The existence of light-like homogeneous geodesics in homogeneous Lorentzian manifolds Zdeněk Dušek Sao Paulo, 2013
Motivation In a previous project, it was proved that any homogeneous affine manifold (and in particular any homogeneous pseudo-riemannian manifold) admits a homogeneous geodesic through arbitrary point. In pseudo-riemannian geometry, null homogeneous geodesics are of particular interest. Plane-wave limits (Penrose limits) of homogeneous spacetimes along light-like homogeneous geodesics are studied. However, it was not known whether any homogeneous pseudo-riemannian or Lorentzian manifold admits a null homogeneous geodesic. An example of a 3-dimensional Lie group with an invariant Lorentzian metric which does not admit light-like homogeneous geodesic was described (G. Calvaruso).
Results In the present project, the affine method is adapted to the pseudo-riemannian case. We show that any Lorentzian homogeneous manifold of even dimension admits a light-like homogeneous geodesic. In the case of a Lie group G = M with a left-invariant metric, the calculation is particularly easy. As an illustration, we apply the method on an example of a Lie group in dimension 3.
Homogeneous geodesics in affine manifolds Lemma Let (M, ) be a homogeneous affine manifold. Then each regular curve which is an orbit of a 1-parameter subgroup g t G on M is an integral curve of an affine Killing vector field on M. Lemma Let (M, ) be a homogeneous affine manifold and p M. There exist n = dim(m) affine Killing vector fields which are linearly independent at each point of some neighbourhood U of p. Lemma The integral curve γ(t) of the Killing vector field Z on (M, ) is geodesic if and only if Zγ(t) Z = k γ Z γ(t) holds along γ. Here k γ R is a constant.
Existence of homogeneous geodesics Theorem Let M = (G/H, ) be a homogeneous affine manifold and p M. Then M admits a homogeneous geodesic through p. Proof. Killing vector fields K 1,..., K n independent near p, basis B = {K 1 (p),..., K n (p)} of T p M. Any vector X T p M, X = (x 1,... x n ) in B, determines a Killing vector field X = x 1 K 1 + + x n K n and an integral curve γ X of X through p. Sphere S n 1 T p M, vectors X = (x 1,..., x n ) with X = 1. Denote v(x ) = X X and t(x ) = v(x ) v(x ), X X, then t(x ) X and X t(x ) defines a vector field on S n 1. If n is odd, according to the Hair-Dressing Theorem for sphere, there is X T p M such that t( X ) = 0. We see v( X ) = k X, hence X X = k X.
Existence of homogeneous geodesics We refine the proof to arbitrary dimension: Recall that X t(x ) defines a smooth vector field on S n 1. Assume now that t(x ) 0 everywhere. Putting f (X ) = t(x )/ t(x ), we obtain a smooth map f : S n 1 S n 1 without fixed points. According to a well-known statement from differential topology, the degree of f is odd (integral degree is deg(f ) = ( 1) n ). On the other hand, we have v(x ) = v( X ) and hence f (X ) = f ( X ) for each X. If Y is a regular value of f, then the inverse image f 1 (Y ) consists of even number of elements, hence deg(f ) is even, which is a contradiction. This implies that there is X T p M such that t( X ) = 0 and again, a homogeneous geodesic exists.
Homogeneous Lorentzian manifolds Proposition Let φ X (t) be the 1-parameter group of isometries corresponding to the Killing vector field X. For all t R, it holds φ X (t)(p) = γ X (t), φ X (t) (X p ) = X γ X (t). The covariant derivative X X depends only on the values of X along γ X (t). From the invariance of g and, we obtain Proposition Along the curve γ X (t), it holds for all t R g p (X, X ) = g γx (t)(x γ X (t), X γ X (t) ), φ X (t) ( X X p ) = X X γx (t).
Proposition Let (M, g) be a homogeneous Lorentzian manifold, p M and X T p M. Then, along the curve γ X (t), it holds X X γx (t) (X γ X (t) ). Proof. We use the basic property g = 0 in the form X g(x, X ) = 2g( X X, X ). (1) According to Proposition 2, the function g(x, X ) is constant along γ X (t). Hence, the left-hand side of the equality (1) is zero and the right-hand side gives the statement.
Theorem Let (M, g) be a homogeneous Lorentzian manifold of even dimension n and let p M. There exist a light-like vector X T p M such that along the integral curve γ X (t) of the Killing vector field X it holds where k R is some constant. Proof. X X γx (t) = k X γ X (t), Killing vector fields K 1,... K n such that {K 1 (p),..., K n (p)} is a pseudo-orthonormal basis of T p M with K n (p) timelike. Any airthmetic vector x = (x 1,..., x n ) R n determines the Killing vector field X = n i=1 x i K i. We identify x with X p and R n T p M. We consider x = ( x, 1), where x S n 2 R n 1. For X, we have g p (X p, X p ) = 0 and the vectors x S n 2 determine light-like directions in R n T p M.
