j. differential geometry 63 (2003) 41-62

EXOTIC NEGATIVELY CURVED STRUCTURES ON CAYLEY HYPERBOLIC MANIFOLDS

C.S. ARAVINDA & F.T. FARRELL

Abstract

We construct examples of closed negatively curved manifolds M which are homeomorphic but not diffeomorphic to Cayley locally symmetric spaces. Given ϵ > 0, we can construct such an M with sectional curvatures all in [−4 − ϵ, −1].

1. Introduction

Margulis [16] discovered a strengthening of Mostow’s strong rigidity theorem [17] to a phenomenon called Archimedean superrigidity valid for lattices in semisimple Lie groups G of real rank bigger than or equal to two. (Here G is assumed to be centerless and to contain no compact normal subgroup other than 1.) Later Corlette [5] proved a version of superrigidity for lattices in the automorphism groups of quarternion hyperbolic spaces or the Cayley hyperbolic plane. It is known that superrigidity fails for other real rank 1 situations; i.e., for lattices in the automorphism groups of the real or complex hyperbolic spaces. Stronger versions of Corlette superrigidity were later proven by Jost and Yau [13] and Mok, Siu and Yeung [19]. A consequence of these superrigidity theorems is that if M and N are homeomorphic closed negatively curved manifolds and the universal cover of M is either a quaternionic hyperbolic space ℍH^{n}, n ≥ 2, or the Cayley hyperbolic plane 𝕆H^{2}, then M and N are isometric up to a scaling of the metric on either of them by a constant (the isometry being the unique harmonic map in the homotopy class of the homeomorphism [6], [10]) under any of the following three extra conditions on N:

This research was supported in part by a grant from the National Science Foundation.

Received 05/08/2002.

1. The curvature operator of N is nonpositive [5].

2. The complexified sectional curvatures of N are nonpositive [19].

3. The sectional curvatures of N are pointwise \frac{1}{4}-pinched; i.e., lie in a closed interval [−4a_{x}, −a_{x}] where a_{x} > 0 and x ∈ N (cf [11] and [24]).

In fact, each of Conditions 1 and 3 independently imply Condition 2.

In [9], homeomorphic pairs of closed negatively curved n-manifolds M and N are constructed where the universal cover M̃ of M is the complex hyperbolic space ℂH^{m} (and n = 2m) but M and N are not diffeomorphic; indeed, given ϵ > 0, such pairs of M and N were constructed so that M̃ = ℂH^{m} and the sectional curvatures of N are “almost \frac{1}{4}-pinched”, i.e., lie in [−4 − ϵ, −1].

It was conjectured in [9] that such examples could be constructed where the universal cover M̃ of M is either the quaternionic hyperbolic plane ℍH^{2} or 𝕆H^{2}. We prove here this conjecture for the case where M̃ = 𝕆H^{2}. The smooth manifolds N are the connected sum M#Σ^{16} where Σ^{16} is the unique smooth manifold homeomorphic but not diffeomorphic to the 16-dimensional round sphere S^{16}. The case when M̃ = ℍH^{n} is treated separately in [2] where we use a different technique to show that the manifolds M#Σ^{4n} admit metrics of negative curvature but get a weaker result without the “almost 1/4-pinched” conclusion. However, we believe that the method used in this paper could be used to get the pinching result for the case M̃ = ℍH^{2}. A corollary of our construction is that Condition 1 or 2 on N in the superrigidity theorems mentioned above is optimal in a sense, i.e., neither of them can be replaced by the condition that the sectional curvatures of N are nonpositive. On the other hand, in view of superrigidity under Condition 3 on N, we note that ϵ cannot be 0 in any of our examples.

We conclude this introduction with an outline of the paper. There are two problems that must be addressed: (1) How to put a negatively curved metric on M#Σ^{16}. (2) How to show that M#Σ^{16} is not diffeomorphic to M. (M#Σ^{16} is clearly homeomorphic to M.) We solve problem (1) in the next section and problem (2) in the final section of the paper. Broadly speaking, we follow the pattern established in [8] and [9]. But the difficulties encountered are more formidable and require substantial modifications to the arguments in [9].

To solve the first problem, we construct a 1-parameter family b_{γ}( , ) of Riemannian metrics on ℝ^{16} indexed by γ ∈ [e, +∞) which satisfy the following properties:

(i) The sectional curvatures of b_{γ} lie in the closed interval [−4 − ϵ(γ), −1] where ϵ(γ) > 0 and ϵ(γ) → 0 as γ → +∞.

(ii) The ball of radius γ about 0 in (ℝ^{16}, b_{γ}) is isometric to a ball of radius γ in real hyperbolic space ℝH^{16}.

(iii) The complement of the ball of radius γ^{2} about 0 in (ℝ^{16}, b_{γ}) is isometric to the complement of a ball of radius γ^{2} in 𝕆H^{2}.

To construct these metrics we make use of the explicit description of the Riemannian curvature tensor for 𝕆H^{2} given in [4]. We use this result together with [9, Lemma 3.18] to put an “almost \frac{1}{4}-pinched” negatively curved Riemannian metric on M#Σ^{16} provided M has sufficiently large injectivity radius. Here M is a closed, orientable Cayley hyperbolic manifold. This injectivity radius condition is satisfied when we pass to sufficiently large finite sheeted covers of M since π_{1}(M) is a residually finite group.

The second problem (i.e., to show that M and M#Σ^{16} are not diffeomorphic) is reduced via Kirby-Siebenmann smoothing theory and using Mostow’s strong rigidity theorem [17] together with its topological analogue [7] to showing that the group homomorphism

θ_{16} = [S^{16}, Top/O] ⟶_{ϕ^{∗}} [M, Top/O]

is monic where ϕ : M → S^{16} is a degree 1 map. Now, a result of Okun [21] shows that ϕ^{∗} is the initial map in a factoring of

θ_{16} = [S^{16}, Top/O] ⟶_{ψ^{∗}} [𝕆P^{2}, Top/O]

where ψ : 𝕆P^{2} → S^{16} is a degree 1 map and 𝕆P^{2} is the Cayley projective plane. Hence it suffices to show that ψ^{∗} is monic. This is done by making delicate use of some calculations of Toda [22] on the stable homotopy groups of spheres.

