2. Metrics as Symmetric Tensors So far we have studied exterior products of 1-forms, which obey the rule called skew symmetry: α β = β α. There is another operation for forming something called the symmetric tensor product, designated here by the symbol, which obey: Rule ST1 (Symmetry). α β = β α for 1-forms α and β. Like the exterior product, the symmetric tensor product is also linear in each slot : Rule ST2 (Bilinearity). (f 1 α 1 + f 2 α 2 ) β = f 1 (α 1 β) + f 2 (α 2 β), and α (g 1 β 1 + g 2 β 2 ) = g 1 (α β 1 ) + g 2 (α β 2 ). Here f 1, f 2, g 2, g 2 are scalar functions (i.e. 0-forms) and α, β, α 1, α 2, β 1, β 2 are 1-forms. Unlike exterior products, there is no general agreement about the notation for symmetric tensor products. Some books use, or even some symbols cannot be produced here, such as a capital S surrounded by a circle. More often a symmetric tensor product of α and β is simply written as αβ, such as dudv instead of du dv and du 2 instead of du du, a custom we shall follow. Watch out that in many books dudv stand for either du dv or du dv depending on the context, which may cause some confusion. The most often used symmetric tensor product is the so-called metric tensor for the Euclidean space R n : dx 2 1 + dx 2 2 + + dx 2 n, which is traditionally denoted by ds 2 ; (strictly speaking, ds 2 is an illegitimate expression). Formally, letting x = (x 1, x 2,..., x n ), we have dx = (dx 1, dx 2,..., dx n ) and dx dx = n k=1 dx k dx k n k=1 dx2 k, which is the metric tensor ds 2 for R n. In case n = 2 or n = 3, we use the notation r = (x, y) or r = (x, y, z) instead, so that the corresponding metrics becomes ds 2 = dr dr = dx 2 + dy 2 or ds 2 = dr dr = dx 2 + dy 2 + dz 2. Example 2.1. Rewrite the metric tensor dx 2 + dy 2 in polar coordinates. Solution: Recall that dx = cos θdr r sin θdθ and dy = sin θdr + r cos θdθ. So dx 2 + dy 2 = (cos θdr r sin θdθ) 2 + (sin θdr + r cos θdθ) 2 = (cos 2 θdr 2 2 cos θ sin θdrdθ + r 2 sin 2 θdθ 2 ) + (sin 2 θdr 2 + 2 sin θ cos θdrdθ + r 2 cos 2 θdθ 2 ) = dr 2 + r 2 dθ 2. 1
The answer is: dx 2 + dy 2 = dr 2 + r 2 dθ. Strictly speaking, the way handling the symbols and putting down this answer is incorrect. To be careful, we should regard the change of coordinates here as a mapping h sending (r, θ) to (x, y) given by x = r cos θ and y = r sin θ and dr 2 + r 2 dθ as the pullback of dx 2 + dy 2 by h, that is, h (dx 2 + dy 2 ) = dr 2 + r 2 dθ. Just like exterior products, symmetric tensor products get along with pullbacks well: for example, we also have h (α β) = (h α) (h β). Now we consider a type of line integrals f ds associated with the metric tensor ds2. Here f is a function defined in the Euclidean space R n, and is an arc in R n given in parametric equation x = x(t), with a t b. To compute this line integral, again we pull back fds by to convert it into an expression of the form g(t)dt and then compute the integral b a g(t)dt. The pull-back of f ds by is (fds) = f ds = f(x(t)) ds, where ds dx 2 1 + + dx2 n is manipulated by replacing each dx k by dx k dt dt: [dx1 ] 2 ds = dt dt + + [ ] 2 dxn dt dt = (recall that v x is the speed). Thus we have fds = b a (f ds) [dx1 ] 2 + + b a dt [ ] 2 dxn dt x (t) dt, dt f(x(t)) x (t) dt. (2.1) Of course the last expression is all we need for computation. This line integral has the following physical interpretation: when f(x) is the linear density of a wire at x, then f ds is the total mass of the wire. In the special case f 1, the integral ds is the arc length of. In practice we often skip the symbol, as shown in the following example. Example 2.2. Find the length of the part of the Archimedian spiral which lies inside the unit circle, given by x = e t cos t and y = e t sin t with 0 t <. Solution: We have dx = ( e t cos t + e t ( sin t))dt, dy = ( e t sin t + e t cos t)dt, dx 2 = e 2t (cos 2 t + 2 cos t sin t + sin 2 t)dt 2, and dy 2 = e 2t (sin 2 t 2 sin t cos t + cos 2 t)dt 2. So ds 2 = dx 2 + dy 2 = 2e 2t dt 2 and ds = 2e t dt. Therefore the required length is ds = 2e t dt = 2( e t ) = 2. 0 0 Alternatively, we can use polar coordinates: r cos θ = x = e t cos t and r sin θ = y = e t sin t, or r = e t and t = θ. So dr = e t dt and dθ = dt. It follows from the previous example that ds 2 = dr 2 + r 2 dθ = ( e t dt) 2 + (e t ) 2 dt 2 = 2e 2t dt 2. The rest is the same. 2
The metric tensor ds 2 = dx 2 1 + +dx 2 n is not the only symmetric tensor that interests us. In special relativity, we consider the so-called Minkowski metric dt 2 dx 2 dy 2 dz 2. Example 2.3. By a boost in the x-direction we mean a Lorentz transformation t = t vx, x = vt + x, ỹ = y, z = z, where v (satisfying 1 < v < 1) is a constant representing the velocity of a second observer, and = 1/ 1 v 2. Show that the Minkowski metric dt 2 dx 2 dy 2 dz 2 sense that d t 2 d x 2 dỹ 2 d z 2 = dt 2 dx 2 dy 2 dz 2. is invariant under this transformation in the Solution: We have d t = dt vdx, d x = vdt + dx, dỹ = dy and d z = dz. So d t 2 d x 2 dỹ 2 d z 2 = 2 (dt 2 2vdtdx + v 2 dx 2 ) 2 (v 2 dt 2 2vdtdx + dx 2 ) dy 2 dz 2 = 2 ((1 v 2 )dt 2 + (v 2 1)dx 2 ) dy 2 dz 2 = dt 2 dx 2 dy 2 dz 2, in view of 2 (1 v 2 ) = 1. We may get the metric of a surface S by restricting the Euclidean metric dx 2 +dx 2 +dz 2 to S. Often S is given by some parametric equations, say σ: x = x(u, v), y = y(u, v) and z = z(u, v) (sometimes we use u 1 and u 2 instead of u and v for parameters), which can be considered as a mapping σ sending a point (u, v) in the plane to a point (x, y, z) in the space and the required metric on S is just the pull-back σ (dx 2 + dy 2 + dz 2 ), which is also (wrongly) denoted by ds 2. Now σ dx = dx(u, v) = x u We have similar expressions for σ dy and σ dz. So du + x v dv x udu + x v dv. ds 2 = (x u du + x v dv) 2 + (y u du + y v dv) 2 + (z u du + z v dv) 2 = (x 2 u + y 2 u + z 2 u) du 2 + 2(x u x v + y u y v + z u z v ) dudv + (x 2 v + y 2 v + z 2 v) dv 2 = r u r v du 2 + 2r u r v dudv + r v r v dv 2 = g 11 du 2 1 + g 12 du 1 du 2 + g 21 du 2 du 1 + g 22 du 2 2, where r = r(u, v) = (x(u, v), y(u, v), z(u, v)) as usual, and g ij = r ui r uj for 1 i, j 2, with u 1 = u and u 2 = v. Notice that g 12 = g 21. Often g ij is (mistakenly) also called the metric for S. The determinant of the symmetric 2 2-matrix [g ij ] is usually denoted by g: g = g 11 g 12 g 21 g 22 = g 11g 22 g12. 2 3
