$QR$-submanifolds of maximal $QR$-dimension in quaternionic projective space
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(2)
π(S 4n1
π(S 4n1
2) × S 4n2
(4)
(5)
(6)
(7)
(8)
(9)
Proof. Sine n = 4m + 3 and p = 4t + 1 for some integers m and t, and since the subspace ν is quaternionic invariant (see also (2.13)), we can take a local orthonormal basis {ξ, ξ a , ξ a∗
ξ 1∗
{g(A a X, Y )ξ a + g(A a∗
+ g(A a∗∗
A a∗
A a∗∗∗
ξ a∗
∇ X F ξ a = −A a∗
∇ X Gξ a = −A a∗∗
∇ X Hξ a = −A a∗∗∗
− A a∗
{s ab∗
(3.8) ϕA a = A a∗
Similarly, from the other equations of (3.7), we have (3.9) ψA a = A a∗∗
Therefore, for any vectors X, Y tangent to M , it is clear from (3.8) that g(A a∗
g(A a∗
A a∗
which and (3.6) imply ϕA a∗
(12)
(4.1) ∇ ′ i1
π(S 4n1
5. An integral formula for the model space π(S 4n1
T := ∇ U U + ∇ V V + ∇ W W − (divU)U − (divV )V − (divW )W and take an orthonormal basis {U, V, W, e a , e a∗
e a∗
g(e i , ∇ ei
{g((∇ ea
− (∇ ea∗∗∗
− (∇ ea∗
− (∇ ea∗∗
(16)
π(S 4n1
for some n 1 , n 2 with 4n 1 + 4n 2 = n − 3, that is, M is a tube over a totally geodesic submanifold QP n1
π(S 4n1
2) × S 4n2
S 4n1
2) × S 4n2
4n ∑1
where 4n 1 + 4n 2 = n − 3 and n = 4s + 3 for some integer s. Then, since S 4n1
2) × S 4n2
S 3 → S 4n1
2) × S 4n2
2) → M n Q1
M n Q1
2) × S 4n2
In this case we can easily see that the geodesic distance from QP n1
where A denote the shape operator of M n Q1
Hence, for M n Q1
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