RIGIDITY OF THE CANONICAL ISOMETRIC IMBEDDING OF THE CAYLEY PROJECTIVE PLANE P^{2}(Cay Cay Cay)

YOSHIO AGAOKA AND EIJI KANEDA

Abstract. In [7], we have proved that P^{2}(Cay Cay Cay) cannot be isometrically immersed into R R R^{25} even locally. In this paper, we investigate isometric immersions of P^{2}(Cay Cay Cay) into R R R^{26} and prove that the canonical isometric imbedding fff_{0} of P^{2}(Cay Cay Cay) into R R R^{26}, which is defined in Kobayashi [17], is rigid in the following strongest sense: Any isometric immersion fff_{1} of a connected open set U(⊂ P^{2}(Cay Cay Cay)) into R R R^{26} coincides with fff_{0} up to a euclidean transformation of R R R^{26}, i.e., there is a euclidean transformation a of R R R^{26} satisfying fff_{1} = afff_{0} on U.

1. Introduction

In the previous paper [7], we investigated the problem of (local) isometric immersions of the quaternion projective plane P^{2}(H H H) and the Cayley projective plane P^{2}(Cay Cay Cay). In particular, we proved the following non-existence theorem of (local) isometric immersions:

Theorem 1. Any open set of the Cayley projective plane P^{2}(Cay Cay Cay) cannot be isometrically immersed into R R R^{25}.

As is well-known, there is an isometric immersion fff_{0} of P^{2}(Cay Cay Cay) into the euclidean space R R R^{26}, which is called the canonical isometric imbedding of P^{2}(Cay Cay Cay) (Kobayashi [17]). This fact together with Theorem 1 implies that R R R^{26} is the least dimensional euclidean space into which P^{2}(Cay Cay Cay) can be (locally) isometrically immersed.

In this paper, we consider (local) isometric immersions of P^{2}(Cay Cay Cay) into R R R^{26} and discuss the rigidity of the canonical isometric imbedding fff_{0}. Concerning the rigidity of fff_{0} Kaneda [15] has shown that the canonical isometric imbedding fff_{0} is of finite type, i.e., the space of local infinitesimal isometric deformations of fff_{0} is of finite dimension. However, it seems to the authors that any further result concerning the rigidity of fff_{0} has not been obtained.

In the present paper, we will show the rigidity of the canonical isometric imbedding fff_{0} in the following strongest form:

Date: April 21, 2002.

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

Key words and phrases. Curvature invariant, isometric immersion, Cayley projective plane, rigidity.

2 YOSHIO AGAOKA AND EIJI KANEDA

Theorem 2. Let fff_{0} be the canonical isometric imbedding of P^{2}(Cay Cay Cay) into the euclidean space R R R^{26}. Then, for any isometric immersion fff_{1} defined on a connected open set U of P^{2}(Cay Cay Cay) into R R R^{26}, there exists a euclidean transformation a of R R R^{26} satisfying fff_{1} = afff_{0} on U.

To prove Theorem 2, we first establish a rigidity theorem for an isometric immersion of a Riemannian manifold. Let M be an n-dimensional Riemannian manifold and let fff_{0} be an isometric immersion of M into the m-dimensional euclidean space R R R^{m}. We will prove that if the Gauss equation in codimension r (= m − n) admits essentially one solution everywhere on M, then fff_{0} is rigid, i.e., for any isometric immersion fff_{1} of M into R R R^{m} there exists a euclidean transformation a of R R R^{m} such that fff_{1} = afff_{0} (see Theorem 5). This theorem may be established by various methods; for example, by combining the results of Nomizu [19] and Szczarba [21], [22] (cf. Agaoka [1]) or by solving a differential system of Pfaff (cf. Bishop–Crittenden [10], Ch. X). In this paper, we will give a simple proof based on a congruence theorem of differentiable mappings, which is easy to understand and gives a clear view on the geometric meaning (see Theorem 6).

Next, we will show that for the Cayley projective plane P^{2}(Cay Cay Cay) the Gauss equation in codimension 10 (= 26 − dim P^{2}(Cay Cay Cay)) admits essentially one solution (see Theorem 10). To show this, we utilize the results obtained in [6] and [7]. Among all, the result concerning pseudo-abelian subspaces (Proposition 8) plays an important role in our proof.

Then, Theorem 2 is a direct consequence of Theorem 5 and Theorem 10.

Throughout this paper we assume the differentiability of class C^{∞}. Notations for Lie algebras are the same that are used in [6] and [7].

2. The Gauss equation

Let M be a Riemannian manifold and T (M) the tangent bundle of M. We denote by g the Riemannian metric of M and by R the Riemannian curvature tensor of type (1, 3) with respect to g.

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 p ∈ M and let S^{2}T_{p}^{∗}(M) ⊗N N N be the space of N N N-valued symmetric bilinear forms on T_{p}(M). We call the following equation on Ψ ∈ S^{2}T_{p}^{∗}(M) ⊗N N N the Gauss equation at p ∈ M:

−g_{p}(R_{p}(x, y)z, w) = ⟨Ψ(x, z), Ψ(y, w)⟩ − ⟨Ψ(x, w), Ψ(y, z)⟩,  (2.1)

where x, y, z, w ∈ T_{p}(M). We denote by G_{p}(N N N) the set of all solutions of (2.1), which is called the Gaussian variety associated with N N N at p ∈ M. As is well-known, G_{p}(N N N) = ∅

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

happens in case the dimensionality r (= dimN N N) is so small, however, G_{p}(N N N) ≠ ∅ if r is sufficiently large (see Cartan [11] or Kaneda–Tanaka [16]).

Let N N N_{1} and N N N_{2} be two euclidean vector spaces and let φ be a linear mapping of N N N_{1} to N N N_{2}. Define a linear map φ̂ of S^{2}T_{p}^{∗}(M) ⊗ N N N_{1} to S^{2}T_{p}^{∗}(M) ⊗ N N N_{2} by

(φ̂ Ψ)(x, y) = φ(Ψ(x, y)),  Ψ ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N_{1}, x, y ∈ T_{p}(M).  (2.2)

Then, we can easily verify

Lemma 3. Let φ be a linear mapping of a euclidean vector space N N N_{1} to a euclidean vector space N N N_{2}. Assume that φ is isometric, i.e., ⟨φ(x), φ(y)⟩_{2} = ⟨x, y⟩_{1} (x, y ∈ T_{p}(M)), where ⟨ , ⟩_{i} (i = 1, 2) denotes the inner product of N N N_{i}. Then φ̂ G_{p}(N N N_{1}) ⊂ G_{p}(N N N_{2}). In particular, if dimN N N_{1} = dimN N N_{2}, then φ̂ G_{p}(N N N_{1}) = G_{p}(N N N_{2}).

In view of Lemma 3, the solvability of the Gauss equation (2.1) substantially depends on the dimensionality of N N N. To emphasize dimN N N we call (2.1) the Gauss equation in codimension r (= dimN N N).

Let N N N be a euclidean vector space and 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}T_{p}^{∗}(M) ⊗ N N N by

(hΨ)(x, y) = h(Ψ(x, y)),

where Ψ ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N, h ∈ O(N N N), x, y ∈ T_{p}(M). We say that two elements Ψ and Ψ′ ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N are equivalent if there is an element h ∈ O(N N N) such that Ψ′ = hΨ. It is easily seen that if Ψ and Ψ′ ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N are equivalent and Ψ ∈ G_{p}(N N N), then Ψ′ ∈ G_{p}(N N N). We say that the Gaussian variety G_{p}(N N N) is EOS if G_{p}(N N N) ≠ ∅ and if it is composed of essentially one solution, i.e., any solutions of the Gauss equation (2.1) are equivalent to each other under the action of O(N N N).

