RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF THE QUATERNION PROJECTIVE PLANE P^{2}(H H H)

YOSHIO AGAOKA AND EIJI KANEDA

Abstract. In this paper, we investigate isometric immersions of P^{2}(H H H) into R R R^{14} and prove that the canonical isometric imbedding fff_{0} of P^{2}(H H H) into R R R^{14}, which is defined in Kobayashi [11], is rigid in the following strongest sense: Any isometric immersion fff_{1} of a connected open set U(⊂ P^{2}(H H H)) into R R R^{14} coincides with fff_{0} up to a euclidean transformation of R R R^{14}, i.e., there is a euclidean transformation a of R R R^{14} satisfying fff_{1} = afff_{0} on U.

1. Introduction

In our previous paper [8], we proved the rigidity of the canonical isometric imbedding of the Cayley projective plane P^{2}(Cay Cay Cay). The purpose of this paper is to investigate a similar problem for (local) isometric immersions of the quaternion projective plane P^{2}(H H H). As we have proved in [7], any open set of the quaternion projective plane P^{2}(H H H) cannot be isometrically immersed into R R R^{13}. On the other hand, there is an isometric immersion fff_{0} of P^{2}(H H H) into the euclidean space R R R^{14}, which is called the canonical isometric imbedding of P^{2}(H H H) (see Kobayashi [11]). Therefore, it follows that R R R^{14} is the least dimensional euclidean space into which P^{2}(H H H) can be (locally) isometrically immersed.

In the present paper, we will show that the canonical isometric imbedding fff_{0} is rigid in the following strongest sense:

Theorem 1. Let fff_{0} be the canonical isometric imbedding of P^{2}(H H H) into the euclidean space R R R^{14}. Then, for any isometric immersion fff_{1} defined on a connected open set U of P^{2}(H H H) into R R R^{14}, there exists a euclidean transformation a of R R R^{14} satisfying fff_{1} = afff_{0} on U.

The proof of this theorem will be given by solving the Gauss equation associated with the isometric imbeddings (immersions) of P^{2}(H H H) into R R R^{14} in the same line of [8] (see Theorem 7). We use the same notations and terminology as those of the previous papers [6], [7] and [8].

Date: April 8, 2004.

2000 Mathematics Subject Classification. 17B20, 53B25, 53C24, 53C35.

Key words and phrases. Curvature invariant, isometric immersion, quaternion projective plane, rigidity, root space decomposition.

2. The quaternion projective plane P^{2}(H H H)

In this section we review the structure of the quaternion projective plane P^{2}(H H H) and prepare several formulas concerning the bracket operation.

As is well-known, P^{2}(H H H) can be represented by P^{2}(H H H) = G/K, where G = Sp(3) and K = Sp(2) × Sp(1). Let 𝔤 (resp. 𝔨) be the Lie algebra of G (resp. K) and let 𝔤 = 𝔨 + 𝔪 be the canonical decomposition of 𝔤 associated with the symmetric pair (G, K). We denote by ( , ) the inner product of 𝔤 given by the (−1)-multiple of the Killing form of 𝔤. As usual, we can identify 𝔪 with the tangent space T_{o}(G/K) at the origin o = {K}. We assume that the G-invariant Riemannian metric g of G/K satisfies

g_{o}(X, Y) = (X, Y ), X, Y ∈ 𝔪.

Then, it is well-known that at the origin o the Riemannian curvature tensor R of type (1, 3) is given by

R_{o}(X, Y)Z = −[[X, Y ], Z], ∀X, Y, Z ∈ 𝔪.

We now take a maximal abelian subspace 𝔞 of 𝔪 and fix it in the following discussions. We note that since rank(P^{2}(H H H)) = 1, we have dim 𝔞 = 1.

For each element λ ∈ 𝔞 we define two subspaces 𝔨(λ) (⊂ 𝔨) and 𝔪(λ) (⊂ 𝔪) by

𝔨(λ) = {X ∈ 𝔨 | [H, [H, X]] = −(λ, H)^{2}X, ∀H ∈ 𝔞} ,
𝔪(λ) = {Y ∈ 𝔪 | [H, [H, Y ]] = −(λ, H)^{2}Y, ∀H ∈ 𝔞} .

Let Σ be the set of all non-zero restricted roots. (An element λ ∈ 𝔞 is called a restricted root if 𝔪(λ) ≠ 0.) As is known, there is a restricted root μ such that Σ = {±μ, ±2μ}. We take and fix such a restricted root μ. For each integer i we set 𝔨_{i} = 𝔨(|i|μ), 𝔪_{i} = 𝔪(|i|μ) (|i| ≤ 2), 𝔨_{i} = 𝔪_{i} = 0 (|i| > 2). Then, we have 𝔪_{0} = 𝔞 = R R Rμ and

𝔨 = 𝔨_{0} + 𝔨_{1} + 𝔨_{2} (orthogonal direct sum),
𝔪 = 𝔪_{0} + 𝔪_{1} + 𝔪_{2} (orthogonal direct sum).

The dimensions of the factors are given by dim 𝔨_{0} = 6, dim 𝔨_{1} = dim 𝔪_{1} = 4 and dim 𝔨_{2} = dim 𝔪_{2} = 3 (precisely, see [7]).

We now show several formulas concerning the bracket operation of 𝔤. By the definition of the subspaces 𝔨_{i} and 𝔪_{i} we easily have

[𝔨_{i}, 𝔨_{j}] ⊂ 𝔨_{i+j} + 𝔨_{i−j}, [𝔪_{i}, 𝔪_{j}] ⊂ 𝔨_{i+j} + 𝔨_{i−j}, [𝔨_{i}, 𝔪_{j}] ⊂ 𝔪_{i+j} + 𝔪_{i−j}.  (2.1)

Moreover, we have

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 3

Proposition 2. Let Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2}, Y_{1}, Y′_{1} ∈ 𝔪_{1}. Then:

[Y_{i}, [Y_{i}, Y′_{j}]] = − (1 + 3δ_{ij}) (μ, μ) {(Y_{i}, Y_{i})Y′_{j} − (Y_{i}, Y′_{j})Y_{i}} , (i, j = 0, 1),  (2.2)
[Y_{i}, [Y′_{i}, Y_{j}]] + [Y′_{i}, [Y_{i}, Y_{j}]] = −2(μ, μ)(Y_{i}, Y′_{i})Y_{j}, (i, j = 0, 1, i ≠ j),  (2.3)
[Y_{i}, [Y_{i}, X_{1}]] = −(μ, μ)(Y_{i}, Y_{i})X_{1}, ∀X_{1} ∈ 𝔨_{1} (i = 0, 1),  (2.4)

where δ_{ij} denotes the Kronecker delta.

Proof. We first prove (2.2). Assume that i = j and Y_{i} ≠ 0. Set Y′′_{i} = Y′_{i}−(Y′_{i}, Y_{i})/(Y_{i}, Y_{i})· Y_{i}. Then, we know that (Y_{i}, Y′′_{i}) = 0 and that Y′′_{i} ∈ 𝔞 + 𝔪_{2} if i = 0 and Y′′_{i} ∈ 𝔪_{1} if i = 1. Hence, by Proposition 10 of [7], we have [Y_{i}, [Y_{i}, Y′′_{i}]] = −4(μ, μ)(Y_{i}, Y_{i})Y′′_{i}. Therefore, we can easily obtain (2.2) in the case i = j. In the case i ≠ j, (2.2) directly follows from Proposition 10 of [7].

We next prove (2.3). Since i ≠ j, it follows that (Y_{i}, Y_{j}) = (Y′_{i}, Y_{j}) = 0. Hence, by (2.2) we have [Y_{i} + Y′_{i}, [Y_{i} + Y′_{i}, Y_{j}]] = −(μ, μ)(Y_{i} + Y′_{i}, Y_{i} + Y′_{i})Y_{j}. This, together with [Y_{i}, [Y_{i}, Y_{j}]] = −(μ, μ)(Y_{i}, Y_{i})Y_{j} and [Y′_{i}, [Y′_{i}, Y_{j}]] = −(μ, μ)(Y′_{i}, Y′_{i})Y_{j}, proves (2.3).

