Béatrice I. Chetard  Last update: November 20, 2017

Saturated graded rings of finite group representations

Contents

1 Classic graded representation rings  1
1.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 R^{∗} and PR^{∗} are not Mackey functors  13
2.1 The graded ring of A_{4} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 The permutation ring PR^{∗}(A_{4}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3 The saturated ring R  16
3.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4 Completed rings  25
4.1 p-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 General finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5 Saturated rings as Tambara functors  27
5.1 Representation rings and tensor induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
5.2 Steenrod operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6 Table of results  30

1 Classic graded representation rings

1.1 Theory

1.1.1 Definitions, first properties

Notation. I write R_{}^{∗} (G) for the graded representation ring associated to G with coefficients in the field . When a (lower case) is an element of the graded ring R_{}^{∗} (G), denote by A (upper case) any lift of a to R_{}(G). The induction of representations from a subgroup H to a bigger group G is denoted Ind_{H}^{G}, or i_{∗} for short (when H, G are obvious from the context). Similarly for the restriction Res_{H}^{G} or i^{∗}.

Definition. The weight of a monomial t = c_{i_{1}}(x_{1}) ⋯c_{i_{n}}(x_{n}), is w = ∑ i_{k}. The degree of t is n.

Definition. (See [Web]).Let R be a commutative ring with a 1 and let R-mod be the category of R-modules. Let G be a finite group. A Mackey functor over R is a function

M : {subgroups of G} −→ R-mod with, for all subgroups K ≤ H ≤ G and for all elements g ∈ G, morphisms:

I_{K}^{H} : M(K) −→ M(H)
R_{K}^{H} : M(H) −→ M(K)
c_{g} : M(H) −→ M(^{g}H)

where^{g}H = gHg^{−1} , such that, for all subgroups J ≤ K ≤ H ≤ G:

(i) I_{H}^{H}, R_{H}^{H}, c_{h} : M(H) → M(H) are the identity morphisms for all subgroups H and for all h ∈ H. (ii) R_{J}^{K}R_{K}^{H} = R_{J}^{H} (iii) I_{K}^{H}I_{J}^{K} = I_{J}^{H} (iv) c_{h}c_{h} = c_{gh} for g, h ∈ G

(v) R_{^{g}K}^{^{g}H}c_{g} = c_{g}R_{K}^{H} (vi) I_{^{g}K}^{^{g}H}c_{g} = c_{g}I_{K}^{H}

(vii) R_{J}^{H}I_{K}^{H} = ∑_{x∈[J\H/K]} I_{J∩^{x}K}^{J}c_{x}R_{J^{x}∩K}^{K}

Proposition 1.1. The graded representation ring R^{∗}(G) is of |G|-torsion.

Proof. To be added.

Proposition 1.2. 1.1 Let G and H be groups with coprime order. Then

R^{∗}(G × H) = R^{∗}(G) ⊗_{}R^{∗}(H)

Proof. Any irreducible representation ρ of G × H is of the form ρ_{G}⊗ ρ_{H} for some irreducible representations ρ_{G} of G and ρ_{H} of H. Thus R(G × H) = R(G) ⊗_{}R(H), and the filtration on R(G × H) is

Γ^{n}(G × H) = ⨁_{i=0}^{n} Γ^{i}(G) ⊗ Γ^{n−i} (H).

Since R^{k}(G) (resp. R^{k}(H)) is of |G| (resp. |H|)-torsion for k > 0, the terms Γ^{i}(G) ⊗ Γ^{n−i} (H) vanish mod Γ^{n+1} (G × H) whenever i ≠ 0, n, so that:

R^{n}(G × H) = \frac{(Γ^{0}(G) ⊗ Γ^{n}(H)) ⊕ (Γ^{n}(G) ⊗ Γ^{0}(H))}{(Γ^{0}(G) ⊗ Γ^{n+1} (H)) ⊕ (Γ^{n+1} (G) ⊗ Γ^{0}(H))} ≅ R^{n}(G) ⊕ R^{n}(H).

Thus:

R^{∗}(G × H) = ⊕ ( ⨁_{i≥1} R^{i}(G) ⊕ R^{i}(H))
= R^{∗}(G) ⊗ R^{∗}(H)

1.1.2 Evaluation of characters, topology

Topological considerations. In the sequel, we view R(G) as a topological ring, with the topology induced by the filtration {Γ^{n}}; that is, a subset U ⊆ R(G) is open if for any x ∈ U there is a t such that x + Γ^{t} ⊆ U.

Proposition 1.3 ([Ati61, Cor. 12.3]). The topology induced by the Γ-filtration coincides with the I-adic topology, where I is the augmentation ideal of R(G).

Proof. To be added.

