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The Second Fundamental Form. Geodesics. the Curvature Tensor

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The Second Fundamental Form. Geodesics. the Curvature Tensor Differential Geometry Lia Vas The Second Fundamental Form. Geodesics. The Curvature Tensor. The Fundamental Theorem of Surfaces. Manifolds The Second Fundamental Form and the Christoffel symbols. Consider a surface x = @x @x x(u; v): Following the reasoning that x1 and x2 denote the derivatives @u and @v respectively, we denote the second derivatives @2x @2x @2x @2x @u2 by x11; @v@u by x12; @u@v by x21; and @v2 by x22: The terms xij; i; j = 1; 2 can be represented as a linear combination of tangential and normal component. Each of the vectors xij can be repre- sented as a combination of the tangent component (which itself is a combination of vectors x1 and x2) and the normal component (which is a multi- 1 2 ple of the unit normal vector n). Let Γij and Γij denote the coefficients of the tangent component and Lij denote the coefficient with n of vector xij: Thus, 1 2 P k xij = Γijx1 + Γijx2 + Lijn = k Γijxk + Lijn: The formula above is called the Gauss formula. k The coefficients Γij where i; j; k = 1; 2 are called Christoffel symbols and the coefficients Lij; i; j = 1; 2 are called the coefficients of the second fundamental form. Einstein notation and tensors. The term \Einstein notation" refers to the certain summation convention that appears often in differential geometry and its many applications. Consider a formula can be written in terms of a sum over an index that appears in subscript of one and superscript of P k the other variable. For example, xij = k Γijxk + Lijn: In cases like this the summation symbol is omitted. Thus, the Gauss formula for xij in Einstein notation is written simply as k xij = Γijxk + Lijn: An important benefit of the use of Einstein notation can be seen when considering n-dimensional manifolds { all the formulas we consider for surfaces generalize to formulas for n-dimensional man- k ifolds. For example, the formula xij = Γijxk + Lijn remains true except that the indices i; j take integer values ranging from 1 to n not just values 1 and 2. If we consider the scalar components in certain formulas as arrays of scalar functions, we arrive to the concept of a tensor. 1 For example, a 2 × 2 matrix with entries gij is considered to be a tensor of rank 2. This matrix is referred to as the metric tensor. The scalar k functions Γij are considered to be the components of a tensor Γ of rank 3 (or type (2,1)). The Ein- stein notation is crucial for simplification of some complicated tensors. Another good example of the use of Einstein notation is the matrix multiplication (students who did not take Linear Algebra can skip this ex- ample and the next several paragraphs that relate to matrices). If A is an n × m matrix and v is a m × 1 (column) vector, the product Av will be a n × 1 column vector. If we denote the elements of i j A by aj where i = 1; : : : ; n, j = 1; : : : ; m and x denote the entries of vector x, then the entries of P i j i j the product Ax are given by the sum j ajx that can be denoted by ajx using Einstein notation. Note also that the entries of a column vector are denoted with indices in superscript and the entries of row vectors with indices in subscript. This convention agrees with the fact that the entries i j of the column vector ajx depend just on the superscript i: Another useful and frequently considered tensor is the Kronecker delta symbol. Recall that the identity matrix I is a matrix with the ij-th entry 1 if i = j and 0 otherwise. Denote these entries by i δj: Thus, 1 i = j δi = j 0 i 6= j i j i In this notation, the equation Ix = x can be written as δjx = x : Recall that the inverse of a matrix A is the matrix A−1 with the property that the products AA−1 −1 ij and A A are both equal to the identity matrix I: If aij denote the elements of the matrix A; a denote the elements of the inverse matrix Aij; the ij-th element of the product A−1A in Einstein ik ik i notation is given by a akj: Thus a akj = δj: ij In particular, let g denote the entries of the inverse matrix of [gij] whose entries are the coeffi- cients of the first fundamental form. The fact that the matrix and its inverse multiply to the identity gives us the following formulas (all given in Einstein notation). kj j ik i gikg = δi and g gkj = δj: The coefficients gij of the inverse matrix are given by the formulas g −g g g11 = 22 ; g12 = 12 ; and g22 = 11 g g g where g is the determinant of the matrix [gij]: Computing the second fundamental form and the Christoffel symbols. The formula k computing the Christoffel symbols can be obtained by multiplying the equation xij = Γijxk + Lijn by xl where k = 1; 2: Since n · xl = 0; and xk · xl = gkl; we obtain that k xij · xl = Γijgkl 2 k To solve for Γij; we have to get rid of the terms gkl from the left side. This can be done by using the inverse matrix gls: ls k ls k s s (xij · xl)g = Γijgklg = Γijδk = Γij Thus, we obtain that the Christoffel symbols can be computed by the formula k lk Γij = (xij · xl)g : k To compute the coefficients of the second fundamental form, multiply the equation xij = Γijxk + Lijn by n: Since xl · n = 0; we have that xij · n = Lijn · n = Lij: Thus, x1×x2 Lij = xij · n = xij · : jx1×x2j While the first fundamental form determines the intrinsic geometry of the surface, the second fun- damental form reflects the way how the surface embeds in the surrounding space and how it curves relative to that space. Thus, the second fundamental form reflects the extrinsic geometry of the surface. The presence of the normal vector n in the formula for Lij reflects this fact also since n \sticks out" of the surface. In contrast, we shall see that the Christoffel symbols can be computed using the first fundamental form only which shows that they are completely intrinsic. Example 1. The Christoffel symbols and the second fundamental form of a plane. Consider the xy-plane z = 0 for simplicity. Thus x(x; y) = (x; y; 0): Recall (or compute again) that x1 = xx = (1; 0; 0); x2 = xy = (0; 1; 0); g11 = g22 = 1; and g12 = 0: Since the first derivatives x1 and x2 are constant, the second derivatives x11; x12 = x21; x22 are all zero. As a result, the second fundamental form Lij are zero. Lij = xij · n = (0; 0; 0) · n = 0 0 0 Represented as a matrix, the second fundamental form is : This means that the surface does 0 0 not deviate from its tangent plane at all and this should not be surprising since the tangent plane of a plane is that plane itself. The Christoffel symbols are also zero. k lk lk Γij = (xij · xl)g = ((0; 0; 0) · xl)g = 0: We shall see later that this means that surfaces with this property are such that their shortest-distance curves (geodesics) are really lines. Example 2. The Christoffel symbols and the second fundamental form of a cylinder. Consider the cylinder (a cos t; a sin t; z) with a circular base of radius a: Recall (or compute again) 2 2 that x1 = (−a sin t; a cos t; 0); x2 = (0; 0; 1) and so g11 = a ; g12 = 0 and g22 = 1: Thus, g = a and so 2 11 g22 1 12 −g12 22 g11 a g = g = a2 ; g = g = 0 and g = g = a2 = 1 so the matrix inverse to the first fundamental 1 0 form is a2 : The unit normal vector is n = (cos t; sin t; 0): 0 1 3 The second derivatives are x11 = (−a cos t; −a sin t; 0); x12 = (0; 0; 0); and x22 = (0; 0; 0): Thus, 2 2 L11 = x11 · n = −a cos t − a sin t = −a; L12 = x12 · n = 0; and L22 = x22 · n = 0: So, the second −a 0 fundamental form is : When comparing the second fundamental form of the cylinder to 0 0 that of the plane from the previous example, we can see that they differ only in the first coefficient. Note also that these two matrices have the same determinant { we shall elaborate on this fact later. k k k 1 Since x12 = x22 = (0; 0; 0); Γ12 = Γ21 = Γ22 = 0: The remaining two Christoffel symbols, Γ11 and 2 Γ11 can be computed as follows. 1 11 21 11 Γ11 = x11 · x1g + x11 · x2g = 0g + 0(0) = 0 and 2 12 22 22 Γ11 = x11 · x1g + x11 · x2g = 0(0) + 0g = 0: Hence, all Christoffel symbols are zero. Example 3. The Christoffel symbols and the second fundamental form of a sphere. Consider the sphere x = (a cos θ cos φ, a sin θ cos φ, a sin φ) of radius a parametrized by geographic coordinates. Recall (or compute again) that x1 = (−a sin θ cos φ, a cos θ cos φ, 0); x2 = (−a cos θ sin φ, 2 2 2 −a sin θ sin φ, a cos φ); g11 = a cos φ, g12 = 0; g22 = a ; and n = (cos θ cos φ, sin θ cos φ, sin φ): Compute that x11 = (−a cos θ; −a sin θ cos φ, 0); x12 = x21 = (a sin θ sin φ, −a cos θ sin φ, 0); and 2 x22 = (−a cos θ cos φ, −a sin θ cos φ, −a sin φ); so that L11 = −a cos φ, L12 = 0 and L22 = −a and −a cos2 φ 0 the second fundamental form is : 0 −a 11 1 12 22 1 Compute then that g = a2 cos2 φ ; g = 0; g = a2 ; so that the inverse of the first fundamental 1 a2 cos2 φ 0 form is 1 : Lastly, compute the dot products of the second and the first derivatives to 0 a2 2 2 be x11 · x1 = 0; x11 · x2 = a sin φ cos φ, x12 · x1 = x21 · x1 = −a sin φ cos φ, x12 · x2 = x21 · x2 = 0; x22 · x1 = 0; x22 · x2 = 0: 1 11 21 11 Thus, the Christoffel symbols are Γ11 = x11 · x1g + x11 · x2g = 0g + x11 · x2(0) = 0; 1 Γ2 = x · x g12 + x · x g22 = 0(0) + a2 sin φ cos φ = sin φ cos φ, 11 11 1 11 2 a2 1 − sin φ Γ1 = Γ1 = x · x g11 + x · x g21 = −a2 sin φ cos φ + 0(0) = 12 21 12 1 12 2 a2 cos2 φ cos φ 2 2 12 22 1 11 21 and Γ12 = Γ21 = x12 · x1g + x12 · x2g = 0 + 0 = 0; Γ22 = x22 · x1g + x22 · x2g = 0 + 0 = 0; and 2 12 22 Γ22 = x22 · x1g + x22 · x2g = 0 + 0 = 0: Thus, all but three Christoffel symbols are zero.
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