LOCAL ISOMETRIC IMBEDDINGS OF P^{2}(H H H) AND P^{2}(Cay Cay Cay)

YOSHIO AGAOKA AND EIJI KANEDA

Abstract. We investigate local isometric imbeddings of the quaternion projective plane P^{2}(H H H) and the Cayley projective plane P^{2}(Cay Cay Cay) into the Euclidean spaces. We prove a non-existence theorem of local isometric imbeddings (see Theorem 2), by which we can conclude that the isometric imbeddings given in Kobayashi [7] are the least dimensional isometric imbeddings of P^{2}(H H H) and P^{2}(Cay Cay Cay).

1. Introduction

In this paper we investigate local isometric imbeddings of the quaternion projective plane P^{2}(H H H) and the Cayley projective plane P^{2}(Cay Cay Cay) into the Euclidean spaces.

In [5], we determined the pseudo-nullity p(G/K) for each compact rank one symmetric space G/K. (For the definition of the pseudo-nullity, see [5].) Utilizing p(G/K), we have obtained the following result concerning the non-existence of isometric imbeddings of the complex projective spaces P^{n}(C C C) (n ≥ 2), the quaternion projective spaces P^{n}(H H H) (n ≥ 2) and the Cayley projective plane P^{2}(Cay Cay Cay) (see Theorem 5.6 of [5]).

Theorem 1. Let G/K be one of the complex projective space P^{n}(C C C) (n ≥ 2), the quaternion projective space P^{n}(H H H) (n ≥ 2) and the Cayley projective plane P^{2}(Cay Cay Cay). Define an integer q(G/K) by setting q(G/K) = 2 dim G/K − p(G/K), i.e.,

q(G/K) = \begin{cases}min\{4n − 2, 3n + 1\}, & if G/K = P^{n}(C C C) (n ≥ 2), \\ min\{8n − 3, 7n + 1\}, & if G/K = P^{n}(H H H) (n ≥ 2), \\ 25, & if G/K = P^{2}(Cay Cay Cay). \end{cases}

Then, any open set of G/K cannot be isometrically imbedded into the Euclidean space R R R^{Q} with Q ≤ q(G/K) − 1.

As is well known, P^{n}(C C C) (resp. P^{n}(H H H), resp. P^{2}(Cay Cay Cay)) can be globally isometrically imbedded into R R R^{n^{2}+2n} (resp. R R R^{2n^{2}+3n}, resp. R R R^{26}) (see Kobayashi [7]). By these facts, it follows that if G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay), then G/K can be isometrically imbedded

Date: August 5, 2002.

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

Key words and phrases. Pseudo-nullity, isometric imbedding, projective plane.

2 YOSHIO AGAOKA AND EIJI KANEDA

into R R R^{q(G/K)+1}. Then a natural question arises: Is there any isometric imbedding of G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay) into the Euclidean space R R R^{q(G/K)} ?

In this paper, we will solve this problem. The main result of this paper is the following

Theorem 2. Let G/K be the quaternion projective plane P^{2}(H H H) or the Cayley projective plane P^{2}(Cay Cay Cay). Then any open set of G/K cannot be isometrically imbedded into the Euclidean space R R R^{q(G/K)}. Accordingly, R R R^{q(G/K)+1} is the least dimensional Euclidean space into which G/K can be locally isometrically imbedded.

2. The Gauss equation

In the following G/K implies the quaternion projective plane P^{2}(H H H) = Sp(3)/Sp(2) × Sp(1) or the Cayley projective plane P^{2}(Cay Cay Cay) = F_{4}/Spin(9).

Let 𝔤 (resp. 𝔨) be the Lie algebra of G (resp. K). Let 𝔤 = 𝔨 + 𝔪 be the canonical decomposition of 𝔤 associated with the Riemannian 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 identify 𝔪 with the tangent space T_{o}(G/K) at the origin o = {K} ∈ G/K. We assume that the G-invariant Riemannian metric g of G/K satisfies g(X, Y) = (X, Y ) (X, Y ∈ 𝔪). Then the curvature tensor R at o is given by

R(X, Y)Z = −[[X, Y ], Z],  ∀X, Y, Z ∈ 𝔪.  (2.1)

Suppose that there is a local isometric imbedding of G/K into the Euclidean space R R R^{Q}, i.e., there is an open set U of G/K and an isometric imbedding fff of U into R R R^{Q}. Because of homogeneity, we may assume that U contains the origin o ∈ G/K. Let N N N be the normal space of fff(U) at fff(o) and let ⟨ , ⟩ be the inner product of N N N induced from the canonical inner product of R R R^{Q}. Then N N N is a vector space with dimN N N = Q−dim G/K and the second fundamental form Ψ of fff at o, which is regarded as an N N N-valued symmetric bilinear form on 𝔪, must satisfy the following Gauss equation:

\begin{aligned}−(R(X, Y)Z, W ) & = ⟨Ψ(X, Z), Ψ(Y, W)⟩ − ⟨Ψ(X, W), Ψ(Y, Z)⟩, \\ & ∀X, Y, Z, W ∈ 𝔪. \end{aligned}  (2.2)

On the contrary, we can prove

Theorem 3. Let G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay). If dimN N N ≤ q(G/K) − dim G/K, then the Gauss equation (2.2) does not admit any solution, i.e., there is no N N N-valued symmetric bilinear form Ψ on 𝔪 satisfying (2.2).