Proposition 4. Let M be a Riemannian manifold and let p ∈ M. Let N N N be an rdimensional euclidean vector space such that G_{p}(N N N) is EOS. Then:

(1) Let Ψ be an arbitrary element of G_{p}(N N N). Then, the vectors Ψ(x, y) (x, y ∈ T_{p}(M)) span the whole space N N N.

(2) Let N N N_{1} be a euclidean vector space. Then:

(2a) G_{p}(N N N_{1}) = ∅ if dimN N N_{1} < r; (2b) G_{p}(N N N_{1}) is EOS if dimN N N_{1} = r; (2c) G_{p}(N N N_{1}) is not EOS if dimN N N_{1} > r.

Proof. Note that if Ψ′ ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N is equivalent to Ψ, then we have |Ψ′(x, y)| = |Ψ(x, y)| for any x, y ∈ T_{p}(M), where |n n n| denotes the norm of n n n ∈ N N N with respect to ⟨ , ⟩.

4 YOSHIO AGAOKA AND EIJI KANEDA

Now, suppose that the vectors Ψ(x, y) (x, y ∈ T_{p}(M)) do not span the whole space N N N. Then, there is a non-zero vector n n n ∈ N N N satisfying ⟨n n n, Ψ(x, y)⟩ = 0 for any x, y ∈ T_{p}(M). Define an element Ψ′ ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N by

Ψ′ = Ψ + (ξ^{∗})^{2} ⊗ n n n,

where ξ^{∗} is a non-zero element of T_{p}^{∗}(M). Then, it is easy to see that Ψ′ ∈ G_{p}(N N N). However, by a simple calculation, we have |Ψ′(x, x)|^{2} = |Ψ(x, x)|^{2} +|n n n|^{2}ξ^{∗}(x)^{2}. Therefore, if we take x ∈ T_{p}(M) such that ξ^{∗}(x) ≠ 0, then we have |Ψ′(x, x)| ≠ |Ψ(x, x)|. This proves that Ψ′ is not equivalent to Ψ and hence G_{p}(N N N) is not EOS. Thus, we obtain (1).

Next we prove (2). First assume dimN N N_{1} = r. Let φ be an isometric linear isomorphism of N N N onto N N N_{1}. Then we have O(N N N_{1}) = φ · O(N N N) · φ^{−1}. Moreover, by Lemma 3 we have φ̂ G_{p}(N N N) = G_{p}(N N N_{1}). Since G_{p}(N N N) is EOS, O(N N N) acts transitively on G_{p}(N N N). Therefore, it is easily seen that O(N N N_{1}) acts transitively on G_{p}(N N N_{1}). This proves that G_{p}(N N N_{1}) is EOS.

We next consider the case dimN N N_{1} < r. Suppose that G_{p}(N N N_{1}) ≠ ∅ and Ψ_{1} ∈ G_{p}(N N N_{1}). Let φ be an isometric linear mapping of N N N_{1} to N N N. Then, we know that φ̂ Ψ_{1} ∈ G_{p}(N N N) and the vectors (φ̂ Ψ_{1})(x, y) (x, y ∈ T_{p}(M)) are contained in the proper subspace φ(N N N_{1}) (⊊ N N N). This contradicts (1). The case dimN N N_{1} > r is similarly dealt with. □

We say that a Riemannian manifold M is formally rigid in codimension r if there is a euclidean vector space N N N with dimN N N = r such that the Gaussian variety G_{p}(N N N) is EOS at each p ∈ M. By virtue of Proposition 4 (2), we know that if M is formally rigid in codimension r, then it is not formally rigid in any other codimension r′ (≠ r).

Remark 1. It should be noted that there is a Riemannian manifold M that is not formally rigid in any codimension r. For example, assume that M is the space of negative constant curvature of dimension n. Let N N N be a euclidean vector space of dimension r. Then, by Ôtsuki’s lemma we have G_{p}(N N N) = ∅ if r < n − 1 (see Ôtsuki [20]). On the other hand, Kaneda [13] proved that if r = n − 1, then G_{p}(N N N) ≠ ∅ and around a suitable Ψ_{0} ∈ G_{p}(N N N), G_{p}(N N N) forms a submanifold of S^{2}T_{p}^{∗}(M)⊗N N N of dimension n(n−1) (see Theorem 3.1 of [13]). Since n(n − 1) > dim O(N N N), G_{p}(N N N) cannot be EOS. If r ≥ n, then by Proposition 4 (2a) we know that G_{p}(N N N) is not EOS. Accordingly, the space of negative constant curvature M is not formally rigid in any codimension r.

Remark 2. For each Riemannian submanifold M ⊂ R R R^{m} listed below, G_{p}(N N N) is known to be EOS at each p ∈ M, where N N N is the normal vector space of M at p in R R R^{m}:

(1) The sphere S^{n} ⊂ R R R^{n+1} (n ≥ 3);

(2) The symplectic group Sp(2) ⊂ R R R^{16} (see Agaoka [1]);

(3) A submanifold M ⊂ R R R^{m} with type number ≥ 3 (see Allendoerfer [9], Kobayashi– Nomizu [18]).

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

Consequently, these submanifolds are formally rigid in our sense and it has been proved that they are actually rigid in R R R^{m} (see [1], [9]).

However, we note that the formal rigidness of M in codimension r does not imply the existence of an isometric immersion of M into R R R^{n+r} (n = dim M). Indeed, Kaneda [14] gave an example of three dimensional Riemannian manifold M that is formally rigid in codimension 1 but cannot be locally isometrically immersed into R R R^{4}.

We will prove in the next section that if a connected Riemannian manifold M is formally rigid in codimension r and if there is an isometric immersion fff of M into R R R^{n+r} (n = dim M), then M (precisely, fff(M)) is actually rigid in R R R^{n+r} (see Theorem 5).

3. Rigidity theorem

In this section, we will prove the following rigidity theorem:

Theorem 5. Let M be an n-dimensional Riemannian manifold and let fff_{0} be an isometric immersion of M into the euclidean space R R R^{m}. Assume:

(1) M is connected;

(2) M is formally rigid in codimension r = m − n. Then, any isometric immersion fff_{1} of M into the euclidean space R R R^{m} coincides with fff_{0} up to a euclidean transformation of R R R^{m}, i.e., there exists a euclidean transformation a of R R R^{m} such that fff_{1} = afff_{0}.

Before proceeding to the proof of Theorem 5, we make some preparations. Let M(m, m′) be the space of real matrices of degree m × m′, where m and m′ are non-negative integers. In what follows we identify M(m, 1) with the m-dimensional euclidean space R R R^{m} in a natural way. Then, we note that the canonical inner product ⟨ , ⟩ of R R R^{m} is given by ⟨vvv,w w w⟩ = ^{t}vvv · w w w for vvv, w w w ∈ R R R^{m}.

Let us define an operation of M(m, m) on R R R^{m} by

M(m, m) × R R R^{m} ∋ (H,vvv) ⟼ H · vvv ∈ R R R^{m},

where · means the usual matrix multiplication.

Let ∇ be the Riemannian connection associated with M. Let fff = ^{t}(f^{1}, …, f^{m}) be a

differentiable map of M into the euclidean space R R R^{m}. By \overbrace{∇ ⋯ ∇}^{k}fff we denote the k-th order covariant derivative of fff, which is defined as follows:

\overbrace{∇_{x_{1}} ⋯ ∇_{x_{k}}}^{k}fff = ^{t}(…, \overbrace{∇_{x_{1}} ⋯ ∇_{x_{k}}}^{k} f^{i}, …) ∈ R R R^{m},

6 YOSHIO AGAOKA AND EIJI KANEDA

where p ∈ M; x_{1}, …, x_{k} ∈ T_{p}(M). (Precisely, see Tanaka [23], Kaneda–Tanaka [16] or Kaneda [14].) It is known that ∇∇fff and ∇∇∇fff satisfy the following integrability conditions:

∇_{x}∇_{y}fff = ∇_{y}∇_{x}fff,  (3.1)
∇_{z}∇_{x}∇_{y}fff = ∇_{x}∇_{z}∇_{y}fff − ∇_{R(z,x)y}fff.  (3.2)

We say that a differentiable map fff of M into R R R^{m} is 2-generic if at each p ∈ M, the whole space R R R^{m} is spanned by the vectors of the form ∇_{x}fff (x ∈ T_{p}(M)), ∇_{y}∇_{z}fff (y, z ∈ T_{p}(M)). It is clear that if fff is 2-generic, then we have the inequality m ≤ \frac{1}{2}n(n + 3). Note that a 2-generic map fff is not necessarily an immersion.

