DIFFERENTIAL GEOMETRY AND APPLICATIONS Satellite Conference of ICM in Berlin, Aug. 10 – 14, 1998, Brno Masaryk University in Brno (Czech Republic) 1999, 73–87

RIEMANNIAN MANIFOLDS WHOSE SKEW–SYMMETRIC

CURVATURE OPERATOR HAS CONSTANT EIGENVALUES II

PETER B. GILKEY

Abstract. A Riemannian metric on a manifold is said to be IP if the eigenvalues of the skew-symmetric curvature operator are pointwise constant, i.e. they depend upon the point of the manifold but not upon the particular 2 plane in the tangent bundle at that point. The IP metrics for manifolds of dimension m ≥ 4 and m ≠ 7, 8 were classified previously; the cases m = 7 and m = 8 are exceptional. In this paper, we use somewhat different techniques to classify IP metrics if m = 8.

0. Introduction

Let R_{g} be the skew-symmetric curvature operator of a Riemannian manifold (M, g) of dimension m ≥ 4. Let {X, Y } be an oriented orthonormal basis for an oriented 2 plane π in the tangent space T_{P} M to M at P. We define a skewsymmetric endomorphism R_{g}(π) := R(X, Y) of T_{P} M which independent of the particular orthonormal basis chosen for π. We say that the metric g is IP at a point P ∈ M if the eigenvalues of R_{g} are constant on the Grassmanian Gr_{2}^{+}(T_{P} M) of oriented 2 planes in T_{P} M; we say that the metric g is globally IP if g is IP at every point of M;the eigenvalues in question may vary with the point P ∈ M.

0.1 Examples of metrics which are globally IP.

(1) Let g be a metric of constant sectional curvature C on a manifold M. The local isometries of (M, g) act transitively on the set of all 2 planes in the tangent bundle of M so g is globally IP.

(2) Let M = I × N be a product manifold where I is a subinterval of ℝ and where ds_{N}^{2} is a metric of constant sectional curvature K on N. Give M the metric ds_{M}^{2} := dt^{2} + f(t)ds_{N}^{2} where f(t) := (Kt^{2} + At + B)/2 > 0 and where A^{2} − 4BK ≠ 0. Then g is globally IP and is not a metric of constant sectional curvature. We refer to [3, 5] for further details.

1991 Mathematics Subject Classification. 53B20.

Key words and phrases. Curvature operator, K theory.

Research of P. Gilkey partially supported by the NSF (USA) and IHES (FRANCE).

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

It is convenient to study a purely algebraic problem. An algebraic curvature tensor R is a 4 tensor which satisfies the usual curvature symmetries:

R_{ijkl} = −R_{jikl}, R_{ijkl} = R_{klij}, and R_{ijkl} + R_{iklj} + R_{iljk} = 0

where R is expressed relative to an orthonormal basis for ℝ^{m}. If R = R_{g} at a point P, then R is an algebraic curvature tensor. Conversely, given an algebraic curvature tensor R and a metric g defined at a single point P of M, there exists an extension of g to all of M with the given curvature tensor at the point in question. We say that algebraic curvature tensor R is IP if R(π) has constant eigenvalues on Gr_{2}^{+}(T_{P} M). We refer to [5] for the proof of the first assertion and to [3] for the proof of the second two assertions in the following classification result. We note the situation if m = 3 is quite different; see [5] for details.

0.2 Theorem.

(1) Let g be a globally IP metric on a 4 dimensional manifold M. Then the metric g on an open dense subset of M is locally isometric to one of the metrics given in Example 0.1.

(2) Let R be an algebraic IP curvature tensor. If m = 5, if m = 6, or if m ≥ 9, then rank(R) ≤ 2.

(3) Let g be a globally IP metric on a manifold M of dimension at least 5. Assume that rank(R_{g}) ≤ 2 at each point of M. Then the metric g is locally isometric to one of the metrics given in Example 0.1.

We note that assertion (2) fails if m = 4; there exists an algebraic IP curvature tensor R of rank 4 if m = 4 see [5, 4] for details; the non-zero entries of R up to the obvious symmetries be given by:

R_{1221} = 1, R_{1243} = −2, R_{3421} = −2, R_{3443} = 1,

R_{1331} = −2, R_{1342} = −1, R_{2431} = −1, R_{2442} = −2,
R_{1441} = −2, R_{1432} = 1, R_{2341} = 1, R_{2332} = −2.

The discussion in [5] uses a Singer-Thorpe basis and applies the methods of classical differential geometry. One classifies the IP algebraic curvature tensors and then uses the second Bianchi identities to derive the required integrality results necessary to complete the proof. The discussion in [3] uses algebraic topology; the proof of (2) is quite topological in flavor. The authors then classify the algebraic IP curvature tensors of rank 2 and proceed to the proof of (3).

The cases m = 7 and m = 8 are left unsettled by the results of [3, 5]. In this paper we extend Theorem 0.2 (2) to the case m = 8; it then follows that any globally IP metric in dimension m is locally isometric to one of the metrics given in Example 0.1. We also obtain some partial results concerning the possible eigenvalue structures of algebraic IP curvature tensors if m = 7. The exceptional geometry of dimensions 7 and 8 play a crucial role in our arguments; the octonians are always lurking in the background.

RIEMANNIAN MANIFOLDS … 75

We introduce the following notational conventions. Let 𝔰𝔬(m) be the space of skew-symmetric m × m real matrices; this is the Lie algebra of the special orthogonal group SO(m). A continuous map R from Gr_{2}^{+}(m) to 𝔰𝔬(m) is said to be admissible if R(−π) = −R(π) and if R has constant rank. We showed [3] that if R is admissible and non-trivial, then rank(R) = 2 if m ≠ 4, m ≠ 7, and m ≠ 8. As noted above, this fails if m = 4. Let R : Gr_{2}^{+}(m) → 𝔰𝔬(m) be admissible. Let

(0.3)  W (λ_{i}) := {(π, ξ) ∈ Gr_{2}^{+}(m) × ℝ^{m} : R(π)^{2}ξ = λ_{i}ξ};

since the eigenvalues of R(π) are constant and R(π)^{2} is self-adjoint, the W (λ_{i}) are vector bundles over Gr_{2}^{+}(m) and we may decompose the trivial bundle m · 1 in the form m·1 = ⊕_{i}W (λ_{i}). We assume that λ_{0} = 0; this eigenspace may be trivial. We order the remaining eigenvalues so that 0 < dim(W_{1}) ≤ dim(W_{2}).... Let n⃗ := (n_{i}) where n_{i} := dim(W_{i}). The main result of this paper is the following:

0.4 Theorem.

(1) Let R : Gr_{2}^{+}(m) → 𝔰𝔬(m) be admissible with rank(R) > 2. 1-a) If m = 7, then n⃗ = (1, 6), n⃗ = (1, 2, 4), or n⃗ = (3, 4). 1-b) If m = 8, then n⃗ = (0, 8), n⃗ = (0, 2, 6), or n⃗ = (2, 6). 1-c) There exist admissible maps with these eigenvalue structures.

(2) If M is an algebraic IP curvature tensor and if m = 8, then rank(R) ≤ 2.

(3) If R is an algebraic IP curvature tensor and if m = 7, then R does not have the eigenvalue structure (1, 6).

In §1, we will use methods of algebraic topology to prove Theorem 0.4 (1). We will also use the exceptional geometries of ℝ^{7} and ℝ^{8} to construct maps R which have these eigenvalue structures; these maps will not satisfy the other curvature symmetries of course. In §2, we eliminate the eigenvalue structure (0, 8). In §3, we will establish some technical results we shall need to deal with the remaining possibilities. In §4, we eliminate the eigenvalue structures (2, 6) and (0, 2, 6), and in §5, we eliminate the eigenvalue structure (1, 6).

The maps constructed in Theorem 0.4 (1) are multilinear. There are similar investigations in the literature dealing with an analogous problem. Let L(m, n, k) be the dimension of the largest linear subspace V in Hom(ℝ^{m}, ℝ^{n}) so that the rank of 0 ≠ v ∈ V is k. Various authors have tried to bound L for given k where k is relatively large. Adams [1] determines L(m, m, m). Lam and Yiu [6] determine L(m, m, m − 1), L(m, m − 1, m − 2), and L(m, m, m − 2). There is an extensive literature on the subject; see the bibliographies in Lam and Yiu [6] and in Meshulam [7] for further references.