We finally prove (2.4). We note that [Y_{1}, 𝔞 + 𝔪_{2}] = 𝔨_{1} holds for any Y_{1} ∈ 𝔪_{1} (≠ 0). In fact, it is easy to see [Y_{1}, 𝔞 + 𝔪_{2}] ⊂ 𝔨_{1} (see (2.1)). Moreover, the map 𝔞 + 𝔪_{2} ∋ Y′_{0} ⟼ [Y_{1}, Y′_{0}] ∈ 𝔨_{1} is bijective, because [Y_{1}, Y′_{0}] ≠ 0 if Y′_{0} ∈ 𝔞 + 𝔪_{2} (Y′_{0} ≠ 0) (recall that rank(P^{2}(H H H)) = 1) and because dim(𝔞 + 𝔪_{2}) = dim 𝔨_{1}. Let X_{1} ∈ 𝔨_{1}. Then, by [Y_{1}, 𝔞 + 𝔪_{2}] = 𝔨_{1} we can take an element Y′_{0} ∈ 𝔞 + 𝔪_{2} such that [Y_{1}, Y′_{0}] = X_{1}. Now, applying ad Y_{1} to the equality [Y_{1}, [Y_{1}, Y′_{0}]] = −(μ, μ)(Y_{1}, Y_{1})Y′_{0} (see (2.2)), we have [Y_{1}, [Y_{1}, X_{1}]] = −(μ, μ)(Y_{1}, Y_{1})X_{1}, proving (2.4) for the case i = 1. Similarly, we can prove (2.4) for the case i = 0. □

Let Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2}. Define a linear mapping L(Y_{0}, Y′_{0}) of 𝔪_{1} to 𝔪 by

L(Y_{0}, Y′_{0})Y_{1} = [Y_{0}, [Y′_{0}, Y_{1}]], Y_{1} ∈ 𝔪_{1}.

Then, we have

Proposition 3. Let Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2}. Then:

(1) L(Y_{0}, Y′_{0})𝔪_{1} ⊂ 𝔪_{1}. The transpose of L(Y_{0}, Y′_{0}) with respect to ( , ) is given by L(Y′_{0}, Y_{0}), i.e., ^{t}L(Y_{0}, Y′_{0}) = L(Y′_{0}, Y_{0}).

(2) Let 111_{𝔪_{1}} be the identity map of 𝔪_{1}. Then:

(2a) L(Y_{0}, Y′_{0}) + L(Y′_{0}, Y_{0}) = −2(μ, μ)(Y_{0}, Y′_{0})111_{𝔪_{1}}; (2b) L(Y_{0}, Y′_{0}) · L(Y′_{0}, Y_{0}) = (μ, μ)^{2}(Y_{0}, Y_{0})(Y′_{0}, Y′_{0})111_{𝔪_{1}}.

Proof. The assertion (1) is clear from (2.1) and the ad 𝔤-invariance of ( , ). Let Y_{1} ∈ 𝔪_{1}. Since [Y_{0}, Y_{1}] ∈ 𝔨_{1}, we have [Y′_{0}, [Y′_{0}, [Y_{0}, Y_{1}]]] = −(μ, μ)(Y′_{0}, Y′_{0})[Y_{0}, Y_{1}] (see (2.4)).

Hence, by applying ad Y_{0} to this equality, we easily have (2b). The equality (2a) directly follows from (2.3). □

Here, we recall the notion of pseudo-abelian subspace of 𝔪. Let Q be a subspace of 𝔪. Q is called pseudo-abelian if it satisfies [Q, Q] ⊂ 𝔨_{0} (see [6]).

Proposition 4. (1) Any subspace Q of 𝔪_{2} is pseudo-abelian.

(2) Let Q be a pseudo-abelian subspace satisfying Q ⊄ 𝔪_{2}. Then, dim Q ≤ 2.

Accordingly, the inequality dim Q ≤ 3 holds for any pseudo-abelian subspace Q, and the equality holds when and only when Q = 𝔪_{2}.

Proof. Since [𝔪_{2}, 𝔪_{2}] ⊂ 𝔨_{0} (see (2.1)), it follows that any subspace of 𝔪_{2} is pseudo-abelian. On the contrary, we already proved in Lemma 5.4 of [6] that for a pseudo-abelian subspace Q with Q ⊄ 𝔪_{2} it holds dim Q ≤ 1 + n(μ), where n(μ) means the local pseudo-nullity of the restricted root μ. (For the definition of the local pseudo-nullity, see §3 of [6].) In the case G/K = P^{2}(H H H), we have n(μ) = 1 (see Theorem 3.2 and Table 3 of [6]). Hence, we have dim Q ≤ 2. □

For later use, we obtain the normal form of a 2-dimensional pseudo-abelian subspace Q with Q ⊄ 𝔪_{2}.

Proposition 5. Let ξ_{1} and η_{1} be elements of 𝔪_{1} satisfying (ξ_{1}, ξ_{1}) = 2(μ, μ), η_{1} ≠ 0 and (ξ_{1}, η_{1}) = 0. Then, the 2-dimensional subspace Q (⊂ 𝔪) defined by

Q = R R R (μ + ξ_{1}) + R R R (η_{1} + \frac{1}{4(μ, μ)^{2}} [μ, [ξ_{1}, η_{1}]])  (2.5)

is pseudo-abelian and Q ⊄ 𝔪_{2}.

Conversely, if Q is a pseudo-abelian subspace of 𝔪 with Q ⊄ 𝔪_{2} and dim Q = 2, then Q can be written in the form (2.5) by utilizing suitable elements ξ_{1} and η_{1} ∈ 𝔪_{1} satisfying (ξ_{1}, ξ_{1}) = 2(μ, μ), η_{1} ≠ 0 and (ξ_{1}, η_{1}) = 0.

Proof. Let ξ_{1} and η_{1} be elements of 𝔪_{1} satisfying (ξ_{1}, ξ_{1}) = 2(μ, μ), η_{1} ≠ 0 and (ξ_{1}, η_{1}) = 0. Then, the subspace Q defined by (2.5) satisfies Q ⊄ 𝔪_{2} and dim Q = 2. Set η_{2} = (1/4(μ, μ)^{2})[μ, [ξ_{1}, η_{1}]]. Then, it is easily verified that η_{2} ∈ 𝔪_{2}. We now show that Q is pseudo-abelian. By (2.3) and (ξ_{1}, η_{1}) = 0, we have [ξ_{1}, [η_{1}, μ]] = −[η_{1}, [ξ_{1}, μ]]. Hence, by the Jacobi identity we have

[μ, [ξ_{1}, η_{1}]] = [[μ, ξ_{1}], η_{1}] + [ξ_{1}, [μ, η_{1}]] = −2[ξ_{1}, [η_{1}, μ]].

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 5

Consequently, we have η_{2} = −(1/2(μ, μ)^{2})[ξ_{1}, [η_{1}, μ]]. Note that [η_{1}, μ] ∈ 𝔨_{1}. Then, by the formula (2.4) and the assumption (ξ_{1}, ξ_{1}) = 2(μ, μ) we have

[ξ_{1}, η_{2}] = −\frac{1}{2(μ, μ)^{2}} [ξ_{1}, [ξ_{1}, [η_{1}, μ]]] = \frac{(ξ_{1}, ξ_{1})}{2(μ, μ)}[η_{1}, μ] = −[μ, η_{1}].

Moreover, since [μ, η_{2}] + [ξ_{1}, η_{1}] ∈ 𝔨 and since

[μ, [μ, η_{2}] + [ξ_{1}, η_{1}]] = −4(μ, μ)^{2}η_{2} + [μ, [ξ_{1}, η_{1}]] = 0,

it follows that [μ, η_{2}] + [ξ_{1}, η_{1}] ∈ 𝔨_{0}. (Note that an element X ∈ 𝔨 belongs to 𝔨_{0} if and only if [μ, X] = 0.) By these relations we have

[μ + ξ_{1}, η_{1} + η_{2}]= [μ, η_{1}] + [ξ_{1}, η_{2}] + [μ, η_{2}] + [ξ_{1}, η_{1}] = 0 + [μ, η_{2}] + [ξ_{1}, η_{1}] ∈ 𝔨_{0}.

Since Q = R R R(μ + ξ_{1}) + R R R(η_{1} + η_{2}), this implies that Q is a pseudo-abelian subspace.