The Cauchy-Schwarz inequality tells us that g 0 and hence we are allowed to take the square root of g. The 2-form α = g du 1 du 2 (2.2) is called the area form for S. Another expression for the area form is α = r u r v dudv, which is often (incorrectly) denoted by ds; see Exercise 5 in 2.2. The area form ds allows us to consider surface integrals of the form fds, defined in the same way as the line σ integral fds: if the map σ is given by equation r = r(u, v), where (u, v) varies in a domain D, then f ds = σ D σ (fds) = D f(r(u, v)) r u r v dudv. Example 2.4. Consider the parametric equations x = cos θ cos φ, y = sin θ cos φ and z = sin φ for the unit sphere S 2 ; (see Example 1.4 in 4.1). Find its metric tensor and its area form in terms of the parameters θ and φ. Solution: r θ = ( sin θ cos φ, cos θ cos φ, 0), So we have cos 2 φdθ 2 + dφ 2. Also, r φ = ( cos θ sin φ, sin θ sin φ, cos φ). r θ r θ = cos 2 φ, r θ r φ = 0, and r φ r φ = 1. Hence the required metric is g = cos2 φ 0 0 1 = cos2 φ, and hence the area form for S 2 is g dθ dφ = cos φ dθ dφ. We give another parametrization of the unit sphere S 2 : x 2 + y 2 + z 2 = 1 via a mapping π N from the uv-plane to the sphere, called stereographic projection, which is used in making maps. The points N = (0, 0, 1) and S = (0, 0, 1) on the sphere are called the north pole and the south pole respectively. The uv-plane overlaps with the xy-plane and a point Q = (u, v) on it is the image of π N of a point P = (x, y, z) on the sphere if three points N = (0, 0, 1), P = (x, y, z) and Q = (u, v, 0) are collinear: the components of NP = (x, y, z 1) are proportional to the components of N Q = (u, v, 1). Thus u/x = v/y = 1/(z 1), which gives Since x 2 + y 2 + z 2 = 1, we have u 2 + v 2 + 1 = u = x/(1 z) and v = y/(1 z). (2.3) x 2 (1 z) 2 + y 2 (1 z) 2 + 1 = x2 + y 2 + 1 2z + z 2 (1 z) 2 = 2 2z (1 z) 2 = 2 1 z 4
and hence 1 z = 2(u 2 + v 2 + 1) 1. (2.4) Thus z = 1 2(u 2 + v 2 + 1) 1 = (u 2 + v 2 1)(u 2 + v 2 + 1) 1. It follows from (2.3) that x = u(1 z) = 2u(u 2 + v 2 + 1) 1 and y = v(1 z) = 2v(u 2 + v 2 + 1) 1. Thus x = 2u u 2 + v 2 + 1, y = 2v u 2 + v 2 + 1, z = u2 + v 2 1 u 2 + v 2 + 1 (2.5) are the parametric equations for the sphere S 2 obtained by spherical projection. Example 2.5. Find the metric for the sphere in terms of parameters u, v above. Solution: The direct computation is rather messy. We start with the simplest relations (2.3) and (2.4) above, which are recast as x = (1 z)u, y = (1 z)v and (1 z)(u 2 +v 2 +1) = 2. Differentiate both sides of each of them to get dx = udz+(1 z)du, dy = vdz + (1 z)dv and (1 z). 2(udu + vdv) (u 2 + v 2 + 1)dz = 0. Thus dx 2 +dy 2 + dz 2 =u 2 dz 2 2u(1 z)dudz + (1 z) 2 du 2 + v 2 dz 2 2v(1 z)dvdz + (1 z) 2 dv 2 + dz 2 =(u 2 + v 2 + 1)dz 2 2(udu + vdv)(1 z)dz + (1 z) 2 (du 2 + dv 2 ) =(1 z) 2 (du 2 + dv 2 ) = (1 + u 2 + v 2 ) 2 (du 2 + dv 2 ). Hence the required metric is ds 2 = du2 + dv 2 (1 + u 2 + v 2 ) 2. (2.6) Using symbols g ij, we have g 11 = g 22 = (1 + u 2 + v 2 ) 2 and g 12 = g 21 = 0. So g (g 11 g 22 g 12 g 21 ) 1/2 = (1 + u 2 + v 2 ) 2. Hence the area form is (1 + u 2 + v 2 ) 2 du dv. Consider the two sheets hyperboloid H 2 : x 2 y 2 + z 2 = 1. The north pole N = (0, 0, 1) is at the bottom of the upper sheet. Again, let the uv-plane overlap with the xy-plane. A point P = (x, y, z) on the lower sheet satisfies