If G has exponent m, then each conjugacy class representative g ∈ G gives rise to a ring morphism:

ϕ_{g} : \{ \begin{matrix}R(G) → [μ_{m}] \\ ρ ↦ χ_{ρ}(g) \end{matrix} ,

where μ_{m} is a choice of primitive m-th root of unity. We are interested in continuity and density questions with respect to p-adic topologies on . Note that our choice of primitive root μ_{m} fixes asn extension of the p-adic valuation to [μ_{m}], for any prime number p. Suppose we are given groups Γ̃^{n} (n ≥ 1) such that:

A. Γ̃^{n+1} ⊆ Γ̃^{n}

B. Γ^{n} = Γ̃^{n} + Γ^{n+1}

(think of Γ̃^{n} as a truncated Γ^{n}). Then by an immediate induction:

Lemma 1.4. For all k ∈ ,

Γ^{n} = Γ̃^{n} + Γ^{n+k}

Definition. Call (Γ̃^{n})_{n} an admissible approximation for (Γ^{n})_{n} if it satisfies the conditions above.

Proposition 1.5. Let p be a prime number, and fix an extension suppose the evaluation morphisms

ϕ_{1}, ⋯ϕ_{k} : R(G) ↦ [μ_{m}]

are continuous with respect to the topology induced by the filtration {Γ^{n}} on R(G), and the p-adic topology on [μ_{m}]. Then for all x ∈ Γ^{n}, there is an element x̃ ∈ Γ̃^{n} such that for all i = 1, ⋯, k:

v_{p}(ϕ_{i}(x)) = v_{p}(ϕ_{i}(x̃))

Proof. Let x ∈ Γ^{n}. Write x = x̃ + r with x̃ ∈ Γ̃^{n} and r ∈ Γ^{N}, choosing N large enough that v_{p}(ϕ_{i}(r)) > v_{p}(ϕ_{i}(x)) for all i. So

v_{p}(ϕ_{i}(x)) ≥ min{v_{p}(ϕ_{i}(x̃)), v_{p}(ϕ_{i}(r))}

with equality if v_{p}(ϕ_{i}(x̃)) ≠ v_{p}(ϕ_{i}(r)). Now, we must have

v_{p}(ϕ_{i}(x̃)) < v_{p}(ϕ_{i}(r))

since otherwise v_{p}(ϕ_{i}(x̃) + ϕ_{i}(r)) ≥ v_{p}(ϕ_{i}(r)) > v_{p}(ϕ_{i}(x)). Thus

v_{p}(ϕ_{i}(x)) = v_{p}(ϕ_{i}(x̃))

Proposition 1.6. Let G be a p-group. Then the morphisms ϕ_{g}, for g ∈ G, are all continuous with respect to the p-adic topology on [μ_{m}].

Proof. Fix an element g ∈ G and let |G| =: p^{n}. By proposition 1.3, it suffices to show that ϕ_{g} is continuous with respect to the I-adic topology on the left, and it is enough to show that for any irreducible character ρ of G,

v_{p}(ϕ_{g}(ρ − ε(ρ))) > 0.

Since G is a p-group, every character is a sum of p^{n}-th roots of unity, so

ϕ_{g}(ρ − ε(ρ)) = ρ(g) − ε(ρ)

= ∑_{l=1}^{ε(ρ)}(μ_{p^{n}}^{i_{l}} − 1),

and each (μ_{p^{n}}^{i_{l}} − 1) has positive valuation.

1.2 Examples

1.2.1 From the definitions

Cyclic groups.

Proposition 1.7 ([GM14, Prop. 3.4]). Let C_{N} be the cyclic group of order N. Whenever is an algebraically closed field of characteristic prime to N,

R_{}^{∗} (C_{N}) = \frac{[z]}{(Nz)}

with |z| = 1 and z = c_{1}(ρ) for some nontrivial, one-dimensional representation ρ of C_{N}.

Proposition 1.8.

R^{∗}(C_{N}, ) = \{ \begin{matrix}\frac{[t]}{(Nt)} & N = 2m + 1 \\ \frac{[t, s]}{(Nt, 2s, s^{2})} & N ≡ 0 (mod 4) \\ \frac{[t, s]}{(Nt, 2s, s^{2} − \frac{N}{2} t)} & N ≡ 2 (mod 4) \end{matrix}

where |t| = 2 and |s| = 1.

Proof. Representations. Let g be a choice of generator of C_{N}. The real representations of C_{N} are obtained by pairing complex representations, and are of the form:

ρ_{k}(g) = ( \begin{matrix}cos ( \frac{2kπ}{N} ) & − sin ( \frac{2kπ}{N} ) \\ sin ( \frac{2kπ}{N} ) & cos ( \frac{2kπ}{N} ) \end{matrix} )

When N = 2m + 1 is odd, we obtain m irreducible representations in this way. When N = 2m, we have (m − 1) such representations, as well as the 1-dimensional representation ε: g ↦ −1. We have the relations:

ρ_{k}ρ_{l} = ρ_{k+l} + ρ_{k−l}  (1.1)
ε^{2} =  (1.2)
ερ_{k} = ρ_{k+m}  (1.3)

where the last two relations, of course, only apply when N = 2m, and indices are taken mod N.

