Lemma 4.3. Let M = (M, g) be a connected Lorentzian manifold and Xa non-trivial lightlike complete Killing vector field. Then Xvanishes nowhere, that is, X(x) = 0 for all x ∈ M. Equivalently, the action of the corresponding one-parameter group ψ :=
{exp(tX)}t∈R ⊆Isom(M) (cf. Proposition 1.1) is locally free.
Proof. Let ψt := exp(tX) and assume, there is x ∈ M such that X(x) = 0. Then ψt(x) =x for all t in R.
gx defines a quadratic form Q : TxM → R. Since ψ preserves the Lorentzian metric g on M, it follows that dψx = {dψtx}t∈R preserves the scalar product gx as well as the quadratic form Q. Thus, dψx is a one-parameter group of isometries of (TxM, gx).
According to Proposition 1.1, it is associated to a complete Killing vector field X on TxM.
4.2. Locally free action 62
The isometry ψt of M is uniquely defined by ψt(x) and dψxt. Since the action of ψ is locally free, it follows that dψxt = idTxM for all t ∈ R\Γ, where Γ is a free subgroup in R generated by at most one element. Thus, the flow dψx is non-trivial and X does not vanish on an open neighborhood of TxM (because the Killing vector field X is determined by X(v) and ∇vX for some v ∈TxM, ∇ being the Levi-Civita connection onM).
Let U ⊂ M be a star-shaped open neighborhood of x, on which the exponential map expx is invertible, and define q :=Q◦exp−1x .
Letv ∈exp−1x (U). Then
γ : [0,1]→M, γ(τ) = expx(τ v), is the unique geodesic with
γ(0) =x, ∂
∂τγ(τ)|τ=0 =v.
ψt is an isometry with fixed point x, so (ψt◦γ) is a geodesic with (ψt◦γ)(0) =x and ∂
∂τ(ψt◦γ)(τ)|τ=0 =dψxt(v).
Thus, we have
ψt(expx(v)) =ψt(γ(1)) = expx(dψtx(v)) for small t. Differentiating with respect to t in t= 0 yields
X(exp x(v)) = (dexpx)v(X(v)). (1)
Forw∈Tv(TxM)∼=TxM, gx((grad Q)(v), w) = ∂
∂tQ(v+tw)|t=0 = ∂
∂tgx(v+tw, v+tw)|t=0 =gx(2v, w).
Hence, grad Q is radial. By the lemma of Gauß, gexp
x(v)((dexpx)v((grad Q)(v)),(dexpx)v(w))
=gx((gradQ)(v), w)
=dQv(w)
=dqexp Choosing w = X(v) and later w = (grad Q)(v) in Equation 2, we obtain successively with the help of Equation 1
It follows together with Equation 4, that
0 = gu((grad q)(u),X(u)) (6)
for all u∈U.
By assumption, 0 = gu(X(u), X(u)) since X is lightlike. Now choose a timelike v ∈ exp−1x (U) such that X(v) = 0 (remember that on any open neighborhood, X does not vanish everywhere). Then because of Equation 1, X(u) = 0 for u = expx(v), but the radial vector (grad Q)(v) and so (grad q)(u) by Equation 5 are timelike. Since (grad q)(u) is in the gu-orthogonal complement of the lightlike vector X(u) by Equa- tion 6, (grad q)(u) is not timelike, contradiction.
Remark. Lemma 4.3 is also true if X is not complete, in this case one consider the local flow ψ defined by X.
4.2. Locally free action 64
Definition. LetG be a Lie group with Lie algebra gacting continuously on a compact topological spaceM. For p∈M denote byGp the subgroup of Gfixing pand gp its Lie algebra.
Remark. Gp is a closed subgroup, especially a Lie subgroup. Thus, its Lie algebra is defined.
Lemma 4.4. Let G be a connected Lie group with Lie algebra g acting continuously on a compact topological space M. Differentiating with respect toτ inτ = 0 yields
Adexp(tX)Y ∈gexp(tX)·x.
for all t∈R. SinceM is compact, we can choose an increasing sequence{tn}∞n=0 of real numbers, such that tn → ∞and
exp(tnX)·x→y∈M as n→ ∞.
Since
Z = lim
n→∞
⎛
⎝Z+
k−1
j=0
k!
j!tk−jn Yj
⎞
⎠= lim
n→∞Ztn, exp(τ Ztn)·(exp(tnX)·x)→exp(τ Z)·y as n→ ∞ for all τ ∈R. But for all n, it holds
exp(τ Ztn)·(exp(tnX)·x) = exp(tnX)·x.
It follows that exp(τ Z)·y =y for all τ ∈R, so Z ∈gy.
