A new proof of a theorem of Petersen
Yi-Hu Yang∗ and Yi Zhang
Abstract
Let M be an n-dimensional complete Riemannian manifold with
Ricci curvature ≥ n − 1. In [4, 5], Tobias Colding, by developing
some new techniques, proved that the following three condtions: 1)
dGH (M, S n ) → 0; 2) the volume of M Vol(M ) → Vol(S n ); 3) the radius
of M rad(M ) → π are equivalent. In [9], Peter Petersen, by developing a different technique, gave the 4-th equivalent condition, namely he
proved that the n + 1-th eigenvalue of M λn+1 (M ) → n is also equivalent to the radius of M rad(M ) → π, and hence the other two. In this
note, we give a new proof of Petersen’s theorem by utilizing Colding’s
techniques.
MSC Classification: 53C20, 53C21, 53C23
Keywords and phrases: radius, eigenvalues, Gromov-Hausdorff distance
1
Introduction
Let M be an n-dimensional complete Riemannian manifold with Ricci curvature ≥ n − 1. In [4, 5], T. Colding proved that the following three conditions
are equivalent: 1) dGH (M, S n ) → 0; 2) the volume of M Vol(M ) → Vol(S n );
3) the radius of M rad(M ) → π. To this end, he developed some new techniques and got some local L2 -estimates of distances and angles (for details, see
§2). On the other hand, by developing a completely different technique, Peter
Petersen later got the 4-th equivalent condition [9], i.e. proved the following
theorem.
Petersen’s Theorem: Let M be an n dimensional complete Riemannian
manifold with Ricci curvature RicM ≥ n − 1. Then the radius rad(M ) of M
is close to π if and only if the n + 1-th eigenvalue λn+1 (M ) of M is close to n.
∗
Supported partially by NSF of China (No.11171253)
1
2
The aim of this note is to give a new proof of Petersen’s theorem by utilizing
Colding’s techniques. As pointed out by Petersen in the introduction of [9]
(also cf. [5], Remark 4.21), Colding may be able to prove something like the
above theorem by using the technique developed by himself. So, this note can
be also considered as a supplement to this. Here, we want to thank Peter
Petersen for friendly pointing out this to us.
In [1, 8], Bertrand and Honda give some (indirect but very natural) generalizations of Petersen’s theorem by showing (roughly speaking) that the sufficient
pinching of the first k (1 ≤ k ≤ n + 1) eigenvalues can result in a GromovHausdorff approximation of the manifolds in question to a (k − 1)-dimensional
suspension of a geodesic subspace of M ; in the case of k = n + 1, this together
with Colding’s theorem implies Petersen’s theorem. Our main proof is directly to show that the sufficient pinching of the first n + 1 eigenvalues implies
the radius of the manifold in question being sufficiently close to that of the
standard sphere S n by completely using Colding’s techniques.
Finally, the first author also thanks Xiaochun Rong for several valuable
talks about this note.
2
Colding’s local L2-estimates of distances and
angles
In this section, we recall Colding’s integral estimates of distances and angles.
We first fix some notation. Let N be a closed Riemannian manifold. For
p ∈ [1, ∞) and f ∈ Lp (N ), set
Z
1
1
|f |p dvol p .
kf kp =
Vol(N ) N
In the sequel, we will always omit ”dvol” in the integral. Let f be a Lipschitz
function and ∇f the gradient of f , then by kf k2,1 denote the (2, 1)-Sobolev
norm of f , i.e.
Z
Z
1
1
1
2
kf k(2,1) =
|f | +
|∇f |2 2 .
Vol(N ) N
Vol(N ) N
For f ∈ C 2 (N ), define the Hessian of f as follows
Hess(f )(u, v) =< ∇u ∇f, v >,
and the laplacian as follows
∆f = TrHess(f ).
So, ∆ is negative semi-definite.
3
Recall also that f is an eigenfunction with eigenvalue λ if ∆f +λf = 0. Let
SN be the unit tangent bundle of N , g t is the geodesic flow and π : SN → N
the corresponding projection. For A ⊂ N , set SA = π −1 (N ). For SN , one has
an obvious Riemannian measure which is induced from that of N . So, we will
integrate on various sets, e.g. N, SN, SA × [0, l], etc.