We next prove the converse. Let Q be a pseudo-abelian subspace with Q ⊄ 𝔪_{2} and dim Q = 2. Then, viewing the proof of Lemma 5.4 of [6], we know that Q ∩ 𝔪_{2} = 0 and dim(Q ∩ (𝔪_{1} + 𝔪_{2})) ≤ n(μ) = 1. Consequently, we have Q ⊄ 𝔪_{1} + 𝔪_{2}, because dim Q = 2. Therefore, there is a basis {ξ, η} of Q written in the form ξ = μ+ξ_{1}+ξ_{2}, η = η_{1}+η_{2}, where ξ_{1}, η_{1} ∈ 𝔪_{1}, ξ_{2}, η_{2} ∈ 𝔪_{2}. Here, we note that η_{1} ≠ 0, because Q ∩ 𝔪_{2} = 0. Subtracting a constant multiple of η from ξ if necessary, we may assume that (ξ_{1}, η_{1}) = 0. Since

[ξ, η] = [μ + ξ_{2}, η_{1}] + [ξ_{1}, η_{2}] + [μ + ξ_{2}, η_{2}] + [ξ_{1}, η_{1}] ∈ 𝔨_{0}

and since [μ + ξ_{2}, η_{1}] + [ξ_{1}, η_{2}] ∈ 𝔨_{1}, [μ + ξ_{2}, η_{2}] + [ξ_{1}, η_{1}] ∈ 𝔨_{0} + 𝔨_{2} and [ξ_{2}, η_{2}] ∈ 𝔨_{0}, it follows that

[μ + ξ_{2}, η_{1}] + [ξ_{1}, η_{2}] = 0,  (2.6)
[μ, η_{2}] + [ξ_{1}, η_{1}] ∈ 𝔨_{0}.  (2.7)

Applying ad μ to (2.7), we have η_{2} = (1/4(μ, μ)^{2})[μ, [ξ_{1}, η_{1}]]. By this equality and the assumption (ξ_{1}, η_{1}) = 0, we can deduce [ξ_{1}, η_{2}] = ((ξ_{1}, ξ_{1})/2(μ, μ))[η_{1}, μ] (see the arguments stated above). Putting this into (2.6), we have

[(1 − \frac{(ξ_{1}, ξ_{1})}{2(μ, μ)})μ + ξ_{2}, η_{1}] = 0.

Since η_{1} ≠ 0 and rank(P^{2}(H H H)) = 1, we have (1 − (ξ_{1}, ξ_{1})/2(μ, μ)) μ + ξ_{2} = 0. This proves (ξ_{1}, ξ_{1}) = 2(μ, μ) and ξ_{2} = 0, completing the proof of the converse. □

3. The Gauss equation

Let N N N be a euclidean vector space, i.e., N N N is a vector space over R R R endowed with an inner product ⟨ , ⟩. Let S^{2}𝔪^{∗} ⊗ N N N be the space of N N N-valued symmetric bilinear forms on 𝔪. We call the following equation on Ψ ∈ S^{2}𝔪^{∗} ⊗ N N N the Gauss equation associated with N N N:

([[X, Y ], Z], W ) = ⟨Ψ(X, Z), Ψ(Y, W)⟩ − ⟨Ψ(X, W), Ψ(Y, Z)⟩,  (3.1)

where X, Y, Z, W ∈ 𝔪. We denote by G(P^{2}(H H H),N N N) the set of all solutions of (3.1), which is called the Gaussian variety associated with N N N.

As in the case of P^{2}(Cay Cay Cay) (Theorem 11 of [8]), we can prove the following

Theorem 6. Let N N N be a euclidean vector space with dimN N N = 6. Let Ψ ∈ S^{2}𝔪^{∗} ⊗ N N N be a solution of the Gauss equation (3.1), i.e., Ψ ∈ G(P^{2}(H H H),N N N). Then:

(1) There are linearly independent vectors A and B ∈ N N N satisfying (i) ⟨A, A⟩ = ⟨B, B⟩ = 4(μ, μ) and ⟨A, B⟩ = 2(μ, μ); (ii) Ψ(Y_{0}, Y′_{0}) = (Y_{0}, Y′_{0})A, ∀Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2}; (iii) Ψ(Y_{1}, Y′_{1}) = (Y_{1}, Y′_{1})B, ∀Y_{1}, Y′_{1} ∈ 𝔪_{1}; (iv) ⟨A, Ψ(μ, 𝔪_{1})⟩ = ⟨B, Ψ(μ, 𝔪_{1})⟩ = 0.

(2) Ψ(Y_{1}, Y_{2}) = −\frac{1}{(μ, μ)^{2}} Ψ(μ, L(μ, Y_{2})Y_{1}), ∀Y_{1} ∈ 𝔪_{1}, ∀Y_{2} ∈ 𝔪_{2}.

(3) ⟨Ψ(μ, Y_{1}), Ψ(μ, Y′_{1})⟩ = (μ, μ)^{2}(Y_{1}, Y′_{1}), ∀Y_{1}, Y′_{1} ∈ 𝔪_{1}.

Let O(N N N) be the orthogonal transformation group of N N N. We define an action of O(N N N) on S^{2}𝔪^{∗} ⊗ N N N by

(hΨ)(X, Y) = h(Ψ(X, Y)),

where Ψ ∈ S^{2}𝔪^{∗} ⊗N N N, h ∈ O(N N N). It is easily seen that G(P^{2}(H H H),N N N) is invariant under this action, i.e., h G(P^{2}(H H H),N N N) = G(P^{2}(H H H),N N N) for any h ∈ O(N N N). We say that the Gaussian variety G(P^{2}(H H H),N N N) is EOS if G(P^{2}(H H H),N N N) ≠ ∅ and if G(P^{2}(H H H),N N N) is consisting of essentially one solution, i.e., for any solutions Ψ and Ψ′ ∈ G(P^{2}(H H H),N N N), there is an element h ∈ O(N N N) satisfying Ψ′ = hΨ (see [8]).

By Theorem 6 we can show

Theorem 7. Let N N N be a euclidean vector space with dimN N N = 6. Then, G(P^{2}(H H H),N N N) is EOS.

Proof. The proof of this theorem is quite similar to that of Theorem 10 in [8].

First we note that G(P^{2}(H H H),N N N) ≠ ∅, because the second fundamental form of the canonical isometric imbedding fff_{0} at the origin o ∈ P^{2}(H H H) satisfies (3.1).

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 7

Let {E_{i} (1 ≤ i ≤ 4)} be an orthonormal basis of 𝔪_{1}. (Note that dim 𝔪_{1} = 4.) Let Ψ ∈ G(P^{2}(H H H),N N N) and let A, B be the vectors of N N N stated in Theorem 6. We define vectors \{F_{i} (1 ≤ i ≤ 6)\} of N N N by setting F_{i} = Ψ(μ, E_{i})/(μ, μ) (1 ≤ i ≤ 4), F_{5} = (A + B)/2\sqrt{3}|μ| and F_{6} = (A − B)/2|μ|. By Theorem 6 we can show that {F_{i} (1 ≤ i ≤ 6)} forms an orthonormal basis of N N N. Now let Ψ′ be another element of G(P^{2}(H H H),N N N). Let A′ and B′ be the vectors stated in Theorem 6 for Ψ′. As in the case of Ψ we can also define an orthonormal basis {F′_{i} (1 ≤ i ≤ 6)} of N N N. Then, there is an element h ∈ O(6) satisfying F′_{i} = hF_{i} (1 ≤ i ≤ 6). Here, we note that A′ = hA, B′ = hB and Ψ′(μ, E_{i}) = hΨ(μ, E_{i}) (1 ≤ i ≤ 4). Set Φ = Ψ′ − hΨ ∈ S^{2}𝔪^{∗} ⊗ N N N. Then, by Theorem 6 (1) we have

Φ(𝔞 + 𝔪_{2}, 𝔞 + 𝔪_{2}) = Φ(𝔪_{1}, 𝔪_{1}) = Φ(𝔞, 𝔪_{1}) = 0.

By Theorem 6 (2) and by the fact L(μ, 𝔪_{2})𝔪_{1} ⊂ 𝔪_{1} we have

Φ(𝔪_{2}, 𝔪_{1}) ⊂ Φ(μ, L(μ, 𝔪_{2})𝔪_{1}) ⊂ Φ(𝔞, 𝔪_{1}) = 0,

which proves Φ(𝔪_{2}, 𝔪_{1}) = 0. Therefore, we have Φ = 0, i.e., Ψ′ = hΨ, completing the proof of Theorem 7. □

By Theorem 7 we know that P^{2}(H H H) is formally rigid in codimension 6 in the sense of Agaoka–Kaneda [8]. Therefore, Theorem 1 can be obtained by Theorem 7 and the rigidity theorem (Theorem 5 of [8]).

Before proceeding to the proof of Theorem 6, we make several preparations.

Let N N N be a euclidean vector space. In what follows we assume dimN N N = 6. Let S^{2}𝔪^{∗}⊗N N N be the space of N N N-valued symmetric bilinear forms on 𝔪. Let Ψ ∈ S^{2}𝔪^{∗} ⊗ N N N and Y ∈ 𝔪. We define a linear map Ψ_{Y} of 𝔪 to N N N by

Ψ_{Y} : 𝔪 ∋ Y′ ⟼ Ψ(Y, Y′) ∈ N N N,

and denote by Ker(Ψ_{Y}) the kernel of Ψ_{Y}. We call an element Y ∈ 𝔪 singular (resp. non-singular) with respect to Ψ if Ψ_{Y} (𝔪) ≠ N N N (resp. Ψ_{Y} (𝔪) = N N N).