Theorem 3 implies that if G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay), then there is no local isometric imbedding of G/K into R R R^{q(G/K)}, proving Theorem 2.

LOCAL ISOMETRIC IMBEDDINGS OF P^{2}(H H H) AND P^{2}(Cay Cay Cay) 3

We now make a preparatory discussion for the proof of Theorem 3.

Take and fix a maximal abelian subspace 𝔞 of 𝔪. Then we have dim 𝔞 = 1, because rank(G/K) = 1. We consider the root space decompositions of 𝔨 and 𝔪 with respect to 𝔞. Let λ ∈ 𝔞. We define subspaces 𝔨(λ) (⊂ 𝔨) and 𝔪(λ) (⊂ 𝔪) by setting

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} .

λ is called a restricted root when 𝔪(λ) ≠ 0. We denote by Σ the set of non-zero restricted roots. In the case where G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay), it is well known that there is a restricted root μ such that Σ = {±μ, ±2μ} and it holds the following root space decompositions:

𝔨 = 𝔨(0) + 𝔨(μ) + 𝔨(2μ) (orthogonal direct sum),  (2.3)
𝔪 = 𝔪(0) + 𝔪(μ) + 𝔪(2μ) (orthogonal direct sum),  (2.4)

where 𝔪(0) = 𝔞 = R R Rμ (see § 5 of [5]). In the following discussions we fix this restricted root μ and decompositions (2.3) and (2.4).

For convenience, for each integer i we set 𝔨_{i} = 𝔨(|i|μ), 𝔪_{i} = 𝔪(|i|μ) (|i| ≤ 2) and 𝔨_{i} = 𝔪_{i} = 0 (|i| > 2). Then we have

Proposition 4. (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}.  (2.5)

(2) Let i = 1, 2. Then it holds dim 𝔨_{i} = dim 𝔪_{i}.

(3) It holds the following table:

G/K dim G/K dim 𝔪_{1} dim 𝔪_{2} q(G/K)
P^{2}(H H H)  8  4  3  13
P^{2}(Cay Cay Cay)  16  8  7  25

Proof. (1) and (2) are clear (see Proposition 2.1 of [5]). (3) is obtained by Table 2 and Table 3 of [5]. □

4 YOSHIO AGAOKA AND EIJI KANEDA

3. Proof of Theorem 3

In this section we prove Theorem 3. Here we suppose that dimN N N = q(G/K)−dim G/K and that there is a solution Ψ of the Gauss equation (2.2).

Let 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.

By Ker(Ψ_{Y})(⊂ 𝔪) we denote the kernel of the linear map Ψ_{Y}. We now show a key proposition, which plays an important role in the following discussion.

Proposition 5. Let Y ∈ 𝔪 (Y ≠ 0) and let k ∈ K satisfy Ad(k)μ ∈ R R RY. Then it holds

Ker(Ψ_{Y}) = Ad(k)𝔪_{2}.  (3.1)

In particular, it holds Ker(Ψ_{μ}) = 𝔪_{2}.

Before proceeding to the proof of Proposition 5, we recall the notion of pseudo-abelian subspaces of 𝔪. Let V be a subspace of 𝔪. Then, V is called pseudo-abelian if it satisfies [V, V ] ⊂ 𝔨_{0} (or equivalently [[V, V ], μ] = 0). By (2.5) we can easily verify that 𝔪_{2} is pseudo-abelian. On the contrary, we have

Lemma 6. Let G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay). Then, any pseudo-abelian subspace V of 𝔪 with dim V > 2 must be contained in 𝔪_{2}.

Proof. Let V be a pseudo-abelian subspace of 𝔪 satisfying V ⊄ 𝔪_{2}. Then by Lemma 5.4 of [5], we obtain dim V ≤ 1 + n(μ), where n(μ) is the local pseudo-nullity associated with μ. (For the definition of the local pseudo-nullity, see § 3 in [5].) Moreover, we have n(μ) = 1 if G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay) (see Table 2 of [5]). Therefore, we get dim V ≤ 2, proving the lemma. □

We now start the proof of Proposition 5. Proof of Proposition 5. We first note that dim Ker(Ψ_{Y}) ≥ dim 𝔪_{2} > 2. In fact, since dimN N N = q(G/K) − dim G/K = dim G/K − dim 𝔪_{2}, we have dim Ker(Ψ_{Y}) ≥ dim G/K − dimN N N = dim 𝔪_{2} > 2 (see Proposition 4 (3)).

