S_{3} × S_{3} acts freely on S^{3} × S^{3}

James F. Davis^{∗} Department of Mathematics Indiana University Bloomington, IN 47405 email: jfdavis@indiana.edu

Abstract

A free action of the direct product of two copies of the symmetric group on 3 elements on the cartesian product of two copies of the 3-sphere is constructed. This nonlinear action is constructed using surgery. The action provides a counterexample to a conjecture of Lewis made in 1968.

Results of P. A. Smith [10] and J. Milnor [5] show that a dihedral group cannot act freely on a sphere. This implies that if a group G acts freely on a sphere, then any element of order 2 must be central. Based on this G. Lewis [3] conjectured^{1} that if a group G acts freely on S^{n} × S^{n}, then any subgroup isomorphic to the Klein 4-group (ℤ_{2} ×ℤ_{2}) must intersect the center non-trivially. (This condition must hold true for a free, linear action.) We will use standard techniques from surgery theory to prove the assertion in the title of this note and thus give a counterexample to Lewis’ conjecture.

R. G. Swan [11] constructed a finite CW -complex X whose fundamental group is the symmetric group S_{3} and whose universal cover X̃ has the homotopy type of the 3-sphere S^{3}. Milnor’s result shows that X does not have the homotopy type of a closed manifold. However, there is the following well-known result, of which we sketch a proof.

∗Partially support by an NSF grant.

1I thank Alejandro Adem bringing this conjecture to my attention and Peter May for providing background for the proof.

Proposition. Swan’s complex X is a Poincaré complex whose Spivak bundle ν : X → BG reduces to a stable vector bundle ν̃ : X → BO.

Proof. Swan’s construction (or a spectral sequence argument) shows H_{3}X̃ → H_{3}X is isomorphic to ℤ ⟶^{×6} ℤ. Hence the transfer H_{3}X → H_{3}X̃ is an isomorphism, X has a fundamental class, and X is a Poincaré complex.

A direct analysis of the J-homomorphism π_{∗}O → π_{∗}G shows π_{i}(G/O) ≅ ℤ, 0, ℤ_{2}, 0, ℤ, 0, ℤ_{2} for i = 0, 1, 2, 3, 4, 5, 6 (see [4] for background on surgery classifying spaces). Thus there is a single obstruction in H^{3}(X; π_{2}(G/O)) = ℤ_{2} to reduction of the Spivak bundle of X. The transfer to the 3-fold cover Y = X̃/(12) of X is an injection on mod 2 cohomology, so it suffices to show that Y has the homotopy type of a closed manifold, namely ℝP^{3}. Proceed as follows. It is easy to find a map f : ℝP^{3} → Y which is an isomorphism on π_{1}. Then f must have odd degree, since f^{∗}(a^{3}) = f^{∗}(a)^{3} ≠ 0, where a ∈ H^{1}(Y; ℤ_{2}) is the generator. By replacing f by

h : ℝP^{3}#S^{3}# … #S^{3} → Y

with the orientation on the S^{3} chosen suitably and the map on the S^{3}’s given by S^{3} ˜→X̃ → Y, one obtains the desired homotopy equivalence h.

We need a computational lemma for the proof of the main theorem.

Lemma. 1. K̃_{0}(ℤ[S_{3} × S_{3}]) = 0.

2. L_{2}^{h}(ℤ[S_{3} × S_{3}]) = L_{2}^{p}(𝔽_{2}[S_{3} × S_{3}]).

Proof. By induction theory [1], it suffices to prove that K̃_{0} vanishes for the maximal hyperelementary subgroups: ℤ_{3} × ℤ_{3} and S_{3} × ℤ_{2}. This result for ℤ_{3} × ℤ_{3} is contained in [8, 8.4]. For S_{3} × ℤ_{2} we use the cartesian square

ℤ[S_{3} × ℤ_{2}] ⟶ ℤ[S_{3}]
↓  ↓
ℤ[S_{3}] ⟶ 𝔽_{2}[S_{3}]

where the top map and left map send the generator of ℤ_{2} to +1 and −1 respectively. Thus there is a Mayer-Vietoris exact sequence [6, Chap. 3]