Let Ψ ∈ G(P^{2}(H H H),N N N) and let Y ∈ 𝔪 (Y ≠ 0). Take an element k ∈ K such that Ad(k)μ ∈ R R RY. Then, as shown in the proof of Proposition 5 of [7], the subspace Q_{Y} = Ad(k)^{−1}Ker(Ψ_{Y}) is a pseudo-abelian subspace of 𝔪.

Proposition 8. Let Ψ ∈ G(P^{2}(H H H),N N N) and let Y ∈ 𝔪 (Y ≠ 0). Then:

(1) dim Ker(Ψ_{Y}) = 2 or 3. Moreover, Y is non-singular (resp. singular) with respect to Ψ if and only if dim Ker(Ψ_{Y}) = 2 (resp. dim Ker(Ψ_{Y}) = 3).

(2) Let k ∈ K satisfy Ad(k)μ ∈ R R RY. Then, Ker(Ψ_{Y}) ⊂ Ad(k)𝔪_{2}. Consequently, Y is non-singular (resp. singular) with respect to Ψ if and only if Ker(Ψ_{Y}) ⊊ Ad(k)𝔪_{2} (resp. Ker(Ψ_{Y}) = Ad(k)𝔪_{2}).

8 YOSHIO AGAOKA AND EIJI KANEDA

Remark 1. Recall that in the case of the Cayley projective plane P^{2}(Cay Cay Cay) the inclusion Ker(Ψ_{Y}) ⊂ Ad(k)𝔪_{2} in Proposition 8 (2) can be proved by a simple discussion. There, the inclusion automatically follows from the fact that any high-dimensional pseudo-abelian subspace must be contained in 𝔪_{2} (see Propositions 8 and 12 of [8]). In contrast, it is not a simple task to show the inclusion Ker(Ψ_{Y}) ⊂ Ad(k)𝔪_{2} in our case P^{2}(H H H). We will prove this inclusion by making use of the normal form of the pseudo-abelian subspaces not contained in 𝔪_{2} (see Proposition 5).

Proof of Proposition 8. Let Y ∈ 𝔪 (Y ≠ 0). Set Q_{Y} = Ad(k)^{−1}Ker(Ψ_{Y}), where k ∈ K is an element satisfying Ad(k)μ ∈ R R RY. Since Q_{Y} is pseudo-abelian, it follows that dim Q_{Y} ≤ 3 (see Proposition 4). Hence, dim Ker(Ψ_{Y}) ≤ 3. On the other hand, since dimN N N = 6 and dim 𝔪 = 8, it follows that dim Ker(Ψ_{Y}) ≥ 2. Therefore, Y is non-singular (resp. singular) with respect to Ψ if and only if dim Ker(Ψ_{Y}) = 2 (resp. dim Ker(Ψ_{Y}) = 3). This proves (1).

To show the first statement of (2) it suffices to prove Q_{Y} ⊂ 𝔪_{2}. Now, let us suppose the contrary, i.e., Q_{Y} ⊄ 𝔪_{2}. Then, we have dim Q_{Y} = 2 (see (1) and Proposition 4 (2)). Hence, there is a basis {ξ, η} of Q_{Y} written in the form ξ = μ+ξ_{1}, η = η_{1}+(1/4(μ, μ)^{2})[μ, [ξ_{1}, η_{1}]], where ξ_{1} and η_{1} are elements of 𝔪_{1} satisfying (ξ_{1}, ξ_{1}) = 2(μ, μ), η_{1} ≠ 0, (ξ_{1}, η_{1}) = 0 (see Proposition 5). Let {ζ_{1}^{1}, ζ_{1}^{2}} be a basis of the orthogonal complement of R R Rξ_{1} + R R Rη_{1} in 𝔪_{1}. Set ζ^{i} = ζ_{1}^{i} + (1/4(μ, μ)^{2})[μ, [ξ_{1}, ζ_{1}^{i}]] (i = 1, 2). Since [μ, [ξ_{1}, ζ_{1}^{i}]] ∈ 𝔪_{2} (i = 1, 2), we know that the vectors ζ^{1} and ζ^{2} are linearly independent. More strongly, they are linearly independent modulo Q_{Y}, i.e., Q_{Y} ∩(R R Rζ^{1} +R R Rζ^{2}) = 0. Moreover, by Proposition 5 we know that the subspace Q^{i} = R R Rξ + R R Rζ^{i} (i = 1, 2) is also pseudo-abelian, because (ξ_{1}, ζ_{1}^{i}) = 0. Consequently, we have [[ξ, ζ^{i}], μ] = 0 (i = 1, 2).

Set X = Ad(k)ξ, Z^{i} = Ad(k)ζ^{i} (i = 1, 2). Then, we have X ∈ Ker(Ψ_{Y}) (X ≠ 0), Ker(Ψ_{Y})∩(R R RZ^{1} +R R RZ^{2}) = 0 and [[X, Z^{i}], Y ] = 0 (i = 1, 2). By the Gauss equation (3.1) we have

0 = ([[X, Z^{i}], Y ], W ) = ⟨Ψ(X, Y), Ψ(Z^{i}, W)⟩ − ⟨Ψ(X, W), Ψ(Z^{i}, Y)⟩, (i = 1, 2),

where W is an arbitrary element of 𝔪. Since Ψ_{Y} (X) = 0, we obtain by this equality ⟨Ψ_{X}(W), Ψ(Z^{i}, Y)⟩ = 0, i.e., ⟨Ψ_{X}(𝔪), Ψ(Z^{i}, Y)⟩ = 0 (i = 1, 2). We note that the vectors Ψ(Z^{1}, Y) and Ψ(Z^{2}, Y) are linearly independent, because Ker(Ψ_{Y}) ∩ (R R RZ^{1} + R R RZ^{2}) = 0. Hence, we have dim Ψ_{X}(𝔪) ≤ dimN N N − 2 = 4, implying dim Ker(Ψ_{X}) ≥ 4. This contradicts the assertion (1). Thus, we have Q_{Y} ⊂ 𝔪_{2}, proving the first statement of (2). The last statement of (2) is now clear. □

As a corollary of Proposition 8 we obtain

Proposition 9. Let Ψ ∈ G(P^{2}(H H H),N N N). Then:

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 9

(1) Let Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0). Then, Ker(Ψ_{Y_{0}}) ⊂ {ξ ∈ 𝔞 + 𝔪_{2} | (ξ, Y_{0}) = 0}. If Y_{0} is singular with respect to Ψ, then Ker(Ψ_{Y_{0}}) = {ξ ∈ 𝔞 + 𝔪_{2} | (ξ, Y_{0}) = 0}.

(2) Let Y_{1} ∈ 𝔪_{1} (Y_{1} ≠ 0). Then, Ker(Ψ_{Y_{1}}) ⊂ {η ∈ 𝔪_{1} | (η, Y_{1}) = 0}. If Y_{1} is singular with respect to Ψ, then Ker(Ψ_{Y_{1}}) = {η ∈ 𝔪_{1} | (η, Y_{1}) = 0}.

Proof. Let Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0). Then, we can take an element k_{0} ∈ K such that Ad(k_{0})μ ∈ R R RY_{0} and Ad(k_{0})(𝔪_{2}) = {ξ ∈ 𝔞 + 𝔪_{2} | (ξ, Y_{0}) = 0} (see Proposition 7 of [7]). This proves (1). Similarly, for Y_{1} ∈ 𝔪_{1} (Y_{1} ≠ 0), we can easily show (2). □

Let Ψ ∈ S^{2}𝔪^{∗} ⊗ N N N. We call a subspace U of 𝔪 singular with respect to Ψ if each element of U is singular with respect to Ψ.

Proposition 10. Let Ψ ∈ G(P^{2}(H H H),N N N). Assume that Y ∈ 𝔪 (Y ≠ 0) is non-singular with respect to Ψ. Then, there is a non-zero vector E ∈ N N N such that

N N N = R R RE + Ψ_{ξ}(𝔪) (orthogonal direct sum)  (3.2)

holds for any ξ ∈ Ker(Ψ_{Y}) (ξ ≠ 0). Consequently, Ker(Ψ_{Y}) is a singular subspace with respect to Ψ.

