LECTURE 25: THE LAPLACIAN AND VOLUME COMPARISON Today we discuss comparison under Ricci curvature condition. We first prove the Laplacian comparison theorem, which can be viewed as an averaged version of the Hessian comparison (under slightly stronger condition). Then we prove the very useful Bishop-Gromov volume comparison theorem, which can be viewed as an “integrated” version of the Laplacian comparison theorem. 1. The Laplacian Comparison Theorem Recall that if (M, g),(M, f g˜) are complete Riemannian manifolds, γ : [0, a] → M and ˜γ : [0, a] → Mf are minimizing normal geodesics with Ke +(t) ≤ K−(t) holds for all t ∈ [0, a]. and if Xq ∈ TqM and Xeq˜ ∈ Tq˜Mf are roughly the same, where q = γ(b), ˜q = ˜γ(b) and 0 < b < a, then we have ∇2 dp(Xq, Xq) ≤ ∇e 2 ˜dp˜(Xeq˜, Xeq˜). Moreover, the equality holds if and only if Ke +(t) = K−(t) for all t ∈ [0, b]. Since ∆ = Tr∇2 , the Hessian comparison theorem will imply a Laplacian comparison ∆dp(q) ≤ ∆e ˜dp˜(˜q). Note that by taking the trace of the Hessian, what we get is (up to a constant) “the average of the Hessian”. As a result, one can anticipate to weaken the comparison condition from sectional curvature to a weaker “averaged version”, namely, a comparison condition on Ricci curvature. Theorem 1.1 (The Laplacian Comparison Theorem). Let (M, g) be a Riemannian manifold, and γ : [0, l] → M a minimizing normal geodesic with γ(0) = p. Suppose Ric( ˙γ(t)) ≥ (m − 1)k. We denote by ∆ek, ˜d, γ˜ etc the corresponding objects in Mm k . Then ∆dp(γ(t)) ≤ ∆e ˜dp˜(˜γ(t)), ∀0 < t < l. Moreover, ∆dp(γ(b)) ≤ ∆e ˜dp˜(˜γ(b)) for some b < l if and only if for any 0 ≤ t ≤ b, Ric( ˙γ(t)) = (m−1)k, and any normal Jacobi field X along γ|[0,b] with X(0) = 0 is almost parallel, i.e. is of the form X = snk(t) snk(b) e(t), where e is a parallel vector field along γ. Proof. Fix b < l. As usual we let {e1(t), · · · , em(t)} be a parallel orthonormal frame along γ with e1(t) = ˙γ(t), and let {e˜1(t), · · · , e˜m(t)} be a parallel orthonormal frame along ˜γ with ˜e1(t) = γ˜˙(t). For any i ≥ 2, let Xi(τ ) be the normal Jacobi field along 1
2 LECTURE 25: THE LAPLACIAN AND VOLUME COMPARISON γ|[0,b] with Xi(0) = 0 and Xi(b) = ei(b), and let Xei(τ ) be the normal Jacobi field along ˜γ|[0,b] with Xei(0) = 0 and Xei(b) = ˜ei(b). Then for q = γ(b) we have ∆dp(q) = Xm i=2 (∇2 dp)q(ei(b), ei(b)) = Xm i=2 I(Xi , Xi) and similarly for ˜q = ˜γ(b), ∆e ˜dp˜(˜q) = Pm i=2 I(Xei , Xei). It remains to prove Xm i=2 I(Xi , Xi) ≤ Xm i=2 I(Xei , Xei). We shall apply the same trick that we played in the proof of the basic index comparison lemma, namely we transplant Xei to γ. For this purpose we first recall that under the condition “ Mf has constant sectional curvature k along ˜γ” (c.f. PSet 4), the normal Jacobi field Xei is given by Xei (t) = snk(t) snk(b) e˜i(t). So for each 2 ≤ i ≤ m we define on γ|[0,b] a vector field X ′ i (t) = snk(t) snk(b) ei(t). Obviously X′ i has the same boundary condition as the Jacobi field Xi . So we get I(Xi , Xi) ≤ I(X′ i , X′ i ). Now the conclusion follows from XI(X ′ i , X′ i ) = XZ b 0 |∇γ˙ X ′ i | 2 + Rm( ˙γ, X′ i , γ, X˙ ′ i ) dt = X i Z b 0 ( sn′ k (t) snk(b) ) 2 − ( snk(t) snk(b) ) 2K( ˙γ, ei) dt = Z b 0 (m − 1)(sn′ k (t) snk(b) ) 2 − ( snk(t) snk(b) ) 2Ric( ˙γ) dt ≤ Z b 0 (m − 1)(sn′ k (t) snk(b) ) 2 − ( snk(t) snk(b) ) 2 (m − 1)k dt = XZ b 0 |∇γ˜˙ Xei | 2 + Rmg(γ, ˜˙ Xei , γ, ˜˙ Xei) dt = XI(Xei , Xei). From the proof we see that the equality holds if and only if “Ric( ˙γ) = (m − 1)k, and Xi = X′ i for all i”. Since any normal Jacobi field X along γ with X(0) = 0 is a linear combination of these Xi , the conclusion follows. □ Remark. As we have seen, (∇2dp)q(Xq, Yq) = ⟨∇X1N, Yq⟩, where N = ∂r is the outward unit normal vector of the geodesic sphere S(p, d(p, q)) at q. In other words, we may replace ∇X1N by the shape vector (∇X1N) ⊥ (c.f. PSet 2 Problem 9) and conclude that ∇2dp is the second fundamental form of the geodesic sphere S(p, d(p, q)). It follows that ∆dp(q) = Tr(∇2dp) is the trace of the second fundamental form, i.e. the ✿✿✿✿✿✿ mean ✿✿✿✿✿✿✿✿✿✿ curvature of the geodesic sphere S(p, d(p, q)) at q
LECTURE 25: THE LAPLACIAN AND VOLUME COMPARISON 3 2. The Bishop-Gromov Volume Comparison Theorem ¶ The volume measure in coordinates. Recall that the Riemannian volume density is defined in a chart (φ, U, V ) as dVg = √ G ◦ φ −1 dx1 · · · dxm, where G = det(gij ) and dx1 · · · dxm is the Lebesgue measure on R m. Now let (M, g) be a complete Riemannian manifold. Although in general there is no global chart, there is a large chart that covers almost the whole of M, namely {exp−1 p , M\Cut(p), Σ(p)}, which is defined except for the measure zero closed subset Cut(p), where Σ(p) = exp−1 (M \ Cut(p)) is an open star-shaped domain in TpM. In particular, on TpM we may use polar coordinates and write dx1 · · · dxm = r m−1 drdΘ, where dΘ is the usual surface measure on S m−1 . Combine this with the chart {exp−1 p , M \Cut(p), Σ(p)} we get dVg = √ G(expp (rΘ))r m−1 drdΘ, rΘ ∈ Σ(p). We denote µp(r, Θ) = (√ G(expp (rΘ))r m−1 , rΘ ∈ Σ(p), 0, rΘ ̸∈ Σ(p). Note that by definition Br(p) = expp (Br(0)) = expp (Br(0) ∩ Σ(p) ). Since Cut(p) is of measure zero in M, we get Vol(Br(p)) = Z Br(0)∩Σ(p) µ(t, Θ)dtdΘ = Z Br(0) µ(t, Θ)dtdΘ. Example. We may calculate the function µp(r, Θ) for the three model spaces, • R m: µ(r, Θ) = r m−1 . • S m: µ(r, Θ) = sinm−1 (r). • Hm: µ(r, Θ) = sinhm−1 (r). ¶ The volume measure via Jacobi fields. Observe that the three functions µ(r, Θ) in the previous example are closely related to the Jacobi fields on the three model spaces. This is not a coincidence:
4 LECTURE 25: THE LAPLACIAN AND VOLUME COMPARISON Proposition 2.1. Given any Θ ∈ SpM and write γ(t) = expp (tΘ). Then for any basis v2, · · · , vm of γ˙(0)⊥, if we let Vj (t) (i ≥ 2) be the normal Jacobi fields along γ with Vj (0) = 0 and ∇γ˙ (0)Vj = vj , then for any point rΘ ∈ Σ(p), µp(r, Θ) = det(V2(r), · · · , Vm(r)) det(∇γ˙ (0)V2, · · · , ∇γ˙ (0)Vm) . Proof. Using v1 = Θ, v2, · · · , vm as a basis one can define a set of global linear coordinates u1, · · · , um on TpM, u ∈ TpM ⇝ u = u1v1 + · · · + umvm. Then we have du1 · · · dum = r m−1 | det(v2, · · · , vm)| drdΘ. Since Vj is a Jacobi field along γ with Vj (0) = 0, we have Vj (t) = t(d expp )tΘ(vj ). Note that under exp−1 p , (u1, · · · , um) also gives a coordinate system near expp (rΘ) for rΘ ∈ Σ(p). With this coordinate system, we have (at expp (rΘ)) ∂1 = ˙γ(r), ∂j = 1 r Vj (r) (j ≥ 2). It follows gij (expp (rΘ)) = ⟨∂i , ∂j ⟩|expp (rΘ) = 1 r 2 ⟨Vi(r), Vj (r)⟩ for i, j ≥ 2. Since γ is a normal geodesic and since each Vj is a normal Jacobi field, we have ⟨∂1, ∂1⟩ = 1 and ⟨∂1, ∂i⟩ = 0 for i ≥ 2. So we get, at expp (rΘ), G = det(gij ) = r −2m+2 det(⟨Vi , Vj ⟩)i,j≥2 = r −2m+2 det(V2(r), · · · , Vm(r))2 . It follows dVg = √ G(expp (rΘ))du1 · · · dum = det(V2(r), · · · , Vm(r)) det(∇γ˙ (0)V2, · · · , ∇γ˙ (0)Vm) drdΘ. This completes the proof. □ ¶ The volume measure v.s. the Laplacian of distance. The crucial observation is Lemma 2.2. Suppose Θ ∈ SpM and rΘ ∈ Σ(p), then µ ′ p (r, Θ) µp(r, Θ) = (∆dp)(expp (rΘ)), where the derivative is taken with respect to r