In § 1 of [2], by considering the Gauss equation (2.2), we have proved

R(Ker(Ψ_{Y}), Ker(Ψ_{Y}))Y = 0.  (3.2)

Because of (2.1), the equality (3.2) means

[[Ker(Ψ_{Y}), Ker(Ψ_{Y})], Y ] = 0.  (3.3)

Applying Ad(k^{−1}) to the both sides of (3.3), we get

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

LOCAL ISOMETRIC IMBEDDINGS OF P^{2}(H H H) AND P^{2}(Cay Cay Cay) 5

(Note that Ad(k^{−1})Y can be written as Ad(k^{−1})Y = cμ for some c ∈ R R R (c ≠ 0).) Hence, Ad(k^{−1})Ker(Ψ_{Y}) is a pseudo-abelian subspace of 𝔪 with dim Ad(k^{−1})Ker(Ψ_{Y}) ≥ dim 𝔪_{2} > 2. Therefore, we have Ad(k^{−1})Ker(Ψ_{Y}) = 𝔪_{2} (see Lemma 6). This proves (3.1). □

Utilizing Proposition 5, we will characterize solutions Ψ of the Gauss equation (2.2). For this purpose we need more informations about the action of the isotropy group Ad(K).

As is well known, any element of 𝔪 is conjugate to an element of R R Rμ(= 𝔞) under the action of Ad(K). More strongly, under our assumption G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay), we have

Proposition 7. (1) Let Y_{0} ∈ 𝔞 + 𝔪_{2} satisfy Y_{0} ≠ 0. Then there is an element k_{0} ∈ K satisfying Ad(k_{0})μ ∈ R R RY_{0} and Ad(k_{0})(𝔞+𝔪_{2}) = 𝔞+𝔪_{2}. Consequently, Ad(k_{0})𝔪_{2} coincides with the orthogonal complement of R R RY_{0} in 𝔞 + 𝔪_{2}, i.e.,

Ad(k_{0})𝔪_{2} = {Y′_{0} ∈ 𝔞 + 𝔪_{2} | (Y′_{0}, Y_{0}) = 0}.  (3.4)

(2) Let Y_{1} ∈ 𝔪_{1} satisfy Y_{1} ≠ 0. Then there is an element k_{1} ∈ K satisfying Ad(k_{1})μ ∈ R R RY_{1} and Ad(k_{1})(𝔞 + 𝔪_{2}) = 𝔪_{1}. Consequently, Ad(k_{1})𝔪_{2} coincides with the orthogonal complement of R R RY_{1} in 𝔪_{1}, i.e.,

Ad(k_{1})𝔪_{2} = {Y′_{1} ∈ 𝔪_{1} | (Y′_{1}, Y_{1}) = 0}.  (3.5)

Under the same setting in Proposition 7 (2), we have

Proposition 8. Let Y_{1} ∈ 𝔪_{1} satisfy Y_{1} ≠ 0. Then there is an element k′_{1} ∈ K satisfying

Ad(k′_{1})μ = \frac{1}{\sqrt{2}} \{μ + \frac{|μ|}{|Y_{1}|}Y_{1}\} ,  (3.6)
Ad(k′_{1})Y_{2} = \frac{1}{\sqrt{2}} \{Y_{2} + \frac{1}{|μ|^{3}|Y_{1}|}[[μ, Y_{1}], Y_{2}]\} ,  ∀Y_{2} ∈ 𝔪_{2}.  (3.7)

Here |v| denotes the norm of v ∈ 𝔪, i.e., |v| = (v, v)^{1/2}.

The proofs of Proposition 7 and Proposition 8 will be given in § 4.

Utilizing Propositions 5, 7 and 8 we first show the following:

Proposition 9. Assume that dimN N N = q(G/K) − dim G/K and that there is a solution Ψ of the Gauss equation (2.2). Then there exist two vectors A and B ∈ N N N satisfying

Ψ(Y_{0}, Y′_{0}) = (Y_{0}, Y′_{0})A,  ∀Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2},  (3.8)
Ψ(Y_{1}, Y′_{1}) = (Y_{1}, Y′_{1})B,  ∀Y_{1}, Y′_{1} ∈ 𝔪_{1},  (3.9)
Ψ(Y_{1}, Y_{2}) = −\frac{1}{(μ, μ)^{2}} Ψ(μ, [[μ, Y_{1}], Y_{2}]),  ∀Y_{1} ∈ 𝔪_{1}, ∀Y_{2} ∈ 𝔪_{2}.  (3.10)

6 YOSHIO AGAOKA AND EIJI KANEDA

Proof. First we prove

Ψ(Y_{0}, Y′_{0}) = 0,  ∀Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2} satisfying (Y_{0}, Y′_{0}) = 0.  (3.11)

We may assume that Y_{0}, Y′_{0} ≠ 0. Then, by Proposition 7 (1), we know that there is an element k_{0} ∈ K satisfying Ad(k_{0})μ ∈ R R RY_{0}. Since (Y_{0}, Y′_{0}) = 0, we have Y′_{0} ∈ Ad(k_{0})𝔪_{2}. Then, by Proposition 5, we know Y′_{0} ∈ Ker(Ψ_{Y_{0}}). Hence Ψ(Y_{0}, Y′_{0}) = 0, completing the proof of (3.11).