Generators and relations. Computing gamma operations explicitly, we have

C_{1}(ρ_{k}) = ρ_{k} − 2
C_{2}(ρ_{k}) = γ^{2}(ρ_{k} − 2) = λ^{2}(ρ_{k} − 1) = λ^{2}(ρ_{k}) − ρ_{k} + 1 = ρ_{k} + 2
C_{1}(ε) = ε − 1

We see immediately that C_{1}(ρ_{k}) = −C_{2}(ρ_{k}) ∈ Γ^{2} and so c_{1}(ρ_{k}) = 0. For the other relations, we use eq. (1.1) above and apply the total Chern class c_{t} to both sides:

c_{t}(ρ_{k}ρ_{l}) = c_{t}(ρ_{k+l} )c_{t}(ρ_{k−l} )  (1.4)

Let y_{i} = c_{2}(ρ_{i}). Expand the right-hand side:

c_{t}(ρ_{k+l} )c_{t}(ρ_{k−l} ) = (1 + y_{k+l} T^{2})(1 + y_{k−l} T^{2})
= 1 + (y_{k+l} + y_{k−l} )T^{2} + y_{k+l} y_{k−l} T^{4}  (1.5)

For the left-hand side, we use the splitting principle. In some extension of R(D_{p}), we can write ρ_{k} = σ_{1} + σ_{2} and ρ_{l} = η_{1} + η_{2} with σ_{i}, η_{i} of dimension 1. Then

c_{t}(ρ_{k}ρ_{l}) = c_{t}((σ_{1} + σ_{2})(η_{1} + η_{2}))
= c_{t}(σ_{1}η_{1})c_{t}(σ_{1}η_{2})c_{t}(σ_{2}η_{1})c_{t}(σ_{2}η_{2})

= (1 + (c_{1}(σ_{1}) + c_{1}(η_{1}))T) · (1 + (c_{1}(σ_{1}) + c_{1}(η_{2}))T) · (1 + (c_{1}(σ_{2}) + c_{1}(η_{1}))T) · (1 + (c_{1}(σ_{2}) + c_{1}(η_{2}))T)

Now, let s_{1}, s_{2} (resp. t_{1}, t_{2}) be the first and second symmetric polynomials in (σ_{1}, σ_{2}) (resp. (η_{1}, η_{2})). Then c_{i}(ρ_{k}) = s_{i} and c_{i}(ρ_{l}) = t_{i}. The last equality can be rewritten:

c_{t}(ρ_{k}ρ_{l}) = 1 + 2(s_{1} + t_{1})T + (t_{1}^{2} + s_{1}^{2} + 3s_{1}t_{1} + 2t_{2} + 2s_{2})T^{2}
+ (s_{1}^{2}t_{1} + s_{1}t_{1}^{2} + 2s_{1}s_{2} + 2t_{1}t_{2} + 2s_{1}t_{2} + 2s_{2}t_{1})T^{3}
+ (t_{2}^{2} + s_{2}^{2} + s_{1}s_{2}t_{1} + s_{1}t_{1}t_{2} + s_{1}^{2}t_{2} + s_{2}t_{1}^{2} − 2s_{2}t_{2})T^{4}

We eliminate all occurrences of s_{1} = t_{1} = 0 to obtain:

c_{t}(ρ_{k}ρ_{l}) = 1 + (2y_{k} + 2y_{l})T^{2} + (y_{k} − y_{l})^{2}T^{4}  (1.6)

Comparing coefficients in eq. (1.5) and eq. (1.6), we obtain:

y_{k+l} + y_{k−l} = 2(y_{k} + y_{l})  (1.7)
y_{k+l} y_{k−l} = (y_{k} − y_{l})^{2}  (1.8)

Now plugging l = k and l = k + 1 in relation (1.7), we get

y_{2k} = 4y_{k}
y_{2k+1} = 2y_{k} + 2y_{k+1} − y_{1}

which by induction gives y_{k} = k^{2}y_{1}. Let t = y_{1}. Since the order N of the group C_{N} annihilates R^{∗}(C_{N}), we have Nt = 0. To show that the order of t is N we look at the complexification morphism ψ : R^{∗}(C_{N}, ) → R^{∗}(C_{N}, ) which sends t to −x^{2}, where x is the degree 1 generator of R^{∗}(C_{N}, ). This incidentally shows that t is not nilpotent. Thus if N is odd, the ring is generated by t alone, and we have indeed:

R^{∗}(C_{N}, ) = \frac{[t]}{(Nt)}

Now if N = 2m is even, we take into account the other generator s = c_{1}(ε). The fact that ε^{2} = 1 implies immediately that 2s = 0, and we only need to apply c_{t} to relation 1.3 to derive:

y_{k+m} = y_{k} + s^{2}

in which we can plug y_{k} = k^{2}t to obtain

s^{2} = m(2k + m)t

for all k = 1, ⋯, m − 1.

