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DIFFERENTIAL GEOMETRY AND APPLICATIONS Proc. Conf., Aug. 28 – Sept. 1, 1995, Brno, Czech Republic Masaryk University, Brno 1996, 93–101

THE INTEGRAL OF MIXED SCALAR

CURVATURE ALONG A LEAF OF FOLIATION

Vladimir Rovenskii

Abstract. By using the method of matrix Riccati ODE, we study Riemannian manifolds with foliations (distributions), whose mixed Ricci or scalar curvature is nonnegative and the norms of integrability tensors are bounded above. We obtain the integral formula for mixed scalar curvature along the complete leaf, rigidity of totally geodesic distribution, splitting of foliation with ”large” dimension.

1. Main results

For two complementary orthogonal distributions T_{1} and T_{2} on T M we define the structural tensors B_{1} : T_{2} × T_{1} → T_{1}, B_{2} : T_{1} × T_{2} → T_{2} by formulae, where P_{i} : T M → T_{i} are orthoprojectors

B_{1}(y, x) := P_{1}(∇_{x}y), B_{2}(x, y) := P_{2}(∇_{y}x).  (1)

Obviously, the equation B_{i} = 0 means that T_{i} is tangent to totally geodesic foliation. The equalities B_{1} = B_{2} = 0 mean by de Rham decomposition theorem that M is locally a product L_{1} × L_{2}, where T L_{1} = T_{1}, T L_{2} = T_{2}. Tensors (1) (as well as Gray’s tensors O, T [Gra2] or O’Neill’s tensors A, T [O’Ne]) contain the whole local information about distributions, for example, their mean curvature vector fields H_{1} ⊂ T_{2}, H_{2} ⊂ T_{1} are given by formulae

(H_{2}, x) = −trace B_{2}(x, ∗), (H_{1}, y) = −trace B_{1}(y, ∗).  (2)

Firstly such so called co-nullity operator was studied by [Ros], [Gra1] for nullity foliation of curvature operator and by [O’NS] for relative nullity foliation on submanifold in space form (see [Fer], [Dom] for totally geodesic foliation).

The second fundamental forms h_{i} : T_{i} × T_{i} → T_{i}^{⊥} and integrability tensors A_{i} : T_{i} × T_{i} → T_{i}^{⊥} are defined by formulae

\begin{matrix}h_{1}(x, u) = P_{2}(∇_{x}u + ∇_{u}x)/2, A_{1}(x, u) = P_{2}(∇_{x}u − ∇_{u}x)/2, \\ h_{2}(y, z) = P_{1}(∇_{y}z + ∇_{z}y)/2, A_{2}(y, z) = P_{1}(∇_{y}z − ∇_{z}y)/2. \end{matrix}  (3)

1991 Mathematics Subject Classification. 53C12.

Key words and phrases. Foliation, distribution, mixed curvature, Riccati equation.

Supported by RFFI, grant No 94-01-00271.

This paper is in final form and no version of it will be submitted for publication elsewhere.

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Note, that h_{i} is a second fundamental form at a point p ∈ M of submanifold, consisting of geodesics in M, tangent to subspace T_{i}(p), the equality A_{i} = 0 means that T_{i} is tangent to foliation.

Definition 1. A distribution (or foliation) T_{i} on a Riemannian manifold M is called (totally) geodesic or minimal if h_{i} = 0 or H_{i} = 0.

Geodesic distribution is characterized by the property: if geodesic in M is tangent to T_{i} at one point, then it stays tangent to T_{i} at every point.