If V is a finite dimensional vector space with an inner product < ·, · > and
ω is a bilinear from on V , then we set
|ω|2 = Σi,j (ω(ei , ej ))2 ,
where {ei } is an orthonormal basis of V .
In this note, unless stated otherwise, we always assume that M is an ndimensional complete Riemannian manifold with the Ricci curvature satisfying
RicM ≥ n − 1, n ≥ 2.
For f ∈ C ∞ (M ) and l > 0, define a function hf ∈ C ∞ (SM × [0, l]) as
follows
hf (v, t) = f (γv (0)) cos t +
f (γv (l)) − f (γv (0)) cos l
sin t,
sin l
(1)
where γv (t) is the geodesic at π(v) along the direction v. The function obviously
satisfies
∂ 2 hf
= −hf
∂t2
(2)
and is also determined uniquely under the conditions: hf (v, 0) = f (γv (0)),
hf (v, l) = f (γv (l)). We remark that if Hess(f ) = −f g (here g is the Riemannian metric on M ), then hf (v, t) = f (γv (t)).
Proposition 1 ([4], Lemma 1.4) Given , k, and l ∈ [ π2 , π], there exists a
positive number δ = δ(, k, l, n) such that if f ∈ C ∞ (M ) with kf k2 ≤ k and
k∆f + nf k2 ≤ δ, then
1
lVol(SM )
Z
l
Z
SM
|f (γv (t)) − hf (v, t)|2 dt < ,
(3)
∂f (γv (t)) ∂hf (v, t) 2
dt < .
−
∂t
∂t
(4)
0
and
1
lVol(SM )
Z
SM
Z
0
l
4
In the sequel of this note, we will use ψ(δ|·, · · · , ·) to denote a certain
positive function depending on δ and some additional parameters such that
when these parameters are fixed, limδ→0 ψ(δ|·, · · · , ·) = 0.
For any f ∈ C ∞ (M ), we define another function gf on SM ×[0, l] as follows
gf (v, t) =< ∇f, v > sin t + f (π(v)) cos t.
(5)
Similar to the hf before, it is easily to see that gf (v, t) is uniquely determined
by
∂ 2 gf
= −gf ,
∂t2
gf (v, 0) = f (π(v)),
∂gf
(v, 0) =< ∇f, v > .
∂t
(6)
(7)
(8)
The following proposition also is essentially due to Colding. We only make
a little bit modification so that we can apply it conveniently in the proof of
the next section.
Proposition 2 ([5], Proposition 4.5) Given , k, Ā, C > 0 and l ∈ [ π2 , π),
there exists a positive number δ = δ(n, k, , l, Ā, C) such that if A ⊂ M with
V ol(A)
≥ Ā and f ∈ C ∞ (M ) with kf k2 ≤ k, k∆f + nf k2 ≤ δ, max |f |, and
V ol(M )
max |∇f | ≤ C, then we have
Z Z l
1
|gf (v, t) − f (γv (t))|2 dt < ,
(9)
l3 Vol(SA) SA 0
and
1
lVol(SA)
Z
SA
Z
0
l
∂gf (v, t) ∂f (γv (t)) 2
dt < .
−
∂t
∂t
(10)
Proof. The first estimate easily follows from the second one by integration.
By the definition of hf and the boundedness of f and |∇f |, for ∀s ∈ [0, l] we
have
|hf (v, s) − f (γv (s))|2 ≤ |hf (v, s) − hf (v, 0)|2 + |f (γv (s)) − f (γv (0))|2
< ψ(|n, C).
By Proposition 1, we also have, for some δ = δ(2 , k, l, n),
Z
Z l
∂hf (v, t) ∂f (γv (t)) 2
1
dt < 2 ,
−
lVol(SM ) π−1 Tl (A) 0
∂t
∂t
5
where π −1 Tl (A) is the -neighborhood of A. In particular, ∃s0 ∈ [0, l] such
that
Z
∂hf (v, s0 ) ∂f (γv (s0 )) 2
1
< .
−
Vol(SM ) π−1 Tl (A)
∂t
∂t
On the other hand, by the definition of hf and gf , we have
0
∂f (γv ) 0
∂gf (g s v, 0)
(s ) =
.
∂t
∂t
Also, for the boundedness of f and |∇f |, we have
∂hf (g s0 v, 0) ∂hf (v, s0 ) < ψ(|n, · · · ).