• If m is itself even, then this can be rewritten as s^{2} = 2m(k + \frac{m}{2})t = 0, as n = 2m kills all elements in R^{∗}(C_{N}, ). Thus we have the relation s^{2} = 0. Moreover, since the complexification morphism ψ sends s = c_{1}(ϵ) to mx ∈ R^{∗}(C_{N}, ), we have ψ(st^{i}) = −mx^{2i=1} ≠ 0, thus

R^{∗}(C_{N}, ) = \frac{[t, s]}{(Nt, 2s, s^{2})}

• If m = 2l + 1 for some l then s^{2} = 2m(k + s)t + mt = mt, and yet another application of ψ yields:

R^{∗}(C_{N}, ) = \frac{[t, s]}{(Nt, 2s, s^{2} − \frac{N}{2} t)}

Elementary abelian groups. Let p be a prime number, and let C_{p} be the cyclic group of order p, generated by g.

Proposition 1.9. Let G = C_{p}^{k}. Then

R_{}^{∗} (G) = \frac{[x_{1}, ⋯, x_{k}]}{(px_{i}, x_{i}^{p}x_{j} = x_{i}x_{j}^{p})}

Proof. Denote by g_{i} the element (1, ⋯, 1, g, 1, ⋯, 1) with g in i-th position. Thus G is generated by g_{1}, ⋯, g_{k}. Let ω = exp^{2iπ/p} . There are p^{k} irreducible complex representations of G, of the form ρ_{v}, with v = (v_{1}, ⋯, v_{k}) ∈ _{≥0}^{k} , and ρ_{v}(g_{i}) = ω^{v_{i}}. These satisfy the two relations:

ρ_{v}ρ_{v′} = ρ_{v+v′}  (1.9)
ρ_{v}^{p} = 1.  (1.10)

Let x_{v} = c_{1}(ρ_{v}), and let ρ_{i} be the representation associated to w = (0, ⋯, 0, 1, 0 ⋯, 0) with a 1 in position i. Then by (1.9), we have

x_{v} = ∑_{i=1}^{k} v_{i} · x_{i}

thus the x_{i} generate R_{}^{∗} (G). The relation (1.10) yields px_{i} = 0 for any i. The relation x_{i}^{p}x_{j} = x_{i}x_{j}^{p} is obtained as follows: let X_{i} be the standard lift ρ_{i} − 1 of x_{i} to R_{}(G). Then (X_{i} + 1)^{p} = 1. Thus

X_{i}^{p} = − ∑_{l=1}^{p−1} \binom{p}{l} X_{i}^{l} = X_{i} ( − ∑_{l=0}^{p−2} \binom{p}{l + 1} X_{i}^{l})
= pX_{i}(−1 + ϕ(X_{i})),

where ϕ(T) ∈ [X] has no constant term. Thus for any i, j, we have

X_{i}^{p}X_{j} (−1 + ϕ(X_{j})) = pX_{i}X_{j} (−1 + ϕ(X_{i})) (−1 + ϕ(X_{j}))
= X_{i}X_{j}^{p} (−1 + ϕ(X_{i}))

That is, quotienting by Γ^{p+2} on each side:

x_{i}^{p}x_{j} = x_{i}x_{j}^{p}.

This shows that the generators of R_{}^{∗} (G) satisfy all the required relations.

We now show that these are the only relations: the graded piece of rank l is generated by monomials of the form

x_{1}^{s_{1}} ⋯x_{k}^{s_{k}}, ∑_{i=1}^{k} s_{i} = l

Now let S_{l} ⊂ _{≥0}^{k} be the set of multi-indices (s_{1}, ⋯, s_{k}) such that ∑ s_{i} = l and only the first nonzero coordinate of each s ∈ S_{l} is (possibly) greater than p − 1. To show that the relations above are the only one, we must show that the monomials x^{s} = x_{1}^{s_{1}} ⋯x_{k}^{s_{k}} are linearly independent. So let us consider a zero linear combination of those:

∑_{s∈S_{l}} a_{s}x^{s} = 0  (1.11)

Consider the restriction ψ : R_{}^{∗} (C_{p}^{k}) → R_{}^{∗} (C_{p}) to the group generated by g_{1}^{i_{1}} ⋯g_{k}^{i_{k}} for some 0 ≤ i_{j} ≤ p − 1. Then ψ(x_{j}) = i_{j}z, where z is the standard one-degree generator of R_{}^{∗} (C_{p}). Then eq. (1.11) becomes:

∑_{s∈S_{l}} a_{s}i_{1}^{s_{1}} ⋯i_{k}^{s_{k}}z^{l} = 0,

that is,

∑_{s∈S_{l}} a_{s}i_{1}^{s_{1}} ⋯i_{k}^{s_{k}} = 0 ∈ _{p},  (1.12)

for all possible strings (i_{1}, ⋯, i_{k}) with 0 ≤ i_{j} ≤ p − 1. In particular, grouping terms by powers of i_{k} in eq. (1.12), we get:

\{ \begin{matrix}a_{(0,···,0,l)}i_{k}^{l} = 0 & when i_{1} = ⋯= i_{k−1} = 0 \\ ∑_{t=0}^{p−1} ( ∑_{s∈S_{l−t}} b_{s}i_{1}^{s_{1}} ⋯i_{k−1}^{s_{k−1}} ) i_{k}^{t} = 0 & otherwise. \end{matrix}  (1.13)

This implies that the coefficient of x_{k}^{l} in eq. (1.11) is zero; more generally, the second equation must be true for all values of i_{k}, from 0 to p − 1. In other words, the ( ∑ b_{s}i_{1}^{s_{1}} ⋯i_{k−1}^{s_{k−1}} ) are the entries of a vector in the

kernel of the Vandermonde matrix (i_{k}^{t})_{i_{k}=1,···,p−1}^{t=1,···,p−1} , which is invertible in _{p}. Therefore

∑_{s∈S_{l−t}} b_{s}i_{1}^{s_{1}} ⋯i_{k−1}^{s_{k−1}} = 0

for all combinations (i_{1}, ⋯, i_{k−1} ). An immediate induction shows that we must have each a_{s} = 0, so the monomials {x^{s}}_{s∈S_{l}} are linearly independent.

(Some) dihedral groups. Let p be an odd prime and consider the dihedral group

D_{p} = ⟨τ, σ|τ^{2}, σ^{p}, τστ = σ^{−1} ⟩ .

There are (p + 1)/2 irreducible representations of D_{p}:

• Two representations of degree 1, the trivial representation 1 and the "signature" ε which sends elements of the form σ^{j} to -1, and elements of the form τσ^{j} to 1.

• And (p − 1)/2 representations χ_{1}, ⋯χ_{(p−1)/2} of degree 2. These are defined over the reals as:

χ_{k}(σ^{j}) = ( \begin{matrix}cos(\frac{2kjπ}{p} ) & − sin(\frac{2kjπ}{p} ) \\ sin(\frac{2kjπ}{p} ) & cos(\frac{2kjπ}{p} ) \end{matrix} )

χ_{k}(τ) = ( \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix})

And over the complex numbers:

χ_{k}(σ^{j}) = ( \begin{matrix}e^{\frac{2ikjπ}{p}} & 0 \\ 0 & e^{− \frac{2ikjπ}{p}} \end{matrix} )

χ_{k}(τ) = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix})

These generate the ring R(D_{p}). For convenience, define χ_{0} = 1 + ε, then we have the following relations:

ε^{2} = 1  (1.14)
ε · χ_{k}= χ_{k}  (1.15)
χ_{k} · χ_{l} = χ_{k+l} + χ_{k−l}  (1.16)

Proposition 1.10. Let c_{1}(χ_{1}) = x and c_{2}(χ_{1}) = y, then

R^{∗}(D_{p}) = \frac{[x, y]}{(2x, py, xy)}

where |x| = 1 and |y| = 2.

Proof. As above, note that 0 = c_{1}(1) = c_{1}(ε^{2}) = 2c_{1}(ε) while c_{1}(χ_{k}) = c_{1}(det χ_{k}) = c_{1}(ε) for any k. So let x = c_{1}(χ_{k}) = c_{1}(ε). For the other relations, we use eq. (1.16) above and apply the total Chern class c_{t} to both sides:

c_{t}(χ_{k}χ_{l}) = c_{t}(χ_{k+l} )c_{t}(χ_{k−l} )  (1.17)

Let y_{i} = c_{2}(χ_{i}). Expand the right-hand side:

c_{t}(χ_{k+l} )c_{t}(χ_{k−l} ) = (1 + xT + y_{k+l} T^{2})(1 + xT + y_{k−l} T^{2})
= 1 + 2xT + (x^{2} + y_{k+l} + y_{k−l} )T^{2} + (xy_{k+l} + xy_{k−l} )T^{3} + y_{k+l} y_{k−l} T^{4}  (1.18)

For the left-hand side, we use the splitting principle. In some extension of R(D_{p}), we can write χ_{k} = ρ_{1} + ρ_{2} and χ_{l} = η_{1} + η_{2} with ρ_{i}, η_{i} of dimension 1. Then

c_{t}(χ_{k}χ_{l}) = c_{t}(ρ_{1} + ρ_{2})(η_{1}) + η_{2}))
= c_{t}(ρ_{1}η_{1})c_{t}(ρ_{1}η_{2})c_{t}(ρ_{2}η_{1})c_{t}(ρ_{2}η_{2})

= (1 + (c_{1}(ρ_{1}) + c_{1}(η_{1}))T) · (1 + (c_{1}(ρ_{1}) + c_{1}(η_{2}))T) · (1 + (c_{1}(ρ_{2}) + c_{1}(η_{1}))T) · (1 + (c_{1}(ρ_{2}) + c_{1}(η_{2}))T)

