Laitinen conjecture

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[edit] 1 Primary problem

Let G be a finite group. A real G-module V is a finite dimensional real vector space with a linear action of G, i.e., the action of G on V is given by a representation G \to GL(V).

Definition 1.1. Two real G-modules U and V are called Smith equivalent if there exists a smooth action of G on a homotopy sphere \varSigma with exactly two fixed points x and y at which the tangent G-modules are isomorphic to U and V, respectively, where the tangent G-modules are determined on the tangent spaces T_x(\varSigma) and T_y(\varSigma) at x and y by taking the derivatives at x and y of the diffeomorphisms \varSigma \to \varSigma, z \mapsto gz considered for all g \in G.

  • Let RO(G) be the representation ring of G, i.e., the Grothendieck ring of the differences U - V of real G-modules U and V. As a group, RO(G) is a finitely generated free abelian group whose rank is the number of real conjugacy classes of elements g\in G. Recall that the real conjugacy class of g \in G is defined by (g)^{\pm} := (g) \cup (g^{-1}). Hereafter, r_G denotes the number of real conjugacy classes (g)^{\pm} represented by elements g \in G not of prime power order.
  • Let PO(G) be the subgroup of RO(G) consisting of the differences U - V of two \mathcal{P}-matched real G-modules U and V, where the \mathcal{P}-matched condition means that U and V are isomorphic when restricted to any prime power order subgroup of G. The group PO(G) is trivial, PO(G) = 0, if and only if r_G = 0. In the case PO(G) \neq 0, PO(G) is a finitely generated free abelian group of rank r_G, \operatorname{rank} PO(G) = r_G.

We shall make use of the notions of Smith set and pimary Smith set of G.

  • The Smith set of G is the subset Sm(G) of RO(G) consisting of the differences U - V \in RO(G) of two Smith equivalent real G-modules U and V.
  • The primary Smith set of G is the subset PSm(G) of RO(G) consisting of the differences U - V \in RO(G) of two \mathcal{P}-matched and Smith equivalent real G-modules U and V.

If two real G-modules U and V with \dim U^G = \dim V^G = 0 are isomorphic, then U and V are Smith equivalent (the sphere \varSigma = S(V \oplus \mathbb{R}) admits the required action of G, where G acts trivially on \mathbb{R} and diagonally on V \oplus \mathbb{R}). Therefore, the sets PSm(G) and Sm(G) both contain the zero 0 = V - V of RO(G). The following problems are motivated by the question of Paul A. Smith posed in 1960, in the article [Smith1960, the footenote on p. 406].

  • Is it true that any two Smith equivalent real G-modules are isomorphic, i.e., is it true that Sm(G) = 0?
  • Is it true that any two \mathcal{P}-matched and Smith equivalent real G-modules are isomorphic, i.e., is it true that PSm(G) = 0?

Let d^G \colon RO(G) \to \mathbb{Z} be the dimension homomorphism, i.e., d^G(U-V) = \dim U^G - \dim V^G for any two real G-modules U and V.

Lemma 1.2 (see [Laitinen&Pawałowski1999]). For a finite group G, the following two statements are true.

  • The group PO(G) \cap \Ker \, d^G is trivial, PO(G) \cap \Ker \, d^G = 0, if and only if r_G = 0 or 1.
  • If r_G \geq 2, PO(G) \cap \Ker \, d^G is a finitely generated free abelian group of rank r_G - 1, \operatorname{rank}(PO(G) \cap \Ker \, d^G) = r_G - 1.

As PSm(G) = PO(G) \cap Sm(G) = PO(G) \cap \Ker \, d^G \cap Sm(G), Lemma 1.2 yields the following corollary.

Corollary 1.3. Let G be a finite group with r_G = 0 or 1. Then any two \mathcal{P}-matched and Smith equivalent real G-modules are isomorphic, i.e., PSm(G) = 0.

Problem 1.4 (Primary problem). For which finite groups G, the following statement is true?

  • Any two \mathcal{P}-matched and Smith equivalent real G-modules are isomorphic if and only if r_G = 0 or 1.

In order to answer the question in Problem 1.4, one shall check for which finite groups G with r_G \geq 2, there exist two \mathcal{P}-matched and Smith equivalent real G-modules that are not isomorphic (cf. Corollary 1.3). For a systematic discussion on the Smith equivalence of real G-modules, we refer to the page Smith equivalence of reresentations. Here, we shall focus only on a related conjecture posed by Erkki Laitinen (cf. [Laitinen&Pawałowski1999, Appendix]).

[edit] 2 Laitinen conjecture

Definition 2.1. Let G be a finite group. Two real G-modules U and V are called Laitinen-Smith equivalent if there exists a smooth action of G on a homotopy sphere \varSigma with exactly two fixed points x and y at which the tangent G-modules are isomorphic to U and V, respectively (cf. Definition 1.1), and the action of G on \varSigma satisfies the Laitinen condition asserting that for any element g \in G of order 2^a for a \geq 3, the fixed point set \varSigma^g = \{z \in \varSigma \ |\ gz = z\} is connected.