1. The proof of Theorem 0.4 (1)

We begin by recalling some facts concerning real projective space ℝℙ^{k}. If U is a real vector bundle of dimension r over ℝℙ^{k}, let [U] denote the corresponding element of the K theory group KO(ℝℙ^{k}) and let w(U) = 1 + w_{1}(U) + ... + w_{r}(U) be the total Stiefel-Whitney class in ℤ_{2} cohomology H^{∗}(ℝℙ^{k}; ℤ_{2}). Recall that w(U) = 1 if U is trivial, that w(U ⊕V) = w(U)w(V), and that w(L) = 1+x where L is the classifying real line bundle over ℝℙ^{k} and where x generates H^{1}(ℝℙ^{k}; ℤ_{2}).

1.1 Lemma. Let k = 5 or k = 6.

(1) We have H^{∗}(ℝℙ^{k}; ℤ_{2}) = ℤ_{2}[x]/x^{k+1} = 0 is a truncated polynomial ring.

(2) We have KO(ℝℙ^{k}) = [1] · ℤ ⊕ ([L] − [1]) · ℤ_{8}.

(3) Expand [U] = dim(U)[1] + a(U)([L] − [1]). Then a(U) is well defined mod 8 and w(U) = (1 + x)^{a(U)}.

(4) If U is isomorphic to U ⊗ L, then 2a(U) ≡ dim(U) mod 8.

Proof. The first assertion is well known; the second follows from the work of Adams [1], and the third follows from the second. If U is isomorphic to U ⊗ L, then

[U] = dim(U)[1] + a(L)([L] − [1])
=[U ⊗ L] = dim(U)[L] + a(L)([1] − [L])
= dim(U)[1] + (dim(U) − a(L))([L] − [1])

so by assertion (2), a(U) ≡ dim(U) − a(U) mod 8; the final assertion follows. □

Suppose that R : Gr_{2}^{+}(m) → 𝔰𝔬(m) is admissible. The eigenbundles of R^{2} define a decomposition of the trivial bundle over Gr_{2}^{+}(m) of the form m·1 = ⊕_{i}W_{i} where the W_{i} are as defined in equation (0.3). We have R(π) preserves the W_{i} and W_{0} = ker(R). Since R^{2}(−π) = R^{2}(π), these bundles descend to bundles V_{i} over the unoriented Grassmanian Gr_{2}(m). Since π_{1}(Gr_{2}(m)) = ℤ_{2}, there is only one non-trivial real line bundle over Gr_{2}(m); denote this bundle by L_{1}. Since R(−π) = −R(π), R descends to define isomorphisms between V_{i} and V_{i} ⊗ L_{1} for i > 0. Let {e_{i}} be the standard basis for ℝ^{m}. The map ξ ↦ Span{ξ, e_{m}} defines a natural embedding of S^{m−2} in Gr_{2}^{+}(m) and ℝℙ^{m−2} in Gr_{2}(m). Let U_{i} be the restriction of the bundles V_{i} to ℝℙ^{m−1}. Note that the restriction of L_{1} to ℝℙ^{m−1} is L. We summarize our discussion so far. There are bundles U_{i} over ℝℙ^{m−1} so

(1.2)  m · 1 = ⊕_{i}U_{i}, n⃗ = (dim U_{0}, dim U_{1}, ...), and U_{i} = U_{i} ⊗ L for i > 0.

To prove Theorem 0.4 1-a), we must show that the eigenvalue decompositions n⃗ = (3, 2, 2) and n⃗ = (1, 2, 2, 2) are impossible in equation (1.2) if m = 7. If one of these cases occurs, we could combine eigenspaces if necessary to construct a decomposition 7 · 1 = U_{0} ⊕ U_{1} ⊕ U_{2} over ℝℙ^{5} with U_{1} = U_{1} ⊗ L, U_{2} = U_{2} ⊗ L, and dim(U_{1}) = dim(U_{2}) = 2. We use Lemma 1.1 to see 2a(U_{1}) ≡ 2 mod 8. If a(U_{1}) ≡ 5 mod 8, then w(U_{1}) = (1 + x)^{5} in ℤ_{2}[x]/x^{6} which is impossible for dimensional reasons. Thus a(U_{1}) ≡ 1 mod 8; similarly a(U_{2}) ≡ 1 mod 8. Thus w(U_{1}) = w(U_{2}) = (1 + x) so 1 = w(U_{0})w(U_{1})w(U_{2}) = w(U_{0})(1 + x)^{2}. This implies w(U_{0}) = (1 + x)^{6}; this is impossible for dimensional reasons.

To prove Theorem 0.4 1-b), we must show that the eigenvalue decompositions n⃗ = (2, 2, 4), n⃗ = (0, 2, 2, 4), n⃗ = (4, 4), n⃗ = (0, 4, 4), n⃗ = (2, 2, 2, 2), and n⃗ = (0, 2, 2, 2, 2) are impossible in equation (1.2) if m = 8. In these 6 cases, we could combine eigenspaces to construct a decomposition of the form 8·1 = U_{0} ⊕U_{1} over ℝℙ^{6} where dim(U_{1}) = 4 and where U_{1} = U_{1} ⊗ L. We use Lemma 1.1 to see 2a(U_{1}) ≡ 4 mod 8. If a(U_{1}) ≡ 2 mod 8, then w(U_{1}) = (1 + x)^{2} so w(U_{0}) = (1 + x)^{6} which is impossible for dimensional reasons. If a(U_{1}) ≡ 6 mod 8, then w(U_{1}) = (1 + x)^{6} which is impossible for dimensional reasons. This completes the proof of Theorem 0.4 i-b).

To prove Theorem 0.4 i-c), we construct admissible maps R with the given eigenvalue decompositions. We refer to Atiyah, Bott, and Shapiro [2] for details concerning Clifford algebras and spinors. Recall that the Clifford algebra Clif(m) is the universal real unital algebra generated by ℝ^{m} subject to the commutation relations v∗w+w∗v = −2(v, w)1. Let Spin(m) be the subset of Clif(m) generated by all products of the form ω = v_{1} ∗ ... ∗ v_{k} where the v_{i} are unit vectors in ℝ^{m} and k is even. This is a smooth submanifold of Clif(m) which has the structure of a Lie group under Clifford multiplication. If {v_{1}, v_{2}} is an oriented orthonormal basis for a 2 plane π, let σ(π) := v_{1} ∗ v_{2} ∈ Spin(m); this is independent of the particular oriented orthonormal basis for π and defines a smooth embedding σ : Gr_{2}^{+}(m) ↦ Spin(m). There is a natural representation ρ : Spin(m) ↦ SO(m) that exhibits Spin(m) as the universal cover of SO(m) for m ≥ 3; ρ(σ(π)) is −1 on π and +1 on π^{⊥}.

Suppose m = 8. The spin representation c gives a representation of Clif(8) on ℝ^{16}. There is a decomposition ℝ^{16} = ℝ_{+}^{8} ⊕ ℝ_{−}^{8} so that Clifford multiplication by a vector v ∈ ℝ^{8} interchanges the two summands, i.e. c(v) : ℝ_{±}^{8} → ℝ_{∓}^{8}. Clifford multiplication by an element ω ∈ Spin(m) preserves the two summands; thus c restricts to define representations, called the half spin representations, c_{±} of Spin(8) on ℝ_{±}^{8}. Let R(π) = c_{+}(σ(π)); R is an admissible map from Gr_{2}^{+}(8) to 𝔰𝔬(8) which has constant rank 8 since R(π)^{2} = −1 and R(−π) = −R(π). This shows the eigenvalue decomposition (0, 8) is possible. Fix 0 ≠ x_{−} ∈ ℝ_{−}^{8}. If {v_{1}, v_{2}} is an orthonormal basis for an oriented 2 plane π, let

E_{0}(π) := Span{c(v_{1})x_{−}, c(v_{2})x_{−}};

E_{0}(π) and E_{0}π)^{⊥} are invariant subspaces of R(π). Decompose R = R_{0} ⊕ R_{2} where R_{0} is the restriction of R to E_{0} and R_{2} is the restriction of R to E_{0}^{⊥}; −R_{0}^{2} is orthogonal projection on E_{0} and −R_{2}^{2} is orthogonal projection on E^{⊥}. The map C_{0}R_{1} ⊕ R_{2} is an admissible map of Gr_{2}^{+}(8) to 𝔰𝔬(8) with n⃗ = (2, 6) if C_{1} = 0 and n⃗ = (0, 2, 6) if C_{1} ≠ 0.

Suppose m = 7. Let {e_{i}} be the standard orthonormal basis for ℝ^{8}. The inclusion ℝ^{7} ⊂ ℝ^{8} embeds Gr_{2}^{+}(7) in Gr_{2}^{+}(8). Let E_{0}(π) be as above and let

E_{1}(π) := Span{c(π)c(e_{8})x_{−}, c(e_{8})x_{−}} ⊂ ℝ_{+}^{8}.