Proof. Let γ : I → M be a T_{i}−geodesic, i.e. γ′ ⊂ T_{i} and ∇_{γ′}γ′ = h_{i}(γ′, γ′) ⊥ T_{i}. If h_{i} = 0, then ∇_{γ′}γ′ = 0, i.e. γ is geodesic in M. □

The symmetric and skew-symmetric with respect to T_{i}^{⊥} components B_{i}^{+}, B_{i}^{−} of structural tensors may be considered as self-adjoint Weingarten tensors associated to h_{i}, and as equivalent terms of integrability tensors for T_{i} in view of identities

(B_{1}^{+}(y, x), u) = −(h_{1}(x, u), y), (B_{2}^{+}(x, y), z) = −(h_{2}(y, z), x),  (4)
(B_{1}^{−}(y, x), u) = −(A_{1}(x, u), y), (B_{2}^{−}(x, y), z) = −(A_{2}(y, z), x).  (5)

Lemma 1 [Rov3]. The following differential equation is true (similar to Codazzi equation (2.4) in [Gra2])

\begin{matrix}((∇_{x_{1}}B_{2})(x_{2}, y), z) & + (B_{2}(x_{2}, B_{2}(x_{1}, y), z) + ((∇_{y}B_{1})(z, x_{1}), x_{2})+ \\ (B_{1}(z, B_{1}(y, x_{1})), x_{2}) + (R(y, x_{1})x_{2}, z) = 0. \end{matrix}  (6)

If γ : I → M is a unit speed T_{1}−geodesic, then the endomorphism field B_{2}(t) = B_{2}(γ′(t), ∗) of T_{2} along γ satisfies the Riccati ODE

B′_{2} + B_{2}^{2} + F (t) = 0, (t ∈ I),  (7)

where (F (t)y, z) = (R(y, γ′)γ′, z) + ((∇_{y}B_{1})(z, γ′), γ′) + (B_{1}(z, B_{1}(y, γ′)), γ′). For totally geodesic foliation F (t) coincides with Jacobi operator R_{γ′}, see [Fer], [Dom].

The expansion θ, sheaf tensor σ and vorticity tensor ω are calculated from B_{2}: θ = trace B_{2}^{+}, σ = B_{2}^{+} − a θ Id, ω = B_{2}^{−}, where a = \frac{1}{dim T_{2}} . Note, that σ^{2} ≥ 0, but ω^{2} ≤ 0, and σ = 0 exactly for conformal operator B_{2}. It is clear, that θ satisfies the following equation, which is also called the Raychaudhuri equation in Relativity theory:

θ′ + aθ^{2} + {trace F (t) + trace (σ^{2}) + trace (ω^{2})} = 0.  (8)

Let {x_{i}} ⊂ T_{1}, {y_{j}} ⊂ T_{2} are local orthonormal bases. The mixed Ricci and scalar curvature are defined by

Ric_{1}(x) = ∑_{j} K(x, y_{j}), Ric_{2}(y) = ∑_{i} K(x_{i}, y), s_{mix} = ∑_{i,j}(R(y_{j}, x_{i})x_{i}, y_{j}).

The method of matrix Riccati ODE (see Lemma 4 below) allows to prove

THE INTEGRAL OF MIXED SCALAR CURVATURE 95

Proposition 1 [BW] ([Abe] for K_{mix} ≥ 0). Let {L} be a totally geodesic foliation with complete leaves, integrable T_{2} = T L^{⊥} and Ric_{1} ≥ 0. Then foliation splits locally.

The contraction of (6) with respect to i, j gives us equation

div(H_{1} + H_{2}) = s_{mix} + ||h_{1}||^{2} + ||h_{2}||^{2} − ||A_{1}||^{2} − ||A_{2}||^{2} − |H_{1}|^{2} − |H_{2}|^{2}, (9)

where we used div_{1}H_{2} = div H_{2} + |H_{2}|^{2}, div_{2}H_{1} = div H_{1} + |H_{1}|^{2},

∑_{i,j}(B_{2}(x_{i}, B_{2}(x_{i}, y_{j})), y_{j}) = ||B_{2}^{+}||^{2} − ||B_{2}^{−}||^{2} = ||h_{2}||^{2} − ||A_{2}||^{2},  (10)

∑_{i,j}(B_{1}(y_{j}, B_{1}(y_{j}, x_{i})), x_{i}) = ||B_{1}^{+}||^{2} − ||B_{1}^{−}||^{2} = ||h_{1}||^{2} − ||A_{1}||^{2}.  (11)