We first show the following congruence theorem:

Theorem 6. Let M be an n-dimensional Riemannian manifold and let fff_{i} (i = 0, 1) be two differentiable maps of M into the euclidean space R R R^{m}. Assume:

(1) M is connected;

(2) fff_{0} is 2-generic;

(3) At each p ∈ M there is an element H(p) ∈ O(m) satisfying

∇_{x}fff_{1} = H(p) · (∇_{x}fff_{0})  ∀x ∈ T_{p}(M),  (3.3)
∇_{y}∇_{z}fff_{1} = H(p) · (∇_{y}∇_{z}fff_{0}),  ∀y, z ∈ T_{p}(M).  (3.4)

Then, fff_{1} coincides with fff_{0} up to a euclidean transformation of R R R^{m}. More precisely, H(p) is identically equal to a constant value H_{0} ∈ O(m) everywhere on M and fff_{1} can be written as fff_{1} = H_{0}fff_{0} + ccc_{0}, where ccc_{0} is a constant vector of R R R^{m}.

Proof. We first note that, since fff_{0} is 2-generic, H(p) satisfying (3.3) and (3.4) is uniquely determined at each p ∈ M and the map H: M ∋ p ⟼ H(p) ∈ O(m) is differentiable. Via the canonical inclusion O(m) ⊂ M(m, m), we can regard H as an M(m, m)-valued function on M satisfying

tHH = I_{m},  (3.5)

where I_{m} denotes the identity matrix of degree m. Differentiate (3.5) covariantly. Then by Leibnitz’ law we get

∇_{x}(^{t}H)H(p) +^{t}H(p)(∇_{x}H) = 0,  ∀x ∈ T_{p}(M).  (3.6)

In this equality, the covariant derivative ∇_{x}H means the element of M(m, m) given by ∇_{x}H = (∇_{x}h_{i}^{j}), where h_{i}^{j} denotes the (i, j)-component of H. By the very definition of ∇_{x}H we have ∇_{x}(^{t}H) = ^{t}(∇_{x}H).

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

Let us define an M(m, m)-valued 1-form L by

L(x) = ^{t}H(p)(∇_{x}H),  x ∈ T_{p}(M).  (3.7)

Then, by (3.6) we have

tL(x) + L(x) = 0,  ∀x ∈ T_{p}(M),  (3.8)

implying that the matrix L(x) ∈ M(m, m) is skew-symmetric.

We now show that the equality L(x) = 0 holds for any x ∈ T_{p}(M). Since fff_{0} is 2-generic, it suffices to prove

L(y) · (∇_{x}fff_{0}) = 0,  ∀x, y ∈ T_{p}(M),  (3.9)
L(z) · (∇_{y}∇_{x}fff_{0}) = 0,  ∀x, y, z ∈ T_{p}(M).  (3.10)

Differentiating (3.3) and (3.4) covariantly, we have

∇_{y}∇_{x}fff_{1} = ∇_{y}H · (∇_{x}fff_{0}) + H(p) · (∇_{y}∇_{x}fff_{0}),  ∀x, y ∈ T_{p}(M),  (3.11)
∇_{z}∇_{y}∇_{x}fff_{1} = ∇_{z}H · (∇_{y}∇_{x}fff_{0}) + H(p) · (∇_{z}∇_{y}∇_{x}fff_{0}), ∀x, y, z ∈ T_{p}(M). (3.12)

Then by (3.4) and (3.11) we have ∇_{y}H · (∇_{x}fff_{0}) = 0 for each x, y ∈ T_{p}(M). Consequently, multiplying ^{t}H(p) from the left, we have (3.9).

We now prove (3.10). Exchanging z and y in (3.12), we have

∇_{y}∇_{z}∇_{x}fff_{1} = ∇_{y}H · (∇_{z}∇_{x}fff_{0}) + H(p) · (∇_{y}∇_{z}∇_{x}fff_{0}),  ∀x, y, z ∈ T_{p}(M). (3.13)

Subtract (3.13) from (3.12). Then, using the integrability condition (3.2) and the equality (3.3), we have

∇_{z}H(∇_{y}∇_{x}fff_{0}) = ∇_{y}H(∇_{z}∇_{x}fff_{0}),  ∀x, y, z ∈ T_{p}(M).  (3.14)

Consequently, multiplying ^{t}H(p) from the left, we get

L(z) · (∇_{y}∇_{x}fff_{0}) = L(y) · (∇_{z}∇_{x}fff_{0}),  ∀x, y, z ∈ T_{p}(M).  (3.15)

Since L(z) is a skew-symmetric matrix, we have

⟨L(z) · (∇_{y}∇_{x}fff_{0}), ∇_{u}fff_{0}⟩ = −⟨(∇_{y}∇_{x}fff_{0}), L(z) · (∇_{u}fff_{0})⟩ = 0.

Therefore, to prove (3.10), we have to show

⟨L(z) · (∇_{y}∇_{x}fff_{0}), ∇_{v}∇_{w}fff_{0}⟩ = 0,  ∀x, y, z, v, w ∈ T_{p}(M).  (3.16)

Define an element X ∈ ⊗^{5}T_{p}^{∗}(M) by

X(z, y, x, v, w) = ⟨L(z) · (∇_{y}∇_{x}fff_{0}), ∇_{v}∇_{w}fff_{0}⟩,  x, y, z, v, w ∈ T_{p}(M).  (3.17)

In the following, we will show X(z, y, x, v, w) = 0 for x, y, z, v, w ∈ T_{p}(M). By the integrability condition (3.1) and by (3.15), we easily know that X(z, y, x, v, w) is symmetric with

8 YOSHIO AGAOKA AND EIJI KANEDA

respect to the pairs {x, y}, {v, w} and {z, y}. Further, since L(z) is a skew-symmetric endomorphism of R R R^{m} (see (3.8)), it follows that

X(z, y, x, v, w) = −X(z, v, w, y, x).  (3.18)

Therefore, X(z, y, x, v, w) is anti-symmetric with respect to the pair {x, w}, because

X(z, y, x, v, w) = −X(z, v, w, y, x) = −X(v, z, w, y, x)
= X(v, y, x, z, w) = X(y, v, x, z, w)
= −X(y, z, w, v, x) = −X(z, y, w, v, x).

Consequently, we get

X(z, y, x, v, w) = −X(z, y, w, v, x) = −X(z, w, y, x, v)
= X(z, w, v, x, y) = X(z, v, w, y, x).

This together with (3.18) proves X(z, y, x, v, w) = 0. Thus we get (3.10).

By the above argument, we know that L(x) = ^{t}H(p)(∇_{x}H) = 0 for any x ∈ T_{p}(M). This implies that H is a locally constant function and hence H is identically equal to an element H_{0} ∈ O(m) on M, because M is connected. Consequently, the difference ccc = fff_{1} − H_{0} · fff_{0} satisfies

∇_{x}ccc = ∇_{x}(fff_{1} − H_{0} · fff_{0}) = ∇_{x}fff_{1} − H_{0} · (∇_{x}fff_{0}) = 0, ∀x ∈ T_{p}(M).