If π ⊂ Gr_{2}^{+}(7), then E_{0}(π) and E_{1}(π) are orthogonal subspaces invariant under R(π). Decompose R = R_{0}⊕R_{1}⊕R_{2} where R_{0} and R_{1} are the restriction of R to E_{0} and to E_{1} and where R_{2} is the restriction of R to (E_{0} ⊕E_{1})^{⊥}. The map c_{1}R_{1} +R_{2} preserves c(e_{8})x_{−}^{⊥} which we may identify with ℝ^{7}; this gives an admissible map of Gr_{2}^{+}(7) to 𝔰𝔬(7) with n⃗ = (1, 6) if C_{1} = 1, n⃗ = (1, 2, 4) if C_{1} = 2, and n⃗ = (3, 4) if C_{1} = 0. This completes the proof of Theorem 0.4 i-c).

Remark. In joint work with Semmelman [4], we have given an alternate description of the construction in dimension 7 using the exceptional Lie group G_{2} and have shown that it does not lead to an algebraic IP curvature tensor of higher rank in dimension 7. It is still, however, an open question of whether there exist algebraic IP curvature tensors of higher rank in dimension 7 which do not arise from the spin representation.

2. The eigenvalue structure (0, 8)

We wish to prove that there does not exist an algebraic IP curvature tensor on ℝ^{8} with an eigenvalue structure (0, 8). Suppose the contrary. By rescaling R, we may suppose R(ξ, η)^{2} = −1 for any orthonormal set {ξ, η}. Fix a unit vector e_{1} ∈ ℝ^{8}. Let ϕ(ξ) := R(e_{1}, ξ)e_{1}. The identity (R(e_{1}, ξ)e_{1}, η) = (R(e_{1}, η)e_{1}, ξ) shows ϕ is symmetric. If ξ ⊥ e_{1}, then |ϕ(ξ)| = |ξ| so ϕ is an isometry of e_{1}^{⊥}. We diagonalize ϕ to find an orthonormal basis e_{i} for ℝ^{8} so that R(e_{1}, e_{i})e_{1} = ϵ_{i}e_{i} for i ≥ 2 for suitably chosen signs ϵ_{i} = ±1. Let c(ξ) = R(e_{1}, ξ) for ξ ⊥ e_{1}. Then c(ξ)^{2} = −|ξ|^{2}. Thus if 2 < i ≤ j, c(e_{i})c(e_{j}) + c(e_{j})c(e_{i}) = −2δ_{ij} so these endomorphisms satisfy the Clifford commutation relations discussed in §1. We use the Bianchi identity to establish the desired contradiction by computing

\begin{matrix}1 = & |R(e_{2}, e_{3})e_{1}| = |R(e_{3}, e_{1})e_{2} + R(e_{1}, e_{2})e_{3}| \\ = & |ϵ_{2}R(e_{3}, e_{1})R(e_{1}, e_{2})e_{1} + ϵ_{3}R(e_{1}, e_{2})R(e_{1}, e_{3})e_{1}| \\ = & |\{−ϵ_{2}c(e_{3})c(e_{2}) + ϵ_{1}c(e_{2})c(e_{3})\}e_{1}| \\ = & |(ϵ_{1} + ϵ_{2})c(e_{2})c(e_{3})e_{1}| = | ± 1 ± 1| = 0 or 2. \end{matrix}

□

3. Technical preliminaries

In this section, we establish some technical results we will use in §4 and §5 to complete the proof of Theorem 0.4. For the remainder of this paper, we assume that R is an algebraic IP curvature tensor with two distinct eigenvalues λ_{1} and λ_{2} on ℝ^{m} for m = 7 or m = 8. Let W_{u} be the corresponding eigen bundles. We suppose dim W_{2} = 6 and normalize the choice of R so that λ_{1} = −1. If λ_{1} = 0, then W_{1} = dim ker R; this is a slight change of notation from previous sections.

It is convenient to distinguish two copies of ℝ^{m}. We shall for the most part use Greek letters ξ, η, etc. for vectors which are arguments of the curvature operator R(·, ·). We shall use Roman letters e, f, etc. for vectors upon which the curvature operator R acts. If {ξ, η} is an orthonormal set, we let W_{u}(ξ, η) be the eigenspace for the eigenvalue λ_{u} defined by R(ξ, η). If {ξ_{i}} is an orthonormal set, we let r_{ij} := R(ξ_{i}, ξ_{j}) and W_{u,ij} := W_{u}(ξ_{i}, ξ_{j}).

3.1 Lemma.

(1) Let e be a unit vector. We have (r_{12}^{2}e, e) ∈ [min(λ_{1}, λ_{2}), max(λ_{1}, λ_{2})]. We have e ∈ W_{u,12} if and only if (r_{12}^{2}e, e) = λ_{u}. If λ_{1} = 0, then r_{12}^{3} = −r_{12}.

(2) Let E := r_{12}r_{13} + r_{13}r_{12}. If e ∈ W_{u,12}, then (Ee, e) = 0 and r_{12}e ⊥ r_{13}e.

(3) If e ∈ W_{u,12} ∩ W_{u,13}, then Ee = 0.

Proof. Since r_{12}^{2} is symmetric and has exactly two eigenvalues, the first assertion follows. Let r(θ) := R(ξ_{1}, cos θξ_{2} + sin θξ_{3}) and let e ∈ W_{u,12}. Expand (r(θ)^{2}e, e) in a Taylor series about θ = 0:

(r(θ)^{2}e, e) =λ_{u} cos^{2} θ + 2 cos θ sin θ(Ee, e) + sin^{2} θ(r_{13}^{2}e, e)
=λ_{u} + 2θ(Ee, e) + O(θ^{2}).

By assertion (1), λ_{u} is an extremal value. This implies that (Ee, e) = 0 and hence r_{12}e ⊥ r_{13}e which proves assertion (2). If e ∈ W_{u,12}∩W_{u,13}, then r(θ)^{2}e = λ_{u}e+Ee. Since (Ee, e) = 0, (r(θ)^{2}e, e) = λ_{u} so r(θ)^{2}e = λ_{u}e for all θ and hence Ee = 0. □

3.2 Lemma. Let ξ_{1}, η_{2}, η_{3} be unit vectors with ξ_{1} ⊥ η_{2} and ξ_{1} ⊥ η_{3}. The following assertions are equivalent: (1) η_{2} = ±η_{3}, (2) W_{1}(ξ_{1}, η_{2}) = W_{1}(ξ_{1}, η_{3}), and (3) W_{2}(ξ_{1}, η_{2}) = W_{2}(ξ_{1}, η_{3}).

Proof. Clearly assertion (1) implies assertion (2). Since W_{2}^{⊥} = W_{1}, assertion (2) is equivalent to assertion (3). Suppose that assertion (3) holds but that assertion (1) fails; we argue for a contradiction. Let ξ_{2} = η_{2} and choose ξ_{3} so that {ξ_{i}} is an orthonormal set and so that η_{3} = cos θξ_{2} + sin θξ_{3} where sin θ ≠ 0. Let e ∈ W_{2}(ξ_{1}, ξ_{2}). By Lemma 3.1, (E(ξ_{1}, ξ_{3})e, e) = 0. Since W_{2}(ξ_{1}, η_{2}) = W_{2}(ξ_{1}, η_{3}),

−1 = (R(ξ_{1}, η_{3})^{2}e, e) = − cos θ^{2} + sin^{2} θ(r_{13}^{2}e, e).

Since sin θ ≠ 0, (r_{13}^{2}e, e) = −1. By Lemma 3.1, e ∈ W_{2,13}. Thus W_{2,12} ⊂ W_{2,13}. Since dim W_{2,12} = dim W_{2.13}, W_{2,12} = W_{2,13}. Let τ(i) := r_{12} and τ(j) := r_{13}. Then τ(i)^{2} = τ(j)^{2} = −1 on W_{2}(ξ_{1}, η_{1}); Lemma 3.1 implies τ(i)τ(j)+τ(j)τ(i) = 0 on W_{2}(ξ_{1}, η_{1}). Thus τ defines a unital representation of the quaternions ℍ on W_{2}(ξ_{1}, η_{1}). This implies dim W_{2}(ξ_{1}, η_{1}) ≡ 0 mod 4 which is false. □

3.3 Lemma. There exists an orthonormal basis {e_{i}} for ℝ^{m} so that:

(1) We have r_{12}e_{1} = e_{2}, r_{12}e_{2} = −e_{1}, r_{12}e_{3} = −e_{4}, r_{12}e_{4} = e_{3}, r_{13}e_{1} = e_{3}, r_{13}e_{2} = e_{4}, r_{13}e_{3} = −e_{1}, r_{13}e_{4} = −e_{2}, and r_{12}r_{13} = −r_{13}r_{12} on the space

S := Span{e_{1}, e_{2}, e_{3}, e_{4}} ⊂ W_{2,12} ∩ W_{2,13}.