Proof. Take an element k ∈ K such that Ad(k)μ ∈ R R RY. Then, since Y is non-singular, we have Ker(Ψ_{Y}) ⊊ Ad(k)𝔪_{2}. Take a non-zero element satisfying Y′ ∈ Ad(k)𝔪_{2} and Y′ ∉ Ker(Ψ_{Y}) and set E = Ψ(Y, Y′) (≠ 0). Let ξ ∈ Ker(Ψ_{Y}) (ξ ≠ 0). Then, by the Gauss equation (3.1) we have

([[ξ, Y′], Y ], W ) = ⟨Ψ(ξ, Y), Ψ(Y′, W)⟩ − ⟨Ψ(ξ, W), Ψ(Y′, Y)⟩,

where W is an arbitrary element of 𝔪. Here, we note that [[ξ, Y′], Y ] = 0, because [[ξ, Y′], Y ] ∈ Ad(k)[[𝔪_{2}, 𝔪_{2}], μ] = 0. Since Ψ(ξ, Y) = 0, we obtain by the above equality ⟨E, Ψ(ξ, W)⟩ = 0. This shows ⟨E, Ψ_{ξ}(𝔪)⟩ = 0 and hence Ψ_{ξ}(𝔪) ≠ N N N. Consequently, ξ is singular with respect to Ψ. Since dim Ker(Ψ_{ξ}) = 3 (see Proposition 8), we have dim Ψ_{ξ}(𝔪) = 5, which proves the decomposition (3.2). □

4. Proof of Theorem 6

In this section, with the preparations in the previous sections, we will prove Theorem 6. We first show

Proposition 11. Let Ψ ∈ G(P^{2}(H H H),N N N). Then, there are singular subspaces U (⊂ 𝔞 + 𝔪_{2}) and V (⊂ 𝔪_{1}) with respect to Ψ satisfying dim U ≥ 2 and dim V ≥ 2.

10 YOSHIO AGAOKA AND EIJI KANEDA

Proof. If 𝔞 + 𝔪_{2} contains no non-singular element with respect to Ψ, then set U = 𝔞 + 𝔪_{2}. On the contrary, if there is a non-singular element Y_{0} ∈ 𝔞 + 𝔪_{2}, then set U = Ker(Ψ_{Y_{0}}). In this case we know that dim U = 2, U ⊂ 𝔞 + 𝔪_{2} and that U is a singular subspace with respect to Ψ (see Proposition 8, Proposition 9 and Proposition 10).

Similarly, we can show that there is a singular subspace V of 𝔪_{1} with respect to Ψ satisfying the desired properties. □

Proposition 12. Let Ψ ∈ G(P^{2}(H H H),N N N). Let U (⊂ 𝔞 + 𝔪_{2}) and V (⊂ 𝔪_{1}) be singular subspaces with respect to Ψ satisfying dim U ≥ 2 and dim V ≥ 2. Then, there are vectors A, B ∈ N N N such that:

(1) ⟨A, A⟩ = ⟨B, B⟩ = 4(μ, μ).

(2) Let ξ ∈ U and η ∈ V. Then:

(2a) Ψ(ξ, Y_{0}) = (ξ, Y_{0})A, ∀Y_{0} ∈ 𝔞 + 𝔪_{2}; (2b) Ψ(η, Y_{1}) = (η, Y_{1})B, ∀Y_{1} ∈ 𝔪_{1}.

(3) Let Y_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Then:

(3a) ⟨A, Ψ_{Y_{0}}(𝔪_{1})⟩ = ⟨B, Ψ_{Y_{0}}(𝔪_{1})⟩ = 0;

(3b) ⟨A, Ψ_{Y_{1}}(𝔞 + 𝔪_{2})⟩ = ⟨B, Ψ_{Y_{1}}(𝔞 + 𝔪_{2})⟩ = 0.

(4) Let ξ ∈ U (ξ ≠ 0) and η ∈ V (η ≠ 0). Then:

(4a) Ψ_{ξ}(𝔪) = R R RA + Ψ_{ξ}(𝔪_{1}) (orthogonal direct sum); (4b) Ψ_{η}(𝔪) = R R RB + Ψ_{η}(𝔞 + 𝔪_{2}) (orthogonal direct sum).

(5) Let Y_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Then:

(5a) ⟨Ψ(Y_{0}, Y_{0}), A⟩ = 4(μ, μ)(Y_{0}, Y_{0}); (5b) ⟨Ψ(Y_{1}, Y_{1}), B⟩ = 4(μ, μ)(Y_{1}, Y_{1}).

(6) Let ξ ∈ U, η ∈ V, Y_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Assume that (ξ, Y_{0}) = (η, Y_{1}) = 0. Then: (6a) ⟨Ψ(Y_{0}, Y_{0}), Ψ_{ξ}(𝔪_{1})⟩ = 0; (6b) ⟨Ψ(Y_{1}, Y_{1}), Ψ_{η}(𝔞 + 𝔪_{2})⟩ = 0.

Proof. The assertions (1), (2) and (3) can be proved in the same manner as in the proof of Proposition 16 of [8]. Hence, we omit their proofs.

Let ξ ∈ U (ξ ≠ 0). By (2a) we easily get Ψ_{ξ}(𝔞 + 𝔪_{2}) = R R RA and hence Ψ_{ξ}(𝔪) = R R RA + Ψ_{ξ}(𝔪_{1}). Since ⟨A, Ψ_{ξ}(𝔪_{1})⟩ = 0 (see (3a)), we have the decomposition (4a). Similarly, we can show (4b).

The assertions (5a) and (6a) are proved as follows: Let Y_{0} ∈ 𝔞 + 𝔪_{2}. Take ξ ∈ U (ξ ≠ 0) such that (ξ, Y_{0}) = 0. Then, we have [[Y_{0}, ξ], Y_{0}] = 4(μ, μ)(Y_{0}, Y_{0})ξ (see (2.2)) and Ψ(ξ, Y_{0}) = 0 (see (2a)). By the Gauss equation (3.1) we have

([[Y_{0}, ξ], Y_{0}], ξ) = ⟨Ψ(Y_{0}, Y_{0}), Ψ(ξ, ξ)⟩ − ⟨Ψ(Y_{0}, ξ), Ψ(ξ, Y_{0})⟩,

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 11

([[Y_{0}, ξ], Y_{0}], Y′_{1}) = ⟨Ψ(Y_{0}, Y_{0}), Ψ(ξ, Y′_{1})⟩ − ⟨Ψ(Y_{0}, Y′_{1}), Ψ(ξ, Y_{0})⟩,

where Y′_{1} is an arbitrary element of 𝔪_{1}. By these equalities we have ⟨Ψ(Y_{0}, Y_{0}), A⟩ = 4(μ, μ)(Y_{0}, Y_{0}) and ⟨Ψ(Y_{0}, Y_{0}), Ψ(ξ, Y′_{1})⟩ = 0. Therefore, we obtain (5a) and (6a). The assertions (5b) and (6b) can be proved in a similar way. □

Remark 2. As seen in the proof of Proposition 11, singular subspaces U and V may not be uniquely determined. However, the vectors A and B in Proposition 8 do not depend on the choice of singular subspaces U and V, which will be clarified at the last part of this section (see Lemma 20).

In the following argument, we take and fix an element Ψ ∈ G(P^{2}(H H H),N N N). We denote by U and V singular subspaces with respect to Ψ satisfying U (⊂ 𝔞 + 𝔪_{2}), V (⊂ 𝔪_{1}), dim U ≥ 2 and dim V ≥ 2. We also denote by A, B the vectors of N N N obtained by applying Proposition 12 to the pair of singular subspaces U and V.

Lemma 13. (1) Let Y_{0} ∈ 𝔞 + 𝔪_{2}. Then:

⟨Ψ_{Y_{0}}(Y_{1}), Ψ_{Y_{0}}(Y′_{1})⟩ = ⟨Ψ(Y_{0}, Y_{0}), Ψ(Y_{1}, Y′_{1})⟩ − (μ, μ)(Y_{0}, Y_{0})(Y_{1}, Y′_{1}), ∀Y_{1}, Y′_{1} ∈ 𝔪_{1}.

(2) Let Y_{0} ∈ 𝔞 + 𝔪_{2} and ξ ∈ U satisfy (ξ, Y_{0}) = 0. Then:

⟨Ψ_{Y_{0}}(Y_{1}), Ψ_{ξ}(Y′_{1})⟩ = (L(Y_{0}, ξ)Y_{1}, Y′_{1}), ∀Y_{1}, Y′_{1} ∈ 𝔪_{1}.

