1 Curvature of submanifolds of Euclidean space

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1 Curvature of submanifolds of Euclidean space by Min Ru, University of Houston 1 Curvature of submanifolds of Euclidean space Submanifold in R N : A C k submanifold M of dimension n in R N means that for every point p M there is a neigborhood U of p and a neightborhood Ω of 0 in R n and a map X C k (Ω, R N ), which is bijection X : Ω M U R N with injective differential. The map is called a local parametrization. Note that there are many local parametrizations, in fact, if φ : Ω Ω is a diffeomorphism then X φ is another. Tangent space: The tangent space T p (M) consists of all tangent vectors to the cuvres in M passing through p. Meric: The metric on M is induced by the standard metric on R N. Let X : Ω M R N be a local parameterization of M. Let (u 1, u,, u n ) be the coordinates on Ω R. Denote by X j = X u j, 1 j n. Then {X j, 1 j n} forms a basis for the tangent space T p (M). Let g ij = X i X j. Then the matrix g = [g ij ] is called the first fundemental form or metric. It is an intrinsic quantity in that it relates to measurements inside the surface. The arc-length s on along the curve α(t) = X(u(t)) is given by ( ) ds = α (t) du i = g ij duj. 1

2 The second fundamental form: Consider a curve α(t) = X(u(t)) in M, let us look at the second derivative of the curve: α (t) = d u j X j(u(t)) + dui du j X jk(u(t)). The first sum is in the tangent plane and if we project it onto the normal space N p (M) (i.e. all vectors orthogonal to T p (M)), we can eliminate this sum. Let h ij be the orthogonal projection of X ij in the normal space N p (M). The sum h ij du i can be thought of as quadratic form on T p (M) with values in N p (M) independent of the particular choice of parameterization of X. It is called the second fundamental form. Clearly in the case of hypersurface n = N 1, it is a function times the normal. It measures how the hypersurface is curving in the surrounding space(recall for surfaces in R 3, h ij = X ij n.) It is the second order of derivatives of the hypersurface measured in a way which is independent of the parametrization, similar to how the curvature of a curve was defined so it did not depend on a particular parametrization. du j Theorem For every C curve on M with the unit tangent vector α (s) = X i du i p = X(u(t)) the sum h ij du i is the curvature of the curve times the orthogonal projection of the principal normal of the curve in N p (M). Christofell symbols, covariant derivative(or connection): Again, we can write du j ds at X ik = l Γ l ikx i + h ik

3 where h ik is the orthogonal projection of X ik in the normal space N p (M). By a miracle one can compute the right hand side by means of the coeffcients of the first fundamental form g ij = X i X j and its derivatives, for we have g ij,k = X ik X j + X i X jk, g jk,i = X ji X k + X j X ki, g ki,j = X kj X i + X k X ij. If we add the first two equations and substract the third, we get Γ ikj := X ik X j = 1 (g ij,k + g jk,i g ki,j ). Note the symmetry in indices i, k, and that g ij,k = Γ ikj + Γ jki. If (g ij ) denotes the inverse of (g ij ) we obtain Γ l ik = g lj Γ ikj = 1 g lj (g ij,k + g jk,i g ki,j ). j The functions Γ l ik are called the Christofell symbols. Note, the term l Γ l ikx i represents the orthonormal projection of X ik to the tangent space of M, which will be called the covariant derivative of X i with respect to the tangent vector X k, we denote it by Xk X i := l Γ l ikx i The operator is called the connection. Geodesic curvature and Geodesic principal normal direction Consider a curve α(t) = X(u(t)) in M, we have shown α (t) = d u j X j(u(t)) + dui du j X jk(u(t)), also α (t) = ( d u j + du i Γj ik du k ) X i (u(t)) + dui du k h ik(u(t)). 3

4 Theorem Suppose that curve α(s) = X(u(s)) is parameterized by arc-length. X (t) = du g i ij duj = 1. Then the curvature times the orthgonal projection of the principal normal in T p (M) is ( d u j + du i Γj ik du k ) X i (u(t)). Thus this vector does not depend on the choice of reparametrization f. It length is called the geodesic curvature and the direction is called the geodesic principal normal direction. Gauss equations and Codazzi-Mainardi equations: Let us now determine the tangential components of derivative of a normal vector field n(u) := n(x(u)): Projection of n i to the tangent space is n k i X k. Since differentiation of the equation n X j = 0 gives that is Multiplying by the inverse g kj gives n i X j + n X ij = 0, n k i g kj = n h ij. n k i = (n h ij )g kj. Actually, if in addition its a hypersurface and n is the unit-normal, then n i n = 0 so that it is already tangetial. It is now easy to calculate the tangetial compotents of X ijk. Since we obtain X ij = Γ l ijx l + h ij, X ijk = ( Γ m ij,k + Γ l ijγ m lk (h ij h kl )g lm) X m mod N p (M). 4

