Positively curved 6-manifolds with simple symmetry groups

Let M be a simply connected compact 6-manifold of positive sectional curvature. If the identity componentoftheisometrygroupcontainsasimpleLiesubgroup, weprovethat M isdiffeomorphic to one of the ﬁve manifolds listed in Theorem A


FUQUAN FANG
In the geometry of positive sectional curvature, 6-dimensional manifolds play very interesting roles.This is because, there are so called ''flag'' manifolds, the homogeneous space SU (2)/T 2 and the biquotient space SU (3)//T 2 on which there exist infinitely many circle bundles whose total spaces (dimension 7) admit positive sectional curvature, the Aloff-Wallach spaces and the Eschenberg spaces (cf.Aloff and Wallach 1975, Eschenberg 1982, 1992).Except the two flag manifolds, so far only S 6 and CP 3 are known to admit positive curvature metrics.All of the four examples admit a symmetry of a simple Lie group.
The purpose of this paper is to study the isometric G-actions on positively curved manifolds of dimension 6 where G is a simple Lie group.We will prove almost a converse holds, except that one more topology type, S 2 × S 4 , maybe occur.
In the below = will indicate manifolds of the same diffeomorphism type.
Theorem A. Let M be a simply connected compact 6-manifold of positive sectional curvature.
If the identity component of the isometry group of M contains a simple Lie subgroup.Then M = S 6 , S 2 × S 4 , CP 3 , SU(3)/T 2 or SU (3)//T 2 .
Note that any simple Lie group contains either SU (2) or SO(3) as a subgroup.Therefore one really needs only to study manifolds with symmetry group G = SU (2) or G = SO(3).Thus Theorem A follows immediately from the following two results.
The rest of the paper is organized as follows: In Section 1, we collect some necessary preliminaries in the proof of Theorems B and C. In Section 2, we prove Theorems B and C.

A. Positive Curvature
For positively curved simply connected manifolds, few general results are known.In our proof of Theorem A, the following results in this subsection are required.
(1.1.1)If n is even and M is orientable, then any orientation preserving isometry φ has a non-empty fixed point set.
(1.1.2)If n is odd, then M is orientable and any orientation reversing isometry φ has a non-empty fixed point set.
then N 1 and N 2 have a non-empty intersection.
The asymptotic index of an immersion f : N → M is defined by ν f = min x∈N ν f (x), where ν f (x) is the maximal dimension of a subspace of T x N on which the second fundamental form vanishes.Clearly, f is a totally geodesic immersion if and only if ν f = dim(N ).
Theorem 1.3 (Fang et al. 2002).Let M m be a closed manifold of positive sectional curvature, and let N j → M m be a closed embedded submanifold of asymptotic index ν j (j = 1, 2).If either N j (j = 1, 2) is minimal or N 1 = N 2 , then the following natural homeomorphisms, Applying to totally geodesic submanifold, where the asymptotic index is exactly the dimension, Theorem 1.3 implies immediately that Theorem 1.4 (Wilking 2002).If M n has positive sectional curvature and if N n−k is an embedded totally geodesic submanifold then the inclusion map i : N n−k → M n is n-2k+1 connected, that is i induces an isomorphism of the homotopy groups up to dimension n − 2k and it maps π n−2k+1 (N ) surjectively onto π n−2k+1 (M).

B. Alexandrov Spaces with Positive Curvature
Recall that an Alexandrov space, X, is a finite Hausdorff dimensional complete inner metric space with a lower curvature bound in distance comparison sense (cf.Burago et al. 1992).In particular, a Riemannian manifold of sectional curvature bounded from below is an Alexandrov space.Typical examples of non-manifold type Alexandrov spaces are given by the following Lemma 1.5 (Burago et al. 1992).Let X be an Alexandrov space with curvature ≥ − .Let G be a compact group of isometries.Then, the quotient space, X/G, is also an Alexandrov space with curvature ≥ − .

PROOFS OF THEOREMS B AND C
In this section we consider simply connected positively curved 6-manifolds with G-symmetry where G = SO(3) or SU (2).Note that the connected symmetry group of SU (3)/T 2 is P SU(3) = SU (3)/Z 3 , where Z 3 is the center of SU (3) (cf.Shankar 2001a).It was pointed out to the author by Shankar (2001b) that the identity component of the symmetry group of the biquotient SU (3)//T 2 is U(2).In both cases, SU ( 2) is a proper subgroup of the isometry groups.

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FUQUAN FANG Lemma 2.1.Let M be a simply connected compact 6-manifold of positive sectional curvature.If M admits an isometric SU (2)-action such that the fixed point set Fix(M, SU (2)) is not empty, then M is either diffeomorphic to S 6 or CP 3 .
Proof.Let X = Fix(M, SU ( 2)).Since an irreducible representation of SU (2) has dimension 4, the fixed point set must be of dimension 2. Therefore the orbifold M/SU ( 2) is an Alexandrov space with boundary X.Thus where E(η) is a 4-disk bundle over X, and H ⊂ SU ( 2) is the isotropy group for some single orbit, m + 3 = dimH + 6 (cf.Grove and Searle 1997).In particular, X is connected, and so X = S 2 (since X is totally geodesic with positive curvature).Note that the boundary of Then the neighborhood SU (2) × H D 4 must be SU (2) × S 1 D 4 , which is equivariant diffeomorphic to S 2 × D 4 with the product action of SU (2), i.e. on the second factor is given by the irreducible representation of SU (2) on D 4 ⊂ C 2 , on the first factor the action is given by the natural action of SU (2) on CP 1 = S 2 .To match the two actions on the boundary, it is easy to see that the gluing map f : S 2 × S 3 → S 2 × S 3 may be written as f (x, y) = (y • x, y), where y ∈ S 3 = SU (2) acts on x factoring through the natural SO(3)-action.It is an easy exercise to show that M = CP 3 .
Lemma 2.2.Let M be a simply connected compact 6-manifold of positive sectional curvature.If M admits an isometric SO(3)-action such that the fixed point set Fix(M, SO(3)) is not empty, then M is diffeomorphic to S 6 .
Proof.First we claim that the fixed point set X = Fix(M, SO(3)) must have positive dimension.In fact, if x ∈ X is an isolated fixed point, the local isotropy representation of SO(3) gives a linear action on S 5 .But a linear action of SO(3) must have all principal orbit dimension 2, and so the action must have a 2-sphere as the fixed point set (cf. Bredon 1972 exercise in Chap. 5).
By the local isotropy representation we also know that X has dimension 3 and so the principal orbits are of dimension 2. Thus the orbit space is 4-dimensional with boundary X.Hence An Acad Bras Cienc (2002) 74 (4) POSITIVELY CURVED 6-MANIFOLDS