Acknowledgments. This research was done when the first named author was a visiting faculty member at the Mathematics Department of SUNY, Binghamton while on sabbatical leave from Chennai Mathematical Institute. He wishes to thank both these institutions for making this collaboration possible.

2. Tapering between 𝕆H^{2} and ℝH^{16}

We begin with a brief description of the Cayley hyperbolic plane 𝕆H^{2}.

The Cayley numbers, denoted by 𝕆, is an 8-dimensional non associative division algebra over the real numbers. It has a multiplicative identity 1 and a positive definite bilinear form ⟨ , ⟩ whose associated norm ∥ ∥ is multiplicative, i.e., ∥uv∥ = ∥u∥∥v∥. Every element u ∈ 𝕆 can be written as α + u_{0} when α is real and ⟨α, u_{0}⟩ = 0. The conjugation map u ↦ u := α − u_{0} is an antiautomorphism, i.e., \overline{(uv)} = vu for all u, v ∈ 𝕆. Moreover, uu = ∥u∥^{2} and one has the following identities which can be checked easily: ⟨uv, w⟩ = ⟨vu, w⟩ = ⟨v, wu⟩ = ⟨v, uw⟩ for u, v, w ∈ 𝕆.

On 𝕆^{2} = 𝕆 × 𝕆, one has the positive definite bilinear form given by ⟨(u_{1}, u_{2}), (v_{1}, v_{2})⟩ = ⟨u_{1}, v_{1}⟩ + ⟨u_{2}, v_{2}⟩ for u_{1}, v_{1}, u_{2}, v_{2} ∈ 𝕆. The set D = {u ∈ 𝕆^{2} | ⟨u, u⟩ < 1} equipped with the metric given by formula (20.4) in [17, p. 144] is a model for the Cayley hyperbolic plane.

It is convenient for us to consider the set 𝕆^{2} itself as the underlying set for the Cayley hyperbolic plane 𝕆H^{2} equipped with the metric gotten by scaling the above metric from D to 𝕆^{2}. This enables one to identify 𝕆^{2} with T_{0}(𝕆H^{2}), the tangent space to 𝕆H^{2} at the origin 0 ∈ 𝕆^{2}.

The Riemannian metric on the distance sphere S^{15} at distance t from the origin can be described as follows. Firstly, one has the Hopf fibering of S^{15} over S^{8} with S^{7} as fiber. This equips S^{15} with complementary distributions η_{1}, η_{2} where Whitney sum η_{1}⊕η_{2} equals the tangent bundle of S^{15}. η_{1} is the 7-dimensional distribution tangent to the S^{7} fibers and η_{2} is the 8-dimensional distribution perpendicular to η_{1} (perpendicular with respect to the round metric on S^{15}). We call the subspace of the tangent space to S^{15} belonging to the distribution η_{1} “the vertical subspace” and the subspace belonging to η_{2} “the horizontal subspace”. The induced Riemannian metric ⟨ , ⟩ on S^{15} is then,

⟨X, X⟩ = b^{2}X · X, ⟨U, U⟩ = a^{2}U · U and ⟨X, U⟩ = 0

where X ∈ η_{1}, U ∈ η_{2}, a = sinh t, b = sinh t cosh t and “·” is the inner product with respect to the round metric on S^{15}. For brevity, we denote the distance sphere with the above metric by S_{a,b}^{15}.

Fix a smooth function ϕ : (0, +∞) → [0, +∞) such that ϕ^{·}(t) ≥ 0 for all t ∈ [0, +∞). We put a Riemannian metric on S^{15} × (0, +∞) using the function ϕ(t) as follows. The foliations S^{15} × t and x × (0, +∞) are required to be perpendicular where x ∈ S^{15} and t ∈ (0, +∞). We set

|N| = 1 where N = \frac{∂}{∂t} and t is the second coordinate variable in the

product structure S^{15} ×(0, +∞). We require that the induced metric on S^{15}×t is S_{a,b}^{15} where a = sinh t and b = sinh t cosh ϕ(t). This Riemannian manifold is denoted S^{15} ×^{ϕ} (0, +∞). Notice that when ϕ(t) = t for all t ∈ (0, +∞), S^{15} ×^{ϕ} (0, +∞) is the punctured Cayley hyperbolic plane 𝕆H^{2} − ∗. Our object is to calculate the sectional curvatures of S^{15} ×^{ϕ} (0, +∞).

Let P be a real 2-plane tangent to S^{15} ×^{ϕ} (0, +∞). Let τ denote the angle made by the plane P with the distance sphere S^{15}. Then P is spanned by vectors {u, cos(τ)v + sin(τ)N} where vectors {u, v} are tangent to the S^{15} × t foliation and satisfy |u| = |v| = 1 and (u · v) = 0 where | | denotes the norm and (·) denotes the inner product with respect to the metric on S^{15}×^{ϕ}(0, +∞). If K(P) denotes the sectional curvature of the plane P in S^{15} ×^{ϕ} (0, +∞), then we have

(1)  K(P) = K(u, cos τv + sin τN)
= cos^{2} τK(u, v) + sin^{2} τK(u, N)
+ 2 sin τ cos τ(R(u, N)v · u)

where R denotes the Riemann curvature tensor in S^{15} ×^{ϕ} (0, +∞).

We shall denote the vectors tangent to S^{15} and lying in the vertical subspace by symbols X, Y and those lying in the horizontal subspace by symbols U, V. If σ is the angle between the vector u and the horizontal subspace and α is the angle between v and the horizontal subspace, let u = sin σX + cos σU and v = sin αY + cos αV where |X| = |Y | = |U| = |V | = 1. (Note σ, α ∈ [0, π/2].) We calculate K(P) by explicitly calculating K(u, v), K(u, N) and (R(u, N)v · u) separately.

Before starting off to compute the above terms, we make the following important observations.