Integration of (9) along compact manifold, or over compact leaf L_{0} of foliation, in view of Green’s Theorem, leads to integral formulae having many interesting global corollaries ([Wal], [Ran], [Roc])

∫_{M}{s_{mix} + ||B_{1}^{+}||^{2} + ||B_{2}^{+}||^{2} − ||B_{1}^{−}||^{2} − ||B_{2}^{−}||^{2} − |H_{1}|^{2} − |H_{2}|^{2}} = 0, (12)

∫_{L_{0}}{s_{mix} + div_{2}H_{1} + ||B_{1}^{+}||^{2} + ||B_{2}^{+}||^{2} − ||B_{2}^{−}||^{2}} = 0.  (13)

In relation with Proposition 1 and Theorems 1, 2 we give

Proposition 2. A compact minimal foliation {L} on a Riemannian manifold M with integrable normal bundle and inequality s_{mix} ≥ 0 splits locally: M = L × L^{⊥}. (In particular, a minimal foliation {L} with integrable normal bundle and inequality s_{mix} > 0 has no compact leaves [Wal]).

Proof. From (13) in view B_{2}^{−} = 0 it follows B_{1}^{+} = B_{2}^{+} = 0, i.e. T L and T L^{⊥} are tangent to totally geodesic foliations. By de Rham theorem foliation splits. □

The consequence ρ(n) = 8b + 2^{c}, where n = (odd) 2^{4b+c}, (b ≥ 0, 0 ≤ c ≤ 3) is well-known in topology: ρ(n) − 1 is number of continuous pointwise linearly independent vector fields on (n−1)−sphere. It is true inequality ρ(n) ≤ 2 log_{2} n+2. Let |A_{i}|(e) = sup{|A_{i}(e, v)| : |v| = 1}, |A_{i}|(p) = sup{|A_{i}|(e) : |e| = 1}.

Theorem 1. Let M be a complete Riemannian manifold equipped with two complementary orthogonal distributions T_{1}, T_{2}, (dim T_{1} = ν, dim T_{2} = n), and T_{1} is totally geodesic. a) Then for every T_{1}−geodesic γ(t) ⊂ M

lim inf_{s→+∞} ∫_{−s}^{s} {Ric_{1}(γ′) − (ν − 1)|A_{1}|^{2}(γ) − n(n − 1)|A_{2}|^{2}(γ)} dt ≤ 0,  (14)

96 VLADIMIR ROVENSKII

and equality holds for every γ iff T_{2} is totally geodesic distribution, and skewsymmetric operators B_{2}, B_{1} have ”constant torsion”:

B_{2}(x, B_{2}(x, y)) = −|A_{2}|^{2}(p)x^{2}y, B_{1}(y, B_{1}(y, x)) = −|A_{1}|^{2}(p)y^{2}x.  (15)

b) If equality holds in (14) for every γ and additionally ν ≥ ρ(n), then T_{2} is also integrable and hence is tangent to Riemannian totally geodesic foliation.

Proof. a) Consider (6) along any unit speed T_{1}−geodesic γ : ℝ → M

(∇_{γ′}B_{2})(γ′, y)+B_{2}(γ′, B_{2}(γ′, y))+(B_{1}(∗, B_{1}(y, γ′)), γ′)+P_{2}(R(y, γ′)γ′) = 0 (16)

with use ((∇_{y}B_{1})(∗, γ′), γ′) = 0, because B_{1} is skew-symmetric for totally geodesic T_{1}. The symmetric for T_{2} component of (16) is

\begin{matrix}(∇_{γ′}B_{2}^{+})(γ′, y)+ & B_{2}^{+}(γ′, B_{2}^{+}(γ′, y)) + B_{2}^{−}(γ′, B_{2}^{−}(γ′, y))+ \\ (B_{1}(∗, B_{1}(y, γ′)), γ′) + P_{2}(R(y, γ′)γ′) = 0. \end{matrix}  (17)