Proof. Putting X = Y_{0}, Y = Y_{1}, Z = Y_{0}, W = Y′_{1} into (3.1), we have

([[Y_{0}, Y_{1}], Y_{0}], Y′_{1}) = ⟨Ψ(Y_{0}, Y_{0}), Ψ(Y_{1}, Y′_{1})⟩ − ⟨Ψ(Y_{0}, Y′_{1}), Ψ(Y_{1}, Y_{0})⟩.

Since [Y_{0}, [Y_{0}, Y_{1}]] = −(μ, μ)(Y_{0}, Y_{0})Y_{1} (see (2.2)), we easily get (1).

Similarly, putting X = ξ, Y = Y_{1}, Z = Y_{0} and W = Y′_{1} into (3.1), we have

([[ξ, Y_{1}], Y_{0}], Y′_{1}) = ⟨Ψ(ξ, Y_{0}), Ψ(Y_{1}, Y′_{1})⟩ − ⟨Ψ(ξ, Y′_{1}), Ψ(Y_{1}, Y_{0})⟩
= ⟨A, Ψ(Y_{1}, Y′_{1})⟩(ξ, Y_{0}) − ⟨Ψ_{ξ}(Y′_{1}), Ψ_{Y_{0}}(Y_{1})⟩.

Since (ξ, Y_{0}) = 0, we have

⟨Ψ_{ξ}(Y′_{1}), Ψ_{Y_{0}}(Y_{1})⟩ = −([[ξ, Y_{1}], Y_{0}], Y′_{1}) = (L(Y_{0}, ξ)Y_{1}, Y′_{1}),

proving (2). □

Let ξ ∈ U (ξ ≠ 0). Since dim Ker(Ψ_{ξ}) = 3 (see Proposition 8) and since dim 𝔪 = 8, we have dim Ψ_{ξ}(𝔪) = 5. Let us denote by E E E_{ξ} the one dimensional orthogonal complement of Ψ_{ξ}(𝔪) in N N N.

Proposition 14. Set C = ⟨A, B⟩ − (μ, μ). Then:

(1) Let ξ ∈ U. Then:

⟨Ψ_{ξ}(Y_{1}), Ψ_{ξ}(η)⟩ = C(ξ, ξ)(Y_{1}, η),  ∀Y_{1} ∈ 𝔪_{1}, ∀η ∈ V.  (4.1)

(2) The inequality 0 < C ≤ 3(μ, μ) holds. The vectors A and B are linearly independent if C ≠ 3(μ, μ) and A = B if C = 3(μ, μ).

(3) Let ξ ∈ U (ξ ≠ 0). Then, Ψ_{Y_{0}}(𝔪_{1}) ⊂ E E E_{ξ} + Ψ_{ξ}(𝔪_{1}), ∀Y_{0} ∈ 𝔞 + 𝔪_{2}.

(4) If C ≠ 3(μ, μ), then:

Ψ_{Y_{0}}(𝔪_{1}) = Ψ_{ξ}(𝔪_{1}),  ∀Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0), ∀ξ ∈ U (ξ ≠ 0);  (4.2)
Ψ(Y_{0}, Y_{0}) ∈ R R RA + R R RB,  ∀Y_{0} ∈ 𝔞 + 𝔪_{2};  (4.3)
Ψ(Y_{1}, Y_{1}) ∈ R R RA + R R RB,  ∀Y_{1} ∈ 𝔪_{1}.  (4.4)

Proof. Put Y_{0} = ξ and Y′_{1} = η into Lemma 13 (1). Then, since Ψ(ξ, ξ) = (ξ, ξ)A and Ψ(Y_{1}, η) = (Y_{1}, η)B, we get (4.1).

In view of Proposition 12 (1), we easily have ⟨A, B⟩ ≤ 4(μ, μ) and hence C ≤ 3(μ, μ). Further, by putting Y_{1} = η (≠ 0) into (4.1) we know C > 0, because Ψ_{ξ}(η) ≠ 0 (see Proposition 9). This shows ⟨A, B⟩ > (μ, μ). Therefore, A and B are linearly independent if ⟨A, B⟩ ≠ 4(μ, μ), i.e., C ≠ 3(μ, μ). It is easy to see that if C = 3(μ, μ), i.e., ⟨A, B⟩ = 4(μ, μ), then A = B.

We next prove (3). Let ξ ∈ U (ξ ≠ 0). By Proposition 12 (4a) we know that the orthogonal complement of R R RA in N N N is given by E E E_{ξ} + Ψ_{ξ}(𝔪_{1}). Hence, by Proposition 12 (3a), we have Ψ_{Y_{0}}(𝔪_{1}) ⊂ E E E_{ξ} + Ψ_{ξ}(𝔪_{1}) for any Y_{0} ∈ 𝔞 + 𝔪_{2}.

Finally, we prove (4). Since C ≠ 3(μ, μ), the subspace R R RA +R R RB forms a 2-dimensional subspace of N N N. Let Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0). Then, by Proposition 12 (3a) we know that Ψ_{Y_{0}}(𝔪_{1}) coincides with the orthogonal complement of R R RA + R R RB in N N N. (Recall that dim Ψ_{Y_{0}}(𝔪_{1}) = 4 and dimN N N = 6.) Let ξ ∈ U (ξ ≠ 0). Since Ψ_{ξ}(𝔪_{1}) is also an orthogonal complement of R R RA + R R RB, it follows that Ψ_{ξ}(𝔪_{1}) = Ψ_{Y_{0}}(𝔪_{1}). If we take ξ ∈ U (ξ ≠ 0) satisfying (ξ, Y_{0}) = 0, then by Proposition 12 (6a) we obtain Ψ(Y_{0}, Y_{0}) ∈ R R RA + R R RB. Similarly, we can prove Ψ(Y_{1}, Y_{1}) ∈ R R RA + R R RB for any Y_{1} ∈ 𝔪_{1}, completing the proof of (4). □

Let Y_{0} ∈ 𝔞 + 𝔪_{2} and ξ ∈ U (ξ ≠ 0). Define a linear mapping Θ_{Y_{0},ξ}: 𝔪_{1} ⟶ N N N by

Θ_{Y_{0},ξ}(Y_{1}) = Ψ_{Y_{0}}(Y_{1}) + \frac{1}{C(ξ, ξ)}Ψ_{ξ}(L(ξ, Y_{0})Y_{1}), Y_{1} ∈ 𝔪_{1}.  (4.5)

Then, we have

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 13

Proposition 15. Let Y_{0} ∈ 𝔞 + 𝔪_{2}, ξ ∈ U (ξ ≠ 0) and Y_{1} ∈ 𝔪_{1}. Assume that (ξ, Y_{0}) = 0 and L(ξ, Y_{0})Y_{1} ∈ V. Then:

(1) Θ_{Y_{0},ξ}(Y_{1}) ∈ E E E_{ξ}. More strongly, if C ≠ 3(μ, μ), then Θ_{Y_{0},ξ}(Y_{1}) = 0.

(2) |Θ_{Y_{0},ξ}(Y_{1})|^{2} = ⟨Ψ(Y_{0}, Y_{0}), Ψ(Y_{1}, Y_{1})⟩ − (μ, μ) {1 + (μ, μ)/C} (Y_{0}, Y_{0})(Y_{1}, Y_{1}).

Proof. By Proposition 14 (3) we know that Θ_{Y_{0},ξ}(Y_{1}) ∈ E E E_{ξ} + Ψ_{ξ}(𝔪_{1}). Here, we note that ⟨E E E_{ξ}, Ψ_{ξ}(𝔪_{1})⟩ = 0, because E E E_{ξ} is orthogonal to Ψ_{ξ}(𝔪). Let Y′_{1} ∈ 𝔪_{1}. Then, by Lemma 13 (2), Proposition 14 (1) and Proposition 3 (2) we have

⟨Θ_{Y_{0},ξ}(Y_{1}), Ψ_{ξ}(Y′_{1})⟩ = ⟨Ψ_{Y_{0}}(Y_{1}), Ψ_{ξ}(Y′_{1})⟩ + \frac{1}{C(ξ, ξ)}⟨Ψ_{ξ}(L(ξ, Y_{0})Y_{1}), Ψ_{ξ}(Y′_{1})⟩
= (L(Y_{0}, ξ)Y_{1}, Y′_{1}) + (L(ξ, Y_{0})Y_{1}, Y′_{1})
= 0,

proving ⟨Θ_{Y_{0},ξ}(Y_{1}), Ψ_{ξ}(𝔪_{1})⟩ = 0. This implies that Θ_{Y_{0},ξ}(Y_{1}) ∈ E E E_{ξ}. In the case where C ≠ 3(μ, μ), we have Θ_{Y_{0},ξ}(Y_{1}) ∈ Ψ_{Y_{0}}(𝔪_{1}) + Ψ_{ξ}(𝔪_{1}) = Ψ_{ξ}(𝔪_{1}) (see (4.2)), which proves Θ_{Y_{0},ξ}(Y_{1}) = 0.