Corollary 4.5. Let M be a compact Lorentzian manifold and S a Lie group with Lie algebra s∼=aff(R) acting isometrically and locally effectively on M. Then this action is locally free.
Proof. Let X, Y ∈s such that [X, Y] = Y. By Lemma 4.1, the subgroup generated by Y has lightlike orbits, and due to Lemma 4.3, it acts locally freely. Especially, Y /∈ sx for all x∈M.
Suppose X+λY ∈sx, λ∈R, for somex∈M. Since
[Y, X+λY] =−Y and [Y,−Y] = 0,
we can apply Lemma 4.4 and obtain −Y ∈sy for somey ∈M, contradiction.
Thus, sx is trivial for all x∈M, that is, S acts locally freely on M.
Corollary 4.6. Let M be a compact Lorentzian manifold and S a Lie group with Lie algebra s∼=sl2(R) acting isometrically and locally effectively on M. Then this action is locally free.
4.2. Locally free action 66
Proof. Without loss of generality, S ⊆ Isom(M). Let e, f, h ∈ s be elements of an sl2-triple, that is, [h, e] = 2e, [h, f] =−2f and [e, f] =h.
Letκ be the induced bilinear form on s.
2κ(e, e) =κ([h, e], e) =κ(h,[e, e]) = 0 because κ is ad-invariant. Analogously,
κ(f, f) = 0.
Because of [e, h] = −2e and [e,−2e] = 0, e generates a non-precompact one-parameter group in Isom(M) by Lemma 3.3. The same is true for f. It follows from Proposi-tion 1.19, that the subgroups generated byeand f have lightlike orbits everywhere. By Lemma 4.3, e, f /∈sx for all x∈M.
Let
0=Y :=αh+βe+γf
for some realα, β, γ. We have to show thatY /∈sx for allx∈M. If α=γ = 0, we have shown this already.
Assume γ = 0 and choose X :=e,Z :=−2γe. Then
[X, Y] =−2αe+γh, [X,[X, Y]] =Z and [X, Z] = 0.
Lemma 4.4 and Z /∈sx for all x yield the required result.
Finally, assume γ = 0, butα= 0. Choose X :=e and Z :=−2αe. Then [X, Y] =Z and [X, Z] = 0,
so we can apply Lemma 4.4 to see thatY /∈sx for all x∈M, sinceZ /∈sx for allx.
Lemma 4.7. Let Gbe a connected nilpotent Lie group acting continuously on a compact manifold M. If the action of the center is locally free, so is the action of G.
Proof. We may assume that G is non-trivial. Let g0 := g, gj := [g,gj−1] for j > 0 be the descending central series of g. Since g is nilpotent, there is an integer k ≥ 0, such that gk ={0}=gk+1.
Since {0}= [g,gk],gk ⊆z(g). Thus, the subgroup generated bygk acts locally freely by assumption. In the following, we prove by induction, that the action of the subgroup generated by gj−1 is locally free, if the action of the subgroup generated by gj is.
Assume the contrary, that is, there is x ∈ M and Y ∈ gj−1 such that Y ∈ gx. Since Y /∈z(g), there isX ∈g such that [X, Y]= 0. g is nilpotent, so there exists an integer l > 0 such that
Z := adlX(Y)= 0 = adl+1X (Y).
By Lemma 4.4, Z ∈gy for some y∈M. But Z ∈gj−1+l ⊆gj, contradiction.
We can now finish the proof of Theorem 4 by showing that the action of the subgroup generated by s is locally free, ifs∼=hed ors∼=heλd.
Corollary 4.8. Let M be a compact Lorentzian manifold and S a Lie group with Lie algebra s ∼= hed, acting isometrically and locally effectively on M. Then this action is locally free.
Proof. By Lemma 4.1, the center of z(s) is κ-isotropic and the subgroup generated by 0=Z ∈z(s) has lightlike orbits. Due to Lemma 4.3, it acts locally free. The result now follows from Lemma 4.7.
Corollary 4.9. Let M be a compact Lorentzian manifold and S a Lie group with Lie algebra s ∼= heλd, acting isometrically and locally effectively on M. Then this action is locally free.
Proof. We identify heλd withs. By Corollary 4.8, the subgroup generated byhed ⊂heλd ∼= s acts locally freely.
LetY ∈heλd\hed. Then there isX ∈hedsuch that [X, Y]= 0 (otherwiseY ∈z(s)⊂hed, contradiction). But [X, Y]∈hed and hed is nilpotent, hence
Z := adkX(Y)= 0 = adk+1X (Y)
for some integer k > 0. The subgroup generated by Z ∈ hed acts locally freely by Corollary 4.8, therefore, for all x∈M,Y /∈sx due to Lemma 4.4.