Consider the Raychaudhuri equation (8), for θ(t) = trace B_{2}^{+}(γ′)

\begin{matrix}θ′ + aθ^{2}+ \\ \{Ric_{1}(γ′) + trace σ^{2}(γ′)− & ∑_{i} |B_{2}^{−}(γ′, y_{i})|^{2} − ∑_{i} |B_{1}(y_{i}, γ′)|^{2}\} = 0. \end{matrix}  (18)

Note that for z_{1} = γ′

\begin{aligned}∑_{i} |B_{1}(y_{i}, γ′)|^{2} = ∑_{j} |A_{1}(γ′, z_{j})|^{2} & ≤ (ν − 1)|A_{1}|^{2}(γ′), \\ ∑_{i} |B_{2}^{−}(γ′, y_{i})|^{2} & ≤ n(n − 1)|A_{2}|^{2}(γ). \end{aligned}  (19)

From Lemma 4 below and (19) it follows (14) and for the equality case:

θ = σ(γ′) = B_{2}^{+}(γ′, ∗) = 0 and |B_{2}^{−}(γ′, ∗)| = |A_{2}|(γ), |B_{1}(∗, γ′)| = |A_{1}|(γ). Applying this to every T_{1}−geodesic we obtain for the equality case: B_{2}^{+} = 0, i.e. T_{2} is totally geodesic and (15) is true.

b) By point a), the distribution T_{2} is totally geodesic and (15) is right. Assume the opposite, that T_{2} in non-involutive somewhere, i.e. by (5) there exist point p ∈ M and unit vectors x_{0} ∈ T_{1}(p), y_{0} ∈ T_{2}(p) with the property B_{2}(x_{0}, y_{0}) ≠ 0. Therefore, |A_{2}|(p) ≠ 0 and in view of (15) B_{2}(x, y) ≠ 0 for each nonzero x ∈ T_{1}(p), y ∈ T_{2}(p). Since skew-symmetric linear operator B_{2}(x, ∗), (x ≠ 0) has no eigenvectors, then following [Fer] we choose a basis {u_{i}} for T_{1}(p) and define ν continuous pointwise linearly independent vector fields {v_{i}} on the unit sphere S^{n−1} ⊂ T_{2}(p) by the rule: v_{i}(y) = B_{2}(u_{i}, y). Thus ν < ρ(n) – a contradiction. □

Combine with the method [Gui], we obtain the inequality for integral of mixed scalar curvature along a leaf.

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Theorem 2. Let {L^{ν}} be a foliation on Riemannian manifold M^{n+ν} with complete minimal leaves, and quadratic form

Ric_{1}(x) + trace B_{2}^{−}(x, B_{2}^{−}(x, ∗)) − trace (h_{1}(x, B_{1}^{+}(∗, x)) + (∇_{∗}h_{1})(x, x))

on a unit tangent bundle SL_{0} of every leaf L_{0} has an integrable positive or negative part. Then

∫_{L_{0}} {s_{mix} − n(n − 1)|A_{2}|^{2}} d vol ≤ 0,  (20)

and equality holds for every leaf iff {L} is a Riemannian totally geodesic foliation. Moreover, if equality in (20) holds for all leaves and ν ≥ ρ(n), then M is locally a product L × L^{⊥}.

Remark 1. If T L^{⊥} in Theorem 2 is integrable, then without inequality for dimension foliation splits. The certain Corollary generalizes Proposition 1. For compact leaf L_{0} see Proposition 2.