z 1. The line joining the point and the north pole intersects a point Q = (u, v) in the uv-plane. Since N, Q, P are collinear, (u, v, 1) is proportional to (x, y, z 1) and, as before, we have u = x/(1 z) and v = y/(1 z). Now we use the relation x 2 y 2 + z 2 = 1 to deduce 1 u 2 v 2 = 1 x 2 (1 z) 2 y 2 (1 z) 2 = 1 2z + z2 x 2 y 2 (1 z) 2 = 2 1 z. Proceed as before, we have the following parametrization of the lower sheet of H 2 : x = 2u 1 u 2 v 2, y = 2v 1 u 2 v 2, z = 1 u2 v 2 1 u 2 v 2. u2 + v 2 < 1). (2.7) 5
The metric for H 2 is defined to be the restriction of the indefinite metric dx 2 + dy 2 dz 2. (The choice of this metric may look unusual. It is suggested by hyperbolic geometry. The main reason for this choice is that it will produce something neat.) We can follow the same computation as that of the above example to obtain the metric and the area form for the hyperbolic plane in Poincarè s disk model D: ds 2 = du2 + dv 2 (1 u 2 v 2 ) 2, α = du dv (1 u 2 v 2 ) 2. (2.8) (The detail for deriving (2.7) and (2.8) is left to the reader as exercises.) From (2.8) we see that when (u, v) approaches to the unit circle, that is, when u 2 + v 2 tends to 1, both ds 2 and α blow up. This is not surprising because (2.7) tells us that in that situation the point (x, y, z) on the hyperboloid runs away to infinity. Example 2.6. Find the length of the arc described by u = 0 and 0 v a, where a is a positive number with a < 1, according the hyperbolic metric given by (2.8). Also find the area of the region R = {(u, v) u 2 + v 2 a 2 } according to this metric. Solution: The required length is l = ds = a 0 [ du2 + dv 2 1 u 2 v 2 ] = a 0 (1 v 2 ) 1 dv = 1 2 log 1 + a 1 a. The required area is R (1 u2 v 2 ) 2 du dv = 2π a 0 0 (1 r2 ) 2 rdrdθ = πa 2 (1 a 2 ) 1. We have to clarify the connection between the metric tensor ds 2 = 2 i,j=1 g ijdu i du j for a surface S and the 1-forms θ i associated with a Darbour frame E i (i = 1, 2, 3) for S (see (2.1) in 4.2). Suppose that E 1, E 2 are tangent to S and E 3 is normal to S. The surface is given by parametric equation r = r(u, v) (x(u, v), y(u, v), z(u, v)), with u = u 1, v = u 2. Then dr = r u du + r v dv. Since both r u and r v are tangent to S, so is dr. Thus, in identity dr = θ 1 E 1 + θ 2 E 2 + θ 3 E 3, we must have θ 3 = 0 and θ i = E i dr = E i r u du + E i r v dv for i = 1, 2. For convenience, write P i = E i r u and Q i = E i r v. Then r u = P 1 E 1 + P 2 E 2 and r v = Q 1 E 1 + Q 2 E 2 are orthogonal decomposition of r u and r v. Hence P 2 1 + P 2 2 = r u r u = g 11, Q 2 1 + Q 2 2 = r v r v = g 22, P 1 Q 1 + P 2 Q 2 = r u r v = g 12 = g 21. Also, θ 1 = P 1 du + Q 1 dv and θ 2 = P 2 du + Q 2 dv. Thus (θ 1 ) 2 + (θ 2 ) 2 = (P 1 du + Q 1 dv) 2 + (P 2 du + Q 2 dv) 2 = (P 2 1 + P 2 2 )du 2 + (Q 2 1 + Q 2 2)dv 2 + 2(P 1 Q 1 + P 2 Q 2 )dudv = g 11 du 2 + 2g 12 dudv + g 22 dv 2 2 6 j,k=1 g jkdu j du k,