Now, let s_{1}, s_{2} (resp. t_{1}, t_{2}) be the first and second symmetric polynomials in (ρ_{1}, ρ_{2}) (resp. (η_{1}, η_{2})). Then c_{i}(χ_{k}) = s_{i} and c_{i}(χ_{l}) = t_{i}. The last equality can be rewritten:

c_{t}(χ_{k}χ_{l}) = 1 + 2(s_{1} + t_{1})T + (t_{1}^{2} + s_{1}^{2} + 3s_{1}t_{1} + 2t_{2} + 2s_{2})T^{2}
+ (s_{1}^{2}t_{1} + s_{1}t_{1}^{2} + 2s_{1}s_{2} + 2t_{1}t_{2} + 2s_{1}t_{2} + 2s_{2}t_{1})T^{3}
+ (t_{2}^{2} + s_{2}^{2} + s_{1}s_{2}t_{1} + s_{1}t_{1}t_{2} + s_{1}^{2}t_{2} + s_{2}t_{1}^{2} − 2s_{2}t_{2})T^{4}

We replace s_{1} = t_{1} = x and eliminate all occurrences of 2x to obtain

c_{t}(χ_{k}χ_{l}) = 1 + (x^{2} + 2y_{k} + 2y_{l})T^{2} + (y_{k} − y_{l})^{2}T^{4}  (1.19)

Comparing coefficients in eq. (1.18) and eq. (1.19), we obtain:

y_{k+l} + y_{k−l} = 2(y_{k} + y_{l})  (1.20)
xy_{k+l} + xy_{k−l} = 0  (1.21)
y_{k+l} y_{k−l} = (y_{k} − y_{l})^{2}  (1.22)

First look at eq. (1.21) with k = l. Note that y_{0} = c_{2}(1 + ε) = 0, and thus eq. (1.21) yields x · y_{2k} = 0 for all k, which is equivalent to x · y_{k} = 0 for all k since indices are understood modulo the odd prime p. We then show is py_{k} = 0 for all k. For this we use that for any group G, the graded ring R^{∗}(G) is |G|-torsion

(this is also in Yohann’s notes, I’ll read the proof an rewrite it here). Here |D_{p}| = 2p, so we multiply the equation

y_{2k} = 4y_{k} by p to obtain that py_{2k} = 0 for all k. Again, this imply that y_{k} = 0 for all k. Finally, consider eq. (1.20), both with l = k and l = k + 1. This gives:

y_{2k} = 4y_{k}
y_{2k+1} = 2(y_{k} + y_{k+1} ) − y_{1}

together, these two relations imply that all y_{k}’s are multiples of y_{1}. We define y = y_{1}.

It remains to show that these are the only relations in R^{∗}(D_{p}). For this we use the standard restriction argument: the representation ϵ restricts to the generator of R^{∗}(C_{2}), whose powers are all nonzero. The restriction of χ_{1} to R(C_{p}, ) is ρ + ρ^{−1} , where ρ sends the a generator σ of C_{p} to e^{2iπ/p} . Thus the induced restriction R^{∗}(D_{p}, ) → R^{∗}(C_{p}, ) sends y to −c_{1}(ρ)^{2}, which is not nilpotent. Finally, the restriction of χ_{1} to R(C_{p}, ) is the representation ρ̂obtained by pairing the complex representations ρ discussed above with its complex conjugate \overline{ρ} = ρ^{−1} . This induced a map R^{∗}(D_{p}, ) → R^{∗}(C_{p}, ) sending y to the (non-nilpotent) generator t of R^{∗}(C_{p}, ). So the powers of y are also nonzero in this case. Note that in all cases, the restriction of ε to C_{p} is the trivial representation, so there cannot be any linear dependency relation between powers of x and powers of y.

Now take the dihedral group D_{4} of order 8. It has four nontrivial irreducible representations:

• In degree 1, the representations ρ : r ↦ −1, s ↦ 1 and η : r ↦ 1, s ↦ −1 and their product ρη.

• In degree 2, the representation Δ, which sends s to ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) and r to ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) .

With the relations:

ρ^{2} = η^{2} = 1  (1.23)
ρΔ = ηΔ = Δ  (1.24)
Δ^{2} = 1 + rho + η + ρη  (1.25)

Then Proposition 1.11. Let c_{1}(ρ) = x; c_{1}(η) = y and c_{2}(Δ) = b. Then

R^{∗}(D_{p}) = \frac{[x, y, b]}{(2x, 2y, 4b, xy, xb − yb)}

where |x| = |y| = 1 and |b| = 2.