Therefore, ccc is also identically equal to a constant vector ccc_{0} ∈ R R R^{m}, completing the proof of the theorem. □

Remark 3. The argument in the proof of the equality X = 0 is essentially the same that is developed in the proof of the uniqueness of the metric connection of the normal bundle associated with an isometric imbedding (see the proof of Theorem 1 of [19]); It is almost the same that is used to calculate the third prolongation of the symbol of the operator L (see Proposition 2.2 of [16]). Here we remark that X = 0 can be proved without assuming the existence of (isometric) immersions.

We are now in a position to prove Theorem 5. Proof of Theorem 5. We show that the map fff_{i} (i = 0, 1) is 2-generic and for each p ∈ M there is an element H(p) ∈ O(m) satisfying the equalities (3.3) and (3.4).

Let i = 0 or 1. Let fff_{i}_{∗}T_{p}(M) (resp. N N N_{i}) be the tangent vector space (resp. normal vector space) of fff_{i}(M) at fff_{i}(p) ∈ R R R^{m}. Then, we have dimfff_{i}_{∗}T_{p}(M) = n and dimN N N_{i} = m − n. We regard fff_{i}_{∗}T_{p}(M) and N N N_{i} as euclidean vector spaces endowed with the inner products induced from the inner product ⟨ , ⟩ of R R R^{m}. By a natural parallel displacement from fff_{i}(p)

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

to the origin o ∈ R R R^{m}, we regard fff_{i}_{∗}T_{p}(M) and N N N_{i} as linear subspaces of R R R^{m}. Since fff_{i} is an isometric immersion, fff_{i}_{∗}T_{p}(M) is spanned by the vectors ∇_{x}fff_{i} (x ∈ T_{p}(M)) and

⟨∇_{x}fff_{i}, ∇_{y}fff_{i}⟩ = g_{p}(x, y),  x, y ∈ T_{p}(M).  (3.19)

The second order derivative ∇∇fff_{i}, which is so called the second fundamental form of fff_{i}, satisfies ∇∇fff_{i} ∈ S^{2}T_{p}^{∗}(M) ⊗ N N N_{i} and ∇∇fff_{i} ∈ G_{p}(N N N_{i}) (see [23], [16]). Since G_{p}(N N N_{i}) is EOS, the vectors ∇_{x}∇_{y}fff_{i} (x, y ∈ T_{p}(M)) span N N N_{i}, implying that fff_{i} is 2-generic (see Proposition 4 (1)). Take an isometric linear isomorphism φ_{2} of N N N_{0} onto N N N_{1}. Since φ̂_{2} ∇∇fff_{0} ∈ G_{p}(N N N_{1}) and since G_{p}(N N N_{1}) is EOS (see Proposition 4 (2b)), there is an element h_{1} ∈ O(N N N_{1}) such that h_{1}(φ̂_{2}∇∇fff_{0}) = ∇∇fff_{1}. On the other hand, in view of (3.19) we also know that there is an isometric linear isomorphism φ_{1} of fff_{0}_{∗}T_{p}(M) onto fff_{1}_{∗}T_{p}(M) satisfying φ_{1}(∇_{x}fff_{0}) = ∇_{x}fff_{1} (x ∈ T_{p}(M)). Define a linear endomorphism H(p) of R R R^{m} satisfying H(p)|_{fff_{0}_{∗}T_{p}(M)} = φ_{1} and H(p)|_{N N N_{0}} = h_{1} · φ_{2}. Then, it is easily seen that H(p) ∈ O(m) and the equalities (3.3) and (3.4) are satisfied.

Therefore, by Theorem 6 we know that fff_{1} can be written as fff_{1} = afff_{0}, where a denotes the euclidean transformation of R R R^{m} defined by R R R^{m} ∋ x x x ⟼ H_{0} · x x x + ccc_{0} ∈ R R R^{m}. Thus, we obtain the theorem. □

4. The Cayley projective plane P^{2}(Cay Cay Cay)

Let M = G/K be a compact Riemannian symmetric space. Let 𝔤 (resp. 𝔨) be the Lie algebra of G (resp. K). We denote by 𝔤 = 𝔨 + 𝔪 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.

Hereafter, we consider the case of the Cayley projective plane P^{2}(Cay Cay Cay). As is wellknown, P^{2}(Cay Cay Cay) can be represented by P^{2}(Cay Cay Cay) = G/K, where G = F_{4} and K = Spin(9). Take a maximal abelian subspace 𝔞 of 𝔪 and fix it in the following discussions. We note that since rank(P^{2}(Cay Cay Cay)) = 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} .

We call λ a restricted root if 𝔪(λ) ≠ 0. Let Σ be the set of all non-zero restricted roots. In the case of P^{2}(Cay Cay Cay), there is a restricted root μ such that Σ = {±μ, ±2μ}. We take and fix such a restricted root μ. Then we have 𝔪(0) = 𝔞 = R R Rμ and

𝔨 = 𝔨(0) + 𝔨(μ) + 𝔨(2μ) (orthogonal direct sum),
𝔪 = 𝔪(0) + 𝔪(μ) + 𝔪(2μ) (orthogonal direct sum).

(For details, see [6], [7].) For simplicity, 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

Proposition 7 ([7]). (1) Let i, j = 0, 1, 2. Then:

[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}.  (4.1)

(2) dim 𝔪 = 16, dim 𝔨_{1} = dim 𝔪_{1} = 8, dim 𝔨_{2} = dim 𝔪_{2} = 7.

In what follows, we recall the results obtained in [7], which will be needed in the proof of Theorem 2. Let V be a subspace of 𝔪. V is called pseudo-abelian if it satisfies [V, V ] ⊂ 𝔨_{0} (or equivalently [[V, V ], 𝔞] = 0). (Precisely, see [6].) As is easily seen, 𝔪_{2} is a pseudoabelian subspace of 𝔪, because [𝔪_{2}, 𝔪_{2}] ⊂ 𝔨_{0} (see (4.1)).

On the contrary, we have

Proposition 8. Let G/K = P^{2}(Cay Cay Cay). Then, any pseudo-abelian subspace V of 𝔪 with dim V > 2 must be contained in 𝔪_{2}.

For the proof, see Lemma 6 of [7]. The following proposition summarizes the results of [7] (see Proposition 7, Proposition 10 and Lemma 17 of [7]).

Proposition 9. (1) Let Y_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Assume that Y_{0} ≠ 0, Y_{1} ≠ 0. Then, there are elements k_{0}, k_{1} ∈ K satisfying

Ad(k_{0})μ ∈ R R RY_{0}, Ad(k_{0})𝔪_{2} = {Y′_{0} ∈ 𝔞 + 𝔪_{2} | (Y′_{0}, Y_{0}) = 0} ,  (4.2)
Ad(k_{1})μ ∈ R R RY_{1}, Ad(k_{1})𝔪_{2} = {Y′_{1} ∈ 𝔪_{1} | (Y′_{1}, Y_{1}) = 0} .  (4.3)

(2) Let Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2}, Y_{1}, Y′_{1} ∈ 𝔪_{1} and X_{1} ∈ 𝔨_{1}. Then:

[Y_{0}, [Y_{0}, Y′_{0}]] = \begin{cases}−4(μ, μ)(Y_{0}, Y_{0})Y′_{0}, & if (Y_{0}, Y′_{0}) = 0, \\ 0, & if Y′_{0} ∈ R R RY_{0}, \end{cases}  (4.4)
[Y_{0}, [Y_{0}, Y_{1}]] = −(μ, μ)(Y_{0}, Y_{0})Y_{1},  (4.5)
[Y_{1}, [Y_{1}, Y_{0}]] = −(μ, μ)(Y_{1}, Y_{1})Y_{0},  (4.6)

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

[Y_{1}, [Y_{1}, Y′_{1}]] = \begin{cases}−4(μ, μ)(Y_{1}, Y_{1}) Y′_{1}, & if (Y_{1}, Y′_{1}) = 0, \\ 0, & if Y′_{1} ∈ R R RY_{1}, \end{cases}  (4.7)

[X_{1}, [X_{1}, Y_{0}]] = −(μ, μ)(X_{1}, X_{1})Y_{0}.  (4.8)

5. Solutions of the Gauss equation

In this and the next sections, we prove

Theorem 10. The projective plane P^{2}(Cay Cay Cay) is formally rigid in codimension 10 (= 26 − dim P^{2}(Cay Cay Cay)).