Now (3.8) can be proved by (3.11) as follows: Let Y_{0} and Y′_{0} be two elements of 𝔞+𝔪_{2} of the same length. Since (Y_{0} + Y′_{0}, Y_{0} − Y′_{0}) = 0, we obtain Ψ(Y_{0} + Y′_{0}, Y_{0} − Y′_{0}) = 0. Hence, we have Ψ(Y_{0}, Y_{0}) = Ψ(Y′_{0}, Y′_{0}). This implies that Ψ(Y_{0}, Y_{0})/(Y_{0}, Y_{0}) (Y_{0} ∈ 𝔞 + 𝔪_{2}, Y_{0} ≠ 0) takes a constant value A(∈ N N N). Therefore, we have Ψ(Y_{0}, Y_{0}) = (Y_{0}, Y_{0})A for any Y_{0} ∈ 𝔞 + 𝔪_{2}. Now (3.8) follows immediately from this equality.

In a similar manner, by applying Proposition 7 (2) we can prove (3.9).

Finally, we prove (3.10). Without loss of generality, we may assume that Y_{1} ≠ 0. Apply Proposition 8 to this Y_{1}(∈ 𝔪_{1}). Then there is an element k′_{1} ∈ K satisfying (3.6) and (3.7). By (3.1) we have

0 = Ψ(Ad(k′_{1})μ, Ad(k′_{1})Y_{2})

= \frac{1}{2}Ψ(μ + \frac{|μ|}{|Y_{1}|}Y_{1}, Y_{2} + \frac{1}{|μ|^{3}|Y_{1}|}[[μ, Y_{1}], Y_{2}]).

Note that [[μ, Y_{1}], Y_{2}] ∈ 𝔪_{1} (see Proposition 4 (1)) and [[μ, Y_{2}], Y_{1}] = 2[[μ, Y_{1}], Y_{2}] (see Lemma 5.3 of [5]). Then, we have

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

Hence by (3.9) we have Ψ(Y_{1}, [[μ, Y_{1}], Y_{2}]) = 0. This together with Ψ(μ, Y_{2}) = 0 proves (3.10). □

To calculate the left hand side of the Gauss equation (2.2), we prepare one more proposition, which will be proved in the last section of this paper.

Proposition 10. (1) Let Y_{0}, Y′_{0} ∈ 𝔞 + 𝔪_{2} and Y_{1} ∈ 𝔪_{1}. Then:

[Y_{0}, [Y_{0}, Y_{1}]] = −(μ, μ)(Y_{0}, Y_{0})Y_{1},  (3.12)
[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}  (3.13)

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

[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}  (3.14)

LOCAL ISOMETRIC IMBEDDINGS OF P^{2}(H H H) AND P^{2}(Cay Cay Cay) 7

[Y_{1}, [Y_{1}, Y_{0}]] = −(μ, μ)(Y_{1}, Y_{1})Y_{0}.  (3.15)

With these preparations, we start the proof of Theorem 3. We first show a series of lemmas by using the Gauss equation (2.2) and Proposition 9.

Lemma 11. ⟨A, A⟩ = ⟨B, B⟩ = 4(μ, μ).

Proof. Take an element Y_{2} ∈ 𝔪_{2} satisfying (Y_{2}, Y_{2}) = 1. Put X = Z = μ and Y = W = Y_{2} into the Gauss equation (2.2). Then, since Ψ(μ, Y_{2}) = 0, we have

([[μ, Y_{2}], μ], Y_{2}) = ⟨Ψ(μ, μ), Ψ(Y_{2}, Y_{2})⟩.

Since Ψ(μ, μ)/(μ, μ) = Ψ(Y_{2}, Y_{2}) = A and ([[μ, Y_{2}], μ], Y_{2}) = 4(μ, μ)^{2}, we have ⟨A, A⟩ = 4(μ, μ).

Next, we prove ⟨B, B⟩ = 4(μ, μ). Take elements Y_{1}, Y′_{1} of 𝔪_{1} satisfying (Y_{1}, Y_{1}) = (Y′_{1}, Y′_{1}) = 1 and (Y_{1}, Y′_{1}) = 0. Put X = Z = Y_{1} and Y = W = Y′_{1} into (2.2). Then, since Ψ(Y_{1}, Y′_{1}) = 0, we have

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

Since Ψ(Y_{1}, Y_{1}) = Ψ(Y′_{1}, Y′_{1}) = B and [[Y_{1}, Y′_{1}], Y_{1}] = 4(μ, μ)Y′_{1} (see (3.14)), we have ⟨B, B⟩ = 4(μ, μ). □

Lemma 12. ⟨A, Ψ_{μ}(𝔪_{1})⟩ = ⟨B, Ψ_{μ}(𝔪_{1})⟩ = 0.