5 Since X kij = X jik, a calculation gives the following: Gauss equations: The first and second fundamental form are related by Rijk m := Γ m ik,j Γ m ij,k + Γ l ikγ m lj Γ l ijγ m lk = (h ik h jl h ij h kl )g lm. Here the first equality is a defintion. If we introduce R lijk = g lm Rijk. m This can be written as R lijk = h ik h jl h ij h kl. Since the second fundamental form is defined invariantly(independent of a particular parametrization) on the tangent space so is the Riemann curvature defined by R(T 1, T, T 3, T 4 ) = R ijkl T i 1T j T k 3 T l 4 = H(T 1, T 3 ) H(T, T 4 ) H(T 1, T 4 ) H(T, T 3 ), where H(T 1, T ) = h ij T i 1T j. We define the Ricci curvature and the scalar curvature by R jl = g ik R ijkl, R = g jl R jl. respectively. These are also invariantly defined. The hypersurface case Suppose now that M is a hypersurface in R N, i.e. N = n + 1 where n is the dimension of M. Then we can write h ij = h ij n where n is a unit normal and h ij is scalar vlaued. Let, for any two vector fields T, S, g(t, S) = g ij T i S j, h(t, S) = h ij T i S j. The quotient h(t, T ) g(t, T ) 5

6 is the curvature of a curve on M with tangent vector T and principal normal in the direction of the normal to the surface, i.e. with vanishing geodesic curvature. We can diagonzlize the first and second fundamental forms simultaneously; h(e k, e m ) = K k δ km, g(e k, e m ) = δ km. In factm since G is a symmetric matrix we can diagonalize it and since it is positive definite, we can take the square root and obtain: G = A T A. Next we find the eigenvalues of H with respect to G : He k = K k e k, which is symmetric so we can diagonalize it with an orthonormal basis X k = Ae k. Then g(e k, e m ) = (A T Ae k ) e m = (Ae k ) (Ae m ) = δ km. The quotient of the second fundamental form divided by the first for a curve with vanishing geodesic curvature hence is maximized by the largest eigenvalue K n and minimized by the smallest. We therefore call K i the principal curvatures and e i the principal curvature directions. It follows from the previous discussion that if n is the unit normal to a hypersurface then n i = h ij g jk X k. Thus (T i n i ) (S k X k ) = h ij T i S j. If we define the Gauss map by p M maps to n(p) S n, then the differential dn : T p (M) T p (M) and the above can be formulated (dn(t )) S = h(t, S). Hence the differential of the Gauss map is the linear transformation that corresponds to the second fundamental form. This is called the shape operator. 6

7 Denote by (S, T ) = g(s, T ) and write n n S = (S, e j )e j, T = (S, e i )e i. j=1 i=1 We get k h(t, S) = K j (S, e j )(T, e j ). j=1 Since h = hn it follows form Gauss equation that R(T 1, T, T 3, T 4 ) = h(t 1, T 3 ) h(t, T 4 ) h(t 1, T 4 ) h(t, T 3 ) = K i K j ((T 1, e i )(T 3, e i )(T, e j )(T 4, e j ) (T 1, e i )(T 4, e i )(T, e j )(T 3, e j )) i j = 1 K i K j ((T 1, e i )(T 3, e i )(T, e j )(T 4, e j ) (T 1, e i )(T 4, e i )(T, e j )(T 3, e j )) i j + 1 K i K j ((T 1, e j )(T 3, e j )(T, e i )(T 4, e i ) (T 1, e j )(T 4, e j )(T, e i )(T 3, e i )) i j Hence R(T 1, T, T 3, T 4 ) = 1 K i K j i j (T 1, e i ) (T 1, e j ) (T, e i ) (T, e j ) (T 3, e i ) (T 3, e j ) (T 4, e i ) (T 4, e j ). If n = we conclude that R(T 1, T, T 3, T 4 ) is K 1 K times the area of the parallelograme spanned by T 1, T times the the area of the parallelograme spanned by T 3, T 4. Hence if T, S are any two orthonormal unit vectors then R(S, T, S < T ) = K 1 K. Since the curvature is determined from the first fundamental form we have proved: Theorema egregium of Gauss The total curvature K 1 K is determined by the first fundamental form. Recall that the curvature tensor R ijkl = h ij h jl h il h jk, h ij = h ji, 7

8 is antisymmetric in the pairs (i, j), (k, l) but symmetric under the exhange of the pairs R ijkl = R jikl = R ijlk = R klij. By expanding in a basis one sees that R(S, T, S, T )(g(s, S)g(T, T ) g(s, T ) ) 1 only depends on the plane spanned by S, T, and is therefore called the sectional curvautre of the two plane. It turns out that the curvature tensor is determined by the sectional curvature. 8