If this theorem is established, then Theorem 2 immediately follows from Theorem 5. On account of homogeneity of P^{2}(Cay Cay Cay), in order to show Theorem 10 we have only to

prove that the Gaussian variety G_{o}(N N N) is EOS at the origin o for any euclidean vector space N N N with dimN N N = 10.

In what follows we assume that M = P^{2}(Cay Cay Cay) and that N N N is a euclidean vector space with dimN N N = 10. We will prove the following theorem:

Theorem 11. Let Ψ ∈ G_{o}(N N N). Then:

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

(2) Ψ(Y_{1}, Y_{2}) + \frac{1}{(μ, μ)^{2}} Ψ(μ, [[μ, Y_{1}], Y_{2}]) = 0, ∀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}.

Before proceeding to the proof of Theorem 11 we make a somewhat lengthy preparation. Let N N N be a euclidean vector space and 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 say that an element Y ∈ 𝔪 is singular (resp. non-singular) with respect to Ψ if Ψ_{Y} (𝔪) ≠ N N N (resp. Ψ_{Y} (𝔪) = N N N). Apparently, 0 (∈ 𝔪) is a singular element for any Ψ ∈ S^{2}𝔪^{∗} ⊗ N N N.

12 YOSHIO AGAOKA AND EIJI KANEDA

Proposition 12. Let Ψ ∈ G_{o}(N N N). Let Y ∈ 𝔪 (Y ≠ 0) and let k be an element of K satisfying Ad(k)μ ∈ R R RY. Then:

(1) Ker(Ψ_{Y}) ⊂ Ad(k)𝔪_{2}. Consequently, dim Ker(Ψ_{Y}) ≤ 7.

(2) Assume that Y is non-singular with respect to Ψ. Then, it holds that dim Ker(Ψ_{Y}) = 6 and Ker(Ψ_{Y}) ⊊ Ad(k)𝔪_{2}.

(3) Assume that Y is singular with respect to Ψ. Then, it holds that Ker(Ψ_{Y}) = Ad(k)𝔪_{2}, dim Ker(Ψ_{Y}) = 7 and dim Ψ_{Y} (𝔪) = 9.

Proof. First, note that dim Ker(Ψ_{Y}) ≥ dim 𝔪−dimN N N = 6. Consequently, it is easy to see that Y is singular (resp. non-singular) with respect to Ψ if and only if dim Ker(Ψ_{Y}) > 6 (resp. dim Ker(Ψ_{Y}) = 6).

Multiplying Y by a non-zero scalar if necessary, we may assume that Y = Ad(k)μ. We have proved in § 1 of [3]

R_{o}(Ker(Ψ_{Y}), Ker(Ψ_{Y}))Y = 0.

In our terminology we have

[[Ker(Ψ_{Y}), Ker(Ψ_{Y})], Y ] = 0.

Applying Ad(k^{−1}) to the both sides of the above equality, we have

[[Ad(k^{−1})Ker(Ψ_{Y}), Ad(k^{−1})Ker(Ψ_{Y})], μ] = 0.

Since 𝔞 = R R Rμ, it follows that Ad(k^{−1})Ker(Ψ_{Y}) is a pseudo-abelian subspace of 𝔪. By Proposition 8 and by the fact dim Ker(Ψ_{Y}) ≥ 6, we have Ad(k^{−1})Ker(Ψ_{Y}) ⊂ 𝔪_{2} and hence Ker(Ψ_{Y}) ⊂ Ad(k)𝔪_{2}, proving (1).

Assume that Y is non-singular with respect to Ψ. Then, as we have stated above, we have dim Ker(Ψ_{Y}) = 6. Since dim 𝔪_{2} = 7 (see Proposition 7 (2)), it follows that Ker(Ψ_{Y}) ⊊ Ad(k)𝔪_{2}, proving (2).

Finally, we assume Y is singular with respect to Ψ. Then, we have dim Ker(Ψ_{Y}) > 6. Since Ker(Ψ_{Y}) ⊂ Ad(k)𝔪_{2} and since dim 𝔪_{2} = 7, we have Ker(Ψ_{Y}) = Ad(k)𝔪_{2} and hence dim Ker(Ψ_{Y}) = 7. This proves (3). □

Corollary 13. Let Ψ ∈ G_{o}(N N N). Let Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0) and Y_{1} ∈ 𝔪_{1} (Y_{1} ≠ 0). Then:

(1) Ker(Ψ_{Y_{0}}) ⊂ {Y′_{0} ∈ 𝔞 + 𝔪_{2} | (Y′_{0}, Y_{0}) = 0}. In particular, if Y_{0} is singular with respect to Ψ, then Ker(Ψ_{Y_{0}}) = {Y′_{0} ∈ 𝔞 + 𝔪_{2} | (Y′_{0}, Y_{0}) = 0}.

(2) Ker(Ψ_{Y_{1}}) ⊂ {Y′_{1} ∈ 𝔪_{1} | (Y′_{1}, Y_{1}) = 0}. In particular, if Y_{1} is singular with respect to Ψ, then Ker(Ψ_{Y_{1}}) = {Y′_{1} ∈ 𝔪_{1} | (Y′_{1}, Y_{1}) = 0}.

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

Proof. Let Y_{0} ∈ 𝔞 + 𝔪_{2} (Y_{0} ≠ 0). By Proposition 9 (1), we know that there is an element k_{0} ∈ K satisfying (4.2). Applying Proposition 12 to Y_{0}, we easily get Ker(Ψ_{Y_{0}}) ⊂ {Y′_{0} ∈ 𝔞 + 𝔪_{2} | (Y′_{0}, Y_{0}) = 0}. Assume that Y_{0} is singular with respect to Ψ. Then, by the equality Ker(Ψ_{Y_{0}}) = Ad(k_{0})𝔪_{2}, we get (1).

The assertion (2) is similarly dealt with. □

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

Proposition 14. Let Ψ ∈ G_{o}(N N N). Let Y ∈ 𝔪 (Y ≠ 0) and let k ∈ K satisfy Ad(k)μ ∈ R R RY. Assume that Y is non-singular with respect to Ψ. Then:

(1) Ker(Ψ_{Y}) is a singular subspace with respect to Ψ.

(2) There is an element Y′ ∈ Ad(k)𝔪_{2} satisfying Ψ(Y, Y′) ≠ 0 and

N N N = R R RΨ(Y, Y′) + Ψ_{Y′′}(𝔪) (orthogonal direct sum),  (5.1)

where Y′′ is an arbitrary non-zero element of Ker(Ψ_{Y}).

Proof. Since Y is non-singular with respect to Ψ, we have Ker(Ψ_{Y}) ⊊ Ad(k)𝔪_{2} (see Proposition 12). Take a non-zero element Y′ ∈ Ad(k)𝔪_{2} such that (Y′, Ker(Ψ_{Y})) = 0. Then, since Y′ ∉ Ker(Ψ_{Y}), we have Ψ(Y, Y′) ≠ 0.

Let Y′′ ∈ Ker(Ψ_{Y}) (Y′′ ≠ 0). Then, by the Gauss equation (2.1) we have

([[Y′, Y′′], Y ], W ) = ⟨Ψ(Y′, Y), Ψ(Y′′, W)⟩ − ⟨Ψ(Y′, W), Ψ(Y′′, Y)⟩,  (5.2)

where W is an arbitrary element of 𝔪. Note that the left hand side of (5.2) vanishes, because

[[Y′, Y′′], Y ] ∈ [[Ad(k)𝔪_{2}, Ad(k)𝔪_{2}], Ad(k)μ] = Ad(k)[[𝔪_{2}, 𝔪_{2}], μ] = 0.

