G-structures on spheres

Abstract

A generalization of classical theorems on the existence of sections of real, complex and quaternionic Stiefel manifolds is proved.

Full text

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arXiv:math/0510149v2 [math.KT] 6 Oct 2006
G-STRUCTURES ON SPHERES
MARTINˇCADEK AND MICHAEL CRABB
Abstract. A generalization of classical theorems on the existence of sections of
real, complex and quaternionic Stiefel manifolds over spheres is proved. We obtain
a complete list of Lie group homomorphisms ρ : G → Gn, where Gn is one of the
groups SO(n), SU(n), Sp(n) and G is one of the groups SO(k), SU(k), Sp(k), which
reduce the structure group Gnin the fibre bundle Gn→ Gn+1→ Gn+1/Gn.
1. Introduction
Consider the fibrations
SO(n) → SO(n + 1) → SO(n + 1)/SO(n) = Sn,
SU(n) → SU(n + 1) → SU(n + 1)/SU(n) = S2n+1,
Sp(n) → Sp(n + 1) → Sp(n + 1)/Sp(n) = S4n+3.
(1.1)
(1.2)
(1.3)
To deal with the three cases we shall write Gn= SO(n), SU(n) or Sp(n) and d = 1, 2
or 4 as appropriate. These principal bundles reduce the structure group O(d(n+1)−1)
of the tangent bundle of the sphere Gn+1/Gn= Sd(n+1)−1to the subgroup Gn.
Let G be a Lie group and ρ : G → Gna homomorphism. We say that the structure
group Gnof a principal fibre bundle τ over a CW-complex X can be reduced to (G,ρ)
if the classifying map of this fibration τ : X → BGncan be factored (up to homotopy)
through Bρ : BG → BGn.
BG
???
?
?
?
?
?
?
?
Bρ
??
X
τ
??
BGn
This paper deals with the problem of determining those groups G and homomor-
phisms ρ to which the structure group Gn in the fibrations above can be reduced.
The problem has been solved in many interesting special cases. Considering stan-
dard inclusions ρ : G = Gk֒→ Gnwe get the famous problem on sections of Stiefel
manifolds over spheres resolved in [1], [3], [5] and [23]. The other standard inclusions
SU(k) ֒→ SO(n), Sp(k) ֒→ SO(n) and Sp(k) ֒→ SU(n) are dealt with in [12], [18] and
[19], respectively. In these cases the question was to find a minimal standard subgroup
to which Gncan be reduced.
In [15] Leonard asked an opposite question: find all maximal proper subgroups to
which Gncan be reduced. He solved it in the cases when G is a reducible maximal
Date: October 4th, 2005.
Research of the first author supported by the grant MSM 0021622409 of the Czech Ministry of
Education.
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2M.ˇCADEK AND M. C. CRABB
subgroup of Gn.
subgroup (G,ρ) if
(1) n is even and Gn= SO(n) or SU(n), unless Gn= SO(6) and G = SU(3);
(2) n ?≡ 11 mod 12 and Gn= Sp(n);
(3) G is a nonsimple irreducible maximal proper subgroup of Gn.
Using similar methods these results were improved in [20]. Nevertheless, the cases
when G is a simple Lie group and ρ : G → Gnis an irreducible representation have
remained unanswered. As a consequence of our main result we will show that if G
is one of the classical Lie groups SO(k), SU(k), Sp(k) such a reduction is impossible
except for the case SU(3) ֒→ SO(6) (and the obvious case Gn= G).
The paper is organized as follows. The main results are described in the next section.
Homotopy theoretical results needed in their proofs are contained in Section 3. The
proofs themselves appear in Section 4, based on statements on dimensions of real
representations of classical Lie groups. The computations proving these statements
are carried out in the following section. In an appendix we give precise self-contained
proofs of several more or less known results which we need and which may be of
independent interest; we also discuss possible generalizations of our main results.
Moreover, he proved that Gn cannot be reduced to any proper
2. Main results
To state our main result we recall several definitions and theorems. For any prime
p let νpstand for the p-adic valuation.
The Hurwitz-Radon number a(r) is the power of 2 given by
ν2(a(r)) = #{i | 1 ≤ i ≤ r − 1 and i ≡ 0,1,2,4 mod 8}.
For n + 1 = (2l + 1)2β+4γwith β ∈ {0,1,2,3} put j(n) = 2β+ 8γ. A well known
result of Adams ([1]) says that the structure group SO(n) in the fibration (1.1) can
be reduced to the standard subgroup SO(k) if and only if k ≥ n − j(n) + 1. This
condition can be expressed in terms of the Hurwitz-Radon numbers as
n + 1 ≡ 0mod a(n + 1 − k).
The complex James number b(r) is the positive integer with
?
0
for all primes p.
Similarly, the quaternionic James number c(r) is the positive integer determined by
νp(b(r)) =
max{i + νp(i) | 1 ≤ i ≤ (r − 1)/(p − 1)}if r ≥ p,
if r < p,
ν2(c(r)) = max{2r − 1, 2i + ν2(i) | 1 ≤ i ≤ r − 1},
νp(c(r)) = νp(b(2r))for all odd primes p.
In the papers [3], [5], [12], [18], [19] and [23] the problem of when the structure
group Gnin one of the fibrations (1.1) – (1.3) can be reduced to a group G = SU(k)
or Sp(k) via a standard inclusion G ֒→ Gnwas solved and the results were expressed
in terms of complex and quaternionic James numbers in a way similar to the result
quoted above. The following theorem can be regarded as a generalization of these
results.