(2) We have e_{6} = r_{12}e_{5}, e_{7} = r_{13}e_{5}, W_{2,12} = Span{e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}}, and

W_{2,13} = Span{e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{7}}.
(3) If m = 7, then W_{1,12} = Span{e_{7}} and W_{1,13} = Span{e_{6}}.

(4) If m = 8, then W_{1,12} = Span{e_{7}, e_{8}}, W_{1,13} = Span{e_{6}, e_{8}}. Furthermore,

r_{12}r_{13}e_{1} = ±|λ_{1}|^{1/2}e_{8}.

Proof. Let V := W_{2,12} ∩ W_{2,13}. Since dim W_{2,12} = dim W_{2,13} = 6 and since m ≤ 8, we have the inequality dim V ≥ 6 + 6 − 8 = 4. Since r_{12} preserves W_{2,12} and since r_{12} is non-singular on W_{2,12}, we have r_{12}V is a 4 dimensional subspace of W_{2,12}. Thus we have the inequality dim(V ∩ r_{12}V) ≥ 4 + 4 − 6 = 2. Choose a unit vector e_{1} in V ∩ r_{12}V. Let e_{2} := r_{12}e_{1}, let e_{3} := r_{13}e_{1}, and let e_{4} := r_{12}r_{13}e_{1}. We use Lemma 3.1 to prove the first assertion:

(1) Since e_{1} ∈ V and e_{2} ∈ V, (r_{12}r_{13} + r_{13}r_{12})e_{1} = 0, (r_{12}r_{13} + r_{13}r_{12})e_{2} = 0,

r_{12}^{2}e_{1} = −e_{1}, r_{13}^{2}e_{1} = −e_{1}, r_{12}^{2}e_{2} = −e_{2}, and r_{13}^{2}e_{2} = −e_{2}.

(2) Since r_{12}^{2}e_{3} = r_{12}^{2}r_{13}e_{1} = −r_{12}r_{13}r_{12}e_{1} = −r_{12}r_{13}e_{2} = r_{13}r_{12}e_{2} = −e_{3}, and r_{13}^{2}e_{3} = r_{13}^{3}e_{1} = −r_{13}e_{1} = −e_{3}, we have e_{3} ∈ V so (r_{12}r_{13} +

r_{13}r_{12})e_{3} = 0.

(3) Since r_{12}^{2}e_{4} = r_{12}^{2}r_{12}e_{3} = −r_{12}e_{3} = −e_{4} and similarly r_{13}^{2}e_{4} = −e_{4}, e_{4} ∈ V

and (r_{12}r_{13} + r_{13}r_{12})e_{4} = 0.

Since r_{12} and r_{13} are skew-symmetric and preserve S, the orthogonal complement T is preserved by r_{12} and r_{13} as well. Let W̃_{2,12} := W_{2,12} ∩ T and W̃_{2,13} := W_{2,13} ∩ T. Since S ⊂ W_{2,12} and S ⊂ W_{2,13}, we can decompose

W_{2,12} = S ⊕ W̃_{2,12} and W_{2,13} = S ⊕ W̃_{2,13}.

We have dim W_{2,12} = dim W_{2,13} = 2. Suppose there exists a unit vector e_{5} in W̃_{2,12} ∩ W̃_{2,13}. Let e_{6} := r_{12}e_{5} and e_{7} := r_{13}e_{5}. Then W̃_{2,12} = Span{e_{5}, e_{6}} and W̃_{2,13} = Span{e_{5}, e_{7}}. By Lemma 3.1, (r_{12}r_{13} + r_{13}r_{12})e_{5} = 0; it now follows that {e_{5}, e_{6}, e_{7}} is an orthonormal subset of T. Since {e_{i}}_{i≤6} is an orthonormal basis for W_{2,12}, it follows that e_{7} ∈ W_{1,12} so e_{7} ∈ W_{1,12}∩W_{2,13}; similarly e_{6} ∈ W_{2,12}∩W_{1,13}. The second assertion of the Lemma now follows if m = 7. If m = 8, we may choose e_{8} to complete the orthonormal basis. We then have e_{8} ⊥ e_{i} for i ≤ 6 so e_{8} ∈ W_{1,12}; similarly e_{8} ∈ W_{1,13} so e_{8} ∈ W_{1,12} ∩ W_{1,13}. Since r_{12}r_{13}e_{5} = r_{12}e_{7} and since e_{7} ∈ W_{1,12}, we have r_{12}e_{7} ∈ W_{1,12}. Since r_{12}e_{7} ⊥ e_{7}, r_{12}e_{7} = ϵe_{8}. Since r_{12}^{2}e_{7} = λ_{1}e_{7}, we have ϵ^{2} = |λ_{1}| and the final assertion of the Lemma follows. Thus to prove the Lemma, we must show

(3.4)  U := W̃_{2,12} ∩ W̃_{2,13} ≠ {0}.

If m = 7, we have dim{W̃_{2,12} ∩ W̃_{2,13}} ≥ 2 + 2 − 3 ≥ 1 and equation (3.4) follows. We may therefore assume that m = 8.

Suppose first that λ_{1} ≠ 0. If f is a unit vector in W_{1,12}, then {f, r_{12}f} spans W_{1,12}. Since r_{13}f ⊥ f and r_{13}f ⊥ r_{12}f, we have r_{13}f ∈ W_{2,12}. Thus r_{13}W_{1,12} ⊂ W̃_{2,12} and r_{13}W̃_{2,12} ⊂ W_{1,12} so r_{13}^{2} preserves W̃_{2,12}. Let μ_{i} be the two eigenvalues of r_{13}^{2} on W̃_{2,12}; each μ_{i} is either λ_{1} or λ_{2}. If μ_{i} = λ_{2} for i = 1 or i = 2, then W̃_{2,12} ∩ W̃_{2,13} ≠ {0} as desired. We therefore suppose μ_{1} = μ_{2} = λ_{1} or equivalently that W̃_{2,12} = W_{1,13}. We have r_{12} and r_{13}/\sqrt{−λ_{1}} define rotations of ±π/2 on W̃_{2,12}. Hence r_{12} = ±r_{13}/\sqrt{−λ_{1}} on this 2 dimensional subspace. Choose θ so that tan θ = ∓\sqrt{−λ_{1}}. Then R(ξ_{1}, cos θξ_{2} + sin θξ_{3}) = cos θr_{12} + sin θr_{13} = 0 on W̃_{2,12}. This is false as R(ξ, η) is invertible for any orthonormal set.

Suppose λ_{1} = 0 so r_{ij}^{3} = −r_{ij}. If e ∈ W̃_{2,12}, then W̃_{2,12} = Span{e, r_{12}e}. Thus r_{13}W̃_{2,12} ⊂ W_{1,12} and ker(r_{13}) = W_{1,13}. Suppose W̃_{2,12} ∩ W_{1,13} = {0}. Then r_{13}e and r_{13}r_{12}e are linearly independent so W_{1,12} = Span{r_{13}e, r_{13}r_{12}e}. We compute

\begin{matrix}(r_{13}^{2}e, r_{13}e) = −(r_{13}^{3}e, e) = −(r_{13}e, e) = 0, and \\ (r_{13}^{2}e, r_{13}r_{12}e) = −(r_{13}^{3}e, r_{12}e) = (r_{13}e, r_{12}e) = 0. \end{matrix}

This shows that r_{13}^{2}e ∈ W̃_{2,12} so r_{13}^{2} preserves W̃_{2,12}. Since r_{13}e ≠ 0, r_{13}^{2}e ≠ 0 and thus W̃_{2,12} = W̃_{2,13} and W_{2,12} = W_{2,13} which contradicts Lemma 3.2.