We also note that Ψ(Y′′, Y) = 0, because Y′′ ∈ Ker(Ψ_{Y}). Consequently, we have ⟨Ψ(Y′, Y), Ψ(Y′′, W)⟩ = 0. This implies that each element of Ψ_{Y′′}(𝔪) is orthogonal to Ψ(Y′, Y). Therefore, Ψ_{Y′′}(𝔪) ≠ N N N, implying that Y′′ is singular with respect to Ψ. Hence, by Proposition 12 (3) we have dim Ψ_{Y′′}(𝔪) = 9, which proves (5.1). □

The following lemma assures that for each Ψ ∈ G_{o}(N N N) there are many high dimensional singular subspaces with respect to Ψ.

Lemma 15. Let Ψ ∈ G_{o}(N N N). Then, there are singular subspaces U and V with respect to Ψ satisfying U ⊂ 𝔞 + 𝔪_{2}, V ⊂ 𝔪_{1}, dim U ≥ 6 and dim V ≥ 6.

14 YOSHIO AGAOKA AND EIJI KANEDA

Proof. If 𝔞 + 𝔪_{2} contains no non-singular element with respect to Ψ, then we can take U = 𝔞+𝔪_{2}. (Note that dim(𝔞+𝔪_{2}) = 8.) On the contrary, if 𝔞+𝔪_{2} contains a non-singular element Y_{0}, then we set U = Ker(Ψ_{Y_{0}}). Then, we know that U ⊂ 𝔞 + 𝔪_{2}, dim U = 6 (see Proposition 12 (2) and Corollary 13 (1)) and that U is a singular subspace with respect to Ψ (see Proposition 14 (2)). Similarly, we can select a singular subspace V ⊂ 𝔪_{1} with dim V ≥ 6. □

Proposition 16. Let Ψ ∈ G_{o}(N N N). Let U and V be arbitrary singular subspaces with respect to Ψ satisfying U ⊂ 𝔞 + 𝔪_{2}, V ⊂ 𝔪_{1} and dim U ≥ 6, dim V ≥ 6. Then there are two vectors A and B ∈ N N N satisfying:

(1) ⟨A, A⟩ = ⟨B, B⟩ = 4(μ, μ);

(2) Ψ(ξ, Y_{0}) = (ξ, Y_{0})A, ∀ξ ∈ U, ∀Y_{0} ∈ 𝔞 + 𝔪_{2};
(3) Ψ(η, Y_{1}) = (η, Y_{1})B, ∀η ∈ V, ∀Y_{1} ∈ 𝔪_{1};
(4) ⟨A, Ψ_{Y_{0}}(𝔪_{1})⟩ = ⟨B, Ψ_{Y_{0}}(𝔪_{1})⟩ = 0, ∀Y_{0} ∈ 𝔞 + 𝔪_{2}.

Proof. Let ξ ∈ U (ξ ≠ 0). Since ξ is singular with respect to Ψ, Ker(Ψ_{ξ}) coincides with the orthogonal complement of R R Rξ in 𝔞 + 𝔪_{2} (see Corollary 13 (1)). Hence, the equality Ψ(ξ, Y_{0}) = 0 holds for each Y_{0} ∈ 𝔞 + 𝔪_{2} satisfying (ξ, Y_{0}) = 0. In particular, we have

Ψ(ξ, ξ′) = 0, ∀ξ, ξ′ ∈ U with (ξ, ξ′) = 0.

Then, applying the same argument as in the proof of Proposition 9 of [7], we can prove that there is a vector A ∈ N N N satisfying

Ψ(ξ, ξ′) = (ξ, ξ′)A,  ∀ξ, ξ′ ∈ U.  (5.3)

Let Y_{0} ∈ 𝔞 + 𝔪_{2} satisfy (Y_{0}, U) = 0. Then, since (ξ, Y_{0}) = 0, we have Ψ(ξ, Y_{0}) = 0 and (ξ, Y_{0})A = 0. This together with (5.3) proves (2). The assertion (3) can be proved in the same way.

We now prove (1). Let ξ, ξ′ ∈ U satisfy (ξ, ξ′) = 0 and (ξ, ξ) = (ξ′, ξ′) = 1. Put X = Z = ξ and Y = W = ξ′ into the Gauss equation (2.1). Then, we have

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

Since [[ξ, ξ′], ξ] = 4(μ, μ)ξ′ (see (4.4)), Ψ(ξ, ξ) = Ψ(ξ′, ξ′) = A and Ψ(ξ, ξ′) = 0, we have ⟨A, A⟩ = 4(μ, μ). Similarly, by (4.7) we can prove ⟨B, B⟩ = 4(μ, μ), proving (1).

Finally, we prove (4). Let Y_{1} ∈ 𝔪_{1} and Y_{0} ∈ 𝔞 + 𝔪_{2}. Take an element ξ ∈ U satisfying (ξ, Y_{0}) = 0 and (ξ, ξ) = 1. Such ξ can exist, because dim U ≥ 6. Put X = Z = ξ, Y = Y_{0} and W = Y_{1} into the Gauss equation (2.1). Then we have

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

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

Since (ξ, Y_{0}) = 0, we have Ψ(ξ, Y_{0}) = 0 and [[ξ, Y_{0}], ξ] = 4(μ, μ)Y_{0} (see (4.4)). Moreover, since Ψ(ξ, ξ) = A and (Y_{0}, Y_{1}) = 0, we have

⟨A, Ψ_{Y_{0}}(Y_{1})⟩ = ⟨Ψ(ξ, ξ), Ψ(Y_{0}, Y_{1})⟩
= ⟨Ψ(ξ, Y_{1}), Ψ(Y_{0}, ξ)⟩ + 4(μ, μ)(Y_{0}, Y_{1})
= 0.

Since Y_{1} is an arbitrary element of 𝔪_{1}, we have ⟨A, Ψ_{Y_{0}}(𝔪_{1})⟩ = 0. In a similar way, the equality ⟨B, Ψ_{Y_{0}}(𝔪_{1})⟩ = 0 can be proved. □

Remark 4. As seen in the proof of Lemma 15, singular subspaces U and V may not be uniquely determined. However, it is noted that the vectors A and B in Proposition 16 do not depend on the choice of U and V. In fact, let U′ and V′ be different singular subspaces with respect to Ψ satisfying U′ ⊂ 𝔞 + 𝔪_{2} and V′ ⊂ 𝔪_{1} with dim U′ ≥ 6, dim V′ ≥ 6. Let A′ and B′ be vectors of N N N satisfying (1) ∼ (4) of Proposition 16. Then, since dim(𝔞 + 𝔪_{2}) = dim 𝔪_{1} = 8, we have U ∩ U′ ≠ 0, V ∩ V′ ≠ 0. Take ξ ∈ U ∩ U′ and η ∈ V ∩ V′ such that (ξ, ξ) = (η, η) = 1. Then we have A = Ψ(ξ, ξ) = A′ and B = Ψ(η, η) = B′, showing our assertion.

In the following discussions, we fix an element Ψ ∈ G_{o}(N N N), singular subspaces U, V and vectors A, B stated in Proposition 16 and prove several lemmas which are indispensable to the proof of Theorem 11.

Lemma 17. Let ξ ∈ U, η ∈ V, Y_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Set C = ⟨A, B⟩ − (μ, μ). Then C > 0 and:

(1) ⟨Ψ_{Y_{0}}(η), Ψ_{Y_{0}}(Y_{1})⟩ = {⟨Ψ(Y_{0}, Y_{0}), B⟩ − (μ, μ)(Y_{0}, Y_{0})} (η, Y_{1});

(2) ⟨Ψ_{ξ}(η), Ψ_{ξ}(Y_{1})⟩ = C(ξ, ξ)(η, Y_{1}).