LECTURE 25: THE LAPLACIAN AND VOLUME COMPARISON 5 Proof. Let γ(t) = expp (tΘ) (0 ≤ t ≤ r) be the normal geodesic starting at p in the direction Θ. Consider a parallel orthonormal frame {ei(t)} along γ with e1(t) = ˙γ(t). Let Vj (t) be a Jacobi field along γ such that Vj (0) = 0 and Vj (r) = ej (r). Then at q = γ(r) we have ∆dp(expp (rΘ)) = Xm i=2 (∇2 dp)q(ei(t), ei(t)) = Xm i=2 I(Vi , Vi). On the other hand, if we denote A(t) = (⟨Vi(t), Vj (t)⟩)i,j≥2, then A(r) = Id, and the derivative of d(t) = det A(t) = (det(V2(t), · · · , Vm(t)))2 is d ′ (t) = d(t)Tr(A −1 (t)A ′ (t)). Thus we get µ ′ p (r, Θ) µp(r, Θ) = 1 2 d ′ (r) d(r) = 1 2 Tr(A ′ (r)) = Xm j=2 ⟨Vj (r), ∇γ˙ (r)Vj (r)⟩ = Xm j=2 I(Vj , Vj ), so the conclusion follows. □ ¶ Comparison of volume elements. In particular if we denote by µk(r) the function µ(r, Θ) for the space Mm k , i.e. µk(r) = sinm−1 ( √ kr), k > 0, r m−1 , k = 0, sinhm−1 ( √ −kr), k 0, (3) if µp(t, Θ)=µk(t) for t ∈ [a, r] and any Θ, then B(p, r) is isometric to Bk(r). Proof. (1) The monotonicity of µp(r,Θ) µk(r) follows from d dt log µp(t, Θ) µk(t) = µ ′ p (t, Θ) µp(t, Θ) − µ ′ k (t) µk(t) ≤ 0. (2) By Vj (r) = r∇γ˙ (0)Vj +O(r 2 ) we get µp(r, Θ) = r m−1 +O(r m). The result follows
6 LECTURE 25: THE LAPLACIAN AND VOLUME COMPARISON (3) If µp(t, Θ)=µk(t) for t ≤ r and any Θ, then (∆dp)(expp (tΘ)) = µ ′ p (t, Θ) µp(t, Θ) = µ ′ k (t) µk(t) = (∆kdk)(t). It follows that Ric( ˙γ(t)) = (m − 1)k, and (since Θ and thus γ are arbitrary) any normal Jacobi field along any geodesic starting at p is almost parallel. By PSet 4, (M, g) has constant sectional curvature, and thus the constant has to be k. Finally by Cartan’s local isometry theorem, B(p, r) is isometric to Bk(r) in Mm k . □ ¶ The Bishop-Gromov volume comparison theorem. Note that for t large, it may happen that tΘ ̸∈ Σ(p). However, in this case µp(t, Θ) = 0. So the monotonicity holds for all t. It is this simple observation that leads to a global comparison instead of a local comparison inside the interior radius. In fact, by integrating the volume density we get Theorem 2.4 (Bishop-Gromov). If (M, g) is a complete Riemannian manifold with Ric ≥ (m−1)k, and p ∈ M is an arbitrary point. Let Sk(r) and Bk(r) be the metric sphere and the metric ball of radius r in Mm k . Then the functions Area(S(p, r)) Area(Sk(r)) and Vol(B(p, r)) Vol(Bk(r)) are non-increasing in r, and both tends to 1 as r → 0+. Moreover, the quotient is a constant for r ∈ [r1, r2] if and only if B(p, r2) is isometric to Bk(r2). Proof. By definition Area(S(p, r)) = Z Sm−1 µp(r, Θ)dΘ, Vol(B(p, r)) = Z r 0 Z Sm−1 µp(r, Θ)dΘdr. Thus if we denote the surface area of the sphere S m−1 ⊂ R m by ωm−1, then Area(S(p, r)) Area(Sk(r)) = R Sm−1 µp(r, Θ)dΘ R Sm−1 µk(r)dΘ = 1 ωm−1 Z Sm−1 µp(r, Θ) µk(r) dΘ is non-increasing and tends to 1 as r → 0. As a consequence, d dr logVol(B(p, r)) Vol(Bk(r)) = Area(S(p, r)) Vol(B(p, r)) − Area(Sk(r)) Vol(Bk(r)) = R r 0 (Area(S(p, r))Area(Sk(t)) − Area(Sk(r))Area(S(p, t)))dt Vol(B(p, r))Vol(Bk(r)) ≤0 and thus Vol(B(p,r)) Vol(Bk(r)) is also non-increasing in r, and tends to 1 as t → 0. □ Corollary 2.5. We have Area(S(p, r)) ≤ Area(Sk(r)), and Vol(B(p, r)) ≤ Vol(B(p, r)) for all r ≥ 0. Moreover, equality holds if and only if B(p, r) is isometric to Bk(r)