Firstly, recall that we identify 𝕆^{2} with the tangent space T_{0}(𝕆H^{2}) where 0 = (0, 0) ∈ 𝕆^{2}. Since 𝕆H^{2} is a homogeneous space, the group G of isometries of 𝕆H^{2} acts transitively on 𝕆H^{2}. Further, G_{0} — the subgroup of G fixing 0 — acts transitively on vectors of unit length in T_{0}(𝕆H^{2}). Therefore one can identify the tangent spaces at other points of 𝕆H^{2} also with 𝕆^{2}. In particular, we identify the vector N, normal to the distance spheres, with (1, 0) ∈ 𝕆^{2}. This would identify 0 × 𝕆 with the horizontal subspace and the subspace of 𝕆 × 0 perpendicular to N with the vertical subspace. Thus the vectors X, Y lying in the vertical subspace are purely imaginary Cayley numbers.

Secondly, we note that any A ∈ O(16) with the property that it induces a permutation of the fibers of the Hopf fibration S^{15} → S^{8} determines an isometry A of S^{15}×^{ϕ}(0, +∞) defined by A(x, t) = (A(x), t). All A ∈ Spin(9) ⊆ O(16) have this property. And since Spin(9) acts transitively on S^{15}, it acts transitively on the fibers of the Hopf fibration. In fact, for any leaf L of Hopf fibration there exists an A ∈ Spin(9) such that A^{2} = I and the fixed set of A (acting on S^{15}) is L. To verify this, it suffices to verify it for L = (𝕆 × 0) ∩ S^{15}. Here we can define A(x, y) = (x, −y) (cf. [4, §§3 and 4]). Consequently, the submanifolds L ×^{ϕ} (0, +∞) are totally geodesic in S^{15} ×^{ϕ} (0, +∞). In particular, L ×^{t} (0, +∞) is totally geodesic in S^{15} ×^{t} (0, +∞) = 𝕆H^{2} − ∗.

Thirdly, the Hopf submersion S^{15} → S^{8} is indeed a Riemannian submersion S^{15}(1) → S^{8}(r) where S^{15}(1) is the sound sphere S^{15} of radius 1 and S^{8}(r) is the round sphere S^{8} of radius r. Therefore, one has a Riemannian submersion from S_{a,b}^{15} → S_{a}^{8} and more generally a Riemannian submersion from S^{15} ×^{ϕ} (0, +∞) → S^{8} ×^{a} (0, +∞) with fibers S_{b}^{7} where a, b are functions of t described earlier and S^{7} = S^{7}(1). Here S^{8} ×^{a} (0, +∞) denotes the product of S^{8}(r) warped over (0, +∞) using a(t) = sinh(t) for the warping function.

Finally, since ϕ is a function from (0, +∞) to [0, +∞), the distance sphere S_{a,b}^{15} where a = sinh ϕ(t) and b = sinh ϕ(t) cosh ϕ(t) is indeed the distance sphere in 𝕆H^{2} at distance ϕ(t) from the origin. Scaling

this metric on S_{a,b}^{15} throughout by the factor s = \frac{sinh t}{sinh ϕ(t)}, we get a

sphere S_{sa,sb}^{15} where sa = sinh t and sb = sinh t cosh ϕ(t). But this is the distance sphere in S^{15} ×^{ϕ} (0, +∞) at distance t from the origin.

Since S^{15} is a Riemannian hypersurface in 𝕆H^{2} and also in S^{15} ×^{ϕ} (0, +∞), in the foregoing calculations, it is important for us to consider the shape operator L of S^{15} ⊂ 𝕆H^{2} and the shape operator L of S^{15} ⊂ S^{15} ×^{ϕ} (0, +∞) corresponding to the normal vector field N on S^{15}. The shape operator is a linear operator acting on each tangent space T_{p}S^{15} at p ∈ S^{15}. We shall compute it by its action on vectors X, U belonging to the vertical and horizontal subspaces respectively. Using formulas from O’Neill [20], we get the following for S^{15} ⊂ S^{15} ×^{ϕ} (0, +∞):

L(X) = (coth t + tanh(ϕ(t))ϕ^{·}(t))X and L(U) = coth tU.

And for S^{15} ⊂ 𝕆H^{2} we get,

L(X) = (coth ϕ(t) + tanh ϕ(t))X and L(U) = coth ϕ(t)U.

Calculation of K(u, N).

K(u, N) = K(sin σX + cos σU, N)
= sin^{2} σK(X, N) + cos^{2} σK(U, N)
+ 2 sin σ cos σ(R(U, N)N · X).

Now X ∈ η_{1} and since the fibers S^{7} ×^{ϕ} (0, +∞) are totally geodesic in S^{15}×^{ϕ}(0, +∞), the vector R(N, X)N ∈ η_{1}. Therefore (R(U, N)N ·X) = −(R(N, U)N ·X) = −(R(N, X)N · U) = 0 since U ∈ η_{2} and η_{1} and η_{2} are complementary distributions on S^{15}. Since S^{7}×^{ϕ}(0, +∞) are totally geodesic in S^{15} ×^{ϕ} (0, +∞) and since X ∈ η_{1}, the curvature K(X, N) in S^{15} ×^{ϕ} (0, +∞) is the same as the curvature of the plane {X, N} in S^{7} ×^{b} (0, +∞). Since the metric on S^{7} ×^{b} (0, +∞) is dt^{2} + b^{2}(dS^{7})^{2}, the

curvature of \{X, N\} is −\frac{b^{··}}{b} where b = sinh t cosh ϕ(t). Therefore,

K(X, N) = −(1 + tanh ϕ(t)ϕ^{··}(t) + (ϕ^{·}(t))^{2} + 2 coth t tanh ϕ(t)ϕ^{·}(t)).

Since U is tangent to S^{15}, [U, N] = 0. And since S^{15}(1) ×^{ϕ} (0; +∞) → S^{8}(r) ×^{a} (0, +∞) is a Riemannian submersion, by O’Neill’s submersion formula [20], K(U, N) is the curvature of the plane spanned by {U, N}

in S^{8}(r) ×^{a} (0, +∞). Thus K(U, N) = −\frac{a^{··}}{a} = −\frac{sinh t}{sinh t} = −1. Piecing together these components we get,

(2)  K(u, N) = −1 − sin^{2} σ(ϕ^{··}(t) tanh ϕ(t) + (ϕ^{·}(t))^{2}
+ 2ϕ^{·}(t) tanh ϕ(t) coth t).