−
∂t
∂t
Thus, we have
∂hf (g s0 v, 0) ∂gf (g s0 v, 0) < ψ(|n, · · · ).
−
∂t
∂t
0
0
Then, the fact that hf (g s vt) and gf (g s v, t) satisfy the same equation f 00 = −f ,
0
0
together with hf (g s v, 0) = gf (g s v, 0) implies
1
lVol(SM )
Z
π −1 Ts0 (A)
Z
0
l
∂gf (g s0 v, t) ∂hf (g s0 v, t) 2
dt < ψ(|n, C...).
−
∂t
∂t
Using the fact that the geodesic flow is volume-preserving, we get
Z Z l
∂gf (v, t) ∂hf (v, t) 2
1
dt < ψ(|n, C...).
−
lVol(SM ) SA 0
∂t
∂t
Then, we have
Z Z l
∂gf (v, t) ∂hf (v, t) 2
1
dt < Vol(SM ) ψ(|n, C...)
−
lVol(SA) SA 0
∂t
∂t
Vol(SA)
= ψ(|n, C, Ā...).
Combining this with Proposition 1, we get the required inequality with replaced by ψ(|k, l, n, C, Ā).
Remark. When the set A in the proposition is sufficiently small and l sufficiently close to π, γv (t) can run over M , but gf (v, t) depends only on f|A . So,
such an f has much more geometric information of M . This is the key point
of the proposition
6
3
3.1
The proof of Petersen’s theorem
λn+1 → n implies rad(M ) → π
We first give some preliminaries. Let p ∈ M . Set r(p) = min{r|Br (p) ⊃ M },
the radius at p of M . Then the radius of M is defined as
rad(M ) = max r(p).
p∈M
We also give the following
Definition 1 For p, q ∈ M and δ, s > 0, set
Cs = {v ∈ SM |π(v) ∈ Bδ (p), exp(sv) ∈ Bδ (q), d(exp(sv), π(v)) = s}.
The following lemma is also due to Colding ([4], Lemma 2.3); for convenience
in the following proofs, we here give it a detailed proof.
Lemma 1 ∀δ > 0 and p, q ∈ M , then there exists an s0 > 0 with
Vol(Cs0 ) ≥
n Vn (δ)Vol(M ) 2
,
πn
ωn
where Vn (δ) denotes the volume of the ball with radius δ in the standard nsphere S n , ωn is the volume of S n .
Proof. Set
C = {v ∈ T M |π(v) ∈ Bδ (p), exp(v) ∈ Bδ (q), d(exp(v), π(v)) = |v|}.
For x ∈ M , by Bishop’s volume comparison, one has that the exponential map
expx at x, when restricted to the set
Cx = {v ∈ Tx M |d(expx (v), x) = |v|},
is volume non-increasing. In particular, for all x ∈ Bδ (p), the Bishop-Gromov
comparison theorem implies
VolRn (Tx M ∩ C) ≥ Vol(Bδ (q)) ≥
Vn (δ)Vol(M )
.
ωn
Thus, one has
Vol(C) ≥
Vn (δ)Vol(M ) 2
.
ωn
On the other hand, one has
Z π
πn
Vol(C) =
Vol(Cs )sn−1 ds ≤
max VolS M (Cs ).
n s>0
0
7
Therefore, there exists s > 0 with
Vol(Cs ) ≥
n Vn (δ)Vol(M 2
.
πn
ωn
The lemma is obtained.
Let n ≤ λ1 ≤ λ2 ≤ · · · ≤ λn+1 be the first n + 1 nonzero eigenvalues of M
and f1 , f2 , · · · , fn+1 be the corresponding eigenfunctions respectively, i.e.
∆fi + λi fi = 0, i = 1, 2, · · · , n + 1;
R
furthermore, we can assume that for i 6= j, M fi fj = 0. We can normalize
each fi to make it satisfy
Z
1
fi2 = 1.
Vol(M ) M
Let a1 , a2 , · · · , an+1 be n + 1 real numbers satisfying a21 + a22 + · · · + a2n+1 = 1.
Set f = a1 f1 + a2 f2 + · · · + an+1 fn+1 . Write λn+1 = n + δ1 . Since we only
1
concern the case of λn+1 near n, so WLOG, we can assume δ1 ≤ n+1
. From
the definition of f , we have
Z
1
f 2 = 1,
Vol(M ) M
and
n+1
∆f + nf = Σn+1
i=1 ai (∆fi + nfi ) = Σi=1 ai (n − λi )fi .