Proof. Putting X = Z = Y_{0}, Y = Y_{1} and W = η into (2.1), we have

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

Since [[Y_{0}, Y_{1}], Y_{0}] = (μ, μ)(Y_{0}, Y_{0})Y_{1} (see (4.5)) and Ψ(Y_{1}, η) = (Y_{1}, η)B, we easily get (1). Putting Y_{0} = ξ ∈ U into (1), we easily have (2). If we set Y_{1} = η ∈ V in (2), we have ⟨Ψ_{ξ}(η), Ψ_{ξ}(η)⟩ = C(ξ, ξ)(η, η). Since Ker(Ψ_{ξ}) ∩ 𝔪_{1} = 0 (see Corollary 13 (1)), we have Ψ_{ξ}(η) ≠ 0 if η ≠ 0. Consequently, we have C > 0. □

Let Y_{0} ∈ 𝔞 + 𝔪_{2}. Let ξ^{0} be a non-zero element of U satisfying (ξ^{0}, Y_{0}) = 0. (Such ξ^{0} exists, because dim U ≥ 6.) We define a linear mapping Θ_{Y_{0},ξ^{0}} : V ⟶ N N N by

Θ_{Y_{0},ξ^{0}}(η) = Ψ_{Y_{0}}(η) + \frac{1}{C(ξ^{0}, ξ^{0})}Ψ_{ξ^{0}}([[ξ^{0}, η], Y_{0}]), η ∈ V.

Then we have

Lemma 18. ⟨A, Θ_{Y_{0},ξ^{0}}(V)⟩ = ⟨Ψ_{ξ^{0}}(V), Θ_{Y_{0},ξ^{0}}(V)⟩ = 0.

Proof. We first note that [[ξ^{0}, η], Y_{0}] ∈ 𝔪_{1} for η ∈ V and note that Θ_{Y_{0},ξ^{0}}(V) ⊂ Ψ_{Y_{0}}(𝔪_{1})+ Ψ_{ξ^{0}}(𝔪_{1}). By Proposition 16 (4), we have ⟨A, Ψ_{Y_{0}}(𝔪_{1})⟩ = ⟨A, Ψ_{ξ^{0}}(𝔪_{1})⟩ = 0 and hence ⟨A, Θ_{Y_{0},ξ^{0}}(V)⟩ = 0.

Let η, η′ ∈ V. Then by putting X = Y_{0}, Y = η′, Z = η and W = ξ^{0} into the Gauss equation (2.1), we have

([[Y_{0}, η′], η], ξ^{0}) = ⟨Ψ(Y_{0}, η), Ψ(η′, ξ^{0})⟩ − ⟨Ψ(Y_{0}, ξ^{0}), Ψ(η′, η)⟩
= ⟨Ψ_{Y_{0}}(η), Ψ_{ξ^{0}}(η′)⟩ − ⟨A, B⟩(Y_{0}, ξ^{0})(η′, η).

Since (Y_{0}, ξ^{0}) = 0, we have

⟨Ψ_{Y_{0}}(η), Ψ_{ξ^{0}}(η′)⟩ = ([[Y_{0}, η′], η], ξ^{0}).  (5.4)

On the other hand, we have ⟨Ψ_{ξ^{0}}([[ξ^{0}, η], Y_{0}]), Ψ_{ξ^{0}}(η′)⟩ = C(ξ^{0}, ξ^{0})([[ξ^{0}, η], Y_{0}], η′) (see Lemma 17 (2)). Therefore,

⟨Θ_{Y_{0},ξ^{0}}(η), Ψ_{ξ^{0}}(η′)⟩ = ⟨Ψ_{Y_{0}}(η) + \frac{1}{C(ξ^{0}, ξ^{0})}Ψ_{ξ^{0}}([[ξ^{0}, η], Y_{0}]), Ψ_{ξ^{0}}(η′)⟩
= ([[Y_{0}, η′], η], ξ^{0}) + ([[ξ^{0}, η], Y_{0}], η′)
= −([Y_{0}, η′], [ξ^{0}, η]) + ([ξ^{0}, η], [Y_{0}, η′])
= 0.

This completes the proof. □

We can further show

Lemma 19. Let η ∈ V. Assume that [[ξ^{0}, η], Y_{0}] ∈ V. Then:

|Θ_{Y_{0},ξ^{0}}(η)|^{2} = [⟨Ψ(Y_{0}, Y_{0}), B⟩ − (μ, μ)(Y_{0}, Y_{0}) {1 + (μ, μ)/C}] (η, η).  (5.5)

Proof. Set η′ = [[ξ^{0}, η], Y_{0}]. By Lemma 18, Lemma 17 and the equality (5.4) we have

⟨Θ_{Y_{0},ξ^{0}}(η), Θ_{Y_{0},ξ^{0}}(η)⟩ = ⟨Ψ_{Y_{0}}(η) + \frac{1}{C(ξ^{0}, ξ^{0})}Ψ_{ξ^{0}}(η′), Θ_{Y_{0},ξ^{0}}(η)⟩

= ⟨Ψ_{Y_{0}}(η), Θ_{Y_{0},ξ^{0}}(η)⟩

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

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

+ \frac{1}{C(ξ^{0}, ξ^{0})}([[Y_{0}, η′], η], ξ^{0}).

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

Since [ξ^{0}, η] ∈ 𝔨_{1}, by (4.8) and (4.5) we have

([[Y_{0}, η′], η], ξ^{0}) = −([Y_{0}, η′], [ξ^{0}, η])
= (Y_{0}, [[ξ^{0}, η], η′])
= (Y_{0}, [[ξ^{0}, η], [[ξ^{0}, η], Y_{0}]])
= −(μ, μ)([ξ^{0}, η], [ξ^{0}, η])(Y_{0}, Y_{0})
= (μ, μ)([ξ^{0}, [ξ^{0}, η]], η)(Y_{0}, Y_{0})
= −(μ, μ)^{2}(ξ^{0}, ξ^{0})(η, η)(Y_{0}, Y_{0}).

Therefore, we obtain (5.5). □

Lemma 20. Let Y_{0} ∈ 𝔞 + 𝔪_{2}. Then:

(1) ⟨Ψ(Y_{0}, Y_{0}), B⟩ = (μ, μ)(Y_{0}, Y_{0}) {1 + (μ, μ)/C}.

(2) Let ξ^{0} be a non-zero element of U satisfying (Y_{0}, ξ^{0}) = 0. Then, Θ_{Y_{0},ξ^{0}}(η) = 0, i.e., the equality

Ψ(Y_{0}, η) + \frac{1}{C(ξ^{0}, ξ^{0})}Ψ(ξ^{0}, [[ξ^{0}, η], Y_{0}]) = 0  (5.6)

holds for each η ∈ V satisfying [[ξ^{0}, η], Y_{0}] ∈ V.