Calculation of K(u, v).

Since the distance sphere in S^{15} ×^{ϕ} (0, +∞) is the distance sphere in

𝕆H^{2} scaled by s = \frac{sinh t}{sinh ϕ(t)}, ∥su∥ = ∥sv∥ = 1 in 𝕆H^{2}. Using Gauss’

equation for the submanifold S^{15} ⊂ S^{15} ×^{ϕ} (0, +∞),

K(u, v) = K_{S^{15}}(u, v) − ((Lu · u)(Lv · v) − (Lu · v)^{2}) (a)

where K_{S^{15}}(u, v) is the curvature of {u, v} in S^{15} ⊂ S^{15} ×^{ϕ} (0, +∞). Similarly, using Gauss’ equation for S^{15} ⊂ 𝕆H^{2},

(b)  K̂(su, sv) = K_{S^{15}}(su, sv)
− (⟨L(su), su⟩⟨L(sv), sv⟩ − ⟨L(su), sv⟩^{2}).

where K̂ is the curvature in 𝕆H^{2} and K_{S^{15}} is the curvature of {su, sv} in S^{15} ⊂ 𝕆H^{2}. Since the metrics on the distance spheres in S^{15}×^{ϕ}(0, +∞) and 𝕆H^{2} differ by the scaling factor s, we have

\frac{1}{s^{2}} K_{S^{15}}(su, sv) = K_{S^{15}}(u, v).

Therefore s^{2} × (a) − (b) and rearranging gives,

(3) K(u, v) = \frac{1}{s^{2}} (⟨L(su), su⟩⟨L(sv), sv⟩ − ⟨L(su), sv⟩^{2})

− ((Lu · u)(Lv · v) − (Lu · v)^{2}) + \frac{1}{s^{2}} K̂(su, sv).

We now calculate the terms on the right-hand side.

A calculation yields,

\frac{1}{s^{2}} (⟨L(su), su⟩⟨L(sv), sv⟩ − ⟨L(su), sv⟩^{2})

= \frac{1}{s^{2}} (coth^{2} ϕ + sin^{2} α + sin^{2} σ + sin^{2} σ sin^{2} α tanh^{2} ϕ

− sin^{2} σ sin^{2} α(coth^{2} ϕ + tanh^{2} ϕ + 2)⟨sX, sY ⟩^{2}

− cos^{2} σ cos^{2} α coth^{2} ϕ⟨sU, sV ⟩^{2}

− 2 sin σ cos σ sin α cos α(1 + coth^{2} ϕ)⟨sX, sY ⟩⟨sU, sV ⟩) and

((L(u) · u)(L(v) · v) − (L(u) · v)^{2})

= coth^{2} t + tanh^{2} ϕ(ϕ^{·})^{2} sin^{2} σ sin^{2} α + coth t tanh ϕϕ^{·}(sin^{2} σ + sin^{2} α)

− sin^{2} σ sin^{2} α(coth t + tanh ϕϕ^{·})^{2}(X · Y)^{2}

− cos^{2} σ cos^{2} α coth^{2} t(U · V)^{2}

− 2 sin σ sin α cos σ cos α coth t(coth t + tanh ϕϕ^{·})(X · Y)(U · V).

Since (u · v) = 0, we get, − sin α sin σ(X · Y) = cos α cos σ(U · V). Using this identity and the relations ⟨sX, sY ⟩ = (X · Y), ⟨sU, sV ⟩ = (U · V), the term

\frac{1}{s^{2}} (⟨L(su), su⟩⟨L(sv), sv⟩ − ⟨L(su), sv⟩^{2})

− ((L(u) · u)(L(v) · v) − (L(u) · v)^{2})

simplifies to

(4) (\frac{1}{s^{2}} − 1) + (sin^{2} σ + sin^{2} α) (\frac{1}{s^{2}} − coth t tanh ϕϕ^{·})

+ sin^{2} σ sin^{2} α tanh^{2} ϕ(1 − (X · Y)^{2}) (\frac{1}{s^{2}} − (ϕ^{·})^{2}) .

Now, to calculate K̂(su, sv) we use the description of the Riemann curvature tensor R̂ of the Cayley hyperbolic plane 𝕆H^{2} in [4]. However, the action of the representation of Spin(9) in [4] is different from the action described in [17]. Indeed, the map (x, y) ↦ (x, y) of 𝕆 × 𝕆 → 𝕆 × 𝕆 sends the Spin(9) action in [17] to that in [4]. In particular, the 𝕆lines in Mostow’s description of 𝕆H^{2} go to 𝕆-lines, i.e., 8-dimensional ℝ-subspaces R of the tangent space to 𝕆H^{2} at 0 such that K̂(P) = −4 for each 2-plane P ⊂ R, in the description in [4] under the above map. Applying this map and using the formula for sectional curvature K̂ in [4] yields,

K̂(su, sv) = −1 − 3 cos^{2} θ (5)

where θ is the angle between the vector sv and the unique 𝕆-line 𝕆u containing the vector su. Since 𝕆u is an 8-dimensional subspace, it is important to note that

θ = min_{\begin{matrix}w∈𝕆u \\ ∥w∥=1 \end{matrix}} < ) (w, sv).

Hence the value cos θ is the maximum for all such angles. Putting (4) and (5) into (3) we get

(6) K(u, v) = (\frac{1}{s^{2}} − 1)

+ (sin^{2} σ + sin^{2} α) (\frac{1}{s^{2}} − coth t tanh ϕ(t) · ϕ^{·}(t))

+ sin^{2} σ sin^{2} α tanh^{2} ϕ(t)(1 − (X · Y)^{2}) (\frac{1}{s^{2}} − (ϕ^{·}(t))^{2})

+ \frac{1}{s^{2}} (−1 − 3 cos^{2} θ).

Calculation of (R(u, N)v · u).

Using the fact that u = sin σX + cos σU and v = sin αY + cos αV, we first expand out (R(u, N)v · u) into 8 terms.

Claim 1. The terms (R (X, N) Y · U), (R (U, N) Y · X) and (R(X, N)V · X) are all zero.