Furthermore, we have
k∆f + nf k2 ≤ δ1 .
So, for all a1 , a2 , · · · , an+1 with a21 + · · · + a2n+1 = 1, we have
k∆f + nf k2 → 0 uniformly, as λn+1 → n.
Moreover, by the standard estimate of PDE of elliptic type, we can assume that
there exists a positive constant C (independent of a1 , · · · , an+1 and depending
only on the dimension n) such that for the above f one has
max |f | ≤ C
M
and
max |∇f | ≤ C.
M
In general, one can not get a universal pointwise estimate of the hessian of
such an f . But the control of k∆f + nf k2 gives a control of kHess(f ) + f gM k2
(cf. [4, 9]) and hence of kHess(f )k2 , here gM is the Riemannian metric of M .
Then, the following segment inequality due to Cheeger and Colding (cf. [3, 2])
gives some necessary (partial) control of |Hess(f )|, which is enough for the
proof of the following Lemma 4.
8
Lemma 2 For y1 , y2 ∈ M , denote γy1 ,y2 is a minimal geodesic connecting y1
and y2 . Then we have
Z
Z d(y1 ,y2 )
Z
2
|Hess(f )(γy1 ,y2 (s))| ds ≤ C(n)
|Hess(f )|2 ,
M ×M
0
M
where C(n) is a universal constant depending only on the dimension n.
In the following, when we mention a smooth function f on M , we always
mean a function like the above, unless stated otherwise.
Using these preliminaries and Colding’s local integral estimates of distances
and angels (Proposition 2), we can now prove the following two main lemmas.
Lemma 3 For arbitrarily given , there exists δ10 = δ10 (), δ20 = δ 0 () and
δ3 = δ3 () such that if λn+1 ≤ n + δ10 , and for some p ∈ M and some α ≤ 8 ,
some smooth function f on M as mentioned before satisfies
Z
1
|∇f |2 < δ20
Vol(Bα (p)) Bα (p)
and
1
Vol(Bα (p))
Z
|f |2 < δ3 ,
Bα (p)
then r(p) > π − .
Proof: Assume r(p) ≤ π − . Then Bπ− (p) ⊃ M . Applying Proposition 2
with A = Bα (p) and l = π − 2 , we can choose a δ10 = δ10 () = δ(n, 2(π−
,π −
)
Vol(S n (α))
,
, C)
2
ωn
2
satisfying
1
(π − 2 )Vol(SBα (p))
Z
SBα (p)
Z
0
π− 2
|gf (v, t) − f (γv (t))|2 ≤ .
2
On the other hand, we have
Z
1
1 =
f2
Vol(M ) M
Z
Z π− 2
(π − 2 )n−1
≤
|f (γv (t))|2
Vol(M )Vol(Bα (p)) SBα (p) 0
Z
Z π− 2
C0
2
2
≤
|f
(γ
(t))
−
g
(v,
t)|
+
|g
(v,
t)|
v
f
f
(π − 2 )Vol(SBα (p)) SBα (p) 0
≤ C 00 ( + δ2 + δ3 ).
2
So, when , δ2 , and δ3 are sufficiently small, this derives a contradiction. The
lemma is obtained.
9
0
2
δ3 Lemma 4 For arbitrarily given and δ3 , set α0 = min{ C80C
, 8 } (we take
1
00
00
00
00
0
C ≤ 10 later). Then there exist δ1 = δ1 (, α0 , δ3 ) and δ2 = δ2 (, δ3 ) satisfying
that if λn+1 ≤ n + δ100 and, for some smooth function f as mentioned before,
p ∈ M and some α ≤ α0 ,
Z
1
|∇f |2 < δ200
Vol(Bα (p)) Bα (p)
and
1
Vol(Bα (p))
Z
|f |2 ≥ δ3 ,
Bα (p)
then r(p) > π − .
