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4.5. Torsion. The corresponding to on a Riemann- ian is called the Levi-Civita connection. Technically speaking, this is a connection on the , TM. Because is locally Euclidean to first order, we can actually compute Lie brackets using the Levi-Civita connection (see the figure). This fact is the torsion-free property

∇uv − ∇vu = [u, v]. (4.5) Note that this condition only makes sense for a connection on TM, not for a con-

"v 2 2 " [u,v] " #uv

2 ! " # v u ! !

"u ! "u

! "v !

FIGURE 3. The torsion-free condition. nection on some other vector! bundle. Remark. For an arbitrary connection on TM, we can define the torsion

T(u, v) := ∇uv − ∇vu − [u, v]. All derivatives of u and v cancel, so this is a tensor. In , i i j k T(u, v) = Tjku v . The torsion-free condition (4.5) is just T = 0. Exercise 4.2. Check that T(u, fv) = f T(u, v). There have been proposed theories of gravity which include torsion, including Einstein’s theory of “distant parallelism”. However, such theories violate the prin- ciple of equivalence, and none have survived experimental tests. A defines a trivialization of the tangent bundle over the coor- dinate patch. The trivial connection is torsion-free. This property is also what allows us to compute exterior derivatives using the Levi-Civita connection or a trivial coordinate connection. 27

4.6. The metric. The basic structure of Riemannian geometry is the metric. Tech- nically, this is a symmetric C (M)-bilinear map g : X(M) × X(M) C (M). In tensor notation, this is ∞ ∞ i j g(u, v) = g(v, u) = giju v . → This allows us to define the magnitude (length) of a vector or vector field as kvk := p g(v, v). We can define a pair of vectors to be orthogonal if g(u, v) = 0. It allows us to convert a vector v into a 1-form g(v, · ). In index notation, vi = j i ij gijv . We can identify the other way by using the inverse of the metric, v = g vj. Most importantly, a metric allows us to define length. Imagine breaking a curve C up into infinitesimal segments. Each of these is described by an infinitesimal vector. The sum of the magnitudes of these vectors is the length of C. Using coor- dinates and a parametrization of C, this is computed as q q i j i j Length(C) = ds = gijdx dx dt = gijx˙ x˙ dt (4.6) ZC ZC ZC Despite the notation, these are not differential forms. The simplest way to write a metric down in coordinates is usually as a line element: 2 i j ds = gijdx dx . For example, the metric for (a patch of) S2 in standard polar coordinates is ds2 = dθ2 + sin2 θ dφ2.

4.7. The Levi-Civita connection. Parallel transport respects the metric. If a vector is parallel transported, then its magnitude remains constant. If two vectors are orthogonal, and they are parallel transported (along the same curve) then they remain orthogonal. The Levi-Civita connection (or the covariant derivative4) is the unique connection which is torsion-free and respects the metric. In a coordinate patch, the Levi-Civita connection can be written as the trivial connection, plus a correction: ∇ = ∂ + Γ. In index notation, this is i i i k ∇jv = v,j + Γjkv . i i We will also use the more concise notation, v|j := ∇jv . Because both the trivial connection and the Levi-Civita connection are torsion-free, i i Γjk = Γkj. Any connection on TM also defines a connection on 1-forms and more general . Let u, v ∈ X(M) and α ∈ Ω1(M). We can pair α and v to make a scalar function hα, vi. We don’t need any connection to compute the derivatives of a function, and the pairing should fit into some Leibniz rule, so

u(hα, vi) = h∇uα, vi + hα, ∇uvi.

4The structure ∇ is “the Levi-Civita connection”. The result of applying ∇ to X is “the of X”. 28

This leads to the explicit formula for the covariant derivative of a 1-form

k ∇jαi = αi,j − Γjiαk.

i... For a general tensor Lj..., there is the following recipe for ∇L. The first term is the partial derivative of L. For every superscript index, there is a Γ term. For every subscript index, there is a −Γ term. The derivative of a scalar is just the partial derivative. Parallel transport respects the metric. This just means that the covariant deriva- tive of the metric is 0 (see Exercise 4.3): l l 0 = ∇kgij = gij,k − gljΓki − gilΓkj. Using this and the torsion-free condition (symmetry of Γ) we can compute

l l gilΓjk = gij,k − gjlΓki l = gij,k − gjk,i + gklΓij l = gij,k − gjk,i + gki,j − gilΓjk 1 = 2 (gij,k − gjk,i + gki,j) and so i 1 il Γjk = 2 g (glj,k + glk,j − gjk,l) . This is known as the Christoffel symbol(s). Note that it is entirely coordinate de- pendent and has no intrinsic significance. It is just a correction to get from the coordinate-dependent trivial connection to the Levi-Civita connection.