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where E(ν) is a disk bundle of dimension 3 over the fixed point set X, and H is a circle subgroup of SO(3) (cf.Grove and Searle 1994).Clearly, X is a spherical form (cf. Hamilton 1982), since it is totally geodesic.In order that the boundary of the disk bundle matches the boundary of SO( 3) × H D 4 the core X must be simply connected.By a straightforward argument one knows that M is a homotopy sphere and therefore M = S 6 .
The q-extent xt q (X), q ≥ 2, of a compact metric space (X, d) is, by definition, given by the following formula: Recall that the 3-extent xt 3 S 2 1 2 = π 3 (Grove and Markvosen 1995).
Lemma 2.3.In the orbit space M * there are at most three isolated singular orbits.
Proof.We prove by contradiction.Assume that there are at least 4 isolated singular orbits.Let A 1 , A 2 , • • • , A 4 be the 4 isolated singular orbits in M * .
For any i = j , let Since the curvature of M * is positive (cf.Lemma 1.5), by the Toponogov's comparison theorem it follows that, for each triple (i, j, k) of distinct integers in [1,4], Summing over all possible choices of the triple (i, j, k) we know that On the other hand, for each A i the isotropy group H i ∼ = S 1 , note that the space of directions at A i , Proof of Theorem B. By Lemma 2.1 we may assume that Fix(M, SU ( 2)) is empty.
An Acad Bras Cienc ( 2002) 74 (4) By the Synge theorem the center Z 2 ⊂ SU (2) must act on M with non empty fixed point (since it preserves the orientation).Consider the fixed point set M Z 2 ⊂ M. Clearly SU (2) acts on M Z 2 .If it has dimension 4, it must be CP 2 or S 4 (cf.Hsiang and Kleiner 1989).But any action of SU (2) on S 4 or CP 2 must have fixed point.A contradiction.Therefore M Z 2 has dimension 2 and consists of the union of some isolated singular orbits of dimension 2. Around every singular orbit, there is a regular neighborhood of the form SU (2) × H D 4 where H is the isotropy group of the singular orbit (a 1-dimensional subgroup of SU ( 2)).In the Alexandrov space M * = M/SU (2), it gives a neighborhood of the singular orbit and homeomorphic to D 4 /H , a cone over S 3 /H .Note that the diameter of S 3 /H is at most π 2 .By Lemma 2.3 above we know that there are at most three such isolated singular orbits.
Observe that the orbit space M * must be a simply connected 3-manifold (therefore M * is a homotopy 3-sphere) with at most three marked singular points.This shows that M may be re-constructed by gluing at most three handles SU (2) × H i D 4 (with H i the isotropy groups of the singular orbits) to SU ( 2 , where D 3 i , i = 1, 2, 3 are disjoint small disks neighborhood around the singular orbits in M * .It is easy to check that the third Betti number b 3 (M) = 0 and, the Euler characteristic χ(M) = 2, 4 or 6, according to 1, 2 or 3 singular orbits.
On the other hand, as we noted in the beginning of the section, SU (3)/T 2 and SU (3)//T 2 both have effective isometric actions by SU (2).The induced actions of SU (2) on the manifolds must have only isolated singular orbits (if not, the manifold must be S 6 or CP 3 by the above argument).
FUQUAN FANG isolated singular orbits.As in the proof of Theorem B, the possible values of χ(M) = 2, 4, 6.In the case χ We claim that χ(M) = 2 and 6 are impossible.This will complete the proof.
Let X = M * −∪ l i=1 int D 3 i with l the number of singular orbits (here l = 1 or 3), where D 3 i is as in Theorem A. Note that M * is again a homotopy 3-sphere.Since all orbits in X is of type SO(3), the principal SO(3)-action on π −1 (X) corresponds to a unique classifying map f : X → BSO(3) for the action.Note that f restricts on each piece of the boundary (a 2-sphere) of X gives the classifying map of the free SO(3) action on SO(3) × H i S 3 = S 2 × S 3 , which is not null homotopic in π 2 (BSO(3)) = Z 2 .Clearly, the sum of the l boundary pieces must be zero homologous (since the map f is a cycle with boundary the sum).Therefore l must be even.This proves that there are only 2 singular orbits.This proves the desired result.

ACKNOWLEDGMENTS
This paper is written based on the author's talks in Brazilian Colloquium in IMPA and in the conference "Foliation and Geometry 2001" in PUC, Brazil.Supported by CNPq of Brazil, NSFC Grant 19741002, RFDP and Qiu-Shi Foundation of China.