Proof of Theorem 2. is based on some facts from ergodic theory, see 2.2. Consider arbitrary leaf L_{0}. By symmetric part of structural tensor (1) we define tensor U on SL_{0}, such that for every x ∈ SL_{0} : U(x) = B_{2}^{+}(x, ∗) is a linear operator on T L_{0}^{⊥}. Then U satisfies the Riccati equation U′ + U^{2} + R̃(x) = 0, see (7), where (R̃(x)y, z) = (R(y, x)x, z) + (B_{2}^{−}(x, B_{2}^{−}(x, y)), z) + (B_{1}^{+}(x, B_{1}^{+}(y, x)), z) + ((∇_{y}B_{1}^{+})(z, x), x), R(x) = R(∗, x)x – the Jacobi operator on T L_{0}^{⊥} and B_{2}^{−}(x, ∗)

– the vorticity tensor. Theorem will follow from the inequality

∫_{SL_{0}} R̃ic_{1}(x) dμ ≤ − ∫_{SL_{0}} trace U^{2}(x) dμ,  (21)

where R̃ic_{1}(x) = Ric_{1}(x)+trace B_{2}^{−}(x, B_{2}^{−}(x, ∗))−trace (h_{1}(x, B_{1}^{+}(∗, x))+(∇_{∗}h_{1})(x, x)). To see this, by Fubini’s theorem and Lemma 6 below, we obtain

∫_{L_{0}} s̃_{mix} d vol = \frac{1}{ω_{n−1}} ∫_{SL_{0}} R̃ic_{1}(x) dμ ≤ − ∫_{SL_{0}} trace U^{2}(x) dμ ≤ 0,

where s̃_{mix} := trace R̃ic_{1} = s_{mix} − ||B_{2}^{−}||^{2} + ||B_{1}^{+}||^{2} in view of H_{1} = 0, s_{mix} = trace Ric_{1} are functions on L_{0}. Note that ||B_{2}^{−}||^{2}(p) ≤ n(n − 1)|A_{2}|^{2}(p). Furthermore, if ∫_{L_{0}} s̃_{mix} d vol = 0, then U^{2} is identically zero and so R̃(x) = 0 on SL_{0} and B_{1} = 0, i.e. L_{0} is totally geodesic submanifold. Since L_{0} is an arbitrary leaf, then distribution T L^{⊥} and foliation {L} are totally geodesic. Moreover, B_{2} has constant torsion in sence (15). From the given inequality for dimension of L_{0} and Theorem 1 b) it follows that T L^{⊥} is also integrable. The de Rham decomposition Theorem completes the proof.

To prove (21) let r_{1}^{+} = (R̃ic_{1})^{+} and r_{1}^{−} = (R̃ic_{1})^{−} – the positive and negative parts of R̃ic_{1}. If ∫_{SL_{0}} r_{1}^{−}dμ = +∞, then ∫_{SL_{0}} R̃ic_{1} dμ = −∞, because in this case we have ∫_{SL_{0}} r_{1}^{+} dμ < ∞. So (21) holds.

98 VLADIMIR ROVENSKII

Hence we may suppose ∫_{SL_{0}} r_{1}^{−} dμ < ∞. Consider nonnegative integrable function f ≤ trace (U^{2}) + r_{1}^{+} on SL_{0}. Thus it suffices to show that

∫_{SL_{0}} f dμ ≤ ∫_{SL_{0}} r_{1}^{−} dμ.  (22)

To prove (22) we will verify that the integral of f − r_{1}^{−} is nonpositive in each one of the sets D, C^{+} and C^{−} \ C^{+}. Note that SL_{0} is the disjoint union SL_{0} = D ∪ C^{+} ∪ (C^{−} \ C^{+}). We integrate the Raychaudhuri equation for u = trace U, which corresponds to Riccati equation U′ + U^{2} + R̃(x) = 0, and obtain

u(s) − u(t) + \frac{1}{n − 1} ∫_{t}^{s} u^{2}(λ) dλ + ∫_{t}^{s} \{trace (U^{2}) + R̃ic_{1}\} dλ = 0,

i.e.

u(s) − u(t) + \frac{1}{n − 1} ∫_{t}^{s} u^{2}(λ) dλ + ∫_{t}^{s}(f − r_{1}^{−})(T_{λ}(v)) dλ = 0  (23)

for every v ∈ SL_{0} and s, t ∈ ℝ. By Lemma 3 below ∫_{t}^{s}(f − r_{1}^{−})(T_{t}(v)) dt ≤ 0 for almost all v ∈ D. Then by Lemma 5 below

∫_{D}(f − r_{1}^{−}) dμ = ∫_{π(D)} ∫_{−∞}^{∞}(f − r_{1}^{−})(T_{t}(v)) dt dμ̃ ≤ 0.