4.7.1. Parallel transport, again. The velocity of a parametrized curve C is abstractly d i i v := dt or in coordinates v := x˙ . The covariant derivative with respect to v of a tensor only depends upon that tensor’s values along C. So, this derivative makes sense even if the tensor is only defined along C. If u is a vector field defined along C (technically u ∈ Γ(C, TM)) then its covariant derivative along C is explicitly given in coordinates by

i i i j k ∇ d u = u˙ + Γjkx˙ u . dt There are similar formulas for derivatives of tensors along C. The vector u is parallel transported along C if and only if

0 = ∇ d u. dt Exercise 4.3. Check that if u is parallel transported along C, then kuk is constant. d 2 Hint: Compute dt kuk . 4.8. . In , it is often said that “the shortest path be- tween two points is a straight line”. Something like this remains true in Riemannian geometry. The shortest path is the straightest path and is called a . The length of a parametrized curve C can be computed by eq. (4.6). In princi- ple, we can find which curves are shortest by applying variational calculus to that expression. However, the square root makes the result rather ugly. 29

Instead, consider the action functional

1 i j S = 2 gijx˙ x˙ dt (4.7) ZC 1 2 = 2 kvk dt ZC Where vi = x˙ i. Unlike the length, this depends upon the parametrization. What parametrized curves minimize this action? In particular, it will be minimal with respect to variations of the parametrization. The Euler-Lagrange equation coming from such special variations implies that the speed kvk is constant, and the parameter t is proportional to the length along the curve (plus a constant). This is called an affine parameter. Now, suppose that C happens to be affinely parametrized. Then the variation of the action

1 2 δS = δ 2 kvk dt = kvk δkvk dt = kvk δkvk dt = kvk δ(Length) ZC ZC ZC is just proportional to the variation of the length. So, a parametrized curve mini- mizing S is affinely parametrized and has minimal length. If C has minimal length, then we can always use an affine parameter so that it minimizes S. On the other hand, what is a straight curve? This means that if we parallel transport a tangent vector (such as v) along C, then it remains tangent to C (is pro- portional to v). If C is affinely parametrized, then this means that v is covariantly constant along C. So, we have the equation

0 = ∇ d v (4.8a) dt i i j j = x¨ + Γjkx˙ x˙ (4.8b) Exercise 4.4. Show that the Euler-Lagrange equations for (4.7) are equivalent to (4.8b). Unless otherwise stated, geodesics will always be affinely parametrized. In Riemannian geometry, the arc length is the obvious affine parameter to use. So what is the purpose of more general affine parameters? When we generalize to spacetime, we will be interested in the paths of light rays. For these, the arc length is always 0, but affine parameters still exist.

4.9. The Riemann tensor. The of the Levi-Civita connection is called i the Riemann tensor. It is denoted R jkl. The definition of curvature and the torsion-free condition give

R(u, v)w := ∇u∇vw − ∇v∇uw − ∇[u,v]w

= ∇u∇vw − ∇v∇uw − ∇∇uvw + ∇∇vuw but i i k l i ∇u∇vw − ∇∇uvw = u v ∇k∇lw and so (in several different notations) i j i i i i i i R jklw = ∇k∇lw − ∇l∇kw = [∇k, ∇l]w = −w|kl + w|lk = −2w|[kl]. (4.9) 30

i Exercise 4.5. Use eq. (4.9) to compute an explicit formula for R jkl in terms of Γ and its first partial derivatives. The analogous formula with a 1-form is j [∇k, ∇l]αi = −R iklαj. Note the minus sign. If we do this with a general tensor, then there is one term for each index, but superscripts and subscripts give opposite signs. 4.9.1. Its symmetries. The definition immediately implies that i i R jkl = −R jlk. (4.10) In fact, this is true for the curvature of any connection. Since the covariant derivative of the metric is 0, so is its second derivative, and