(where u 1 = u and u 2 = v) which is the metric. Next, we verify θ 1 θ 2 g dudv, where g = g 11 g 22 g12. 2 Indeed, θ 1 θ 2 = (P 1 du + Q 1 dv) (P 2 du + Q 2 dv) = (P 1 Q 2 P 2 Q 1 )du dv. On the other hand, g = g 11 g 22 g12 2 = (P1 2 + P2 2 )(Q 2 1 + Q 2 2) (P 1 Q 1 + P 2 Q 2 ) 2 = P1 2 Q 2 1 + P1 2 Q 2 2 + P2 2 Q 2 1 + P2 2 Q 2 2 (P1 2 Q 2 1 + 2P 1 Q 1 P 2 Q 2 + P2 2 Q 2 2) = P1 2 Q 2 2 + P2 2 Q 2 1 2P 1 Q 2 P 2 Q 1 = (P 1 Q 2 P 2 Q 1 ) 2. Thus the area form is α = g du dv = P 1 Q 2 P 2 Q 1 du dv. The possible discrepancy of the sign between θ 1 θ 2 and α is due to our negligence of the orientation of the surface. But let us not worry about this subtle matter here and conclude Rule ST3. (θ 1 ) 2 + (θ 2 ) 2 = metric tensor, θ 1 θ 2 = area form. Exercises 1. In each of the following parts, compute the symmetric tensor product α β for the given 1-forms α and β: (a) α = dx + dy, β = dx dy, (b) α = β = dx + dy, (c) α = ydx + ydy, β = xdy xdx, (d) α = β = xdy + ydz + zdx. 2. In each of the following parts, find the length of the given parametric arc lying in R 2 or R 3 : (a) : x = t 2, y = t 1 3 t3 (0 t 1), (b) : x = t, y = cosh t (0 t 1), (c) : x = cos t, y = sin t, z = t (0 t π), (d) : x = t, y = t 2, z = 2 3 t3 (0 t 1). (e) : x = cos t + t sin t, y = sin t t cos t (0 t π/2). (f) One loop of cycloid: : x = t sin t, y = 1 cos t (0 t 2π). 3. In each of the following parts, find the line integral fds for the given arc and the given function f: (a) f(x, y) = x and : x = t, y = t 2 (0 t 1), 7
(b) f(x, y) = y and : x = cos t, y = sin t (0 t π/2), (c) f(x, y, z) = 2x + 6xy + 3z and : x = t, y = t 2, z = t 3 (0 t 1). 4. Find the stereographic projection π S from the South pole to parametrize the unit sphere S 2. Compute π S (dx2 + dy 2 + dz 2 ) and compare it with (2.6). Find the transformation π 1 S π N of the uv-plane. 5. Verify (2.7) and (2.8). 6. Consider a surface of revolution given by the parametric equations x = f(u) cos v, y = f(u) sin v, z = g(u). Compute the metric tensor ds 2 and the area form ds in two ways and compare their results. First way: use the connection θ 1 and θ 2 obtained in Exercise 8 of 4.1 and the identities ds 2 = (θ 1 ) 2 +(θ 2 ) 2 and ds = θ 1 θ 2. Second way: compute g ij r ui r uj and ds = g du 1 du 2. (u 1 = u, u 2 = v) and use the identities ds 2 = g ij du i du j 7. Given the parametric surface σ : x(u, v) = 1 2 u2, y(u, v) = 1 2 uv, z(u, v) = 1 2 v2, (u 2 + v 2 1), find (a) the metric tensor ds 2 for this surface, (b) the area form ds, and (c) the surface area of σ. 8. Find the surface integral zds, where σ is the upper unit sphere parametrized by σ (a) x = cos θ cos φ, y = sin θ cos φ, z = sin φ (b) the stereographic projection (2.5) with 1 u 2 + v 2 <, (0 θ 2π, 0 φ π/2), and 9. Consider a parametric surface S given by σ: r = r(u 1, u 2 ) with the metric tensor ds 2 = 2 i,j=1 g ij du i du j. Suppose that u 1 and u 2 are functions of v 1 and v 2 : u 1 = u 1 (v 1, v 2 ), u 2 = u 2 (v 1, v 2 ) so that S can also be parametrized by variables v 1, v 2 as τ: r = r(u 1 (v 1, v 2 ), u 2 (v 1, v 2 )). Let ds 2 = 2 k,l=1 G kl dv k dv l be the metric tensor of S corresponding to the parametrization τ. Verify that G kl = 2 i,j=1 g ij where g ij is evaluated at (u 1 (v 1, v 2 ), u 2 (v 1, v 2 )). u i v k u j v l, 9. Check that, by introducing complex variable w = u + iv, the metric ds 2 given in (2.8) can be rewritten as Ω Ω, where Ω = dw/(1 w 2 ) and Ω = d w/(1 w 2 ). Check that the metric ds 2 is invariant under a Möbius transform T (w) = e iθ (w w 0 )/(1 w 0 w) in the sense that T (ds 2 ) = ds 2. 8