Proof. Note that c_{1}(ρη) = x + y and c_{1}(Δ) = c_{1}(detΔ) = c_{1}(ρη). So the graded ring is indeed generated by x, y, b. We have 2x = 2y = 0 from the relations above, and XB = YB = XY, so XY ∈ Γ^{3} and xy = 0, and xb = yb. Finally, applying c_{t} to eq. (1.25), we get that 5c_{1}(Δ)^{2} + 4b = x^{2} + 3xy + y^{2} = x^{2} + y^{2}, and since c_{1}(Δ) = x + y, we obtain 4b = 0. To see these are the only relations, we use the computation of R^{∗}(D_{4}) ⊗ _{2} = \frac{[x,y,b]}{(xy,xb−yb)} from [GM14]: tensoring with _{2} shows that none of x, y, b is nilpotent. Finally, restriction to C_{4} =< r > show that 2b is non-nilpotent, so that any power of b is 4-torsion.

1.2.2 Topological method

The quaternion group of order 8. Let G = Q_{8} = ⟨ i, j, k | i^{2} = j^{2} = k^{2} = ijk ⟩. We prove:

Theorem 1.12.

R_{}^{∗} (Q_{8}) = \frac{[x, y, u]}{(2x, 2y, 8u, x^{2}, y^{2}, xy − 4u)}

where |x| = |y| = 1 and |u| = 2

The group Q_{8} has 5 conjugacy classes: {1}, {−1}, {±i}, {±j}, {±k} so 5 irreducible representations on :

• In dimension 1, the trivial representation , ρ_{1} \{ \begin{matrix}i ↦ 1 \\ j ↦ −1 \end{matrix} , ρ_{2} = −ρ_{1} and ρ_{3} = ρ_{1}ρ_{2}.

• In dimension 2, the representation Δ, which is in matrix form:

Δ(−1) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) , Δ(i) = ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) , Δ(j) = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , Δ(k) = ( \begin{matrix}0 & −i \\ −i & 0 \end{matrix} )

These representations satisfy the relations

ρ_{i}^{2} = 1  (1.26)
ρ_{3} = ρ_{1}ρ_{2}  (1.27)
Δρ_{i} = Δ  (1.28)
Δ^{2} = + ρ_{1} + ρ_{2} + ρ_{3}  (1.29)

Let us first take a look at the generators and relations of Q_{8}:

Lemma 1.13. The graded ring R_{}^{∗} (Q_{8}) is generated by the elements

x := c_{1}(ρ_{1}), y := c_{1}(ρ_{2}), u := c_{2}(Δ)

and those satisfy the relations in theorem 1.12, that is

2x = 2y = 8u = 0, x^{2} = y^{2} = 0, xy = 4u.

Proof. First we eliminate the redundant generators: c_{1}(ρ_{3}) = c_{1}(ρ_{1}ρ_{2}) = c_{1}(ρ_{1}) + c_{1}(ρ_{2}), and c_{1}(Δ) = c_{1}(det(Δ)) = c_{1}( ) = 0. So R_{}^{∗} (Q_{8}) is indeed generated by x, y and u. Now since ρ_{1}^{2} = ρ_{2}^{2} = , we have 2x = 2y = 0, and the order of Q_{8} kills R_{}^{∗} (Q_{8}) so 8u = 0. For the relations in degree 2, we apply the total class c_{t} to the relation Δρ_{i} = Δ, splitting the 2-dimensional representation Δ into σ_{1} + σ_{2} where σ_{i} has dimension 1. On the left-hand side we have:

c_{t}(Δρ_{i}) = c_{t}(σ_{1}ρ_{i})c_{t}(σ_{2}ρ_{i})

= 1 + [c_{1}(σ_{1}) + c_{1}(σ_{2}) + 2c_{1}(ρ_{i})] T + [c_{1}(σ_{1})c_{1}(σ_{2}) + c_{1}(σ_{1})c_{1}(ρ_{i}) + c_{1}(σ_{2})c_{1}(ρ_{i}) + c_{1}(ρ_{i})^{2}] T^{2}

While on the right-hand side:

c_{t}(Δ) = 1 + [c_{1}(σ_{1}) + c_{1}(σ_{2})] T + [c_{1}(σ_{1})c_{1}(σ_{2})] T^{2}.

In degree 2, this yields the relation:

c_{1}(ρ_{i})(c_{1}(σ_{1}) + c_{1}(σ_{2})) + c_{1}(ρ_{i})^{2} = 0.

Bearing in mind that c_{1}(σ_{1}) + c_{1}(σ_{2}) = c_{1}(Δ) = 0, we obtain x^{2} = y^{2} = 0. The relation xy = 4u is obtained by applying c_{t} to the relation Δ^{2} = + ρ_{1} + ρ_{2} + ρ_{1}ρ_{2} and identifying the terms in degree 2, which yields

5c_{1}(Δ)^{2} + 4u = x^{2} + y^{2} + 3xy,

that is, 4u = xy.

What remains to be proven is that these are the only relations satisfied by the generators; thus we want to check that we have no extra nilpotency or torsion conditions on u, and that the products xu^{i} and yu^{i} are nonzero for any i. For this, we look at the 2-valuation of the characters of Q_{8}: we first define an admissible approxiamtion (Γ̃^{n}) that only takes into accounts lifts of our generators. This allows us to restrict ourselves when we compute the 2-valuations of our evaluation morphisms, which we use to obtain information about torsion in R^{∗}.