Thus we may suppose λ_{1} = 0 and W̃_{2,12} ∩ W_{1,13} ≠ {0}; let e_{6} be a unit vector in this intersection. Let e_{5} = −r_{12}e_{6} ∈ W̃_{2,12}; r_{12}e_{5} = e_{6} so r_{13}r_{12}e_{5} = r_{13}e_{6} = 0. Since r_{13}e_{5} ⊥ e_{5} and r_{13}e_{5} ⊥ e_{12}e_{5} = e_{6}, we have r_{13}e_{5} ∈ W_{1,12} so r_{12}r_{13}e_{5} = 0 as well. Let r(θ) = R(ξ_{1}, cos θξ_{2} + sin θξ_{3}); r(θ)^{3} = −r(θ). We compute:

r(θ)^{3}e_{5} = {cos^{3} θr_{12}^{3} + cos^{2} θ sin θ(r_{12}^{2}r_{13} + r_{12}r_{13}r_{12} + r_{13}r_{12}^{2})

+ cos θ sin^{2} θ(r_{12}r_{13}^{2} + r_{13}r_{12}r_{13} + r_{13}^{2}r_{12}) + sin^{3} θr_{13}^{3}}e_{5}
= {− cos^{3} θr_{12} − cos^{2} θ sin θr_{13} + cos θ sin^{2} θr_{12}r_{13}^{2} − sin^{3} θr_{13}}e_{5}
= {−r(θ) + cos θ sin^{2} θ(r_{12}r_{13}^{2} + r_{12})}e_{5} = −r(θ)e_{5}.

This shows that r_{12}r_{13}^{2}e_{5} = r_{12}e_{5}. Since |r_{12}e_{5}| = 1 and since r_{12} is a linear operator of norm 1, we conclude |r_{13}^{2}e_{5}| ≥ 1. Since 1 is an extremal value, we have e_{5} ∈ W_{2,13} and hence W_{2,13} ∩ W_{2,13} ≠ {0}. □

3.5 Lemma. If |ξ| = 1, let the axis 𝔄_{u}(ξ) := ∩_{η⊥ξ,|η|=1}W_{u}(R(ξ, η)).

(1) If ξ is any unit vector and if ξ ∈ 𝔄_{u}(ξ), then λ_{u} = 0.

(2) We have ∩_{ξ}𝔄_{u}(ξ) = {0}.

Proof. Suppose assertion (1) is false. Thus λ_{u} ≠ 0 and we may choose a unit vector ξ_{1} ∈ 𝔄_{1}(ξ_{1}). Then if {ξ_{1}, η} is an orthonormal set, we have ξ_{1} ∈ W_{u}(R(ξ_{1}, η)) so |R(ξ_{1}, η)ξ_{1}|^{2} = −λ_{u}. Thus the map ϕ : η ↦ R(ξ_{1}, η)ξ_{1}/\sqrt{−λ_{u}} is an isometry from ξ_{1}^{⊥} to itself. The curvature symmetry R(ξ_{1}, η_{1}, ξ_{1}, η_{2}) = R(ξ_{1}, η_{2}, ξ_{1}, η_{1}) implies that (ϕ(η_{1}), η_{2}) = (ϕ(η_{2}), η_{1}). Thus ϕ is a symmetric isometry. We can therefore choose a basis ξ_{i} for ξ_{1}^{⊥} so that ϕξ_{i} = ϵ_{i}ξ_{i} where ϵ_{i} = ±1, i.e. r_{1i}ξ_{1} = ϵ_{i}\sqrt{−λ_{u}}ξ_{i} for i ≥ 2. By Lemma 3.1, (r_{1i}r_{1j} + r_{1j}r_{1i})ξ_{1} = 0 for i ≠ j. Let η_{ij} := r_{1i}r_{1j}e_{1}. Let k ≥ 2 be distinct from i and j. We use the Bianchi identity to compute:

\begin{matrix}0 = & (\sqrt{−λ_{u}})^{2}\{(r_{1i}ξ_{j}, ξ_{k}) + (r_{1j}ξ_{k}, ξ_{i}) + (r_{1k}ξ_{i}, ξ_{j})\} \\ = & ϵ_{j}ϵ_{k}(r_{1i}r_{1i}ξ_{1}, r_{1k}ξ_{1}) + ϵ_{k}ϵ_{i}(r_{1j}r_{1k}ξ_{1}, r_{1i}ξ_{1}) + ϵ_{i}ϵ_{j}(r_{1k}r_{1i}ξ_{1}, r_{1k}ξ_{1}) \\ = & (ϵ_{j}ϵ_{k} + ϵ_{k}ϵ_{i} + ϵ_{i}ϵ_{j})(r_{1i}r_{1j}ξ_{1}, r_{1k}ξ_{1}). \end{matrix}

Since (±1 ± 1 ± 1) ≠ 0, η_{ij} ⊥ r_{1k}ξ_{1} so η_{ij} ⊥ ξ_{k}. We have η_{ij} ⊥ r_{1i}e_{1} and η_{ij} ⊥ r_{1j}e_{1} so η_{ij} ⊥ ξ_{i} and η_{ij} ⊥ ξ_{j}. Since e_{1} ∈ W_{u,1i} ∩ W_{u,1j}, η_{ij} ⊥ e_{1} by Lemma 3.1. Thus η_{ij} ⊥ ξ_{k} for all k so η_{ij} = 0 for all i, j. Since λ_{u} ≠ 0 and e_{1} ∈ W_{u,1i}, f_{i} := r_{1i}e_{1} ≠ 0 for 2 ≤ i. Furthermore f_{i} ⊥ f_{j} for i ≠ j. Since {f_{3}, f_{4}, f_{5}} ⊂ ker r_{12}, we see dim W_{1,12} ≥ 3 which is false.

Suppose assertion (2) fails. Choose a unit vector ξ_{1} in ∩_{ξ}𝔄_{u}(ξ). Then in particular, ξ_{1} ∈ 𝔄_{1}(ξ_{1}) and hence λ_{u} = 0 by assertion (1). This implies u = 1. Let {ξ_{i}} be an orthonormal basis for ℝ^{m}. Since ξ_{1} ∈ W_{1}(ξ_{i}, ξ_{j}), we have R(ξ_{i}, ξ_{j})ξ_{1} = 0 for i ≠ j; trivially R(ξ_{i}, ξ_{i})ξ_{1} = 0. Then: (R(ξ_{1}, ξ_{2})ξ_{j}, ξ_{k}) = (R(ξ_{j}, ξ_{k})ξ_{1}, ξ_{2}) = 0 and thus R(ξ_{1}, ξ_{2})ξ_{j} = 0 so R(ξ_{1}, ξ_{2}) vanishes identically which is false. □

We can now show 𝔄_{2} is non-trivial if m = 7 and that 𝔄_{1} is non-trivial if m = 8.

3.6 Lemma.

(1) Suppose that m = 8. Let {ξ_{1}, ξ_{2}, ξ_{3}, ξ_{4}} be an orthonormal set. Then 1-a) If W_{1}(ξ_{1}, ξ_{2}) ∩ W_{1}(ξ_{1}, ξ_{3}) ∩ W_{1}(ξ_{1}, ξ_{3}) ≠ {0}, then dim 𝔄_{1}(ξ_{1}) = 1.

1-b) If ξ is any unit vector, then dim 𝔄_{1}(ξ) = 1.

1-c) The map Φ : ξ ↦ 𝔄_{1}(ξ) is a continuous map from S^{7} to ℝℙ^{7}. 1-d) We have W_{1}(ξ_{1}, ξ_{2}) ∩ W_{1}(ξ_{1}, ξ_{3}) ∩ W_{1}(ξ_{2}, ξ_{3}) = {0}.

(2) Suppose that m = 7. 2-a) If ξ is a unit vector, then dim 𝔄_{2}(ξ) = 1. 2-b) The map Φ : ξ ↦ 𝔄_{2}(ξ) is a continuous map from S^{6} to ℝℙ^{6}. 2-c) There exists a unit vector ξ so ξ ∈ 𝔄_{2}(ξ).

