Riemannian foliations of spheres
Abstract.
We show that a Riemannian foliation on a topological sphere has leaf dimension or unless and the Riemannian foliation is given by the fibers of a Riemannian submersion to an dimensional sphere. This allows us to classify Riemannian foliations on round spheres up to metric congruence.
1991 Mathematics Subject Classification:
53C12, 57R301. Introduction
We are going to prove the final piece of the following theorem:
Theorem 1.1.
Suppose is a Riemannian foliation by dimensional leaves of a compact manifold which is homeomorphic to . We assume . Then one of the following holds

and the foliation is given by an isometric flow, with respect to some Riemannian metric.

, and the generic leaves are diffeomorphic to or .

, and is given by the fibers of a Riemannian submersion where is homeomorphic to and the fiber is homeomorphic to .
Furthermore all these cases can occur.
A big part of the Theorem follows by putting together various pieces in the literature: Ghys ([Ghy84]) showed that the generic leaves of a Riemannian foliation of a homotopy sphere are closed, unless the leave dimension is and the foliation is given by an isometric flow with respect to a possibly different Riemannian metric. Furthermore, the generic leaves are rational homotopy spheres. Haefliger ([Hae84]) observed that for any Riemannian foliation of a complete manifold with closed leaves, one can find a space homotopically equivalent to such that is the total space of a fiber bundle, where the fibers are homeomorphic to the generic leaves of the foliation (see section 2 for further details). If is a sphere then the fibers are contractible in . Spanier and Whitehead observed ([SW54]) that for any such fibration the fiber must be an space. Furthermore, closed manifolds which are spaces and rational homotopy spheres were classified by Browder ([Bro62]): they are homotopically equivalent to , , , or . With Perelman’s solution of the geometrization conjecture one can improve the statement further to diffeomorphic if .
We are left to consider dimensional foliations of homotopy spheres. Our strategy will be reduce the situation first to the case of and then show that the foliation is simple, i.e. given by the fibers of a Riemannian submersion. By a result of Browder ([Bro62]) this automatically rules out the possibility of an foliation.
To see that all examples can occur, we can again appeal to the literature for the only nonclassical case: the existence of foliations on . It was shown by Oliver ([Oli79]), that contrary to a previous conjecture, there are almost free smooth actions of on for . The actions of Oliver extend to fixed point free smooth actions on the disc , different actions were later exhibited by Grove and Ziller ([GZ00]).
Our topological result allows us to classify Riemannian foliations of the round sphere up to metric congruence. We recall that Gromoll and Grove ([GG88] classified Riemannian foliations of the sphere up to leave dimension 3. Moreover, due to [Wil01], a Riemannian submersion of the round with dimensional fibers is metrically congruent to the Hopf fibration. Combining this work with Theorem 1.1 gives
Corollary 1.2.
Let be a Riemannian foliation on a round sphere with leaf dimension . Then, up to isometric congruence, either is given by the orbits of an isometric action of or with discrete isotropy groups or it is the Hopf fibration of with fiber .
As has been pointed out by Gromoll and Grove a real representation induces an almost free action of on the unit sphere if and only if all irreducible subrepresentations are even dimensional.
The paper is structured as follows. In Section 2 we recall the results stated after Theorem 1.1 and study the fibration from a homotopy sphere to the resolution of the orbifold . The fiber of the fibration is , the principal leaf of , and we only need to consider the cases and . From this fibration we compute the cohomology of . These computations show . Moreover, the cohomology of at all primes but is concentrated in dimensions . In the two subsequent sections, we exclude the possibility that the orbifold is not a manifold. Here we use the local data of the orbfold to find nontrivial cohomology classes of that cannot exist by the previous computations. We rely on the fact that all isotropy groups of act freely on a dimensional sphere or a projective space, a sever restriction on the possible group structure. In Section 3, we use the computation of the cohomology of at the prime , to deduce that all isotropy groups are cyclic of odd order. Here we detect the forbidden classes by loooking at single points of , i.e., by finding the nontrivial restrictions of the cohomology classes to the classifying spaces of the isotropy groups. In Section 4, we exclude the possibility that the set of points with nontrivial isotropy is nonempty, otherwise detecting forbidden cohomology classes by their nontrivial restriction to a component of .