In order to compute ∫_{C^{+}}(f − r_{1}^{−})dμ, let δ > 0 and f_{0} > 0 be integrable on SL_{0}. Define g = f − r_{1}^{−} − δf_{0}. From (23) we have

δ ∫_{0}^{s} f_{0}(T_{t}(v)) dt − u(0) + u(s) + \frac{1}{n − 1} ∫_{0}^{s} u^{2}(t) dt + ∫_{0}^{s} g(T_{t}(v)) dt = 0

for every v ∈ SL_{0} and s ∈ ℝ. If v ∈ C^{+}, using Lemma 2 a) below we conclude that lim inf_{s→+∞} g(T_{t}(v) dt = −∞. Then by Theorem 3 below we obtain ∫_{C^{+}}(f − r_{1}^{−} −

δf_{0}) dμ = ∫_{C^{+}} g dμ and therefore ∫_{C^{+}}(f − r_{1}^{−}) dμ ≤ 0, by letting δ → 0. In order to compute ∫_{C^{+}\C^{−}}(f − r_{1}^{−}) dμ we derive from (23) that

δ ∫_{−s}^{0} f_{0}(T_{t}(v))dt + u(0) − u(s) + \frac{1}{n−1} ∫_{−s}^{0} u^{2}(t)dt + ∫_{−s}^{0} g(T_{t}(v))dt ≤ 0 for every v ∈ SL_{0} and s ∈ ℝ. Hence

δ ∫_{0}^{s} f_{0}(T_{−t}(v))dt + u(0) − u(−s) + \frac{1}{n−1} ∫_{0}^{s} u^{2}(−t)dt + ∫_{0}^{s} g(T_{−t}(v))dt ≤ 0. (24)

If v ∈ C^{−}, then lim inf_{s→+∞} g(T_{−t}(v)) dt = −∞ by (31) and Lemma 2 a) below. Using

Theorem 3 below we obtain ∫_{C^{−}\C^{+}}(f −r_{1}^{−}−δf_{0}) dμ = ∫_{C^{−}\C^{+}} g dμ. Hence letting δ → 0 leads to ∫_{C^{−}\C^{+}}(f − r_{1}^{−}) dμ ≤ 0. □

As from Theorem 1 so from Theorem 2 it follows

Corollary 1 [KT] (rigidity). Let {L} be a Riemannian foliation on a complete Riemannian manifold M. If for each geodesic γ orthogonal to leaves we have Ric_{2}(γ′) ≥ (n − 1)|A_{2}|^{2}(γ′), then {L} is totally geodesic.

Corollary 2 [Rov2] (splitting). Let M, {L} be a foliated Riemannian manifold with complete totally geodesic leaves and Ric_{1}(x) ≥ n(n−1)|A_{2}|^{2}(p), (x ∈ T_{p}M). If ν ≥ ρ(n), then M is locally a product L × L^{⊥}.

THE INTEGRAL OF MIXED SCALAR CURVATURE 99

2. Necessary facts

2.1 Ferus’s scheme. Ferus [Fer] proved the most effective estimate dim L < ρ(codim L) of dimension of totally geodesic foliation {L} with complete leaves and K_{mix} = const > 0. The proof is based on the following scheme, which was used later by many geometers.

1. O’Neill and Steil [O’NS] studied relative nullity foliations on submanifolds in space forms and have used the following property of matrix Riccati equation B′ + B^{2} + kE = 0 with K_{mix} = k = const:

the eigen-vectors of solution B(t) do not depend on t, and eigen-function λ(t) satisfies the scalar Riccati equation λ(t)′ + λ^{2}(t) + k = 0, and for k > 0 can’t be continued on the whole real line.