Definition 2.2. Let G be a finite group. Two real G-modules U and V are said to be \mathcal{P}_2-matched if U and V are isomorphic when restricted to any cyclic subgroup of G of order 2^a for a \geq 3.

In Definition 2.1, the Laitinen condition implies that the two Smith equivalent real G-modules U and V are \mathcal{P}_2-matched. As any two \mathcal{P}_2-matched and Smith equivalent real G-modules U and V are \mathcal{P}-matched, Corollary 1.3 yields the next corollary.

Corollary 2.3. Let G be a finite group with r_G = 0 or 1. Then any two Laitinen-Smith equivalent real G-modules are isomorphic.

In 1996, Erkki Laitinen ([Laitinen&Pawałowski1999, Appendix]) posed the following conjecture (cf. Problem 1.4).

Problem 2.4 (Laitinen conjecture). For a finite Oliver group G, any two Laitinen-Smith equivalent real G-modules are isomorphic if and only if r_G = 0 or 1.

In order to prove that the Laitinen conjecture holds for a finite Oliver group G, it sufficies to restrict attention to the case where r_G \geq 2, and to check that there exist two Laitinen-Smith equivalent real G-modules that are not isomorphic (cf. Corollary 2.3).

[edit] 3 Results so far

  • Laitinen and Pawałowski [Laitinen&Pawałowski1999] prove that the Laitinen conjecture holds for any finite (non-trivial) perfect group G.
  • Pawałowski and Solomon [Pawałowski&Solomon2002] prove that the Laitinen conjecture holds under either of the following condition:
    • G is a finite Oliver group of odd order (where always r_G \geq 2).
    • G is a finite Oliver group with a quotient isomorphic to \mathbb{Z}_{pq} for two distinct odd primes p and q (where always r_G \geq 2).
    • G is a finite non-solvable gap group not isomorphic to P\varSigma L_2(\mathbb{F}_{27}) := PSL_2(\mathbb{F}_{27}) \rtimes \operatorname{Aut}(\mathbb{F}_{27}), the splitting extension of PSL_2(\mathbb{F}_{27}) by the group \operatorname{Aut}(\mathbb{F}_{27}) of automorphism of the field \mathbb{F}_{27}.
  • Morimoto [Morimoto2008] obtains the first counterexample to the Laitinen conjecture by proving that the Smith set Sm(G) = 0 and r_G = 2 for G = \operatorname{Aut}(A_6).
  • Pawałowski and Sumi [Pawałowski&Sumi2009] compute the primary Smith set PSm(G) for some finite solvable Oliver groups, to the effect that:
    • PSm(G) = Sm(G) = 0 and r_G = 1 for G = (\mathbb{Z}_3 \times A_4) \rtimes \mathbb{Z}_2, confirming the Latinen conjecture.
    • PSm(G) = Sm(G) = 0 and r_G = 2 for G = S_3 \times A_4, contrary to the Laitinen conjecture.
    • PSm(G) = Sm(G) = 0 and r_G = 2 for G = (\mathbb{Z}^2_2 \rtimes \mathbb{Z}_3)^2 \rtimes \mathbb{Z}_2, contrary to the Laitinen conjecture.
    • PSm(G) = Sm(G) = 0 and r_G = 2 for G = \operatorname{Aff}_2(\mathbb{F}_3), contrary to the Laitinen conjecture.
    • PSm(G) = Sm(G) = 0 and r_G = 3 for G = (A_4\times A_4) \rtimes \mathbb{Z}^2_2, contrary to the Laitinen conjecture.
    • PSm(G) = Sm(G) = \mathbb{Z} and r_G = 3 for G = \mathbb{Z}_3 \times S_4, and they prove that any element of PSm(G) is the difference of two Laitinen-Smith equivalent real G-modules, confirming the Laitinen conjecture.
  • Morimoto [Morimoto2010] checks that PSm(G) = \mathbb{Z} for G = P\varSigma L_2(\mathbb{F}_{27}), where r_G = 2, and he proves that any element of PSm(G) is the difference of two Laitinen-Smith equivalent real G-modules, confirming the Laitinen conjecture.
  • Pawałowski and Sumi [Pawałowski&Sumi2010] confirm the Laitinen conjecture for any finite non-solvable group G not isomorphic to \operatorname{Aut}(A_6) or P\varSigma L_2(\mathbb{F}_{27}), and more generally, for any finite Oliver group G satisfying the Nil-Condition.

[edit] 4 Further discussion

Summarizing the results of [Laitinen&Pawałowski1999], [Pawałowski&Solomon2002], [Morimoto2008], [Morimoto2010], and [Pawałowski&Sumi2010], one obtains the following theorem.

Theorem 4.1. For a finite non-solvable group G not isomorphic to \Aut(A_6), the following two statements are true.

  • Any two \mathcal{P}-matched and Smith equivalent real G-modules are isomorphic if and only if r_G = 0 or 1 (cf. Problem 1.4).
  • Any two Laitinen-Smith equivalent real G-modules are isomorphic if and only if r_G = 0 or 1 (cf. Problem 2.4).

[edit] 5 References

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