Proof. Suppose there exists a unit vector f_{1} in W_{1,12} ∩W_{1,13} ∩W_{1,14}. Let {f_{1}, f_{i}} be an orthonormal basis for W_{1,1i} for i = 2, 3, 4. We use Lemma 3.3 to see:

\begin{matrix}f_{1} ∈ W_{1,12} ∩ W_{1,13} ∩ W_{1,14}, & f_{2} ∈ W_{1,12} ∩ W_{2,13} ∩ W_{2,14}, \\ f_{3} ∈ W_{2,12} ∩ W_{1,13} ∩ W_{2,14}, & f_{4} ∈ W_{2,12} ∩ W_{2,13} ∩ W_{1,14}. \end{matrix}

Thus {ξ_{1}, ξ_{2}, ξ_{3}, ξ_{4}} is an orthonormal set. Extend this set to an orthonormal basis {ξ_{1}, ξ_{2}, ξ_{3}, ξ_{4}, ξ_{5}, ξ_{6}, ξ_{7}, ξ_{8}} for ℝ^{8}. By Lemma 3.3, dim(W_{1,1i}∩W_{1,1j}) = 1 for i ≠ j. Choose unit vectors g_{ik} = a_{ik}f_{1} +b_{ik}f_{i} ∈ W_{1,1i} ∩W_{1,1k} for 2 ≤ i ≤ 4 < k. If all the b_{ik} are non-zero, then the g_{ik} are 3 linearly independent vectors in W_{1,1k}; this is not possible as dim W_{1,1k} = 2. Thus some b_{ik} = 0 so f_{1} ∈ W_{1,1k}. Thus f_{1} ∈ W_{1,1i} for 2 ≤ i. Let η = Σ_{i≥2}u_{i}ξ_{i} be a unit vector perpendicular to ξ_{1}. We complete the proof of assertion 1-a) by showing f_{1} ∈ W_{2}(ξ_{1}, η). By Lemma 3.1:

R(ξ_{1}, η)^{2}f_{1} = Σ_{i}u_{i}^{2}r_{1i}^{2}f_{1} + Σ_{i<j}u_{i}u_{j}(r_{1i}r_{1j} + r_{1j}r_{1i})f_{1} = λ_{1}f_{1}.

Suppose that assertion 1-b) fails. Since dim W_{1}(ξ, η) = 2, dim 𝔄_{1}(ξ) ≤ 2. If dim 𝔄_{1}(ξ) = 2, then 𝔄_{1}(ξ) = W_{1}(ξ, η) for any η ⊥ ξ; this contradicts Lemma 3.2. Thus assertion 1-b) fails implies 𝔄_{1}(ξ) = {0} for some ξ = ξ_{1}. Let {ξ_{1}, ξ_{2}, ξ_{3}, η} be an orthonormal set. By assertion 1-a), W_{1,12}∩W_{1,13}∩W_{1}(ξ_{1}, η) = {0}. For i = 2, 3, let {f_{1}, f_{i}} be an orthonormal basis for W_{1,1i}; {f_{1}, f_{2}, f_{3}} is an orthonormal set. Let g_{i} = a_{i}f_{1}+b_{i}f_{i} be a unit vector in the 1 dimensional space W_{1,1i}∩W_{1}(R(ξ_{1}, η)). If g_{2} = ±g_{3}, then g_{i} = ±f_{1} and W_{1,12} ∩ W_{1,13} ∩ W_{1}(R(ξ_{1}, η)) is non-trivial which is false. Thus g_{2} and g_{3} are linearly independent so

W_{1}(R(ξ_{1}, η)) = Span{g_{2}, g_{3}} ⊂ W_{1,12} + W_{1,13}.

Note that dim(W_{1,12} + W_{1,13}) = 3. We use the equation above to see that the map Φ : η ↦ W_{1}(R(ξ_{1}, η)) is a smooth injective map from ℝℙ{Span{ξ_{1}, ξ_{2}, ξ_{3})^{⊥}} to Gr_{2}(W_{1,12} + W_{1,13}); this is not possible as dim ℝℙ{Span{ξ_{1}, ξ_{2}, ξ_{3})^{⊥}} = 4 and dim Gr_{2}(W_{1,12} + W_{1,13}) = 2. Since 𝔄_{1}(ξ_{1}) = W_{1}(ξ_{1}, ξ_{2}) ∩ W_{1}(ξ_{1}, ξ_{3}) for any orthonormal set {ξ_{1}, ξ_{2}, ξ_{3}}, and since the map (ξ, η) ↦ W_{1}(ξ, η) is continuous, assertion 1-c) follows.

Suppose assertion (1-d) fails. Let f be a unit vector in W_{1,12} ∩ W_{1,13} ∩ W_{1,23}. Then f ∈ 𝔄_{1}(ξ_{1}) ∩ 𝔄_{1}(ξ_{2}). Let {ξ_{1}, ξ_{2}, η} be an orthonormal set. Then f belongs to W_{1}(ξ_{1}, η) ∩ W_{1}(ξ_{2}, η) implies f belongs to 𝔄_{1}(η). Given a unit vector ζ, we can choose η_{1} and η_{2} so that {ξ_{1}, ξ_{2}, η_{1}, η_{2}} is an orthonormal set and so {η_{1}, η_{2}, ζ} is an orthonormal set. Since {ξ_{1}, ξ_{2}, η_{i}} is an orthonormal set, the previous argument shows that f belongs to 𝔄_{1}(η_{1}) ∩ 𝔄_{1}(η_{2}). Since {η_{1}, η_{2}, ζ} is an orthonormal set, the same argument also implies that f ∈ 𝔄_{1}(ζ). Thus there is a common axis f which contradicts Lemma 3.5.

Suppose m = 7. Let {ξ_{i}} be an orthonormal set. Let f_{i} be unit vectors spanning W_{1,1i} for 2 ≤ i ≤ 7. We use Lemma 3.3 to see f_{i} ⊥ f_{j} for i ≠ j so {f_{2}, ..., f_{7}} is an orthonormal set. Complete this set to an orthonormal basis {f_{1}, ..., f_{7}} for ℝ^{7}. Then f_{1} ⊥ f_{i} implies f_{1} ⊥ W_{1,1i} so f_{1} ∈ W_{2,1i} for 2 ≤ i ≤ 7. Thus r_{1i}^{2}f = −f and (r_{1i}r_{1j} + r_{1j}r_{1i})f = 0 for i and j distinct indices between 2 and 7. If η ⊥ ξ_{1}, we may expand η = Σ_{i}u_{i}ξ_{i} and compute R(ξ_{1}, η)f_{1} = Σ_{i>1}iu_{i}^{2}r_{1i}^{2}f_{1} = −f_{1} so f_{1} belongs to W_{2}(ξ_{1}, η) for any unit vector η perpendicular to e_{1}. This proves assertion 2-a). Let Φ : ξ ↦ 𝔄_{2}(ξ) ∈ ℝℙ^{6}; Φ is a well defined map from S^{6} to ℝℙ^{6}. On a small open neighborhood of ξ in S^{6}, we can choose the f_{i} = f_{i}(ξ) described above to vary continuously with ξ; it now follows that Φ is continuous which proves assertion 2-b). Since 𝔄_{2}(−ξ) = 𝔄_{2}(ξ), Φ descends to a continuous map from ℝℙ^{6} to ℝℙ^{6}. The Lefschetz fixed point formulas imply any map of ℝℙ^{6} to itself has a fixed point; this means that there exists ξ so ξ ∈ 𝔄_{2}(ξ). □

Let m = 8. By Lemma 3.6, dim 𝔄_{1}(ξ) = 1 for any unit vector ξ. The map ξ ↦ 𝔄_{1}(ξ) defines a smooth non-constant map from S^{7} to ℝℙ^{7}. We use covering space theory to see that we may lift this map to take values in S^{7}. This defines a smooth map β on S^{7} so that β(ξ) ∈ 𝔄_{1}(ξ) for all unit vectors ξ. Note that the distinguished vector e_{8} of Lemma 3.3 is ±β(ξ_{1}).

3.7 Lemma.

(1) Let m = 8. There exist universal choices of sign K = ±\sqrt{−λ_{1}} and ϵ = ±1 so that if {ξ_{1}, ξ_{2}, ξ_{3}} is any orthonormal set: 1-a) We have {β(ξ_{1}), β(ξ_{2})} is an orthonormal basis for W_{1}(ξ_{1}, ξ_{2}). 1-b) We have R(ξ_{1}, ξ_{2})β(ξ_{1}) = Kβ(ξ_{2}). 1-c) We have β(ξ_{3}) = ϵR(ξ_{1}, ξ_{2})R(ξ_{1}, ξ_{3})β(ξ_{2}). 1-d) The map β extends to a linear isometry of ℝ^{8}.

(2) Let m = 7. There exists a linear map β : Λ^{2}ℝ^{7} → ℝ^{7} so that if {ξ_{i}} is any orthonormal set, then 2-a) We have β(ξ_{1}, ξ_{2}) is a unit vector which spans W_{1,12}. 2-b) We have β(ξ_{1}, ξ_{2}) = ±R(ξ_{1}, ξ_{3})R(ξ_{1}, ξ_{2})β(ξ_{1}, ξ_{3}).