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G-STRUCTURES ON SPHERES3
Theorem 2.1. Let Gnbe one of the groups SO(n), SU(n) or Sp(n) and let G be one
of the groups SO(k) with k ≥ 4, SU(k) with k ≥ 2 or Sp(k) with k ≥ 2. Suppose that
the dimension of the sphere Gn+1/Gnis at least 8 and that dimG < dimGn. Then the
structure group Gnof the principal fibre bundle
(2.1)Gn→ Gn+1→ Gn+1/Gn
can be reduced to G via a homomorphism ρ : G → Gnif and only if one of the following
cases occurs.
(A) Gn = SO(n), G = SO(k), n = m − 1, m ≡ 0 mod a(m − k) and, up to
conjugation, ρ is the standard inclusion SO(k) ֒→ SO(n).
(B) Gn= SO(n), G = SU(k), n = 2m−1, m ≡ 0 mod 2ν2(b(m−k))and, up to con-
jugation, ρ is the composition of the standard inclusions SU(k) ֒→ SO(2k) ֒→
SO(n) or the composition SU(4) → SO(8)×SO(6) ֒→ SO(15) where the first
homomorphism is given on the first factor by the standard inclusion and on the
second factor by the double covering SU(4)∼= Spin(6) → SO(6).
(C) Gn= SO(n), G = Sp(k), n = 4m −1, m ≡ 0 mod 2ν2(c(m−k))and, up to con-
jugation, ρ is the composition of the standard inclusions Sp(k) ֒→ SO(4k) ֒→
SO(n) or the exterior square Sp(3) → SO(15).
(D) Gn = SU(n), G = SU(k), n = m − 1, m ≡ 0 mod b(m − k) and, up to
conjugation, ρ is the standard inclusion SU(k) ֒→ SU(n).
(E) Gn= SU(n), G = Sp(k), n = 2m − 1, m ≡ 0 mod c(m − k) and, up to con-
jugation, ρ is the composition of the standard inclusions Sp(k) ֒→ SU(2k) ֒→
SU(n).
(F) Gn = Sp(n), G = Sp(k), n = m − 1, m ≡ 0 mod c(m − k) and, up to
conjugation, ρ is the standard inclusion Sp(k) ֒→ Sp(n).
As a consequence of Theorem 2.1 we get a partial answer to Leonard’s question from
[15].
Corollary 2.2. Under the assumptions of Theorem 2.1 there is no irreducible repre-
sentation ρ : G → Gnsuch that the structure group Gnof the principal fibre bundle
(2.1) can be reduced to (G,ρ).
Remark 2.3. As for G = Spin(k) or an exceptional simple Lie group, using Propo-
sition 3.1 and similar statements for SU(n) and Sp(n) one can easily prove that the
structure group Gnof (2.1) cannot be reduced to G via any homomorphism ρ with
the possible exception of finitely many cases in which a certain dimension condition
(the same as or similar to that in Proposition 3.1) is not satisfied.
Remark 2.4. In [6] and [24] it was shown that every stably parallelizable manifold
of dimension n either is parallelizable or has the same span as Sn. This suggests the
possibility of extending Theorem 2.1 for Gn= SO(n) from the case of the tangent
bundle over Snto the case of a stably trivial, but non-trivial, n-dimensional vector
bundle over a stably parallelizable n-manifold. We say a little more about this question
in the final section.