Proof. We first show that there is a non-zero element η^{0} ∈ V satisfying Θ_{Y_{0},ξ^{0}}(η^{0}) = 0 and [[ξ^{0}, η^{0}], Y_{0}] ∈ V. Let D D D be the orthogonal complement of R R RA + Ψ_{ξ^{0}}(V) in N N N and let V′ be the orthogonal complement of V in 𝔪_{1}. By Lemma 18, we easily have Θ_{Y_{0},ξ^{0}}(V) ⊂ D D D. Therefore, to obtain η^{0} satisfying the above condition, it suffices to find a non-zero solution η = η^{0} ∈ V of the system of linear equations

⟨Θ_{Y_{0},ξ^{0}}(η),D D D⟩ = ([[ξ^{0}, η], Y_{0}], V′) = 0.  (5.7)

Since Ker(Ψ_{ξ^{0}}) ∩ 𝔪_{1} = 0 (see Corollary13 (1)) and ⟨A, Ψ_{ξ^{0}}(𝔪_{1})⟩ = 0 (see Proposition 16 (4)), we have dim(R R RA + Ψ_{ξ^{0}}(V)) = 1 + dim V ≥ 7. (Recall that we are assuming V ⊂ 𝔪_{1} and dim V ≥ 6.) Hence, we have dimD D D ≤ dimN N N − 7 = 3. Moreover, we have dim V′ = 8 − dim V ≤ 2. Consequently, (5.7) is the system of linear equations on V (dim V ≥ 6) composed of at most 5 linearly independent equations. Therefore, we can find a non-zero solution η^{0} ∈ V of (5.7). Putting η = η^{0} into (5.5), we obtain the equality (1). Further, putting (1) into (5.5), we have Θ_{Y_{0},ξ^{0}}(η) = 0 for any η ∈ V satisfying [[ξ^{0}, η], Y_{0}] ∈ V. □

Lemma 21. The vectors A and B are linearly independent and ⟨A, B⟩ = 2(μ, μ), C = (μ, μ).

18 YOSHIO AGAOKA AND EIJI KANEDA

Proof. Let ξ ∈ U with (ξ, ξ) = 1. Since Ψ(ξ, ξ) = A (see (5.3)), by putting Y_{0} = ξ into the equality in Lemma 20 (1), we easily have ⟨A, B⟩ = (μ, μ) {1 + (μ, μ)/C}. Since C = ⟨A, B⟩ − (μ, μ), it immediately follows that C^{2} = (μ, μ)^{2}. Since C > 0, we get C = (μ, μ) and hence ⟨A, B⟩ = 2(μ, μ). This together with Proposition 16 (1) proves that A and B are linearly independent. □

These being prepared, we show Theorem 11.

Proof of Theorem 11. First we show that μ is singular with respect to any element Ψ ∈ G_{o}(N N N). Suppose that there is an element Ψ_{0} ∈ G_{o}(N N N) such that μ is non-singular with respect to Ψ_{0}. Then, Ker((Ψ_{0})_{μ}) is a singular subspace with respect to Ψ_{0} and it satisfies dim Ker((Ψ_{0})_{μ}) = 6 and Ker((Ψ_{0})_{μ}) ⊂ 𝔪_{2} (see Proposition 12 and Proposition 14).

Now, set Ψ = Ψ_{0} and U = Ker((Ψ_{0})_{μ}) in Proposition 16. Let A, B be the vectors of N N N satisfying (1)–(4) of Proposition 16. Let ξ ∈ U = Ker((Ψ_{0})_{μ}) with ξ ≠ 0. First, we show B ∈ (Ψ_{0})_{ξ}(𝔪). In fact, there is a non-zero element Y_{2}^{0} ∈ 𝔪_{2} satisfying Ψ_{0}(μ, Y_{2}^{0}) ≠ 0 and N N N = R R RΨ_{0}(μ, Y_{2}^{0}) + (Ψ_{0})_{ξ}(𝔪) (orthogonal direct sum) (see Proposition 14). By Lemma 20 (1) and by the relation

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

we easily have ⟨Ψ_{0}(μ, Y_{2}^{0}), B⟩ = 0, which proves B ∈ (Ψ_{0})_{ξ}(𝔪). Since (Ψ_{0})_{ξ}(𝔪) = R R RA + (Ψ_{0})_{ξ}(𝔪_{1}) (orthogonal direct sum) and ⟨B, (Ψ_{0})_{ξ}(𝔪_{1})⟩ = 0 (see Proposition 16 (2), (4)), we have B ∈ R R RA. This contradicts Lemma 21. Accordingly, we can conclude that μ is singular with respect to any element Ψ ∈ G_{o}(N N N).

Now we show that any element of 𝔪 is singular with respect to any Ψ ∈ G_{o}(N N N). Let Y be a non-zero element of 𝔪. Take an element k ∈ K such that Ad(k)μ ∈ R R RY and define Ψ′ ∈ S^{2}𝔪^{∗} ⊗ N N N by

Ψ′(Y′, Y′′) = Ψ(Ad(k)Y′, Ad(k)Y′′), Y′, Y′′ ∈ 𝔪.

Then, it is easily seen that Ψ′ ∈ G_{o}(N N N). Applying the arguments developed above, we know that μ is also singular with respect to Ψ′. Note that Ψ′_{μ}(𝔪) = Ψ_{Ad(k)μ}(Ad(k)𝔪) = Ψ_{Y} (𝔪). Then, since Ψ′_{μ}(𝔪) ≠ N N N, we have Ψ_{Y} (𝔪) ≠ N N N, implying that Y is singular with respect to Ψ.

Accordingly, in Proposition 16 and in the discussion after it, we may allow to put U = 𝔞 + 𝔪_{2} and V = 𝔪_{1}. Therefore, by Proposition 16 and Lemma 21, we get (1) of Theorem 11. Further, putting Y_{0} = Y_{2} ∈ 𝔪_{2}, ξ^{0} = μ and η = Y_{1} into (5.6), we get (2) of Theorem 11. The assertion (3) of Theorem 11 follows from Lemma 17 (2) and Lemma 21. This completes the proof of the theorem. □

6. Proof of Theorem 10

Let {E_{i} (1 ≤ i ≤ 8)} be an orthonormal basis of 𝔪_{1}. (Note that dim 𝔪_{1} = 8.) Let Ψ ∈ G_{o}(N N N) and let A, B be the vectors of N N N stated in Theorem 11. We define vectors {F_{i} (1 ≤ i ≤ 10)\} of N N N by setting F_{i} = Ψ(μ, E_{i})/(μ, μ) (1 ≤ i ≤ 8), F_{9} = (A + B)/2\sqrt{3}|μ| and F_{10} = (A − B)/2|μ|. We now show that {F_{i} (1 ≤ i ≤ 10)} forms an orthonormal basis of N N N. By Theorem 11 (3) we have ⟨F_{i}, F_{j}⟩ = δ_{ij} (1 ≤ i, j ≤ 8), where δ_{ij} denotes Kronecker’s delta. Moreover, since ⟨A, F_{i}⟩ = ⟨B, F_{i}⟩ = 0 (1 ≤ i ≤ 8) (see Theorem 11 (1d)), we have ⟨F_{9}, F_{i}⟩ = ⟨F_{10}, F_{i}⟩ = 0 (1 ≤ i ≤ 8). The equalities ⟨F_{9}, F_{9}⟩ = ⟨F_{10}, F_{10}⟩ = 1 and ⟨F_{9}, F_{10}⟩ = 0 immediately follow from Theorem 11 (1a).

Now let Ψ′ be another element of G_{o}(N N N). Let A′ and B′ be the vectors stated in Theorem 11 for Ψ′. As in the case of Ψ we can also define an orthonormal basis {F′_{i} (1 ≤ i ≤ 10)} of N N N. Then, there is an element h ∈ O(10) satisfying F′_{i} = hF_{i} (1 ≤ i ≤ 10). Here we note that A′ = hA, B′ = hB and Ψ′(μ, E_{i}) = hΨ(μ, E_{i}) (1 ≤ i ≤ 8). Set Φ = Ψ′ − hΨ ∈ S^{2}𝔪^{∗} ⊗ N N N. Then, by Theorem 11 (1) we have

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

By the fact [[μ, 𝔪_{1}], 𝔪_{2}] ⊂ 𝔪_{1} and Theorem 11 (2), we have

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

which proves that Φ(𝔪_{2}, 𝔪_{1}) = 0. Therefore, we have Φ = 0, i.e., Ψ′ = hΨ. This implies that the Gaussain variety G_{o}(N N N) is EOS. This completes the proof of Theorem 10. □

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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) Department of International 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