Proof. The vectors X, Y belong to the vertical subspace tangential to the fiber S^{7} of the distance sphere S^{15}. And since S^{7}×^{ϕ}(0, +∞) is totally geodesic in S^{15}×^{ϕ}(0, +∞) we conclude that the vectors R(X, N)Y, R(Y, X)N and R(X, N)X are tangent to S^{7} ×^{ϕ} (0, +∞). Since the vectors U and V belong to the horizontal space, the terms (R(X, N)Y · U), (R(Y, X)N · U) and (R(X, N)X · V) are zero and this completes the proof of Claim 1. q.e.d.

To analyze the remaining terms we use the Codazzi-Mainardi equation for the submanifold S^{15} of S^{15} ×^{ϕ} (0, +∞). For this we recall the shape operator L acting on the vectors tangent to S^{15}. We have

L(X) = (coth t + tanh ϕϕ^{·})X for vectors X in vertical subspace and L(U) = (coth t)U for vectors U in the horizontal subspace.

For vectors α, β, γ ∈ T_{p}S^{15}, the Codazzi-Mainardi equation gives (R(α, β)γ · N) = −(Tor_{L}(α, β) · γ) where Tor_{L}(α, β) = D_{α}L(β) − D_{β}L(α) − L([α, β]) where D is the Riemannian connection on S^{15} ×^{ϕ} (0, +∞). Applying this equation for the remaining 5 terms gives (R(X, N)X · Y) = 0, (R(U, N)U · V) = 0 and (R(U, N)U · Y) = 0. For the

remaining two terms we get (R(U, N)V · X) = (tanh ϕ)ϕ^{·}(D_{V} U · X) and

(R(X, N)V · U) = (tanh ϕ)ϕ^{·}((D_{V} U − D_{U}V) · X). Thus (R(u, N)v ·

u) = sin σ cos σ cos α(tanh ϕ)ϕ^{·}((2D_{V} U − D_{U}V) · X). Since the distance spheres in S^{15} ×^{ϕ} (0, +∞) are gotten by scaling the metric on the dis-

tance spheres in 𝕆H^{2} by a factor s = \frac{sinh t}{sinh ϕ(t)} they have the same

affine connection. We therefore have, for vectors U, V, X ∈ T_{p}S^{15}, that tangential part of D_{V} U = tangential part of D̂_{V} U where D̂ denotes the Riemannian connection on 𝕆H^{2}. Hence (D_{V} U · X) = (D̂_{V} U · X) = s^{2}⟨D̂_{V} U, X⟩ where (·) and ⟨ , ⟩ denote the Riemannian metrics on S^{15} ×^{ϕ} (0, +∞) and on 𝕆H^{2} respectively.

On the other hand, proceeding exactly as above while simplifying (R(u, N)v · u), we can show that

⟨R̂(su, sN)sv, su⟩ = sin σ cos σ cos α(tanh ϕ)s^{4}⟨2D̂_{V} U − D̂_{U}V, X⟩.

Now, using relations (D_{V} U · X) = s^{2}(D̂_{V} U, X⟩ and (D_{U}V · X) = s^{2}⟨D̂_{U}V, X⟩ deduced above we get the following:

(R(u, N)v · u) = \frac{ϕ^{·}}{s^{2}} ⟨R̂(su, sN)sv, su⟩.

We then wish to calculate the term on the right-hand side using the formula for the curvature tensor for the Cayley hyperbolic plane 𝕆H^{2} described in [4]. To be able to do so we must as before first transform our description of 𝕆H^{2} to the description in [4] via the map f : (x, y) ↦ (x, y) of 𝕆^{2} → 𝕆^{2}. Also our curvature operator R̂ is negative of that in [4]. Making these necessary changes and using the formula for the curvature operator R̂ in [4, page 52], a calculation yields,

\frac{ϕ^{·}}{s^{2}} ⟨R̂(su, sN)sv, su⟩ = −\frac{3ϕ^{·}}{s} sin σ cos σ cos α⟨s^{2}XU, sV ⟩

= −\frac{3ϕ^{·}}{s} sin σ cos σ cos α⟨s^{2}UX, sV ⟩.

This together with the fact that

⟨s^{2}uX, sv⟩ = cos σ cos α⟨s^{2}UX, sV ⟩

yields

(7) (R(u, N)v · u) = −\frac{3ϕ^{·}}{s} sin σ⟨s^{2}uX, sv⟩ = −\frac{3ϕ^{·}}{s} sin σ cos ω

where ω is the angle between the unit length vectors s^{2}uX and sv. Finally, putting together the calculations (2), (6) and (7) into (1) gives,

K(P) = cos^{2} τ (\frac{1}{s^{2}} − 1

+ (sin^{2} σ + sin^{2} α) (\frac{1}{s^{2}} − coth(t)(tanh ϕ(t))ϕ^{·}(t))

+ sin^{2} σ sin^{2} α tanh^{2} ϕ(t)(1 − (X · Y)^{2}) (\frac{1}{s^{2}} − ϕ^{·}(t)^{2})

+\frac{1}{s^{2}} (−1 − 3 cos^{2} θ))

+ sin^{2} τ(−1 − sin^{2} σ(ϕ^{··}(t) tanh ϕ(t) + ϕ^{·}(t)^{2}

+ 2ϕ^{·}(t)(tanh ϕ(t)) coth(t))

− 6 sin τ cos τ sin σ \frac{ϕ^{·}}{s} cos ω.

Combining and regrouping the above term we get

K(P) (8)

= −1 − 3 (\frac{cos τ cos ω}{s} + sin τ sin σϕ^{·}(t))^{2}

− \frac{3 cos^{2} τ}{s^{2}} (cos^{2} θ − cos^{2} ω)

+ cos^{2} τ(sin^{2} σ + sin^{2} α) (\frac{1}{s^{2}} − coth(t)(tanh ϕ(t))ϕ^{·}(t))

− sin^{2} τ sin^{2} σϕ^{··}(t) tanh ϕ(t)

− 2 sin^{2} τ sin^{2} σϕ^{·}(t)(tanh ϕ(t) coth(t) − ϕ^{·}(t))

+ cos^{2} τ sin^{2} σ sin^{2} α tanh^{2} ϕ(t)(1 − (X · Y)^{2}) (\frac{1}{s^{2}} − ϕ^{·}(t)^{2}) .