2. Topology
2.1. Recollection
Let be as in Theorem 1.1 and assume that the leaves have dimension . Due to [Ghy84], all leaves of are closed. This in turn is equivalent to saying that is a generalized Seifert fibration on , i.e., the space of leaves carries the natural structure of a smooth Riemannian orbifold, such that the induced Riemannian distance corresponds to the distance between leaves in . Due to [Ghy84], the regular leaf of is a rational homology sphere. Following Haefliger, we consider the bundle over given by all oriented horizontal frames in . Then the Riemannian foliation induces a fiber bundle structure on with the fibers being diffeomorphic to and with the base space being the oriented frame bundle of the orbifold . Furthermore, the natural fiber bundle map is equivariant.
Thus one also gets a fiber bundle with total space given by with fiber and with base space , . Clearly is homotopically equivalent to and is the so called resolution (or classifying space) of the orbifold . Its cohomology is the so called orbifold cohomology of . As has been oberserved by Haefliger the natural projection is a rational homotopy equivalence.
Since the fiber is a dimensional manifold and is connected, we see that the fiber is contractible in . Therefore is an space ([SW54]). Since is a rational homology sphere, we may apply [Bro62] and deduce that is homotopy equivalent to , , or . An application of the geometrization conjecture proves Theorem 1.1 for . Thus we only need to consider the case .
Thus is either homeomorphic to or it is homotopy equivalent to and its doublecover is homeomorphic to . We call the first case the spherical case and the second case the projective case.
2.2. Gysin sequence and dimension
Let be any ring with unit. In the projective case we assume in addition that is invertible in , e.g., or . Then . Thus we find the Gysin sequence of the fibration with coefficients in . The Euler class is a generator. Moreover, the cup product is an isomorphism, if .
Since has finite rational cohomology, we use this isomorphism for to see that , for some positive integer .
2.3. Reduction to .
We want to show . Assume on the contrary . Then, due to the above isomorphism, we have in degrees . To obtain a contradiction, we first show:
Lemma 2.1.
Under the assumptions above there exists a space and an element such that the cohomology ring equals the polynomial ring in degrees .
Proof.
For , one could just take . In general, let be the mapping cylinder of , which is a fiber bundle over with fiber being the cone over . Let be the Thom space of the fibration , which is obtained from by identifying all points on the boundary of . For the subbundle of the bundle , we can apply [Hat02], Theorem 4.D.8. Using the fact that the bundle is orientable, we deduce that there is an element (the Thom class of the fibration), such that induces an isomorphism between and the reduced cohomology .
The claim follows from this isomorphism and the structure of . ∎
We now get a contradiction to the following application of Steenrod powers, cf. [Hat02], Theorem 4.L.9.
Lemma 2.2.
Let be a topological space. Assume that . Then there is no element , with .
Proof.
Consider the Steenrod operations . We have . On the other hand, by the Adem relations, is a linear combination of and , which must be zero, since the corresponding cohomology groups are trivial. Thus . ∎
The contradiction shows , hence . Thus has dimension and has the rational homology of .
2.4. Cohomology of
From the Gysin sequence of the fibration we deduce:
Lemma 2.3.
Let be a prime number, with in the projective case. Then either is an homology sphere, or the cohomology ring of has the form
where has degree and has degree .
We will need:
Lemma 2.4.
.
Proof.
In the spherical case is connected. In the projective case, we know and for and . Hence the canonical map from to the EilenbergMacLane space induces isomorphisms on all cohomologies in all degrees . The result follows from the computations of the cohomology groups of (for instance, cf. [Cle02]). ∎
The last result about the cohomology of which we extract from the fibre bundle is the following application of the transgression theorem of Borel ([Bor53], Theorem 13.1). This theorem applies (cf. [Bro62], last paragraph on p. 370), since in the projective case, the fiber has the cohomology of .
Lemma 2.5.
Assume that is homotopy equivalent to . Then the cohomology ring up to degree is freely generated by elements of degree and respectively. In particular, we have and .
3. Isotropy groups are cyclic groups of odd order
In this section we use characteristic classes to see that any Sylow subgroup of any isotropy group is cyclic of order at most . With different methods we then use this to show that all isotropy groups are cyclic groups of odd order.
Consider as the quotient space , where is the bundle of oriented frames of with canonical action of . Recall that the space is nothing else but the Borel construction . We will often consider the canonical dimensional vector bundle (the tangent bundle of the orbifold)
over .
Lemma 3.1.
Let be a vector bundle over . Then the StiefelWhitney classes and vanish.
Proof.
We first assume and prove that this implies .