0 = gij|kl − gij|lk = Rijkl + Rjikl. (4.11) So it is skew symmetric on the first 2 indices as well. The torsion-free condition implies that i i i 0 = R jkl + R klj + R ljk i (4.12) = R [jkl]. Exercise 4.6. Prove eq. (4.12) by using the torsion-free condition (4.5) and the Jacobi identity (2.3d) to show that R(u, v)w + R(v, w)u + R(w, u)v = 0. Together, eqs. (4.10), (4.11), and (4.12) imply that 3  0 = 2 Ri[jkl] − Rj[ikl] − Rk[lij] + Rl[kij] = Rijkl − Rklij. (4.13) Exercise 4.7. Check eq. (4.13). 4.9.2. Ricci. The Ricci tensor is also denoted by an R, but has fewer indices k Rij := R ikj Note that this is defined directly from the “natural” form of the Riemann tensor without using the metric again. Equation (4.13) implies that this is symmetric Rij = Rji. Using the metric, we can get rid of all indices and define the i ij R = R i = R ij. This is the crudest measure of the curvature of a Riemannian manifold. Example. In 2 , the Riemann tensor has only one independent compo- nent. It can always be written as5 i i i R jkl = κ(δkgjl − δlgjk). (4.14)

5The symbol is

i 1 : i = j δj := 0 : i 6= j, i j i so that δjv = v . In other words, it is just the unit . 31

For example, a 2-sphere of radius r has κ = r−2. The Ricci tensor and scalar curva- ture are

Rij = κgij R = 2κ. Exercise 4.8. Check that (4.14) has all the correct symmetries for a Riemann tensor. 4.10. Geodesic deviation. On a sphere S2 of radius r, the geodesics are the great circles. Any great circle passing through a given point must also pass through the antipodal point (which is a distance πr away). So, if two great circles intersect at x ∈ S2, then the separation between them increases at first, but then decreases and returns to 0 after a distance of πr. This is an effect of curvature called conjugate points. In general, we can ask how the separation between two geodesics on a Riemann- ian manifold varies along their length. To make this precise, we should consider two geodesics that are infinitesimally separated. The separation is then described by a vector, called the deviation. To simplify the calculation, let’s consider a large family of geodesics. Let v ∈ X(M) be a vector field whose integral curves are (affinely parametrized) geodesics. This means that is satisfies

0 = ∇vv. So, we have a geodesic through every point of M. Now suppose that u ∈ X(M) is a vector field whose flow preserves this family of geodesics. If x, y ∈ M are two points on one of our geodesics, then x + εu and y + εu are on the same geodesic in this family. The infinitesimal vector field εu is the separation between (some) neighboring geodesics in our family. This implies that the Lie bracket vanishes [u, v] = 0.

"v

"u "u ! "u ! ! "v ! FIGURE 4. Deviation between nearby geodesics, and why [u, v] = 0. ! We are interested in how u changes along these geodesics. The first derivative is