Proof. Assume r(p) ≤ π − . Let f be a smooth function mentioned before,
p ∈ M and α ≤ α0 . For any v, v 0 ∈ SM , t, t0 ∈ [0, π − 4 ], we have
|f (γv (t)) − gf (v, t)|
= |f (γv0 (t0 )) − gf (v 0 , t0 ) − gf (v, t) + gf (v 0 , t0 ) − f (γv0 (t0 )) + f (γv (t))|
≥ |gf (v, t) − gf (v 0 , t0 )| − |f (γv0 (t0 )) − gf (v 0 , t0 )| − |f (γv0 (t0 )) − f (γv (t))|,
so, we have
|f (γv (t)) − gf (v, t)| + |f (γv0 (t0 )) − gf (v 0 , t0 )|
≥ |gf (v, t) − gf (v 0 , t0 )| − |f (γv0 (t0 )) − f (γv (t))|.
(11)
R
Since Vol(B1α (p)) Bα (p) |f |2 ≥ δ3 , there exists a q1 ∈ Bα (p) such that |f (q1 )| ≥
δ3 . From this and the fact that |∇f | ≤ C, ∃r0 = r0 (, δ3 , C) ≤ 32 such that
for any r ≤ r0 and q2 ∈ Br (q1 ), |f (q2 )| ≥ δ23 .
We first derive a lower bound of the term |gf (v, t) − gf (v 0 , t0 )| for some
suitable v, v 0 , t, t0 . By the segment ineguality, it is clear that, for any given
1 > 0, there exists a subset SBr0 (q1 ) ⊂ SBr (q1 ) 1 satisfying Vol(SBr0 (q1 )) >
(1 − )Vol(SBr (q1 )) and that for any v ∈ SBr0 (q1 ) there exists SBβ0 (π(v)) ⊂
SBβ (π(v)) (β will be fixed in the following) satisfying that Vol(SBβ0 (π(v))) >
(1 − )Vol(SBβ (π(v))) and for any v 0 ∈ SBβ0 (π(v))
|f (π(v 0 ))| ≥
δ3
4
and
||∇f (π(v))| − |∇f (π(v 0 ))|| ≤ C(1 )β,
Here, by SBr0 (q1 ) and the following SBβ0 (π(v)) and SBr00 (q1 ), etc., we mean some subsets
which can generate some minimal geodesics on which the hessian of f has uniform control.
1
10
here C(1 ) is some constant depending only on ||Hess(f )||2 and 1 . If β is
sufficiently small, by combining these estimates with |∇f | ≤ C, we then have
|gf (v, t) − gf (v 0 , t0 )|
≥ |f (π(v))|| cos t − cos t0 | − |f (π(v)) − f (π(v 0 ))| − |∇f (π(v))| − |∇f (π(v 0 ))|
≥ |f (π(v))|| cos t − cos t0 | − Cβ − C(1 )β − 2|∇f (π(v))|.
Taking t ∈ [π − 2 , π − 4 ] and t0 ∈ [0, π −
3
],
4
we then have
|gf (v, t) − gf (v 0 , t0 )| ≥ C 0 δ3 2 − Cβ − C(1 )β − 2|∇f (π(v))|,
here C 0 ≤
1
.
10
0
2
C δ3 Take β ≤ min{ 20 max{C,C(
, }. We then have
1 )} 32
9 0 2
C δ3 − 2|∇f (π(v))|.
10
|gf (v, t) − gf (v 0 , t0 )| ≥
Similar to the above, we can also find sufficient large subsets SBα0 (p) ⊂ SBα (p)
and SBr00 (q1 ) ⊂ SBr (q1 ) with the same volume condition as above so that we
have
|Hess(f )(v, v)| ≤ C(1 )
R
for any v ∈ SBα0 (p)∪SBr00 (q1 ). On the other hand, since Vol(B1α (p)) Bα (p) |∇f |2 ≤
δ200 (δ200 will be fixed in the following), so for any v ∈ SBr00 (q1 ), we have
|∇f (π(v))| ≤ C(1 )r + 2C(1 )α + δ200 .
Taking δ200 ≤
C 0 δ3 2
80
0
2
C δ3 and r0 ≤ min{ 80C(
, }, we then have
1 ) 32
|∇f (π(v))| ≤
C 0 δ3 2
.
20
Consequently, we have
|gf (v, t) − gf (v 0 , t0 )| ≥
8 0 2
C δ3 ,
10
(12)
for any (v, t) ∈ SBr000 (q1 )×[π− 2 , π− 4 ] and any (v 0 , t0 ) ∈ SBβ0 (π(v))×[0, π− 34 ],
where SBr000 (q1 ) = SBr0 (q1 ) ∩ SBr00 (q1 ) satisfying
Vol(SBr000 (q1 )) > (1 − 21 )Vol(SBr (q1 )).