By stabilizing with a trivial bundle we may assume that the rank of is at least . Let be the classifying map of the bundle . In particular, the StiefelWhitney classes of are given by pull backs of StiefelWhitney classes of the universal bundle over . Since , can be lifted to a map . Suppose now on the contrary that . Then
is not zero. Since there is a natural map to the EilenbergMacLane space that induces an isomorphism on th cohomology with integral coefficients. Since this map is connected it also induces an isomorphism on th cohomology with coefficients. By composing the map with we get a map which induces a nontrivial map on th cohomology with coeffcients. On the other hand, the homotopy classes of maps are classified by (see Lemma 2.4) and thus any map is null homotopic – a contradiction.
Assume now . Then as well (cf. Lemma 2.5). Consider the bundle . Then the total StiefelWhitney classes satisfy . Since is simply connected, . We deduce and . Applying the previous observation to the bundle , we deduce . This contradicts .
∎
Lemma 3.2.
Let be an isotropy group. Then any element of order is given by . The Sylow group of is a cylic group of order at most .
Proof.
Let be a point in the inverse image of such that is the isotropy group of the action on at . Notice that the image of under the natural projection can be naturally identified with the classifying space of the isotropy group . If we restrict the canonical bundle over to , we get an bundle which is isomorphic to where is acting by the canonical representation on . Let be a subgroup. If we pull back via the covering map , we thus get a bundle which is isomorphic to over . By Lemma 3.1, the second and the fourth StiefelWhitney classes of vanish.
Suppose now that and suppose the nonzero element has times the eigenvalue . Then is a bundle over which decomposes as the sum of canonical line bundles and trivial line bundles. Thus the total StiefelWhitney class is given by , where is the generator of and is the generator of .
If is odd, we get and if , we see that . Since and we obtain a contradiction. This only leaves us with the possibility that has times the eigenvalue and thus .
Since there is at most one order element, it follows that a Sylow subgroup does not contain any abelian noncyclic subgroup. This implies that is either cyclic or generalized quaternionic ([Wol67], [Wal10]). In order to prove that is cyclic it suffices to rule out the possibility that we can realize the quaternion group with elements as a subgroup of an isotropy group . Suppose on the contrary we can. As before, the bundle over can be seen as a pull back bundle of the canonical bundle over . By Lemma 2.4, and thus the first Pontryagin class of vanishes, .
The embedding of is determined by the fact that the center of is mapped to . The representation of decomposes into two equivalent dimensional subrepresentations of . Thus is isomorphic to the sum of two copies of the dimensional bundle , where acts by its unique dimensional irreducible representation on . Since admits a complex structure, we have and thus the first Pointryagin class is additive: . In other words, . If we pull back the bundle to via the natural covering we get a bundle which decomposes into two dimensional subbundles, whose Euler classes are generators of . This in turn implies that is given by the order two element in . On the other hand is given by the image of under the natural homomorphism – a contradiction since any homomorphism has in its kernel.
Thus the Sylow group is cyclic. It remains to rule out that there are elements of order . Suppose on the contrary that is cyclic group of order and fix a generator . Let be a primitive th root and choose numbers such that () are the eigenvalues of counted with multiplicity. Since we know , all are odd.
The bundle over decomposes into four orientable dimensional subbundles whose Euler classes are given by () where is a generator of .
It follows that the first Pontryagin class of the bundle is given by . As before and since is a generator of , this implies . But for any odd number we have – a contradiction.
∎
Lemma 3.3.
Any isotropy group is either cyclic or isomorphic to a semidirect product , where acts on the cyclic group of odd order by an automorphism of order . Moreover, if has even order it has a nontrivial periodic cohomology.
Proof.
Let be a (not necessarily proper) subgroup of an isotropy group. Recall that or a extension of acts freely on and thus has periodic cohomology (cf. [Wal10], [Wol67] for this fact and subsequent results about groups with periodic cohomology). Thus for all odd , the Sylow groups are cyclic. By Lemma 3.2, the Sylow group is cyclic as well.
A classical theorem of Burnside implies that such a group is metacyclic, that is, is isomorphic to a semidirect product where and are relatively prime.
It remains to check that the homomorphism does not contain any elements of odd prime order . In fact then Lemma 3.2 implies that the image of has order at most .
We argue by contradiction and assume that is a minimal counterexample. The minimality easily implies that is a prime and that is a prime power , where are both odd.