Next, we show (2). By Lemma 13 and by the equality ⟨Θ_{Y_{0},ξ}(Y_{1}), Ψ_{ξ}(𝔪_{1})⟩ = 0, we have

⟨Θ_{Y_{0},ξ}(Y_{1}), Θ_{Y_{0},ξ}(Y_{1})⟩ = ⟨Θ_{Y_{0},ξ}(Y_{1}), Ψ_{Y_{0}}(Y_{1})⟩

= ⟨Ψ_{Y_{0}}(Y_{1}), Ψ_{Y_{0}}(Y_{1})⟩ + \frac{1}{C(ξ, ξ)}⟨Ψ_{ξ}(L(ξ, Y_{0})Y_{1}), Ψ_{Y_{0}}(Y_{1})⟩

= ⟨Ψ(Y_{0}, Y_{0}), Ψ(Y_{1}, Y_{1})⟩ − (μ, μ)(Y_{0}, Y_{0})(Y_{1}, Y_{1})

+ \frac{1}{C(ξ, ξ)}(L(ξ, Y_{0})Y_{1}, L(Y_{0}, ξ)Y_{1}).

On the other hand, by Proposition 3 we have

(L(ξ, Y_{0})Y_{1}, L(Y_{0}, ξ)Y_{1}) = (L(ξ, Y_{0})L(ξ, Y_{0})Y_{1}, Y_{1})
= −(L(Y_{0}, ξ)L(ξ, Y_{0})Y_{1}, Y_{1})
= −(μ, μ)^{2}(ξ, ξ)(Y_{0}, Y_{0})(Y_{1}, Y_{1}).

Therefore, we get the assertion (2). □

With these preparations we begin with the proof Theorem 6. First, we consider the case dim V = 2.

Lemma 16. Assume that dim V = 2. Then, C ≠ 3(μ, μ). Accordingly, the vectors A and B ∈ N N N are linearly independent.

14 YOSHIO AGAOKA AND EIJI KANEDA

Proof. Take non-zero elements ξ, ξ′ ∈ U satisfying (ξ, ξ′) = 0. Then, by Proposition 3 (2) it follows that L(ξ, ξ′) = −L(ξ′, ξ) and L(ξ, ξ′) gives an isomorphism of 𝔪_{1} onto itself. Let Y_{1} ∈ L(ξ, ξ′)V. Then, by Proposition 3 (2b) we have L(ξ, ξ′)Y_{1} ∈ V. Hence, by Proposition 15 (1) we have Θ_{ξ′,ξ}(Y_{1}) ∈ E E E_{ξ}. Since dim L(ξ, ξ′)V = dim V = 2 and dimE E E_{ξ} = 1, it is possible to take a non-zero element Y_{1} ∈ L(ξ, ξ′)V satisfying Θ_{ξ′,ξ}(Y_{1}) = 0. Therefore, by Proposition 15 (2) and Proposition 12 (2a) we have

0 = |Θ_{ξ′,ξ}(Y_{1})|^{2} = [⟨Ψ(Y_{1}, Y_{1}), A⟩ − (μ, μ) {1 + (μ, μ)/C} (Y_{1}, Y_{1})] (ξ′, ξ′).

Since (ξ′, ξ′) ≠ 0, we have

⟨Ψ(Y_{1}, Y_{1}), A⟩ = (μ, μ) {1 + (μ, μ)/C} (Y_{1}, Y_{1}).  (4.6)

Now, we suppose the case C = 3(μ, μ). Then, by (4.6) we have ⟨Ψ(Y_{1}, Y_{1}), A⟩ = \frac{4}{3} (μ, μ)(Y_{1}, Y_{1}). On the other hand, by Proposition 12 (5b) we have ⟨Ψ(Y_{1}, Y_{1}), A⟩ = 4 (μ, μ)(Y_{1}, Y_{1}), because A = B in case C = 3(μ, μ) (see Proposition 14 (2)). Hence, we have (Y_{1}, Y_{1}) = 0, which contradicts the assumption Y_{1} ≠ 0. Therefore, we have C ≠ 3(μ, μ) and hence A and B are linearly independent. □

Lemma 17. Assume that dim V = 2. Then, V can be extended to a 3-dimensional singular subspace contained in 𝔪_{1}, i.e., there is a singular subspace V̂ (⊂ 𝔪_{1}) such that V ⊂ V̂ and dim V̂ = 3.

Proof. Let F ∈ R R RA + R R RB be a unit vector which is orthogonal to B. Then, for any η ∈ V we have ⟨F, Ψ_{η}(𝔪)⟩ = 0, because ⟨F, Ψ_{η}(𝔪)⟩ = ⟨F,R R RB + Ψ_{η}(𝔞 + 𝔪_{2})⟩ = 0 (see Proposition 12 (4b) and (3b)).

Now, define a symmetric bilinear form χ on 𝔪_{1} by setting

χ(Y_{1}, Y′_{1}) = ⟨Ψ(Y_{1}, Y′_{1}), F⟩, Y_{1}, Y′_{1} ∈ 𝔪_{1}.

Since Ψ(Y_{1}, Y′_{1}) ∈ R R RB+R R RF (see Proposition 14 (4)) and ⟨Ψ(Y_{1}, Y′_{1}), B⟩ = ⟨B, B⟩(Y_{1}, Y′_{1}) for Y_{1}, Y′_{1} ∈ 𝔪_{1} (see Proposition 12 (5)), we have

Ψ(Y_{1}, Y′_{1}) = (Y_{1}, Y′_{1})B + χ(Y_{1}, Y′_{1})F,  Y_{1}, Y′_{1} ∈ 𝔪_{1}.  (4.7)

Let V^{⊥} be the orthogonal complement of V in 𝔪_{1}. Then, we have dim V^{⊥} = 2. (Recall that dim 𝔪_{1} = 4 and dim V = 2.) Let {Y_{1}, Y′_{1}} be an orthonormal basis of V^{⊥}. Then, putting X = Z = Y_{1} and Y = W = Y′_{1} into the Gauss equation (3.1), we have

\begin{aligned}([[Y_{1}, Y′_{1}], Y_{1}], Y′_{1}) & = ⟨B, B⟩(Y_{1}, Y_{1})(Y′_{1}, Y′_{1}) \\ & + χ(Y_{1}, Y_{1})χ(Y′_{1}, Y′_{1}) − χ(Y_{1}, Y′_{1})χ(Y′_{1}, Y_{1}). \end{aligned}

Since ([[Y_{1}, Y′_{1}], Y_{1}], Y′_{1}) = ⟨B, B⟩(Y_{1}, Y_{1})(Y′_{1}, Y′_{1}) (see (2.2)), we have

χ(Y_{1}, Y_{1})χ(Y′_{1}, Y′_{1}) − χ(Y_{1}, Y′_{1})χ(Y′_{1}, Y_{1}) = 0.

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 15

This implies that χ is degenerate on V^{⊥}. Therefore, there is a non-zero vector ζ ∈ V^{⊥} such that χ(ζ, V^{⊥}) = 0, i.e., ⟨F, Ψ_{ζ}(V^{⊥})⟩ = 0.

Let us show that the subspace V̂ = R R Rζ + V (⊂ 𝔪_{1}) is singular with respect to Ψ. Note that ⟨F, Ψ_{ζ}(𝔞 + 𝔪_{2})⟩ = 0 (see Proposition 12 (3b)). Then, since 𝔪 = 𝔞 + 𝔪_{2} + V + V^{⊥} and Ψ_{ζ}(V) ⊂ R R RB, it follows that

⟨F, Ψ_{ζ}(𝔪)⟩ = ⟨F, Ψ_{ζ}(𝔞 + 𝔪_{2}) + Ψ_{ζ}(V) + Ψ_{ζ}(V^{⊥})⟩ ⊂ 0 + ⟨F,R R RB⟩ + 0 = 0.

Hence, we have ⟨F, Ψ_{aζ+η}(𝔪)⟩ = 0 for any a ∈ R R R and η ∈ V. Consequently, Ψ_{aζ+η}(𝔪) ≠ N N N, which implies that aζ + η ∈ V̂ is singular with respect to Ψ. □

Now, we assume that dim V = 2 and denote by V̂ be the singular subspace stated in the above lemma. Let Â and B̂ be the vectors obtained by applying Proposition 12 to the pair of singular subspaces U and V̂. Then, by Proposition 12 (2) we can easily see that Â = A and B̂ = B. Therefore, we know that all the statements in Proposition 12 and hence the arguments developed after Proposition 12 are also true if we simply replace V by V̂. Accordingly, without loss of generality we can assume that dim V ≥ 3.