2. The key for the existence of eigen-vector for structural tensor B of totally geodesic foliation {L} is the following [Fer]:

in case dim L ≥ ρ(codim L) for any point m ∈ L there exist nonzero vectors x ∈ T_{m}L, y ∈ T_{m}L^{⊥} and the real number λ with the property B(x, y) = λy.

The inequality with dimensions in local Theorem [Fer] is exact, but is wrong without additional conditions for totally geodesic foliation with K_{mix} > 0 (in Riccati equation kE replaced by non-constant symmetric matrix R(t) > 0) [Rov1].

Lemma 2 [EK]. Let ∫_{0}^{+∞} f(t)dt and ∫_{−∞}^{0} f(t)dt converge conditionally and there exists a solution λ(t) of the Riccati equation λ′ + aλ^{2} + f = 0, (t ∈ ℝ). Then a)

lim_{s→+∞}λ(t) = lim_{s→−∞}λ(t) = 0,

b) ∫_{−∞}^{+∞} f(t)dt = −a ∫_{−∞}^{+∞} λ^{2}(t)dt ≤ 0 where equality holds iff λ = f = 0.

Lemma 3 [Gui]. Let λ : [0, +∞) → ℝ and μ : [−∞, +∞) → ℝ be continuous functions and a > 0. Then a) lim sup_{s→+∞} (λ(s) + a ∫_{0}^{s} λ^{2}(t)dt) ≥ 0 with equality iff

λ = 0, b) lim sup_{s→+∞} (μ(s) − μ(−s) + a ∫_{−s}^{s} μ^{2}(t)dt) ≥ 0 with equality iff μ = 0.

Lemma 4. Let a = const > 0, R(t) – a symmetric continuous (n × n)−matrix, and the Riccati equation B′ +aB^{2} +R(t) = 0, (t ∈ ℝ) admits a solution B(t) with a symmetric B(0). Then lim inf_{s→+∞} ∫_{−s}^{s} {trace R(t) + trace (σ^{2})} dt ≤ 0 and equality

holds iff trace B(t) = trace R(t) + trace (σ^{2}) ≡ 0. Moreover, if trace R(t) ≥ 0, then B(t) = R(t) ≡ 0.

Remark 2. Without the symmetry for B(0) Lemma 4 is wrong [Rov1]. Application of Lemma 4 for manifolds without conjugate points see in [EK].

Proof. Since B^{−}(0) = 0, then B(t) is symmetric for all t [Abe]. Then from (8) assuming f(t) = trace R(t)+trace (σ^{2}) and θ(t) = trace B(t) we have θ(s)−θ(−s)+

a ∫_{−s}^{s} θ^{2}(t)dt = − ∫_{−s}^{s} f(t)dt. Hence, by Lemma 3 b) lim sup_{s→+∞} (− ∫_{−s}^{s} f(t)dt) =

−lim inf_{s→+∞} ∫_{−s}^{s} f(t)dt ≥ 0. If equality holds, then θ = 0, and hence f = 0 from (8).

From trace R(t) ≥ 0 it follows trace R(t) = σ = 0, i.e. B = aθ = 0. Finally from given matrix Riccati equation we obtain R = 0. □

100 VLADIMIR ROVENSKII

2.2. Ergodic theory. Let SL_{0} – a unit tangent bundle of a leaf L_{0}, T_{t} – a geodesic flow on SL_{0}, μ – Liuville measure on the Borel sets of SL_{0}, which is σ−finite and invariant under T_{t}.

(Maximal Ergodic) Theorem 3 [Gui]. Let f be an integrable function on SL_{0} and let D ⊂ SL_{0} be T_{t}−invariant Borel set. Define E(f) = {v ∈ D : sup_{s>0} ∫_{0}^{s} f(T_{t}(v)) dt > 0}. Then ∫_{E(f)} f dμ ≥ 0. If we set T = T_{1}, and if A is a T −invariant Borel set and E[f] = {v ∈ A : sup_{n≥0} ∑_{j=0}^{n} f(T^{j}(v)) > 0} , then ∫_{E[f]} f dμ ≥ 0.