∇vu = ∇uv because of the vanishing Lie bracket. 32

The definition of curvature gives

R(u, v)v = ∇u∇vv − ∇v∇uv − ∇[u,v]v = −∇v∇uv = −∇v∇vu. This is the second (covariant) derivative of u along the geodesics. This result is still true if we have just two, infinitesimally separated geodesics. d Suppose that C is a geodesic with tangent vector v = dt . Let u be a deviation vector along C so that εu is the separation between C and a “neighboring” geodesic. Then u satisfies the geodesic deviation equation 2 0 = ∇ d u + R(u, v)v. (4.15) dt Example. In Euclidean space, the Riemann tensor is 0, and the geodesic deviation equation implies that u changes linearly. This reflects the fact that the separation between two lines changes linearly. Example. If the “neighboring” geodesic is just the same geodesic with a different affine parameter, then u must be tangent to C. This means that it can be written as u = f v. By the skew symmetry of the Riemann tensor, R(v, v) = 0. So, the geodesic deviation equation becomes 2 2 d f 0 = d (fv) + f R(v, v)v = v. ∇ 2 dt dt The general solution is f = a + bt. This reflects the fact that any other affine parameter is related to t by rescaling and adding a constant. In general, the affine parametrization of the neighboring geodesic can always be adjusted so that u is orthogonal to C. In any case, only the orthogonal part of u is interesting. Example. Suppose that the Riemann tensor has the form (4.14), C is a geodesic d parametrized by arc length (so that v := dt is a unit vector), and u is orthogonal to C. In that case, i i j k l i j i j i [R(u, v)v] = R jklv u v = κ(u vjv − v ujv ) = κu . So, the geodesic deviation equation simplifies to 2 0 = ∇ d u + κu. dt If κ is constant, then this looks very much like a harmonic oscillator with period √2π κ . In fact, we can compare different values of u(t) by parallel transport, and it √π does oscillate. If u = 0 at t, then u = 0 at t + κ . This means that two intersecting geodesics must periodically intersect again.

Example. Let v, w ∈ TxM be two orthogonal unit vectors at x ∈ M. These span a 2-plane in TxM. We can use the geodesic deviation equation to analyze a thin triangle tangent to this plane. Let C be the geodesic through v at x, parametrized by arc length s from x. Com- pare this with the neighboring geodesic through v + εw (with ε infinitesimal). Let εu be the deviation between these geodesics. These geodesics and the vector εu(s) form a thin triangle with angle ε. 33

w

" "u ! v

FIGURE 5. A thin triangle ! ! ! By construction, u(0) = 0 and ∇ d u(0) = w. The geodesic deviation equation ds gives 2 ∇ d u(0) = 0 ds and 3 ∇ d u(0) = −R(w, v)v. ds So, to order s3, u(s) is approximately 1 3 sw − 6 s R(w, v)v parallel transported along C. The length of the base of the triangle is 1 2 kεu(s)k ≈ εs(1 − 6 s κ) i j k l where κ = Rijklv w v w is called the through v and w. The sectional curvature only depends upon the plane spanned by v and w. If v and w are not necessarily unit vectors or orthogonal to each other, then the sec- tional curvature is R viwjvkwl κ = ijkl . 2 2 m 2 kvk kwk − (vmw ) We can construct a small “circle” around x as the set of points a distance ` away along geodesics tangent to the v-w plane. The circumference of this circle is ap- proximately 1 2 2π`(1 − 6 ` κ). 4.11. Bianchi identities. Now consider the derivative of the Riemann tensor. This can be computed by i j i j i j R jkl|mv = (R jklv )|m − R jklv|m i = [∇m, [∇k, ∇l]]v using the commutator notation, [∇k, ∇l] := ∇k∇l−∇l∇k. The commutator satisfies the Jacobi identity,

0 = [∇m, [∇k, ∇l]] + [∇k, [∇l, ∇m]] + [∇l, [∇m, ∇k]] and so we have the Bianchi identity i i i 0 = R jkl|m + R jlm|k + R jmk|l i (4.16) = R j[kl|m]. 34

Exercise 4.9. Suppose that the sectional curvature of M is the same in all directions. Then the Riemann tensor can be written in terms of κ ∈ C (M) by eq. (4.14). Show that if dim M ≥ 3, then the Bianchi identity implies that κ∞is constant. More generally, suppose that K is the curvature tensor of a connection on a vector bundle E, but that we also have a connection on TM. Then, by exactly the same argument, α 0 = K β[ij|k] Note that this looks a bit like an . This does not really depend upon the connection on TM. We can think of K as a 2-form, valued in End(E), the vector bundle of matrices over E. This general Bianchi identity states that the “covariant exterior derivative” of K vanishes. 4.11.1. The contracted Bianchi identity. Contracting two pairs of indices in eq. (4.16) gives jk jk jk 0 = R jk|i + R ij|k + R ki|j j = R|i − 2R i|j 4.11.2. The . This suggests that we can construct a combination from the Ricci and scalar curvature for which the Bianchi identity is particularly simple. This is the Einstein tensor, 1 Gij := Rij − 2 Rgij. As the name suggests, this is important for . The contracted Bianchi identity becomes simply, ij 0 = G |j.