We consider the normal covering whose deck transformation group is generated by an element of order . Since the order is prime to , the induced map is injective and its image is given by the fixed point set of where is the induced map on cohomology. Clearly acts on by an element of order . This in turn implies that is fixed by if and only is divisible by . Hence the minimal period of is divisible by – a contradiction since we know that has periodic cohomology. Thus is cyclic and has periodic cohomology, or , where acts by an automorphism of order two on . To see that in the latter case has periodic cohomology we construct a free linear action of on . Let be the fixed point set of . Since has order , the numbers and are relatively prime. In particular . We can now embed into by mapping the factor injectively to a central subgroup of and by mapping injectively to a subgroup of . Clearly, the induced action on is free and thus has periodic cohomology. The cohomology of can not be trivial as . ∎
Lemma 3.4.
The isotropy groups are cyclic groups of odd order.
Proof.
By Lemma 3.3 it suffices to show the isotropy groups have odd order. By Lemma 3.2 the subset of points whose isotropy groups have even order is finite . Let denote the isotropy groups. Suppose on the contrary that is not empty.
Let denote the inverse image of in the frame bundle and the corresponding subset in the Borel construction . By assumption there is a tubular neighbourhood of in which is homeomorphic to the normal bundle of in . By excision and Thom isomorphism the relative cohomology group is given by . Furthermore, the cohomology of coincides with the cohomology of and thus is zero in degrees above . Since has nontrivial periodic cohomology we can combine all this with the exact sequence of the relative cohomology of the pair to see that has nontrival periodic cohomology in all degrees .
Remark 3.1.
Once one has established that any order two element in an isotropy group is given by , one can also proceed differently to rule out isotropy groups of even order altogether: As above, there are only finitely many points whose isotropy groups have even order. Moreover, the Sylow group of is either cyclic or generalized quaternionic. By a theorem of Swan [Swa60] this implies that the cohomology of is nontrivial and periodic. One can then directly pass to the proof of Lemma 3.4.
4. All isotropy groups are trivial
We have seen in the last section that all isotropy groups are cyclic groups of odd order, Lemma 3.4. We fix an odd prime . In this section we plan to prove that the order of any isotropy group is not divisible by . We argue by contradiction and assume the set of points in whose isotropy group has torsion is not empty.
In any isotropy group with there is a unique normal subgroup of which is isomorphic to . This implies that is a smooth suborbifold of . Let denote a component of .
Let denote the inverse image of in the frame bundle and the corresponding subset in the Borel construction . By assumption there is a tubular neighbourhood of in which is homeomorphic to the normal bundle of in .
Lemma 4.1.
The image contains the kernel of , where denotes the unit normal bundle of in . If the normal bundle is orientable and denotes its Euler class then the kernel of the latter map is given by the image of , .
Proof.
Consider the Mayer Vietoris sequence of
Since is homotopy equivalent to and is homotopy equivalent to the first statement follows. The second statement is an immediate consequence of the exactness of the Gysin sequence. ∎
We will use that the cohomology is given by for all odd and by for all even positive . It is generated by elements in degree and . Furthermore where has degree and has degree .
We distinguish among three cases.
4.1. Case 1. The normal bundle of is orientable.
Let be a point and let be the fiber of with respect to the natural projection .
Then there is a unique normal subgroup and there are natural maps . Consider the induced map . The Euler class of the normal bundle of is mapped to the Euler class of the bundle , where denotes the natural representation. The representation decomposes into dimensional irreducible subrepresentation and, by construction, each of these dimensional subrepresentations is effective. This in turn implies that the Euler class of the bundle is a generator of , where is the codimension of . Hence is not zero. By Lemma 4.1, this nonzero element lies in the image of . We deduce that does not vanish for all . Combining with Lemma 2.3 this gives . Thus is a single point and .
4.2. Case 2. and the normal bundle of is not orientable.
Since is an orientable orbifold this assumption implies that is a nonorientable orbifold and, in particular, is not a point. Thus .
We consider the twofold cover such that the pull back of the normal bundle is orientable. The map is injective and its image is given by the fixed point set of , where is the map induced by the nontrivial deck transformation of .
By assumption, we know that the Euler class of the pull back bundle satisfies . As before we deduce that the image of in does not vanish. Therefore, is a nontrivial element in the kernel of for . If is even is the pull back of an element , with . Clearly, is in kernel of and, by Lemma 4.1, for all even . Since , this is a contradiction to Lemma 2.3.
4.3. Case 3. and the normal bundle of is not orientable.
This case is technically more
involved and we subdivide its discussion into several steps.
Step 1. Each normal space of a point decomposes into
two inequivalent dimensional subrepresentations of .
It is clear that decomposes into two subrepresentations of .
If the two representations would be equivalent
then each element would naturally induce a complex structure on the normal space,
and up to the sign the complex structure would not depend on the choice of . Since induce the same
orientation this would imply that is orientable – a contradiction.