Lemma 18. ⟨Ψ(Y_{0}, Y_{0}), B⟩ = (μ, μ) {1 + (μ, μ)/C} (Y_{0}, Y_{0}), ∀Y_{0} ∈ 𝔞 + 𝔪_{2}.

Proof. As in the proof of Lemma 16, we can prove that C ≠ 3(μ, μ). Let Y_{0} ∈ 𝔞+𝔪_{2} (Y_{0} ≠ 0). Take ξ ∈ U (ξ ≠ 0) such that (ξ, Y_{0}) = 0, which is possible because dim U ≥ 2. Then, by Proposition 3 (2) it follows that L(ξ, Y_{0}) = −L(Y_{0}, ξ) and that the map L(ξ, Y_{0}) gives an isomorphism of 𝔪_{1} onto itself. Now, take η ∈ V (η ≠ 0) such that L(ξ, Y_{0})η ∈ V. This is also possible because dim L(ξ, Y_{0})V = dim V ≥ 3 and dim (V ∩ L(ξ, Y_{0})V) ≥ 2. (Note that dim 𝔪_{1} = 4.) Then, by Proposition 15 and Proposition 12 (2b) we have

0 = |Θ_{Y_{0},ξ}(η)|^{2} = [⟨Ψ(Y_{0}, Y_{0}), B⟩ − (μ, μ) {1 + (μ, μ)/C} (Y_{0}, Y_{0})](η, η).

Since (η, η) ≠ 0, we get the lemma. □

Lemma 19. C = (μ, μ), i.e., ⟨A, B⟩ = 2(μ, μ).

Proof. Take ξ ∈ U (ξ ≠ 0). Then, by Lemma 18 and Ψ(ξ, ξ) = (ξ, ξ)A (see Proposition 12 (2a)), we have ⟨A, B⟩ = (μ, μ) {1 + (μ, μ)/C}. Since C = ⟨A, B⟩ − (μ, μ), we easily have C^{2} = (μ, μ)^{2}. Moreover, since C > 0 (see Proposition 14 (2)), it follows that C = (μ, μ), i.e., ⟨A, B⟩ = 2(μ, μ). □

Now, we show

Lemma 20. (1) Ψ(Y_{0}, Y′_{0}) = (Y_{0}, Y′_{0})A, ∀Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2}.

(2) Ψ(Y_{1}, Y′_{1}) = (Y_{1}, Y′_{1})B, ∀Y_{1}, Y′_{1} ∈ 𝔪_{1}.

16 YOSHIO AGAOKA AND EIJI KANEDA

Proof. On account of an elementary fact concerning symmetric bilinear forms, we have only to show Ψ(Y_{0}, Y_{0}) = (Y_{0}, Y_{0})A and Ψ(Y_{1}, Y_{1}) = (Y_{1}, Y_{1})B for any Y_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}.

Let Y_{0} ∈ 𝔞 + 𝔪_{2}. Then, by Lemma 18 and Lemma 19 we have ⟨Ψ(Y_{0}, Y_{0}), B⟩ = ⟨A, B⟩(Y_{0}, Y_{0}). Moreover, by Proposition 12 (1) and (5a) we have ⟨Ψ(Y_{0}, Y_{0}), A⟩ = ⟨A, A⟩(Y_{0}, Y_{0}). Since Ψ(Y_{0}, Y_{0}) ∈ R R RA + R R RB (see (4.3)), it follows that Ψ(Y_{0}, Y_{0}) = (Y_{0}, Y_{0})A, which proves (1).

We next prove (2). Let Y_{1} ∈ 𝔪_{1} (Y_{1} ≠ 0). Take elements ξ ∈ U (ξ ≠ 0) and η ∈ V (η ≠ 0) such that (η, Y_{1}) = 0. Set Y_{0} = [Y_{1}, [ξ, η]]. Then, it is easy to see that [ξ, η] ∈ 𝔨_{1} and Y_{0} ∈ 𝔞 + 𝔪_{2} (see (2.1)). Further, we have (ξ, Y_{0}) = 0 and L(ξ, Y_{0})Y_{1} ∈ V, because

(ξ, Y_{0}) = (ξ, [Y_{1}, [ξ, η]]) = −([ξ, [ξ, η]], Y_{1}) = (μ, μ)(ξ, ξ)(η, Y_{1}) = 0,
L(ξ, Y_{0})Y_{1} = [ξ, [[Y_{1}, [ξ, η]], Y_{1}]] = (μ, μ)(Y_{1}, Y_{1})[ξ, [ξ, η]]
= −(μ, μ)^{2}(ξ, ξ)(Y_{1}, Y_{1})η ∈ V

(see (2.2) and (2.4)). Thus, by Proposition 15 (2), Lemma 19 and Ψ(Y_{0}, Y_{0}) = (Y_{0}, Y_{0})A (see (1)), we have

0 = |Θ_{Y_{0},ξ}(Y_{1})|^{2} = [⟨A, Ψ(Y_{1}, Y_{1})⟩ − 2(μ, μ)(Y_{1}, Y_{1})] (Y_{0}, Y_{0}).

Here, we note that Y_{0} ≠ 0, because L(ξ, Y_{0})Y_{1} ≠ 0. Hence, by the above equality and Lemma 19, we have ⟨Ψ(Y_{1}, Y_{1}), A⟩ = ⟨B, A⟩(Y_{1}, Y_{1}). On the other hand, by Proposition 12 (1) and (5b) we have ⟨Ψ(Y_{1}, Y_{1}), B⟩ = ⟨B, B⟩(Y_{1}, Y_{1}). Consequently, it follows that Ψ(Y_{1}, Y_{1}) = (Y_{1}, Y_{1})B, because Ψ(Y_{1}, Y_{1}) ∈ R R RA + R R RB (see (4.4)). This proves (2). □

We are now in a final position of the proof of Theorem 6. Let Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0). Then, by Lemma 20 (1) we have Ker(Ψ_{Y_{0}}) ⊃ {Y′_{0} ∈ 𝔞 + 𝔪_{2} | (Y_{0}, Y′_{0}) = 0}. This shows dim Ker(Ψ_{Y_{0}}) ≥ 3 and hence Y_{0} is singular with respect to Ψ (see Proposition 9 (1)). Accordingly, 𝔞 + 𝔪_{2} is a singular subspace. Similarly, by Lemma 20 (2) we can show that 𝔪_{1} is also a singular subspace.

Now, let us put into Proposition 12 U = 𝔞 + 𝔪_{2} and V = 𝔪_{1}. Then, by Lemma 20 we know that the vectors A and B are not altered by this change of singular subspaces. Therefore, all the statements in Proposition 12 and the arguments developed after Proposition 12 are also true under our setting U = 𝔞 + 𝔪_{2} and V = 𝔪_{1}. Consequently, by Proposition 12 (1), (2), (3) and Lemma 19 we get the assertion (1) of Theorem 6. We also obtain by Proposition 14 and C = (μ, μ) (see Lemma 19) the assertion (3) of Theorem 6.

RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF P^{2}(H H H) 17

Finally, we prove the assertion (2) of Theorem 6. Let Y_{2} ∈ 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Then, since C ≠ 3(μ, μ) and (μ, Y_{2}) = 0, we have

Θ_{Y_{2},μ}(Y_{1}) = Ψ_{Y_{2}}(Y_{1}) + \frac{1}{(μ, μ)^{2}} Ψ_{μ}(L(μ, Y_{2})Y_{1}) = 0

(see Proposition 15). Here we note that the conditions μ ∈ U and L(μ, Y_{2})Y_{1} ∈ V in Proposition 15 have no significance, because U = 𝔞 + 𝔪_{2} and V = 𝔪_{1}. Accordingly, we obtain the assertion (2). This completes the proof of Theorem 6. □

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(Yoshio AGAOKA) Faculty of Integrated Arts and Sciences, Hiroshima University 1-7-1 Kagamiyama, Higashi-Hiroshima City, Hiroshima, 739-8521, Japan

E-mail address: agaoka@mis.hiroshima-u.ac.jp

(Eiji KANEDA) Faculty of Foreign Studies, Osaka University of Foreign Studies 8-1-1 Aomadani-Higashi, Minoo City, Osaka, 562-8558, Japan

E-mail address: kaneda@osaka-gaidai.ac.jp