Proof. Let Y_{1} be an arbitrary element of 𝔪_{1}. Take an element Y_{2} ∈ 𝔪_{2} satisfying (Y_{2}, Y_{2}) = 1. Put X = Z = Y_{2}, Y = μ and W = Y_{1} into (2.2). Then, since Ψ(μ, Y_{2}) = 0, we have

([[Y_{2}, μ], Y_{2}], Y_{1}) = ⟨Ψ(Y_{2}, Y_{2}), Ψ(μ, Y_{1})⟩.

Since Ψ(Y_{2}, Y_{2}) = A and [[Y_{2}, μ], Y_{2}] = 4(μ, μ)μ (see (3.13)), we have ⟨A, Ψ(μ, Y_{1})⟩ = 4(μ, μ)(μ, Y_{1}) = 0. Since Y_{1} is an arbitrary element of 𝔪_{1}, we have ⟨A, Ψ_{μ}(𝔪_{1})⟩ = 0.

Next, let Y_{1} be an arbitrary element of 𝔪_{1}. Take an element Y′_{1} ∈ 𝔪_{1} satisfying (Y′_{1}, Y_{1}) = 0 and (Y′_{1}, Y′_{1}) = 1. Put X = Z = Y′_{1}, Y = μ and W = Y_{1} into (2.2). Then, since Ψ(Y_{1}, Y′_{1}) = 0, we have

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

Since Ψ(Y′_{1}, Y′_{1}) = B and [[Y′_{1}, μ], Y′_{1}] = (μ, μ)μ (see (3.15)), we have ⟨B, Ψ(μ, Y_{1})⟩ = (μ, μ)(μ, Y_{1}) = 0. Since Y_{1} is an arbitrary element of 𝔪_{1}, we have ⟨B, Ψ_{μ}(𝔪_{1})⟩ = 0. □

Viewing Proposition 4 (3), we have dimN N N = dim 𝔪_{1} + 1. Since Ker(Ψ_{μ}) ∩ 𝔪_{1} = 𝔪_{2} ∩ 𝔪_{1} = 0, we have dim Ψ_{μ}(𝔪_{1}) = dim 𝔪_{1} = dimN N N − 1. Consequently, by Lemma 12 and Lemma 11, we easily have B = ±A. More strongly, we can show

Lemma 13. A = B.

8 YOSHIO AGAOKA AND EIJI KANEDA

Proof. By the above discussion, it suffices to prove ⟨A, B⟩ > 0. Let Y_{1} ∈ 𝔪_{1} satisfy (Y_{1}, Y_{1}) = 1. In (2.2), we put X = Z = μ and Y = W = Y_{1}. Then, we have

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

Since Ψ(μ, μ) = (μ, μ)A, Ψ(Y_{1}, Y_{1}) = B and [[μ, Y_{1}], μ] = (μ, μ)^{2}Y_{1}, we have

(μ, μ)⟨A, B⟩ = (μ, μ)^{2}(Y_{1}, Y_{1}) + ⟨Ψ(μ, Y_{1}), Ψ(μ, Y_{1})⟩ ≥ (μ, μ)^{2}.

This proves ⟨A, B⟩ > 0. □

Utilizing Lemma 13, we have

Lemma 14. Let Y_{1}, Y′_{1} ∈ 𝔪_{1}. Then:

⟨Ψ(μ, Y_{1}), Ψ(μ, Y′_{1})⟩ = 3(μ, μ)^{2}(Y_{1}, Y′_{1}).  (3.16)

Proof. Put X = Z = μ, Y = Y_{1} and W = Y′_{1} into (2.2). Then we have

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

Since Ψ(μ, μ) = (μ, μ)A, Ψ(Y_{1}, Y′_{1}) = (Y_{1}, Y′_{1})B and A = B, the first term of the right hand side becomes ⟨Ψ(μ, μ), Ψ(Y_{1}, Y′_{1})⟩ = 4(μ, μ)^{2}(Y_{1}, Y′_{1}) (see Lemma 11). Therefore, by [[μ, Y_{1}], μ] = (μ, μ)^{2}Y_{1}, we have

⟨Ψ(μ, Y_{1}), Ψ(μ, Y′_{1})⟩ = 4(μ, μ)^{2}(Y_{1}, Y′_{1}) − (μ, μ)^{2}(Y_{1}, Y′_{1})
= 3(μ, μ)^{2}(Y_{1}, Y′_{1}).
□

We are now in a position to complete the proof of Theorem 3. Let Y_{1} ∈ 𝔪_{1} (Y_{1} ≠ 0) and Y_{2} ∈ 𝔪_{2} (Y_{2} ≠ 0). Note that [Y_{1}, Y_{2}] ∈ 𝔨_{1} (see Proposition 4 (1)). We also note that [Y_{1}, Y_{2}] ≠ 0. In fact, if [Y_{1}, Y_{2}] = 0, then the 2-dimensional subspace generated by Y_{1} and Y_{2} forms an abelian subspace of 𝔪, which contradicts rank(G/K) = 1. Now, set Y′_{1} = [[Y_{1}, Y_{2}], μ]. Then it is clear that Y′_{1} ∈ 𝔪_{1} (see Proposition 4 (1)). Moreover, we have Y′_{1} ≠ 0, because [μ, Y′_{1}] = (μ, μ)^{2}[Y_{1}, Y_{2}] ≠ 0.