Next, we want to derive an upper bound of the term |f (γv (t)) − gf (v, t)| for
some suitable (v, t) ∈ SBr (q1 ) × [π − 2 , π − 4 ]. Setting l = π − 4 and applying
Proposition 2 on A = B 4 (p), we then have
Z
Z π− 4
1
|gf (v, t) − f (γv (t))|2 dt < ψ0 (δ100 |),
(π − 4 )3 Vol(SB 4 (p)) SB (p) 0
4
11
here ψ0 (δ100 |) → 0 as δ100 → 0 (δ100 will be fixed in the following).
We also remark that α ≤ α0 ≤ 8 and r ≤ r0 ≤ 32 , and hence SBr000 (q1 ) ⊂
SBr (q1 ) ⊂ SB 4 (p) with Vol(SBr000 (q1 )) > (1 − 21 )Vol(SBr (q1 )). So, by the
volume comparison theorem, we have
Z
Z π− 4
1
(13)
|gf (v, t) − f (γv (t))|2 dt < ψ1 (δ100 |).
000
Vol(SBr (q1 )) SBr000 (q1 ) π− 2
Thus, for sufficiently small δ100 and some (v, t) ∈ SBr000 (q1 ) × [π − 2 , π − 4 ], we
have
p
1
|f (γv (t)) − gf (v, t)| < ψ1 (δ100 |) ≤ C 0 δ3 2 .
(14)
10
Combining this with (11) and (12), we have, for sufficiently small δ100 ,
|f (γv0 (t0 )) − gf (v 0 , t0 )| + |f (γv0 (t0 )) − f (γv (t))| ≥
7 0 2
C δ3 ,
10
(15)
for some (v, t) ∈ SBr000 (q1 )×[π− 2 , π− 4 ] and any (v 0 , t0 ) ∈ SBβ0 (π(v))×[0, π− 34 ].
Now, we discuss the terms |f (γv0 (t0 )) − gf (v 0 , t0 )| and |f (γv0 (t0 )) − f (γv (t))|
in (15). To this end, we need Lemma 1. For convenience, we first set
C(v, t) = {(v̄, s) ∈ SM × [0, π] | π(v̄) ∈ Bβ (π(v)), exp(sv̄) ∈ Bβ (γv (t)),
and d(exp(sv̄), π(v̄)) = s};
0
C (v, t) = {v̄ ∈ T M | π(v̄) ∈ Bβ (π(v)), exp(v̄) ∈ Bβ (γv (t)),
and d(exp(v̄), π(v̄)) = |v̄|};
C(s; v, t) = {v̄ ∈ SM | π(v̄) ∈ Bβ (π(v)), exp(sv̄) ∈ Bβ (γv (t)),
and d(exp(sv̄), π(v̄)) = s}.
By the definition of C(v, t) and the assumption r(p) ≤ π − , we have, if
(v̄, s) ∈ C(v, t),
s≤π−+α+r+β+β ≤π−
3
25
<π− .
32
4
So, C(v, t) ⊂ SBβ (π(v)) × [0, π − 34 ] ⊂ SB 4 (p) × [0, π − 34 ]. Similarly, for
v̄ ∈ C 0 (v, t), |v̄| ≤ π − 34 ; and for s ≥ π − 34 , C(s; v, t) = φ. We also have
Z
Vol(C(v, t)) =
π
Z
π− 3
4
Vol(C(s; v, t))ds =
Vol(C(s; v, t))ds
0
0
and
0
Z
Vol(C (v, t)) =
π
Vol(C(s; v, t))s
0
n−1
Z
ds =
0
π− 3
4
Vol(C(s; v, t))sn−1 ds.
12
On the other hand, by the proof of Lemma 1, we have
Vn (β)Vol(M ) 2
.
ωn
Vol(C 0 (v, t)) ≥
So, we have
Z
π− 3
4
Vol(C(v, t)) =
Vol(C(s; v, t))ds
0
Z
1
≥
(π −
3 n−1
)
4
1
=
(π −
(π −
3 n−1
)
4
Vol(C(s; v, t))sn−1 ds
0
3 n−1 Vol(C
)
4
1
≥
π− 3
4
0
(v, t))
Vn (β)Vol(M ) 2
.