For x = ( x, 1) R n T p M, we denote Y x = X X p. With respect to the basis B = {K 1 (p),..., K n (p)}, we denote the components of the vector Y x as y(x) = (y 1,..., y n ). Using Proposition, we see that y(x) x. We define the new vector t x as t x = y(x) y n x. Because x is light-like vector, it holds also t x x. In components, we have t x = ( t x, 0), where t x R n 1. We see that t x x, with respect to the positive scalar product on R n 1 which is the restriction of the indefinite scalar product on R n. The assignment x t x defines a smooth tangent vector field on the sphere S n 2. If n is even, it must have a zero value. There exist a vector x S n 2 such that for the corresponding vector x = ( x, 1) it holds t x = 0. For this vector x, it holds y(x) = k x and X X γx (t) = k X γ X (t) is satisfied.
Corollary Let (M, g) be a homogeneous Lorentzian manifold of even dimension n and let p M. There exist a light-like homogeneous geodesic through p. Proof. We consider the vector X T p M which satisfies Theorem. The integral curve γ X (t) through p of the corresponding Killing vector field X is homogeneous geodesic.
Invariant metric on a Lie group Let M = G be a Lie group with a left-invariant metric g. For any tangent vector X T e M and the corresponding Killing vector field X, we consider the vector function X γ X (t) along the integral curve γ X (t) through e. It can be uniquely extended to the left-invariant vector field L X on G. Hence, along γ X, we have L X γ X (t) = X γ X (t). (2) At general points q G, values of left-invariant vector field L X do not coincide with the values of the Killing vector field X, which is right-invariant. As we are interested in calculations along the curve γ X (t), we can work with respect to the moving frame of left-invariant vector fields and use formula (2).
Proposition Let {L 1,..., L n } be a left-invariant moving frame on a Lie group G with a left-invariant pseudo-riemannian metric g and the induced pseudo-riemannian connection. Then it holds Li L j = n γij k L k, i, j = 1,..., n, k=1 where γ k ij are constants. Proof. It follows from the invariance of the affine connection.
An example of a Lie group Now we illustrate the affine method of the previous section with an example of the 3-dimensional Lie group E(1, 1) with an inv. Lorentzian metric which has no light-like homogeneous geodesic. We choose one of the examples described by G. Calvaruso using the geodesic lemma for reductive pseudo-riemannian homogeneous manifolds. We construct explicitly the vector field t x, which has no zero value in this case.
The group E(1, 1) can be represented by the matrices e w 0 u 0 e w v 0 0 1 Hence, M = E(1, 1) can be identified with R 3 [u, v, w]. The left-inv. vector fields are U = e w u, V = e w v, W = w.. We choose the new moving frame {E 1, E 2, E 3 } given as E 1 = U V, E 2 = W, E 3 = 1/2(U + V ). The pseudo-riemannian metric g such that the basis determined by the above frame at any point p M is pseudo-orthonormal basis of T p M with E 3 timelike.
The above metric g in coordinates is ds 2 = 1 4 (3e2w du 2 + 3e 2w dv 2 + 10dudv 4dw 2 ) and the nonzero Christoffel symbols are Γ 3 11 = 3 4 e2w, Γ 1 13 = 9 16, Γ2 13 = 15 16 e2w, Γ 3 22 = 3 4 e 2w, Γ 2 23 = 9 16, Γ1 23 = 15 16 e 2w. In the frame {E 1, E 2, E 3 }: g(e 1, E 1 ) = g(e 2, E 2 ) = 1, g(e 3, E 3 ) = 1, g(e i, E j ) = 0 and nonzero covariant derivatives (which satisfy the Proposition): E1 E 2 = 3 4 E 3, E1 E 3 = 3 4 E 2, E2 E 3 = 5 4 E 1, E2 E 1 = 5 4 E 3, E3 E 1 = 3 4 E 2, E3 E 2 = 3 4 E 1. We will perform all calculations in this moving frame, or with respect to the corresponding pseudo-orthonormal basis B = {E 1 (e), E 2 (e), E 3 (e)} of the tangent space T e M R 3.
Any x = (x 1, x 2, x 3 ) R 3 determines L X = x 1 E 1 + x 2 E 2 + x 3 E 3. Light-like vectors X T e M are x = (sin(ϕ), cos(ϕ), 1), x = (sin(ϕ), cos(ϕ)) S 1. For L X, it holds L X L X = 2 cos(ϕ)e 1 3 2 sin(ϕ)e 2 + 1 2 sin(ϕ) cos(ϕ)e 3, y(x) = (2 cos(ϕ), 3 2 sin(ϕ), 1 ) 2 sin(ϕ) cos(ϕ). We see that y(x) x. The projection t x is t x = y(x) 1 sin(ϕ) cos(ϕ) x = [ 2 = 2 1 ] ( ) 2 sin2 (ϕ) cos(ϕ), sin(ϕ), 0. t x x and t x x x t x defines the smooth vector field on S 1, which is nonzero everywhere. There is not any vector X T e G which satisfies Main Theorem.
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