Now, put X = Y_{1}, Y = Y_{2}, Z = μ and W = Y′_{1} into (2.2). Since Ψ(Y_{2}, μ) = 0, we have

([[Y_{1}, Y_{2}], μ], Y′_{1}) = ⟨Ψ(Y_{1}, μ), Ψ(Y_{2}, Y′_{1})⟩.  (3.17)

By (3.10) and (3.16), the right hand side of (3.17) becomes

⟨Ψ(Y_{1}, μ), Ψ(Y_{2}, Y′_{1})⟩ = −⟨Ψ(μ, Y_{1}), Ψ(μ, [[μ, Y′_{1}], Y_{2}])⟩/(μ, μ)^{2}
= −3(Y_{1}, [[μ, Y′_{1}], Y_{2}])
= 3([Y_{1}, Y_{2}], [μ, Y′_{1}])
= 3([[Y_{1}, Y_{2}], μ], Y′_{1}).

LOCAL ISOMETRIC IMBEDDINGS OF P^{2}(H H H) AND P^{2}(Cay Cay Cay) 9

Putting this equality into (3.17), we have ([[Y_{1}, Y_{2}], μ], Y′_{1}) = 0, which contradicts our assumption ([[Y_{1}, Y_{2}], μ], Y′_{1}) = (Y′_{1}, Y′_{1}) ≠ 0.

As we have shown above, starting from the assumption that the Gauss equation (2.2) admits a solution Ψ, we finally arrive at a contradiction. Accordingly, we can conclude that if G/K = P^{2}(H H H) or P^{2}(Cay Cay Cay), then the Gauss equation (2.2) does not admit any solution in case dimN N N = q(G/K) − dim G/K. This completes the proof of Theorem 3. □

4. The action of the isotropy group Ad(K)

In this section we prove Propositions 7, 8 and 10, which are needed in the proof of Theorem 3.

Lemma 15. Let X_{i} ∈ 𝔨_{i} (i = 1, 2). Then it holds

[X_{i}, [X_{i}, μ]] = −i^{2}(μ, μ)(X_{i}, X_{i})μ.  (4.1)

Proof. By (2.5) we have [X_{i}, [X_{i}, μ]] ∈ 𝔞 + 𝔪_{2i}. By the Jacobi identity we have

[μ, [X_{i}, [X_{i}, μ]]] = [[μ, X_{i}], [X_{i}, μ]] + [X_{i}, [[μ, X_{i}], μ]] = 0,

because [[μ, X_{i}], μ] ∈ R R RX_{i}. Therefore, we have [X_{i}, [X_{i}, μ]] ∈ 𝔞. Since 𝔞 = R R Rμ, there is a scalar c ∈ R R R satisfying [X_{i}, [X_{i}, μ]] = cμ. Then we have c = −i^{2}(μ, μ)(X_{i}, X_{i}), because

c(μ, μ) = ([X_{i}, [X_{i}, μ]], μ) = (X_{i}, [[X_{i}, μ], μ]) = −(iμ, μ)^{2}(X_{i}, X_{i}).
□

By the above lemma, we obtain

Lemma 16. Let X_{i} ∈ 𝔨_{i} (i = 1, 2) satisfy X_{i} ≠ 0. Then it holds

Ad(exp(tX_{i}))μ = cos(i|μ||X_{i}|t)μ + \frac{sin(i|μ||X_{i}|t)}{i|μ||X_{i}|}[X_{i}, μ],  ∀t ∈ R R R.  (4.2)

Proof. Let n be a non-negative integer. By induction of n, we can easily show

(ad X_{i})^{2n}μ = (−1)^{n}(i|μ||X_{i}|)^{2n}μ,
(ad X_{i})^{2n+1}μ = (−1)^{n}(i|μ||X_{i}|)^{2n}[X_{i}, μ].

Consequently, for all t ∈ R R R we have

Ad(exp(tX_{i}))μ = ∑_{n=0}^{∞} \{\frac{t^{2n}}{(2n)!}(ad X_{i})^{2n}μ + \frac{t^{2n+1}}{(2n + 1)!}(ad X_{i})^{2n+1}μ\}

= ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n)!} (i|μ||X_{i}|t)^{2n}μ

+ \frac{1}{i|μ||X_{i}|} ∑_{n=0}^{∞} \frac{(−1)^{n}}{(2n + 1)!}(i|μ||X_{i}|t)^{2n+1}[X_{i}, μ]

= cos(i|μ||X_{i}|t)μ + \frac{sin(i|μ||X_{i}|t)}{i|μ||X_{i}|}[X_{i}, μ].