ωn
Again applying Proposition 2 with A = M and l = π − 4 , we have
Z
1
Vol(SM )
SM
π− 4
Z
|gf (v̄, s) − f (γv̄ (s))|2 ds < ψ2 (δ100 |).
0
Since C(v, t) ⊂ SBβ (π(v)) × [0, π −
1
Vol(C(v, t))
Z
0
π− 3
4
Z
3
]
4
⊂ SM × [0, π − 4 ], we have
|gf (v̄, s) − f (γv̄ (s))|2 ds < ψ3 (δ100 |).
C(s;v,t)
So, for sufficiently small δ100 (which now can be fixed), there exists an (v 0 , t0 ) ∈
C(v, t) ∩ SBβ0 (π(v)) × [0, π − 34 ] ⊂ SBβ (π(v)) × [0, π − 34 ] satisfying
|gf (v 0 , t0 ) − f (γv0 (t0 ))|2 <
p
1
ψ3 (δ100 |) ≤ C 0 δ3 2 .
10
(16)
Since (v 0 , t0 ) ∈ C(v, t), (by the definition of C(v, t)) we have γv0 (t0 ) ∈ Bβ (γv (t)),
so
|f (γv0 (t0 )) − f (γv (t))| ≤ Cβ ≤
1 0 2
C δ3 .
10
(17)
Combining (17) with (15) and (16), we derive a contradiction. The lemma is
obtained.
Clearly, Lemma 3 and Lemma 4 together imply the following
13
Theorem 1 For arbitrarily given > 0, there exist δ1 = δ1 () > 0, and
δ2 = δ2 () > 0 such that if λn+1 ≤ n + δ1 , and for some p ∈ M , some smooth
function f mentioned before, and some α = α() > 0
Z
1
|∇f |2 < δ2 ,
Vol(Bα (p)) Bα (p)
then r(p) > π − .
Proof of 3.1. Take > 0 (sufficiently small) and p ∈ M arbitrarily. Consider the gradient vector ∇f1 , ∇f2 , · · · , ∇fn+1 at p of the
of M .
Peigenfunctions
n+1 2
=
1
satisfying
There
exist
n
+
1
real
numbers
a
,
a
,
·
·
·
,
a
with
a
n+1
i=1 i
Pn+1
P1n+12
a
∇f
=
0
at
p.
Set
f
=
a
f
.
So,
(∇f
)(p)
=
0. Thus, for the δ2
i
i=1 i
√ i=1 i i
δ2
in Theorem 1, we have, for α ≤ C ,
√
|∇f | < δ 2 , on Bα (p).
So, we have
1
Vol(Bα (p))
Z
|∇f |2 < δ2 .
Bα (p)
We remark that the choose of δ2 and α are independent of p. So, by Theorem
1, as λn+1 is sufficiently close to n, r(p) > π − , and hence rad(M ) > π − .
The proof is finished.
3.2
rad(M ) → π implies λn+1 → n
Colding’s theorem [4] says that rad(M ) → π is equivalent to dGH (M, S n ) → 0.
So, we just need to prove that dGH (M, S n ) → 0 implies λn+1 → n. To do this,
we need the following result of Colding ([4], Lemma 1.10).
Lemma 5 ∀ > 0, ∃ δ = δ(, n) > 0, such that if there exist some p, q ∈
M with d(p, q) > π − δ, then there exists an f ∈ C ∞ (M ) with kf k2 ≤ 1,
k∆f + nf k2 < , and kf − gk2,1 < , here g(x) = cos d(p, x).
Remark.
Actually, for the f in the above lemma, we can further assume
R
f
=
0.
In
fact, for the above f , we have
M
Z
Z
1
1 1
f ≤
(4f + nf )
Vol(M ) M
n Vol(M ) M
1
≤
k∆f + nf k2 < .
n
n
R
R
1
Set f¯ = f − Vol(M
f with M f¯ = 0. Then kf¯k2 ≤ 1 + , k∆f¯ + nf¯k2 < 2,
) M
and kf¯ − gk2,1 < 2. So, we can use f¯ to replace f .
14
Proof of 3.2. Let {p0i } be n + 1 points in the standard n-sphere S n , and {qi0 }
the corresponding anti-podal points satisfying that for i 6= j, d(p0i , p0j ) = π2 .
