) XuXv (1.12) XvXv (1.13)


detQ =

detQ = EG F .

EG-F2. (1.14)