□

With these preparations, we proceed to the proof of Proposition 7. Let Y_{0} ∈ 𝔞 + 𝔪_{2}. If Y_{0} ∈ 𝔞, then we have only to set k_{0} = e, where e is the identity element of K.

Now we assume that Y_{0} ∉ 𝔞 and write Y_{0} = cμ + Y_{2} (c ∈ R R R, Y_{2} ∈ 𝔪_{2}, Y_{2} ≠ 0). Set X_{2} = [Y_{0}, μ]. Then we easily have X_{2} = [Y_{2}, μ] ∈ 𝔨_{2} and [X_{2}, μ] = −4(μ, μ)^{2}Y_{2}. Moreover, we have |X_{2}| = 2|μ|^{2}|Y_{2}|, because

(X_{2}, X_{2}) = ([Y_{2}, μ], [Y_{2}, μ]) = −([[Y_{2}, μ], μ], Y_{2}) = 4(μ, μ)^{2}(Y_{2}, Y_{2}).

Putting this X_{2} into Lemma 16, we have

Ad(exp(tX_{2}))μ = cos(4|μ|^{3}|Y_{2}|t)μ − \frac{|μ|}{|Y_{2}|} sin(4|μ|^{3}|Y_{2}|t)Y_{2}, ∀t ∈ R R R.

Take t_{0} ∈ R R R satisfying cos(4|μ|^{3}|Y_{2}|t_{0}) = c(|μ|/|Y_{0}|) and sin(4|μ|^{3}|Y_{2}|t_{0}) = −|Y_{2}|/|Y_{0}|. Let us set k_{0} = exp(t_{0}X_{2}). Then we have k_{0} ∈ K and

Ad(k_{0})μ = Ad(exp(t_{0}X_{2}))μ = \frac{|μ|}{|Y_{0}|}(cμ + Y_{2}) = \frac{|μ|}{|Y_{0}|}Y_{0}.

Thus we get Ad(k_{0})μ ∈ R R RY_{0}. By (2.5) we immediately have [X_{2}, 𝔞+𝔪_{2}] ⊂ 𝔞+𝔪_{2}. Hence, we have Ad(k_{0})(𝔞 + 𝔪_{2}) = 𝔞 + 𝔪_{2}. Since Ad(k_{0}) is an orthogonal transformation of 𝔪, we know that Ad(k_{0})𝔪_{2} coincides with the orthogonal complement of R R RY_{0} in 𝔞 + 𝔪_{2}. This finishes the proof of Proposition 7 (1).

To prove Proposition 7 (2), we first show

Lemma 17. Let X_{1} ∈ 𝔨_{1}. Then it holds

[X_{1}, [X_{1}, Y_{2}]] = −(μ, μ)(X_{1}, X_{1})Y_{2}, ∀Y_{2} ∈ 𝔪_{2}.

Proof. By (4.1), we have

[X_{1}, [X_{1}, μ]] = −(μ, μ)(X_{1}, X_{1})μ.  (4.3)

Let Y_{2} be a non-zero element of 𝔪_{2}. Then, as in the proof of Proposition 7 (1), we know that there is a scalar t_{0} ∈ R R R such that the element k_{0} = exp(t_{0}X_{2}) ∈ K satisfies Ad(k_{0})μ ∈ R R RY_{2}, where we set X_{2} = [Y_{2}, μ] ∈ 𝔨_{2}. Then, we have Ad(k_{0})𝔨_{1} = 𝔨_{1}, because [X_{2}, 𝔨_{1}] ⊂ 𝔨_{1} (see Proposition 4 (1)).

Now, applying Ad(k_{0}) to the both sides of (4.3), we have

[Ad(k_{0})X_{1}, [Ad(k_{0})X_{1}, Y_{2}]] = −(μ, μ)(X_{1}, X_{1})Y_{2}

LOCAL ISOMETRIC IMBEDDINGS OF P^{2}(H H H) AND P^{2}(Cay Cay Cay) 11

= −(μ, μ)(Ad(k_{0})X_{1}, Ad(k_{0})X_{1})Y_{2}.

Writing X_{1} instead of Ad(k_{0})X_{1} ∈ 𝔨_{1}, we get the lemma. □

Now we return to the proof of Proposition 7 (2). Set X_{1} = [Y_{1}, μ]. In the same way as in the proof of (1), we can easily prove X_{1} ∈ 𝔨_{1}, [X_{1}, μ] = −(μ, μ)^{2}Y_{1} and |X_{1}| = |μ|^{2}|Y_{1}|. Applying Lemma 16 to this X_{1}, we have

Ad(exp(tX_{1}))μ = cos(|μ|^{3}|Y_{1}|t)μ − \frac{|μ|}{|Y_{1}|} sin(|μ|^{3}|Y_{1}|t)Y_{1},  ∀t ∈ R R R.  (4.4)

Let Y_{2} ∈ 𝔪_{2}. By Lemma 17, we have

(ad X_{1})^{2n}Y_{2} = (−1)^{n}(|μ||X_{1}|)^{2n}Y_{2},
(ad X_{1})^{2n+1}Y_{2} = (−1)^{n}(|μ||X_{1}|)^{2n}[X_{1}, Y_{2}].