K_{1}(ℤ[S_{3}])⊕K_{1}(ℤ[S_{3}]) → K_{1}(𝔽_{2}[S_{3}]) ⟶^{∂} K̃_{0}(ℤ[S_{3}×ℤ_{2}]) → K̃_{0}(ℤ[S_{3}])⊕K̃_{0}(ℤ[S_{3}])

By [11] or [9], K̃_{0}(ℤ[S_{3}]) = 0. Now 𝔽_{2}[S_{3}] = 𝔽_{2}[ℤ_{2}] × M_{2}(𝔽_{2}), so K_{1}(𝔽_{2}[S_{3}]) has order two, generated by (12). This is in the image of the left hand map, so K̃_{0}(ℤ[S_{3} × ℤ_{2}]) = 0. Thus K̃_{0}(ℤ[S_{3} × S_{3}]) = 0 and L_{2}^{h}(ℤ[S_{3} × S_{3}]) = L_{2}^{p}(ℤ[S_{3} × S_{3}]).

The rational representation theory for S_{3} × S_{3} is particularly simple: ℚ[S_{3} × S_{3}] = ℚ[S_{3}] ⊗_{ℚ} ℚ[S_{3}] is a product of matrix rings over the rational numbers. It follows that L_{2}^{p}(ℤ[S_{3} × S_{3}]) = L_{2}^{p}(ℤ̂_{2}[S_{3} × S_{3}]) where ℤ̂_{2} is the 2-adic integers (see [2, Section 4]). Since ℤ̂_{2}[S_{3} ×S_{3}] is a complete semilocal ring with 2 in its Jacobson radical, L_{2}^{p}(ℤ̂_{2}[S_{3} ×S_{3}]) = L_{2}^{p}(𝔽_{2}[S_{3} ×S_{3}]) (see [13]).

Theorem. S_{3} × S_{3} acts freely and smoothly on S^{3} × S^{3}.

Proof. By the proposition, X is a Poincaré complex with a reducible Spivak bundle, so a transversality argument (see [4, Theorem 2.23]) gives a degree one normal map

ν_{M} ⟶^{f̂} ξ
↓  ↓
M^{3} ⟶^{f} X.

There is a surgery obstruction [12, 17D],

σ_{∗}(f × f, f̂× f̂) ∈ L_{6}^{h}(ℤ[S_{3} × S_{3}]) = L_{6}^{p}(𝔽_{2}[S_{3} × S_{3}])

which vanishes if and only if f × f is normally bordant to a homotopy equivalence. By the composition formula and product formula [7]

σ_{∗}(f × f, f̂× f̂) = σ_{∗}(Id_{M} × f) + σ_{∗}(f × Id_{M})
= σ_{∗}(f, f̂) ⊗ (σ^{∗}M + σ^{∗}X) ∈ L_{6}^{p}(𝔽_{2}[S_{3} × S_{3}])

with σ_{∗}(f, f̂) ∈ L_{3}^{p}(𝔽_{2}[S_{3} × S_{3}]) and the symmetric signatures σ^{∗}M, σ^{∗}X ∈ L_{p}^{3}(𝔽_{2}[S_{3} × S_{3}]). But L_{3}^{p}(𝔽_{2}[S_{3} × S_{3}]) = 0 since the odd projective L-theory of a complete semilocal ring vanishes. Thus the integral surgery obstruction is zero and X × X has the homotopy type of a closed, smooth manifold N with fundamental group S_{3} × S_{3} and universal cover homotopy equivalent to S^{3} × S^{3}.

The surgery exact sequence

L_{7}(ℤ) → S(S^{3} × S^{3}) → [S^{3} × S^{3}, G/O] → L_{6}(ℤ)

shows that S(S^{3} × S^{3}) = ∗, since L_{7}(ℤ) = 0 and [S^{3} × S^{3}, G/O] → L_{6}(ℤ) is isomorphic to ℤ_{2} ⟶^{Id} ℤ_{2}, since there is a framed 6-manifold with non-trivial Kervaire invariant (coincidently S^{3} × S^{3} with the Lie invariant framing). Thus the universal cover of N is diffeomorphic to S^{3} × S^{3} and the deck transformations give the desired action.

References

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