Remarks on the Dorfmeister-Neher theorem on isoparametric hypersurfaces
Reiko Miyaoka
Acknowledgment. The author cordially appreciate U. Abresch and A.
Siffert’s questions and indication.
1 Introduction
§7 and §8 of [M] are the heart of the paper, but a lack of clear argument causes some questions, although the statement is true. The purpose of the present paper is to make it clear. We follow the notation and the argument in [M].
2 Dim E = 2 ( § 7 [M])
We call a vector fieldv(t) along L6 parametrized byp(t), even when v(t+ π) =v(t), and odd whenv(t+π) =−v(t). Note thatEconsists of∇ke6e3(t), k = 0,1, . . . which are odd or even at the same time, and W consists of
∇ke6∇e3e6(t) of which evenness and oddness is the opossite ofE, sinceL(t+ π) =−L(t).
Proposition 7.1 [M] dimE= 2 does not occur at any point ofM+. Proof : dimE = 2 implies dimW = 1, and so W consists of even vector (∇e3e6 never vanish by Remark 5.3 of [M]). ThusE consists of odd vectors.
For X1, Z1, X2, Z2 in p.709, X1 is parallel to ∇e6e3 at p0 =p(0) and p(π), and so has opposite sign atp(0) andp(π). Note thatZ1 ∈W is a constant unit vector parallel to ∇e3e6(t). Also, span{X2, Z2} is parallel since this is the orthogonal complement of E ⊕W. Because D1(π) = D5(0) and
1
D2(π) =D4(0) etc. hold, four cases occur;
(e1+e5)(π) = (e1+e5)(0) and (e2+e4)(π) = (e2+e4)(0), (e1+e5)(π) = (e1+e5)(0) and (e2+e4)(π) =−(e2+e4)(0), (e1+e5)(π) =−(e1+e5)(0) and (e2+e4)(π) = (e2+e4)(0), (e1+e5)(π) =−(e1+e5)(0) and (e2+e4)(π) =−(e2+e4)(0).
In the first case,α(π) =−α(0) andβ(π) =−β(0) follow. ThenX2 becomes even andZ2 becomes odd, contradicts that span{X2, Z2}, is parallel. In the second case,α(π) =−α(0) andβ(π) =β(0) hold, and soX2 is odd, andZ2 is even, again a contradiction. Other cases are similar. ✷
3 Dim E = 3 ( § 8 [M])
When dimE = 3,e3(t) is an even vector, sinceE is parallel alongL6. Using Proposition 8.1 [M], we extend e1, e2, e4, e5 as follows: Taking the double cover ˜c(t) of c(t), i.e., t ∈ [0,4π), if necessary, we choose a differentiable frame ei(t) as follows: First take e1(t), e2(t) continuously for t ∈ [0,4π).
Then we definee5(t) =e1(t+π) ande4(t) =e2(t+π) fort∈[0,3π). Thus we have a differentiable frame ei(t) for t ∈ [0,3π), though we only need t∈[0,2π].
With respect to this frame, we can take a differentiable orthonormal frame of E and E⊥ by
e3(t), X1 =α(t)(e1+e5)(t) +β(t)(e2+e4)(t) X2(t) = 1
pσ(t)
≥β(t)
√3(e1−e5)(t)−√
3α(t)(e2−e4)(t)¥ (1) and
Z1(t) = 1 pσ(t)
≥√
3α(t)(e1−e5)(t) +β(t)√
3(e2−e4)(t)¥ Z2(t) =β(t)(e1+e5)−α(t)(e2+e4)(t),
(2) whereα(t),β(t),σ(t) are differentiable for t∈[0,3π], satisfying
α2(t) +β2(t) = 1/2, σ(t) = 2(3α2(t) +β2(t)/3). (3) Note thatσ(t) =σ(t+π) holds, sinceσ(t) is an eigenvalue of T(t) =tRR(t) (see (45) [M] and the statement after it).
Proposition 8.2 [M] σ(t) is constant and takes values 1/3 or 3.
Remark. We need not specialize the caseσ = 1 in the proof.
2
Proof : From (3), the conclusion follows if we show α(t)β(t) ≡ 0. Suppose α(t)β(t)6≡0. By definition, we have
e1(π) =e5(0), e2(π) =e4(0). (4) We must be careful for
e5(π) =e1(2π) =≤1e1(0), e4(π) =e2(2π) =≤2e2(0), where≤i =±1. However, sincee3 is even and by (4), we obtain
≤:=≤1 =≤2. [Case 1]≤= 1. In this case, we have
X1(π) =α(π)(e1(π) +e5(π)) +β(π)(e2(π) +e4(π))
=α(π)(e5(0) +e1(0)) +β(π)(e4(0) +e2(0)), (5) which belongs to E, and is orthogonal toe3(0) and X2(0). Thus we obtain X1(π) = ¯≤X1(0), namely,α(π) = ¯≤α(0), β(π) = ¯≤β(0), (6) where ¯≤=±1. On the other hand, we have
X2(π) = 1 pσ(π)
µβ(π)
√3 (e1(π)−e5(π))−√
3α(π)(e2(π)−e4(π))
∂
= 1
pσ(0)
µβ(π)
√3 (e5(0)−e1(0))−√
3α(π)(e4(0)−e2(0))
∂ ,
(7)
where we use σ(π) =σ(0). Thus from (6), we obtain X2(π) =−¯≤X2(0).
However, because E is parallel, X1 and X2 should be even or odd at the same time, a contradiction.
[Case 2] ≤=−1. In this case, we have
X1(π) =α(π)(e1(π) +e5(π)) +β(π)(e2(π) +e4(π))
=α(π)(e5(0)−e1(0)) +β(π)(e4(0)−e2(0)), (8) which belongs to E, and is orthogonal toe3(0) and X1(0). Thus we obtain
X1(π) = ¯≤X2(0), namely,α(π) =−¯≤ β(0)
p3σ(0), and β(π) = ¯≤
√3α(0) pσ(0), (9)
3
for ¯≤=±1. On the other hand, we see that X2(π) = 1
pσ(π)
µβ(π)
√3 (e1(π)−e5(π))−√
3α(π)(e2(π)−e4(π))
∂
= 1
pσ(0)
µβ(π)
√3 (e5(0) +e1(0))−√
3α(π)(e4(0) +e2(0))
∂ (10)
where we use σ(π) = σ(0). Because it belongs to E and is orthogonal to e3(0) and X2(0), and further because (X1(0), X2(0)) 7→ (X1(π), X2(π)) should be orientation preserving, we obtain,
X2(π) =−¯≤X1(0), namely, β(π)
p3σ(0) =−¯≤α(0) and −
√3α(π)
pσ(0) =−¯≤β(0).
(11) However, then (9) and (11) have no solution.
These contradiction is caused by the assumption α(t)β(t) 6≡ 0. Thus α(t)β(t)≡0 follows. Now, by the argument in §9 [M], we obtain
Theorem 3.1 [DN], [M] Isoparametric hypersurfaces with (g, m) = (6,1) are homogeneous.
References
[DN] J. Dorfmeister and E. Neher, Isoparametric hypersufaces, case g = 6, m= 1, Comm. in Alg. 13(1985), 2299-2368.
[M] R. Miyaoka, The Dorfmeister-Neher theorem on isoparametric hyper- surfaces, Osaka J. Math. 46, (2009) 695–715.
Mathematical Institute, Graduate School of Sciences Tohoku University, Sendai, 980-8578/JAPAN [email protected]
4