Proof. Let m = 8. β_{i} := β(ξ_{i}) where i = 1, 2, 3. Let {β_{1}, e_{i}} be an orthonormal basis for W_{1,1i} where i = 2, 3. Expand β_{i} = a_{i}β_{1} +b_{i}e_{i} where i = 2, 3. If β_{2} = ±β_{3}, since {β_{1}, e_{2}, e_{3}} is an orthonormal set we have b_{i} = 0 so β_{2} = ±β_{1}. This would imply that β_{1} ∈ W_{1,12} ∩ W_{1,13} ∩ W_{1,23} and would contradict Lemma 3.6. Thus the β_{i} are linearly independent and b_{i} ≠ 0. Since β_{i} ∈ W_{1,23} for i = 2, 3, we have

(3.8)  W_{1,23} = Span{β_{2}, β_{3}} ⊂ W_{1,12} + W_{1,13}.

Since b_{i} ≠ 0, W_{1,12} = Span{β_{1}, e_{2}} = Span{a_{2}β_{1} + b_{2}e_{2}, −b_{2}β_{1} + a_{2}e_{2}}. Since β_{2} = a_{2}β_{1} + b_{2}e_{2} ∈ 𝔄(ξ_{2}), we have a_{2}β_{1} + b_{2}e_{2} ∈ W_{1,23}. We use Lemma 3.3 to see the complementary vector satisfies −b_{2}β_{1} + a_{2}e_{2} ⊥ W_{1,23}. Similarly we have −b_{3}β_{1} + a_{3}e_{3} ⊥ W_{1,23}. Since dim W_{1,23} = 2, dim{(W_{1,12} + W_{1,13}) ∩ W_{1,23}^{⊥}} = 1 by equation (3.8). Consequently −b_{2}β_{1} + a_{2}e_{2} = ± − b_{3}β_{1} + a_{3}e_{3}. That implies that a_{2} = a_{3} = 0 so β_{2} = e_{2} and β_{3} = e_{3}. Assertions 1-a,b) now follow from Lemma 3.3; continuity shows the sign is in assertion 1-b) is universal.

We use the decomposition of Lemma 3.3. We have β_{1} = ±e_{8}. Since {β_{1}, β_{2}} is an orthonormal basis for W_{1,12} and {β_{1}, β_{3}} is an orthonormal basis for W_{1,13}, we have that β_{2} = ±e_{7} = ±r_{13}e_{5} and β_{3} = ±e_{6} = ±r_{12}e_{5}. This implies that r_{12}r_{13}β_{2} = ±r_{12}r_{13}r_{13}e_{5} = ±r_{12}e_{5} = ±β_{3}. The sign in question is universal and independent of the choice of ξ_{i}; assertion 1-c) now follows.

Let {η_{1}, η_{2}} be an orthonormal set. Let a_{1}^{2} +a_{2}^{2} = 1. Choose ξ_{i} so {ξ_{1}, ξ_{2}, η_{1}, η_{2}} is an orthonormal set. Then

\begin{matrix}β(a_{1}η_{1} + a_{2}η_{2}) = & ϵR(ξ_{1}, ξ_{2})R(ξ_{1}, a_{1}η_{1} + a_{2}η_{2})β(ξ_{2}) \\ = & a_{1}ϵR(ξ_{1}, ξ_{2})R(ξ_{1}, η_{1})β(ξ_{2}) + a_{2}ϵR(ξ_{1}, ξ_{2})R(ξ_{1}, η_{2})β(ξ_{2}) \\ = & a_{1}β(η_{1}) + a_{2}β(η_{2}). \end{matrix}

Let m = 7. The map {ξ, η} → W_{1}(ξ, η) defines a continuous map from Gr_{2}(ℝ^{7}) to ℝℙ^{6} and which lifts to a continuous map from Gr_{2}^{+}(ℝ^{7}) to S^{6}. If {ξ_{1}, ξ_{2}, ξ_{3}} is an orthonormal set, we consider the adapted orthonormal frame of Lemma 3.3. Then β(ξ_{1}, ξ_{2}) = ±r_{13}e_{5} and β(ξ_{1}, ξ_{2}) = ±r_{12}e_{5} so β(ξ_{1}, ξ_{2}) = ±r_{13}r_{12}β(ξ_{1}, ξ_{3}). It now follows that β can be extended bilinearly and the desired conclusion follows.□

4. The eigenvalue structures (2, 6) and (0, 2, 6)

4.1 Normal form for R. Throughout §4, let R be an algebraic IP curvature tensor on ℝ^{8} with an eigenvalue structure of (2, 6) or (0, 2, 6). Let β be the linear isometry of ℝ^{8} defined in Lemma 3.7. If {ξ, η} is an orthonormal set, let

p(ξ, η) : e ↦ (β(ξ), e)β(η) − (β(η), e)β(ξ).

Since {β(ξ), β(η)} is an orthonormal basis for W_{1}(ξ, η), p is a rotation through an angle of \frac{π}{2} on W_{1}(ξ, η) and vanishes on W_{2}(ξ, η). Thus there exists a universal constant K = ±\sqrt{−λ_{1}} so that R = Kp on W_{1}. Let

c(ξ, η) := R(ξ, η) + (1 − K)p(ξ, η).

Then c(ξ, η) agrees with R on W_{2}(ξ, η) and with p on W_{1}(ξ, η). Consequently p and c commute so β(η) = p(ξ, η)β(ξ) = c(ξ, η)β(ξ), c(ξ, η)^{2} = −1, and R = c+(K−1)p. Thus c(ξ_{1}, η_{1})c(ξ_{1}, η_{2}) + c(ξ_{1}, η_{2})c(ξ_{1}, η_{2}) = −2(η_{1}, η_{2}) if ξ_{1} ⊥ η_{i} so c satisfies the Clifford commutation relations.

4.2 Lemma. Let R be an algebraic IP curvature tensor on ℝ^{8} with an eigenvalue structure of type (0, 2, 6) or (0, 6). If |ξ| = 1, let σ := ξ + (K − 1)(ξ, β(ξ))β(ξ) and let Z(ξ) := {η : (ξ, η) = 0, (β(ξ), η) = 0, (ξ, β(η)) = 0}.

(1) If β(ξ) ≠ ±ξ, there an orthonormal set {ξ, Ξ} so that β(ξ) = {a+bc(ξ, Ξ)}ξ

and so that Z(ξ) = {η : (ξ, η) = 0, (β(ξ), η) = 0, (Ξ, η) = 0}.

(2) If β(ξ) = ±ξ, then Z(ξ) = {η : (ξ, η) = 0}.

(3) If η ∈ Z(ξ), then R(ξ, η)ξ = c(ξ, η)σ.

(4) We have ϕ : η ↦ R(ξ, η)ξ is a symmetric map preserving Z(ξ).

(5) We have σ(ξ) = 0.

Proof. If ξ ≠ ±β(ξ), expand β(ξ) = aξ + bζ where {ξ, ζ} is an orthonormal set, where a^{2} +b^{2} = 1 and where b ≠ 0. The map ξ → c(ξ, Ξ)ξ is an isometry of ξ^{⊥} and hence we can choose Ξ so that c(ξ, Ξ)ξ = ζ. We prove assertion (1) by computing:

a) If ξ ⊥ η, then c(ξ, η)c(ξ, Ξ) + c(ξ, Ξ)c(ξ, η) = −2(Ξ, η).

b) If ξ ⊥ η, then (β(η), ξ) = (c(ξ, η)β(ξ), ξ) = (c(ξ, η){a + bc(ξ, Ξ)}ξ, ξ) = \frac{b}{2}(\{c(ξ, Ξ)c(ξ, η) + c(ξ, η)c(ξ, Ξ)\}ξ, ξ) = −b(Ξ, η)(ξ, ξ) = −b(Ξ, η).

c) We have (ξ, η) = 0 and (ξ, β(η)) = 0 if and only if (ξ, η) = 0 and (η, Ξ) = 0. Suppose next that β(ξ) = ±ξ. If η ⊥ ξ, then β(η) = c(ξ, η)ξ ⊥ ξ so η ∈ Z(ξ). We prove assertion (3) by computing:

\begin{matrix}R(ξ, η)ξ = & (K − 1)\{(ξ, β(ξ))β(η) − (ξ, β(η))β(ξ)\} + c(ξ, η)ξ \\ = & (K − 1)(ξ, β(ξ))c(ξ, η)β(ξ) + c(ξ, η)ξ \\ = & c(ξ, η)\{ξ + (K − 1)(ξ, β(ξ))β(ξ)\}. \end{matrix}

The curvature symmetry (R(ξ, η_{1})ξ, η_{2}) = (R(ξ, η_{2})ξ, η_{1}) shows that ϕ is symmetric. If ξ = ±β(ξ), then R(ξ, η)ξ ⊥ ξ so ϕ preserves Z(ξ). We therefore suppose ξ ≠ ±β(ξ). Let η ∈ Z(ξ). Since R is skew-symmetric, (R(ξ, η)ξ, ξ) = 0. We show that R(ξ, η)ξ ⊥ β(ξ) by using assertion (3) to compute

\begin{matrix}0 = & (ξ, β(η)) = (ξ, c(ξ, η)β(ξ)) = −(c(ξ, η)ξ, β(ξ)) \\ 0 = & (c(ξ, η)ξ, β(ξ)) + (ξ, β(ξ))(K − 1)(c(ξ, η)β(ξ), β(ξ)) \\ = & (c(ξ, η)σ, β(ξ)) = (R(ξ, η)ξ, β(ξ)). \end{matrix}

Since R(ξ, η)ξ ⊥ ξ and R(ξ, η)ξ ⊥ β(ξ), by assertion (1) to show R(ξ, η)ξ ∈ Z(ξ), we need only show that R(ξ, η)ξ ⊥ Ξ. Since b ≠ 0,

Span{ξ, β(ξ)} = Span{ξ, c(ξ, Ξ)ξ} = Span{β(ξ), c(ξ, Ξ)β(ξ)},
η ⊥ ξ, η ⊥ β(ξ), η ⊥ c(ξ, Ξ)ξ, and η ⊥ c(ξ, Ξ)β(ξ) = β(Ξ).