From these equalities, it follows

Ad(exp(tX_{1}))Y_{2} = cos(|μ|^{3}|Y_{1}|t)Y_{2} + \frac{sin(|μ|^{3}|Y_{1}|t)}{|μ|^{3}|Y_{1}|}[[Y_{1}, μ], Y_{2}], ∀t ∈ R R R.  (4.5)

Let us take t_{1} ∈ R R R satisfying |μ|^{3}|Y_{1}|t_{1} = −π/2 and set k_{1} = exp(t_{1}X_{1}). Then we can easily show that k_{1} ∈ K, Ad(k_{1})μ = (|μ|/|Y_{1}|)Y_{1} ∈ 𝔪_{1} and

Ad(k_{1})Y_{2} = −\frac{1}{|μ|^{3}|Y_{1}|}[[Y_{1}, μ], Y_{2}].  (4.6)

Hence, we have Ad(k_{1})μ ∈ R R RY_{1} and Ad(k_{1})𝔪_{2} ⊂ [[Y_{1}, μ], 𝔪_{2}]. Since [[Y_{1}, μ], 𝔪_{2}] ⊂ 𝔪_{1} (see Proposition 4 (1)), we have Ad(k_{1})(𝔞+𝔪_{2}) ⊂ 𝔪_{1}. Therefore, we have Ad(k_{1})(𝔞+𝔪_{2}) = 𝔪_{1}, because dim(𝔞 + 𝔪_{2}) = dim 𝔪_{1} (see Proposition 4 (3)). Since Ad(k_{1}) is an orthogonal transformation of 𝔪, we know that Ad(k_{1})𝔪_{2} coincides with the orthogonal complement of R R RY_{1} in 𝔪_{1}. This completes the proof of Proposition 7 (2). □

Next we prove Proposition 8. Under the same situation as in the proof of Proposition 7 (2), let us set k′_{1} = exp(t_{1}X_{1}/2). Then by the equalities (4.4) and (4.5) we easily obtain (3.6) and (3.7). □

Finally, we prove Proposition 10. First we show Proposition 10 (1). If Y_{0} ∈ 𝔞, then there is nothing to prove. Hence we may assume that Y_{0} ∉ 𝔞. Applying Proposition 7 (1), we have an element k_{0} ∈ K satisfying Ad(k_{0})μ ∈ R R RY_{0} and Ad(k_{0})(𝔞+𝔪_{2}) = 𝔞+𝔪_{2}. Then, it is easily seen that Ad(k_{0})𝔪_{1} = 𝔪_{1}. If we write Ad(k_{0})μ = cY_{0} (c ∈ R R R), then we have c^{2} = (μ, μ)/(Y_{0}, Y_{0}). Let Y_{i} be an element of 𝔪_{i} (i = 1, 2). Apply Ad(k_{0}) to the both sides of the equality [μ, [μ, Y_{i}]] = −i^{2}(μ, μ)^{2}Y_{i} (i = 1, 2). Then, since c^{2} = (μ, μ)/(Y_{0}, Y_{0}), we have

[Y_{0}, [Y_{0}, Ad(k_{0})Y_{i}]] = −i^{2}(μ, μ)(Y_{0}, Y_{0})Ad(k_{0})Y_{i}, i = 1, 2.

Now, (3.12) and (3.13) follow immediately from the above equality. (Note the equality (3.4) and the fact Ad(k_{0})𝔪_{1} = 𝔪_{1}.)

By applying Proposition 7 (2), Proposition 10 (2) can be also shown in a similar manner. Details are left to the readers. □

Thus, we have completed the proofs of Propositions 7, 8 and 10.

References

[1] Y. Agaoka and E. Kaneda, On local isometric immersions of Riemannian symmetric spaces, Tôhoku Math. J. 36 (1984), 107–140.

[2] , An estimate on the codimension of local isometric imbeddings of compact Lie groups, Hiroshima Math. J. 24 (1994), 77–110.

[3] , Local isometric imbeddings of symplectic groups, Geometriae Dedicata 71 (1998), 75–82.

[4] , Strongly orthogonal subsets in root systems, Hokkaido Math. J. 31 (2002), 107–136.

[5] , A lower bound for the curvature invariant p(G/K) associated with a Riemannian symmetric space G/K, (preprint).

[6] , Local isometric imbeddings of Grassmann manifolds, (in preparation).

[7] S. Kobayashi, Isometric imbeddings of compact symmetric spaces, Tôhoku Math. J. 20 (1968), 21–25.

(Yoshio AGAOKA) Faculty of Integrated Arts and Sciences, Hiroshima University 1-7-1 Kagamiyama, Higashi-Hiroshima, 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, Osaka, 562-8558, Japan

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