If dGH (M, S 2 ) < 3δ , from the definition of Gromov-Hausdorff distance, we can
then find pi , qi ∈ M, i = 1, 2, · · · , n + 1, satisfying
|d(pi , pj ) −
π
| < δ, for i 6= j,
2
and
|d(pi , qi ) − π| < δ.
Set gi (x) = cos d(pi , x), i = 1, 2, · · · , n + 1. Then, from Lemma 4, we can find
n + 1 functions {fi , i = 1, 2, · · · , n + 1} satisfying kfi k2 ≤ 1,
k∆fi + nfi k2 < ψ1 (δ),
(18)
kfi − gi k2,1 < ψ2 (δ),
(19)
and
where ψi (δ) satisfy
limδ→0 ψi (δ) = 0, i = 1, 2. By the previous remark, we can
R
assume that M fi = 0, i = 1, 2, · · · , n + 1. If {fi } are linearly independent,
then by applying the minimax principle of eigenvaluesR of the laplacian (cf. e.g.
[7], Chapter 4) to the space H0 = {f ∈ C ∞ (M ) : M f = 0}, the result is
obtained. So, we only need to prove that {fi } are linearly independent as δ is
sufficiently small.
Assume that {fi } are not linearly
Then
there exist n + 1 real
Pn+1
Pindependent.
2
=
1
and
a
numbers a1 , a2 , · · · , an+1 satisfying n+1
i=1 i
i=1 ai fi = 0. Set
g(x) =
n+1
X
0
0
ai gi (x) and g (x ) =
i=1
n+1
X
ai cos d(p0i , x0 ).
i=1
From (19), we have
kgk2,1 = k
n+1
X
i=1
ai gi −
n+1
X
ai fi k2,1 < ψ2 (δ).
(20)
i=1
On the other hand, by the definition of g 0 , it is an eigenfunction of S n with
eigenvalue being n; so, there exists q 0 ∈ S n such that g 0 (x0 ) = cos d(q 0 , x0 ). It
is clear that for s < π3 and any q 00 ∈ Bs (q 0 ) ⊂ S n ,
1
g 0 (q 00 ) > .
2
15
From the choose of pi and p0i , we also know that if dGH (x, x0 ) < 3δ , then
|g(x) − g 0 (x0 )| < δ.
We now choose a q ∈ M satisfying dGH (q, q 0 ) ≤ 3δ , then for any q̄ ∈ B 2s (q),
we have g(q̄) ≥ 13 . But, on the other hand, the above (20) implies that for
sufficiently small δ, there must exist a q̃ ∈ B 2s (q) satisfying g(q̃) < 16 . This is
a contradiction. The proof is completed.
References
[1] J. Bertrand, Pincement spectral en courbure de Ricci positive, Comment.
Math. Helv. 82(2007), 323-352.
[2] J. Cheeger, Degeneration of Riemannian metrics under Ricci curvature
bounds, Pubblicazioni della Classe di Scienze della Scuola Normale Superiore, Lezioni Fermiane No. 12, Pisa, 2001.
[3] J. Cheeger and T. H. Colding, Lower bounds on the Ricci curvature and
the almost rigidity of warped products, Ann. of Math., 144(1996), 189-237.
[4] T. H. Colding, Shape of manifolds with positive Ricci curvature, Invent.
Math., 124(1996), 175-191.
[5] T. H. Colding, Large manifolds with positive Ricci curvature, Invent.
Math., 124(1996), 193-214.
[6] T. H. Colding, Ricci curvature and volume convergence, Ann. Math.,
145(1997), 477-501.
[7] S. Gallot, D. Hulin, & J. Lafontaine, Riemannian Geometry, 3rd edition,
Springer.
[8] S. Honda, Ricci curvature and almost spherical multi-suspension, Tohoku
Mathematical Journal, 61(2009), 499-522.
[9] P. Petersen, On eigenvalue pinching in positive Ricci curvature, Invent.
Math., 138(1999), 1-21.
Yi-Hu Yang: Department of Mathematics, Shanghai Jiao Tong University,
Shanghai 200240, China.
Email: [email protected]
16
Yi Zhang: Department of Mathematics, Tongji University, Shanghai 200092,
China.
Email: [email protected]