Again, instead of working directly with we go to a suitable cover .
This time we consider a fourfold cover in which
the pullback of the bundle is orientable and decomposes into the sum of two
orientable dimensional subbundles determined by the first step above.
We summarize the properties of this cover , which are intuitively rather clear,
but whose exact derivation requires some tedious considerations:
Step 2. There is a normal cover of whose group of deck transformation is generated
by one element of order , such that the following holds true:

The pullback bundle of to is orientable and sum of two orientable dimensional subbundles. The map exchanges the subbundles and the map changes the orientation of each of them.

The unit bundle has vanishing cohomology in degrees with coefficients in .

is the total space of a fiber bundle with fiber and connected structure group.

The restrictions of both dimensional subbundles of to a fiber have Euler classes which generate .
Moreover, .
Proof.
As before denotes the inverse image of in the frame bundle of . Let be a point, with isotropy group . Let be the unique normal subgroup of isomorphic to .
We have seen above that acts on as the sum of two inequivalent representations and a trivial fourdimensional representation. Therefore, the normalizer of which is contained in has connected component . Moreover, coincides with the centralizer of . We see that has either two or four connected components.
Let be a fixed point component of , whose projection to is surjective. Then is invariant. If is not invariant, or if has only two connected components, then we could make a continuous choice of pairs along . We can then argue, similarly to the first paragraph of Case 3, that the normal bundle of is orientable, in contradiction to the assumption.
We deduce that has elements. Thus is isomorphic to where acts effectively on and acts on as . Moreover, acts on as the group . In particular, mod because otherwise does not contain elements of order .
The generator of this group exchanges the dimensional invariant subspaces of . The square preserves the subspaces and changes the orientation on each of them.
Since all isotropy groups of points in with respect to the action on are contained in and the orbit through any point of intersects , we see that is equivariantly diffeomorphic to . This in turn shows that is homotopy equivalent to .
We now consider the fold cyclic cover of with the group of deck transformations . Note that the normal bundle decomposes as a sum of invariant orientable dimensional subbundles. Hence, the bundle decomposes as a sum of orientable dimensional subbundles. But this bundle is just the pullback to of the normal bundle of .
The description the action of on above finishes the proof of (1).
The unit bundle is a covering of the unit bundle . The latter space is homotopy equivalent to the resolution of a dimensional orbifold without isotropy. This implies (2).
In order to see (3), observe that lies in the kernel of the action of on . Thus we have a canonical action of (which is isomorphic to ) on . Consider now the canonical action of on and via on . Then, for the diagonal action of on , we see that is homotop to . The canonical projection of this space to is a fiber bundle with fiber . Moreover, the structure group of this bundle is the connected group .
The restriction of each of the dimensional subbundles to the fiber is given by where and are the two inequivalent faithful representations mentioned at the beginning. This proves (4). ∎
The last statement, namely , implies that any endomorphism of order on any finitedimensional vector space is diagonalizable with eigenvalues satisfying . In particular, it is true for the endomorphism of .
If denotes the Euler class (with any coefficients) of the bundle
and denote the Euler classes of the two dimensional subbbundles, then
the first statement of the above lemma reads as follows: ;
preserves the set of four elements ;
and , for .
Step 3. Let denote the graded subalgebra of
that consists of invariant elements
divisible by the Euler class of .
Then and for .
The natural map
is injective and as in Case 2 its image is given by the invariant elements.
The subalgebra is thus isomorphic to the kernel of
.
Combining Lemma 4.1
and Lemma 2.3 Step 3 follows.
Step 4. .
Otherwise, choose a nonzero
eigenvector of .
In the subspace spanned by and we can find an
eigenvector of which is not in kernel of the restriction to .
Of course, the Euler class satisfies .
Since restricts to a generator of , we see that
with mod .
We claim that for all . We choose a circle in the base (see Step 2 (3)) such that restricts to a nonzero element in the first cohomology group of the inverse image of in . We get a fiber bundle and since the structure group is connected this bundle must be trivial. Since and restrict to nonzero elements of the cohomology of the fiber the claim follows.
Depending on the eigenvalue of , we can choose
some such that is fixed by .
The existence of this nonzero element of with
contradicts Step 3.
Step 5. For all , we have .
Proof.
By the previous step . The group is finite without torsion, thus does not have torsion either.
Let be the ring obtained by localizing at , i.e.
From the universal coefficient theorem and , for some . Let