We now proceed to choose functions ϕ so that given an ϵ > 0, the curvature K(P) satisfies −4 − ϵ ≤ K(P) ≤ −1 + ϵ for all plane sections P in S^{15} ×^{ϕ} (0, +∞).

Following [9] first fix a smooth function ψ : ℝ → [0, 1] such that

ψ^{·}(t) ≥ 0 for all t ∈ [1, 2]
ψ^{−1}(0) = (−∞, 1) and
ψ^{−1}(1) = [2, +∞).

For each c ≥ 1, let ϕ_{c}(t) = ψ (\frac{ln t}{c}) t for all t > 0. Therefore ϕ_{c}(t) = 0

for t ∈ (0, e^{c}] and ϕ_{c}(t) = t for t ∈ [e^{2c}, +∞). As in [9] observe that the following limits hold uniformly in t:

lim_{c→+∞} |ϕ^{··}_{c}(t)| = 0,

lim sup_{c→+∞} ϕ^{·}_{c}(t) ≤ 1,

lim sup_{c→+∞} (\frac{1}{s^{2}} − (ϕ^{·})^{2}) ≤ 0

lim_{c→+∞} ϕ^{·}_{c}(t)(tanh ϕ_{c}(t) coth t − 1) = 0.

(9)

(The 3rd inequality is a bit different from the corresponding inequality posited in [9, (2.22)].) Now, for the angles θ and ω as in (8) we have the following:

Lemma 1. | cos ω| ≤ cos θ.

Proof. Recall that ω is defined by the relation cos ω = ⟨s^{2}uX, sv⟩. Consider the vector u^{X} := (s^{2} sin σ, s^{2} cos σUX) = s^{2}uX. It is easy to see that ±u^{X} ∈ 𝕆u where 𝕆u is the unique 𝕆-line containing the vector su. (Note that X = −X and hence (UX)X = U(XX) = U.) And obviously ⟨±u^{X}, sv⟩ = ± cos ω. Since θ = min_{\begin{matrix}w∈𝕆u \\ |w|=1 \end{matrix}} < ) (w, sv), we conclude

that | cos ω| ≤ cos θ. q.e.d.

Lemma 1 together with formulas (8) and (9) yield the following result when ϕ(t) is one of the functions ϕ_{c}(t).

Lemma 2. If t ∈ (0, e^{c}] and ϕ = ϕ_{c}, then K(P) = −1. Moreover, for all t > 0, the following limit holds uniformly in t:

lim sup_{c→+∞} K(P) = −1

where ϕ = ϕ_{c}. Also S^{15} ×^{0} (0, +∞) is ℝH^{16} less a point and S^{15} ×^{t} (0, +∞) is 𝕆H^{2} less a point. Hence S^{15} ×^{ϕ_{c}} (0, e^{c}] can be identified with a closed ball of radius e^{c} in ℝH^{16} with its center deleted. And S^{15} ×^{ϕ_{c}} [e^{2c}, +∞) can be identified with 𝕆H^{2} from which an open ball of radius e^{2c} is deleted.

To obtain a lower bound for K(P) we need the following lemma.

Lemma 3.

(i) The maximum value of |C cos y cos z + D sin y sin z| is max{|C|, |D|}, as both y and z vary over ℝ.

(ii) The maximum value of

B cos^{2} τ cos^{2} θ + (1 − B) cos^{2} τ \frac{a}{3}

+ 2\sqrt{B} cos τ sin τ sin σ cos ω + sin^{2} τ sin^{2} σ

is 1, where τ ∈ ℝ, B ∈ [0, 1], α, σ ∈ [0, π_{2}] and angles θ and ω are as in (8). And a = sin^{2} σ + sin^{2} α + sin^{2} σ sin^{2} α.

Proof. We skip the proof of (i) which can be proved by elementary calculus and proceed directly to prove (ii).

(ii) It is convenient to set

f = B cos^{2} τ cos^{2} θ + (1 − B) cos^{2} τ \frac{a}{3}

+ 2\sqrt{B} cos τ sin τ sin σ cos ω + sin^{2} τ sin^{2} σ.

We first observe that f is quadratic in sin τ and cos τ. Hence, letting

M = (\begin{matrix}B cos^{2} θ + (1 − B)\frac{a}{3} & \sqrt{B} sin σ cos ω \\ \sqrt{B} sin σ cos ω & sin^{2} σ \end{matrix})

we see that f = (cos τ sin τ)M (\begin{matrix}cos τ \\ sin τ \end{matrix}). By a linear algebra argument

it follows that f ≤ 1 for all τ ∈ ℝ if and only if the maximum eigenvalue of M is less than or equal to 1; i.e., f ≤ 1 for all τ ∈ ℝ if and only if g = Trace(M) − Determinant(M) ≤ 1.

Now

g = B cos^{2} θ + (1 − B)\frac{a}{3} + sin^{2} σ

− (B cos^{2} θ + (1 − B)\frac{a}{3}) sin^{2} σ + B sin^{2} σ cos^{2} ω.

Since g is linear in B, thinking of g as a function of B and fixing θ, ω, σ and α, it is easy to see that the maximum value of g occurs at either B = 0 or at B = 1. Therefore, to prove the lemma, it is sufficient to show that g|_{B=0} ≤ 1 and g|_{B=1} ≤ 1. It is easy to see that g|_{B=0} ≤ 1. To show g|_{B=1} ≤ 1, we show equivalently that for all τ ∈ ℝ, f|_{B=1} ≤ 1. This fact follows easily by observing that the formula for K(P) in (8) for values of t ≥ e^{2c} (in which case S^{15} ×^{ϕ_{c}} [e^{2}, +∞) is 𝕆H^{2} less an open ball of radius e^{2c}) reduces to −3f|_{B=1} − 1 from which it follows that f|_{B=1} ≤ 1 for all τ ∈ ℝ. q.e.d.

Lemma 4. lim inf_{c→+∞}K(P) = −4.