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4M.ˇCADEK AND M. C. CRABB
3. Auxiliary results
In this section we will summarize the results which will be needed for the proofs
of Theorem 2.1 and Corollary 2.2 in the next section. In what follows we will not
distinguish between maps and their homotopy classes.
Proposition 3.1. Let τ be a principal SO(n)-bundle over the suspension ΣX of a
pointed finite complex X. Suppose that the structure group SO(n) of τ can be reduced
to (G,ρ), where ρ : G → SO(n) is a homomorphism from a Lie group G of dimension
less than n−j, 1 ≤ j < n. Then the structure group of τ can be reduced to the standard
subgroup SO(n − j) of SO(n).
Proof. We denote the classifying map X → SO(n) of a principal fibre bundle τ by
the same letter. Suppose that SO(n) structure can be reduced to (G,ρ). Since the
standard inclusion ι : SO(n − j) → SO(n) is an (n − j − 1)-equivalence and dimG ≤
n − j − 1, the map ρ : G → SO(n) can be factored as a composition of a map
η : G → SO(n−j) and ι. Then τ can be factored through the standard inclusion ι as
shown by the following diagram.
SO(n − j)
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ι
??
X
??
??
τ
??? ? ?
? ? ? ? ? ?
G
η
ρ
??
SO(n)
?
Lemma 3.2. Let 1 ≤ k < n and let n ≥ 9 be odd. Consider a homomorphism
ρ : Gk→ O(n). Suppose that there is a map ρ′′: BG∞→ BO such that ρ′′◦ιk≃ ιn◦Bρ.
A choice of homotopy induces a diagram of fibrations:
G∞/Gk
ρ′
??
??
O/O(n)
??
BGk
Bρ
??
ιk
??
BO(n)
ιn
??
BG∞
ρ′′
??BO
Let ξ ∈ πn(BGk) be a homotopy class such that τ = Bρ ◦ ξ ∈ πn(BO(n)) classifies
the fibration (1.1). Then there is an element ξ′∈ πn(BGk) such that Bρ ◦ ξ′= τ and
ιk◦ ξ′= 0.
Proof. We deal separately with the three cases (a) Gk= SO(k), (b) Gk= SU(k), (c)
Gk= Sp(k).
Case (a).We consider the only non-trivial case: n ≡ 1 mod 8. Let α denote the
generator of πn(BO) = Z/2. Then πn(BO(n)) = Z/2 ⊕ Z/2 is generated by τ and a

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G-STRUCTURES ON SPHERES5
class β such that
ιnβ = α,ιnτ = 0
in πn(BO).
We deal first with the case that k ≥ 6. Suppose that ξ ∈ πn(BSO(k)) satisfies
Bρ ◦ ξ = τ andιkξ = α.
According to [11] (see also the Appendix, Proposition 6.2) α ∈ πn(BO) can be
factored as a composition ι6η where η ∈ πn(BSO(6)) and ι6: BSO(6) → BO is the
standard inclusion. This gives the diagram
Sn
η
??
ζ
??
α
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???
? ?
?
?
?
BO
BSO(k)
???
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?
?
ιk
??
BSO(6)
ι6
??
ι
in which ι : BSO(6) → BSO(k) is the standard inclusion and ζ = ιη.
If we show that Bρ ◦ ζ = 0 ∈ πn(BO(n)), then ξ′= ξ + ζ will satisfy the required
conditions, since
Bρ ◦ ξ′= Bρ ◦ (ξ + ζ) = Bρ ◦ ξ + Bρ ◦ ζ = τ + 0 = τ,
ιkξ′= ιk(ξ + ζ) = α + α = 0.
Write Bρ ◦ ζ = bβ + cτ with b,c ∈ Z/2. Then
bα = ιn(bβ + cτ + τ) = ιnBρ(ζ + ξ) = ρ′′ιk(ζ + ξ) = ρ′′(α + α) = 0,
which implies that b = 0. Suppose that Bρ ◦ ζ = τ. Then τ = (Bρ) ◦ ι ◦ η in the
diagram:
Sn
τ
ζ
?
?
?
?
η
??
??
???
?
BSO(k)
?
?
?
Bρ
??BO(n)
BSO(6)
ι
??
Consequently, the structure group O(n) of τ can be reduced to SO(6). Since
dimSO(6) = 15 < n − j(n),for n ≡ 1mod 8, n ≥ 17,
it follows from Proposition 3.1 that n = 9. However, the only possible homomorphism
SO(6) → O(9) is (up to conjugation) the standard inclusion. In this case ρ′′= id,
which leads to the contradiction:
α = ρ′′ι6η = ρ′′ιkιη = ιn(Bρ)ιη = ιnτ = 0.
Finally, if k < 6, then according to [11], (ιk)∗ : πn(BO(k)) → πn(BO) always
vanishes and we may take ξ′= ξ.
Case (b).In the complex case the assertion is trivial, because πn(BSU) = 0.