Let f_{0} be an integrable function on SL_{0}. Then the Borel sets

D^{+} = {v ∈ SL_{0} : ∫_{0}^{∞} f_{0}(T_{t}(v)) dt < ∞}, C^{+} = SL_{0} \ D^{+}  (25)

are T_{t}−invariant; they are independent of f_{0} in the following sence [Gui]: if f_{1} is another integrable function, then the intersection of set D_{1}^{+} = {v ∈ SL_{0} : ∫_{0}^{∞} f_{1}(T_{t}(v)) dt = ∞} with D^{+} has measure zero.

The components D^{+} and C^{+} are called, respectively, the dissipative and the conservative parts of the E.Hopf decomposition SL_{0} = D^{+} ∪ C^{+}. Let SL_{0} = D^{−} ∪ C^{−} be the decomposition of SL_{0} associated to the inverse geodesic flow {T_{−t}}, (t ∈ ℝ), i.e. D^{−} = {v ∈ SL_{0} : ∫_{−∞}^{0} f_{0}(T_{t}(v)) dt < ∞}, C^{−} = SL_{0} \ D^{−}. If D = D^{+} ∩ D^{−}, then for every integrable function f on SL_{0}, the Lebesgue integral

∫_{−∞}^{∞} f(T_{t}(v)) dt (26)

exists for almost all v ∈ D. Also, if g_{0} > 0 is integrable, then the T −invariant sets

D[T] = {v ∈ SL_{0} : ∑_{−∞}^{∞} g_{0}(T^{j}(v)) < ∞}, C(T) = SL_{0} \ D(T) are independent of g_{0}, up to subset of measure zero. Moreover, if g ≥ 0 is integrable, then ∑_{−∞}^{∞} g(T^{j}(v)) < ∞ for almost all v ∈ D(T).

Next we consider a measure on the space of orbits. Let ∼ be the equivalence relation in SL_{0} defined by v ∼ w ⟺ ∃j ∈ ℤ : T^{j}(v) = w. Let S̃L_{0} be the space of orbits, and denote by π : S̃L_{0} → SL_{0} the natural projection π(v) = [v]. Consider in S̃L_{0} the σ−algebra β̃ induced by π. Then π(E) is measurable for every Borel set E, because π^{−1}(π(E)) = ⋃_{−∞}^{∞} T^{j}(E).

Given a function f integrable on SL_{0}, then by (26) x ∈ D → ∫_{−∞}^{∞} f(T_{t}(x)) dt defines a measurable function on the Borel sets of D. Furthermore, it is constant in each orbit on D. Thus [x] ∈ π(D) → ∫_{−∞}^{∞} f(T_{t}(x)) dt defines a μ̃−measurable function on π(D).

Lemma 5 [Gui]. If f is an integrable function on SL_{0}, then

∫_{D} f dμ = ∫_{π(D)} ∫_{−∞}^{∞} f(T_{t}(v)) dt dμ̃.

Lemma 6. Let S^{n−1} ⊂ ℝ^{n} be a unit sphere, ω_{n−1} – its volume, A – a quadratic form in ℝ^{n}. Then \frac{1}{ω_{n−1}} ∫_{S^{n−1}} A(x) dx = trace A.

Proof. The function f(x) = A(x)−(trace A)x^{2} is, obviously, harmonic in ℝ^{n} and by the mean value Theorem the integral of f(x) over S^{n−1} is zero. □

THE INTEGRAL OF MIXED SCALAR CURVATURE 101

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Chair of Geometry, Mathematical Department Pedagogical State University Lebedeva st. 89, Krasnoyarsk-49, 660049, RUSSIA

E–mail : roven@edk.krasnoyarsk.su