Proof. It suffices, because of Lemma 2, to show that lim inf_{c→+∞}K(P) ≥

−4. Because of (8) and (9), this is equivalent to showing that lim sup_{c→+∞} v ≤

3, where

v = 3 (\frac{cos^{2} τ cos^{2} θ}{s^{2}} + sin^{2} τ sin^{2} σ(ϕ^{·}_{c}(t))^{2}

+\frac{2}{s} sin τ cos τ sin σ cos ωϕ^{·}_{c}(t))

+ cos^{2} τ(sin^{2} σ + sin^{2} α) (coth t tanh ϕ_{c}(t)ϕ^{·}_{c}(t) − \frac{1}{s^{2}} )

+ 2 sin^{2} τ(sin^{2} σ)ϕ^{·}_{c}(t)(tanh ϕ_{c}(t) coth t − ϕ^{·}_{c}(t))

+ cos^{2} τ sin^{2} σ sin^{2} α tanh^{2} ϕ_{c}(t)(1 − (X · Y)^{2}) ((ϕ^{·}_{c}(t))^{2} − \frac{1}{s^{2}} ) .

Let B = \frac{1}{s^{2}} and x = ψ (\frac{ln t}{c}) and define v_{1} by

v_{1} = 3(B cos^{2} τ cos^{2} θ + sin^{2} τ(sin^{2} σ)x^{2}
+ 2\sqrt{B} sin τ cos τ sin σ(cos ω)x)
+ cos^{2} τ(sin^{2} σ + sin^{2} α)(x − B) + 2 sin^{2} τ(sin^{2} σ)x(1 − x)
+ cos^{2} τ sin σ sin^{2} α tanh^{2} ϕ_{c}(t)(1 − (X · Y)^{2})(x^{2} − B).

Using (9), it is easy to see that v_{1} − v converges to 0 uniformly as c → +∞. Therefore, it suffices to show that lim sup_{c→+∞} v_{1} ≤ 3. Since the maximum values of tanh^{2} ϕ_{c}(t) and (1−(X ·Y)^{2}) is 1, it suffices to show that lim sup_{c→+∞} v_{2} ≤ 3 where

v_{2} = 3(B cos^{2} τ cos^{2} θ + sin^{2} τ(sin^{2} σ)x^{2}
+ 2\sqrt{B} sin τ cos τ sin σ(cos ω)x)
+ cos^{2} τ(sin^{2} σ + sin^{2} α)(x − B) + 2 sin^{2} τ(sin^{2} σ)x(1 − x)
+ cos^{2} τ sin^{2} σ sin^{2} α(x^{2} − B).

Now, define

v_{3} = 3(B cos^{2} τ cos^{2} θ + sin^{2} τ sin^{2} σ + 2\sqrt{B} sin τ cos τ sin σ cos ω)
+ cos^{2} τ(sin^{2} σ + sin^{2} α + sin^{2} σ sin^{2} α)(1 − B).

Since v_{2} is a continuous function of B, τ, σ, α, θ, ω and x, we have lim_{x→1}(v_{2} − v_{3}) = 0 uniformly in B ∈ [0, 1], τ, ω ∈ ℝ and σ, α, θ ∈ [0, π/2].

Since v_{3} ≤ 3 by Lemma 3(ii), it follows therefore that, given an ϵ > 0, there exists a δ > 0 such that x > 1 − δ implies v_{2} ≤ 3 + ϵ. Hence to complete the proof of the Lemma, we may assume that x ≤ 1 − δ. This, together with the fact that ϕ_{c}(t) = xt and the specific choice of functions ϕ_{c}(t) can be used to show that

(10)  lim_{c→+∞} B = 0 uniformly in t.

Now (10), x ≤ 1 − δ and Lemma 3(i) together imply that lim sup_{c→+∞} v_{2} ≤ 3

which completes the proof of the Lemma 4. q.e.d.

3. Detecting exotic smooth structures

The purpose of this section is to prove the following result.

Theorem. Let M^{16} be any closed locally Cayley hyperbolic manifold. Given ϵ > 0 then there exists a finite sheeted cover N^{16} of M^{16} such that the following is true for any finite sheeted cover N^{16} of N^{16}.

(a) N^{16} is not diffeomorphic to N^{16}#Σ^{16}.

(b) N^{16}#Σ^{16} supports a negatively curved Riemannian metric whose sectional curvatures are contained in the interval [−4 − ϵ, −1].

Here Σ^{16} is the unique closed, oriented smooth 16-dimensional manifold which is homeomorphic but not diffeomorphic to the sphere S^{16}. The existence and uniqueness of Σ^{16} is a consequence of the following result which is implicit in [14]. However, for the reader’s convenience, we derive it here from the Sullivan-Wall surgery exact sequence.

Proposition. The group of smooth homotopy spheres θ_{16} is cyclic of order 2.

Proof. We have the following surgery exact sequence from [23]:

0 → θ_{16} → π_{16}(F/O) → L_{16}(O) = ℤ.

This sequence together with the fact that θ_{16} is a finite group show that θ_{16} can be identified with the subgroup S of π_{16}(F/O) consisting of all elements having finite order. Next consider the exact sequence

π_{16}(O) ⟶^{J} π_{16}(F) → π_{16}(F/O) → π_{15}(O) = ℤ.

This sequence and the fact that π_{16}(F) = π_{16}^{s} is a finite group show that S can be identified with cokernel of J. Recall now that Adams [1] proved that J is monic. This result together with the facts that π_{16}(O) = ℤ_{2} and π_{16}^{s} = ℤ_{2} ⊕ ℤ_{2} (cf. [22]) show that θ_{16} = ℤ_{2}. q.e.d.

Now, the proof the theorem posited in the beginning of this section follows the pattern established in [8] and [9]. The main result of Okun’s thesis [21, Theorem 5.1] gives a finite sheeted cover N^{16} of M^{16} and a tangential map f : N^{16} → 𝕆P^{2}. And we can arrange that N^{16} has arbitrarily large preassigned injectivity radius r by taking larger covers since π_{1}(M^{16}) is residually finite. Once r is determined, then this is the manifold N^{16} posited in the theorem. The argument in [9, pp. 69–70] is now easily adapted to yield the following lemma since the other ingredients — Mostow’s strong rigidity theorem [17], its topological analogue [7] and Kirby-Siebenmann smoothing theory [14, pp. 25 and 194] — remain valid.