We complete the proof of assertion (4) by computing:

\begin{matrix}0 = & (K − 1)\{(β(ξ), ξ)(β(Ξ), η) − (β(ξ), η)(β(Ξ), ξ)\} + (c(ξ, Ξ)ξ, η) \\ = & (R(ξ, Ξ)ξ, η) = (R(ξ, η)ξ, Ξ). \end{matrix}

Suppose assertion (5) fails so that σ ≠ 0. For any unit vector η ∈ Z(ξ), we have |R(ξ, η)ξ| = |c(ξ, η)σ| = |σ|. Thus ϕ/|σ| is a symmetric isometry of Z(ξ) and we can diagonalize ϕ. This decomposes Z = Z_{+} ⊕ Z_{−} where ϕ = ±|σ| on Z_{±}. Choose a unit vector η_{1} ∈ Z_{±}. Then ϕη_{1} = ±|σ|η i.e. c(ξ, η_{1})σ = ±|σ|η_{1}. Since dim(Z) ≥ 5, at least one of the eigenspaces has dimension 3 so we can choose a unit vector η_{2} in Z so (η_{1}, η_{2}) = 0, (β(η_{1}), η_{2}) = 0 and c(ξ, η_{2})σ = ±|σ|η_{2}. Then

\begin{matrix}(η_{1}, β(η_{2})) = ±|σ|^{−1}(c(ξ, η_{1})σ, c(ξ, η_{2})β_{1}) \\ = & ∓ |σ|^{−1}(c(ξ, η_{2})σ, c(ξ, η_{1})β_{1}) = ∓(η_{2}, β(η_{1})) = 0. \end{matrix}

To summarize: we have an orthonormal set {ξ, η_{1}, η_{2}} so that

ξ ⊥ β(η_{1}), ξ ⊥ β(η_{2}), R(η_{1}, η_{2})ξ = c(η_{1}, η_{2})ξ,
η_{1} ⊥ β(ξ), η_{1} ⊥ β(η_{2}), R(ξ, η_{2})η_{1} = c(ξ, η_{2})η_{1}, η_{1} = ±|σ|^{−1}c(ξ, η_{1})σ
η_{2} ⊥ β(ξ), η_{2} ⊥ β(η_{1}), R(ξ, η_{1})η_{2} = c(ξ, η_{1})η_{2}, η_{2} = ±|σ|^{−1}c(ξ, η_{2})σ.

Since c(η_{2}, ξ) anti-commutes with c(η_{1}, ξ), R(η_{1}, η_{2})ξ = −R(η_{2}, ξ)η_{1} − R(ξ, η_{1})η_{2} by the Bianchi identities so c(η_{1}, η_{2})ξ = |σ|^{−1}{±c(η_{2}, ξ)c(ξ, η_{1})±c(ξ, η_{1})c(ξ, η_{2})}σ. We take norms to see 1 = | ± 1 + ±1| which is false. □

Proof of Theorem 0.4 (2). We use Lemma 4.2 to see that σ(ξ) = 0 for every unit vector ξ. This means that: ξ + (K − 1)(ξ, β(ξ))β(ξ) = 0. Consequently β(ξ) = ±ξ and K = 0 so λ_{1} = 0. By replacing β by −β if necessary, we may suppose that β(ξ) = ξ for all ξ. Let {ξ_{i}} be an orthonormal set. Adopt the notation of Lemma 3.3. Since {β_{1}, β_{2}} is an orthonormal basis for W_{1,12} and since {β_{1}, β_{3}} is an orthonormal basis for W_{1,13}, we have β_{1} = ±e_{8}, β_{2} = ±e_{7}, and β_{3} = ±e_{6}. Consequently r_{13}β_{2} = ±r_{12}β_{3} = ±e_{5}. Similarly, r_{23}β_{1} = ±r_{12}β_{3}. Thus 0 = r_{12}ξ_{3} + r_{23}ξ_{1} + r_{31}ξ_{2} = ±r_{12}β_{3} ± r_{13}β_{2} ± r_{23}β_{1} = (±1 ± 1 ± 1)e_{5} by the Bianchi identities; this yields the final contradiction. □

5. The proof of Theorem 0.4 (3)

Proof of Theorem 0.4 (3). Choose ξ_{1} ∈ S^{6} so that ξ_{1} ∈ 𝔄_{2}(ξ_{1}). If η ⊥ ξ_{1}, then ξ_{1} belongs to W_{2}(η, ξ) so |R(ξ_{1}, η)ξ_{1}| = 1. Since R(ξ_{1}, η)ξ_{1} ⊥ ξ_{1} and since we have the curvature symmetry (R(ξ_{1}, η_{1})ξ_{1}, η_{2}) = (R(ξ_{1}, η_{2})ξ_{1}, η_{1}), the map ϕ : η ↦ R(ξ_{1}, η)ξ_{1}) is a self-adjoint isometry of ξ_{1}^{⊥}. We may therefore decompose ξ_{1}^{⊥} = V_{+}⊕V_{−} into the ±1 eigenspaces of ϕ. By replacing R by −R, we may suppose dim V_{+} ≤ dim V_{−}. If V_{+} ≠ {0}, choose a unit vector ξ_{2} ∈ V_{+}; otherwise choose ξ_{2} ∈ V_{−}. The map β discussed in Lemma 3.7 is bilinear. The decomposition V_{+} ⊕ V_{−} is orthogonal. We have dim V_{−} ≥ 3; if ξ_{2} ∈ V_{−}, then dim V_{−} ≥ 4. Therefore we may choose a unit vector ξ_{3} ∈ V_{−} so (ξ_{3}, ξ_{2}) = 0 (this is automatic if ξ_{2} ∈ V_{+}), so (β(ξ_{2}, ξ_{3}), ξ_{1}) = 0, and so (β(ξ_{1}, ξ_{2}), ξ_{3}) = 0. We complete the proof:

(1) We have {ξ_{1}, ξ_{2}, ξ_{3}} is an orthonormal set.
(2) Since (β(ξ_{1}, ξ_{2}), ξ_{3}) = 0, ξ_{3} ∈ W_{2,12} so |r_{12}ξ_{3}| = 1.
(3) Since (β(ξ_{2}, ξ_{3}), ξ_{1}) = 0, ξ_{1} ∈ W_{2,23} so |r_{23}ξ_{1}| = 1.
(4) Since ξ_{1} ∈ W_{1}(ξ_{1}), ξ_{1} ∈ W_{1,12} ∩ W_{1,13} so (r_{12}r_{13} + r_{13}r_{12})ξ_{1} = 0.
(5) The Bianchi identities imply −r_{23}ξ_{1} = r_{21}ξ_{3} + r_{13}ξ_{2}.
(6) Since ξ_{i} ∈ V_{±} for i = 2, 3, ξ_{i} = ±r_{1i}ξ_{i}.
(7) We have 1 = |r_{23}ξ_{1}| = |r_{13}ξ_{2} +_{12} ξ_{3})| = |(±r_{13}r_{12}ξ_{1} ± r_{12}r_{13})ξ_{1}|

= | ± 1 ± 1| · |r_{12}r_{13}ξ_{1}| = | ± 1 ± 1| · |r_{12}ξ_{3}| = | ± 1 ± 1| which is false.□

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Mathematics Department, University of Oregon Eugene Or 97403 USA E-mail: gilkey@math.uoregon.edu