Lemma 0. Let N^{16} be any finite sheeted cover of N^{16}. If N^{16}#Σ^{16} is diffeomorphic to N^{16}, then 𝕆P^{2}#Σ^{16} is concordant to 𝕆P^{2}.

The octave projective plane 𝕆P^{2} is the mapping cone of the Hopf map p : S^{15} → S^{8}. Let ϕ : 𝕆P^{2} → S^{16} be the collapsing map obtained by identifying S^{16} with 𝕆P^{2}/S^{8} in an orientation preserving way.

Lemma 1. The homomorphism ϕ^{∗} : [S^{16}, Top/O] → [𝕆P^{2}, Top/O] is monic.

Proof. Recall that [X, Top/O] is the zeroth cohomology group of X in an extraordinary cohomology theory. By considering the long exact sequence in this theory determined by the pair (𝕆P^{2}, S^{8}) and using the identification of 𝕆P^{2} with the mapping cone of ϕ, it is seen that ϕ^{∗} is monic if and only if

(Σp)^{∗} : [S^{9}, Top/O] → [S^{16}, Top/O]

is the zero homomorphism. Here

Σp : S^{16} → S^{9}

denotes the suspension of p.

To show that (Σp)^{∗} is the zero homomorphism consider the following commutative ladder of groups and homomorphisms:

θ_{16} = [S^{16}, Top/O] ⟶^{α} [S^{16}, F/O] ⟵ [S^{16}, F] ⟵^{J} [S^{16}, O]
(Σp)^{∗}↑  ↑^{(Σp)^{∗}}  ↑^{(Σp)^{∗}}
θ_{9} = [S^{9}, Top/O] ⟶ [S^{9}, F/O] ⟵_{β} [S^{9}, F] .

The horizontal homomorphisms in this ladder are induced by the natural maps Top/O → F/O, F → F/O, and O → F. Now the following three facts used in conjunction with a simple “diagram chase” show that (Σp)^{∗} is the zero homomorphism thus proving Lemma 1. q.e.d.

Fact 1. α is monic.

Fact 2. β is an epimorphism.

Fact 3. Image (Σp)^{∗} ⊆ Image J where (Σp)^{∗} : [S^{9}, F] → [S^{16}, F] and J : [S^{16}, O] → [S^{16}, F] is the classical J-homomorphism.

It remains to verify these Facts. Fact 1 is due to Kervaire and Milnor [14]. A more modern proof is given by observing that α is a homomorphism in Sullivan’s surgery exact sequence

⋯ ⟶ L_{17}(0) ⟶ θ_{16} ⟶^{α} π_{16}(F/O) ⟶ …

and that L_{17}(0) = 0. Fact 2 is due to Adams [1] who showed that

J : π_{8}(O) → π_{8}(F)

is monic. (Now consider the homotopy exact sequence for the fibration O → F → F/O.) Fact 3 is a more special result which we proceed to prove.

During this proof we will use Toda’s notation [22, p. 189] for special elements in the stable stems G_{n} = π_{n}^{s}. Recall that G equal the direct sum of the G_{n} is an anti-commutative graded ring with respect to composition as multiplication. First note that the homotopy class

[Σp] = aσ + x ∈ π_{7}^{s}

where a ∈ ℤ and x ∈ π_{7}^{s} has odd order. Using this together with the fact that π_{16}^{s} has order 4, we see that Image (Σp)^{∗} is generated by the following three elements:

v^{3} ∘ aσ, μ ∘ aσ, η ∘ ϵ ∘ aσ.

And Theorem 14.1 (ii, iii) [22, p. 190] yields that

v^{3} ∘ aσ = a(v^{2} ∘ (v ∘ σ)) = 0,
η ∘ ϵ ∘ aσ = a(η ∘ (σ ∘ ϵ)) = 0, and
μ ∘ aσ = a(μ ∘ σ) = −a(σ ∘ μ) = −a(η ∘ ρ) = a(ρ ∘ η).

Consequently, Image (Σp)^{∗} is contained in the subgroup of π_{16}^{s} generated by ρ ∘ η. Hence in order to complete the verification of Fact 3 it suffices to show that

ρ ∘ η ∈ Image J.

To do this let η_{15} : S^{16} → S^{15} represent the element η ∈ π_{1}^{s} and notice that the following digram commutes:

[S^{15}, F] ⟶^{η_{15}^{∗}} [S^{16}, F]
J′↑  ↑^{J}
[S^{15}, O] ⟶_{η_{15}^{∗}} [S^{16}, O]

where J′ denotes the J-homomorphism in dimension 15. We recall that Kervaire and Milnor showed (cf. [18, p. 284]) that Image J′ is a cyclic group of order 480. Using this fact together with Toda’s calculation of π_{15}^{s} in [22, p. 189] it is easily seen that

either ρ ∈ Image J′ or ρ + η ∘ k ∈ Image J′.

Consequently the above commutative diagram shows that

either ρ ∘ η ∈ Image J or ρ ∘ η + η ∘ k ∘ η ∈ Image J.

But Theorem 14.1(i) in [22, p. 190] yields that

η ∘ k ∘ η = η^{2} ∘ k = 0

and consequently ρ ∘ η ∈ Image J.

But Lemma 1 implies that 𝕆P^{2}#Σ^{16} is not concordant to 𝕆P^{2} since the concordance classes of smooth structures on a smooth manifold X are in bijective correspondence with [X, Top/O] provided dim X > 4. Thus assertion (a) of the Theorem is a direct consequence of Lemmas 0 and 1.

Now combining the construction of the previous section with [9, Lemma 3.18] it is seen that there exists a number r_{16} > 0 (independent of M^{16}) such that if the injectivity radius of N^{16} is chosen to be larger than r_{16}, then assertion (b) of the Theorem is also true. Since Borel [3] has constructed closed Riemannian manifolds M^{16} whose universal cover M̃^{16} is 𝕆H^{2}, Theorem produces the examples claimed in the Introduction.

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Chennai Mathematical Institute, India SUNY, Binghamton