Let X, Y, U be standard lifts of x, y, u. We consider the approximation:

Γ̃^{n} = ⟨X^{ϵ_{1}}Y^{ϵ_{2}}U^{k}⟩, 2k + ϵ_{1} + ϵ_{2} ≥ n, 0 ≤ ϵ_{1} + ϵ_{2} ≤ 1  (1.30)

Lemma 1.14. The approximation (Γ̃^{n}) is admissible.

Proof. Obviously we have Γ̃^{n+1} ⊂ Γ̃^{n}, so (Γ̃^{n}) satisfies item A in the definition of an admissible filtration. To check B, let Z = ρ_{3} − 1 and T = Δ − 2 be lifts of c_{1}(ρ_{3}), c_{1}(Δ), respectively. Then let

α = X^{ϵ_{1}}Y^{ϵ_{2}}Z^{ϵ_{3}}U^{k}T^{l} ∈ Γ^{n}, 2k + l + ϵ_{1} + ϵ_{2} + ϵ_{3} ≥ n

If ϵ_{1} ≥ 2, then α contains a factor X^{2}, but the relation x^{2} = 0 implies that X^{2} ∈ Γ^{3}, so in that case α ∈ Γ^{n+1} . The same goes for ϵ_{2}, so we can restrict ourselves to monomials such that ϵ_{1} + ϵ_{2} ≤ 1. Similarly, since c_{1}(Δ) = 0 we have T ∈ Γ^{2}, which means that l ≠ 0 forces α ∈ Γ^{n+1} . We proceed similarly for all factors and obtain

Γ^{n} = Γ̃^{n} + Γ^{n+1}

Now, there are 4 nontrivial evaluation morphisms on R_{}(Q_{8}):

• ϕ_{−1} : ρ_{i} ↦ 1, Δ ↦ −2.

• ϕ_{i} : ρ_{1} ↦ 1, ρ_{2}, ρ_{3} ↦ −1, Δ ↦ 0 • ϕ_{j} : ρ_{2} ↦ 1, ρ_{1}, ρ_{3} ↦ −1, Δ ↦ 0

• ϕ_{k} : ρ_{3} ↦ 1, ρ_{1}, ρ_{2} ↦ −1, Δ ↦ 0

Now we apply the morphisms to the standard lifts of the Chern classes and obtain

• ϕ_{−1} : X, Y ↦ 0, T ↦ −4, U ↦ 4

• ϕ_{i} : X ↦ 0, Y, T ↦ −2, U ↦ 2

• ϕ_{j} : X, T ↦ −2, Y ↦ 0, T ↦ 2

• ϕ_{k} : X, Y, T ↦ −2, T ↦ 2

Lemma 1.15. The morphisms ϕ_{g} are continuous with respect to the 2-valuations on , for all g ∈ Q_{8}.

Proof. Consider the value of ϕ_{−1} on a generic monomial α ∈ Γ^{n}. Write α = X^{ϵ_{1}}Y^{ϵ_{2}}Z^{ϵ_{3}}U^{k}T^{l} with ϵ_{1} + ϵ_{2} + ϵ_{3} + k + l ≥ n. Reading the values of ϕ_{−1} , we see that ϕ_{−1} (α) ≥ 2^{n/2}, which means that ϕ_{−1} is continuous. Similarly for the other ϕ_{g}’s.

We can now wrap up:

Proof of theorem 1.12.We want to show that R_{}^{2n} = ⟨u^{n}⟩ = /8 , and R_{}^{2n+1} = ⟨xu^{n}, yu^{n}⟩ = ( /2 )^{2}. We first look at R_{}^{2n}, where we need to show that 4u^{n} ≠ 0, that is, 4U^{n} ∉ Γ^{2n+1} . We have ϕ_{g}(4U^{n}) = 2^{n+2} , for any g = i, j, k. Suppose that 4U^{n} ∈ Γ^{2n+1} ; then by proposition 1.5 there is an element x̃ ∈ Γ̃^{2n+2} satisfying

n + 2 = v_{2}(ϕ_{g}(4U^{n})) = v_{2}(ϕ_{g}(x̃))

so that v_{2}(ϕ_{g}(x̃)) = n + 2. Write

x̃ = a_{1}U^{n+1} + a_{2}XU^{n} + a_{3}YU^{n} + U^{n+2} P(X, Y, U) + XU^{n+1} Q(X, Y, U) + YU^{n+1} R(X, Y, U)

where P, Q, R are polynomials with integer coefficients. We then apply ϕ_{g} for g = i, j, k to this equation, divide the 3 relations thus obtained by 2^{n+1} and look at the result mod 2. The three ϕ_{g} give us a system of three equations:

0 = a_{1} + a_{3}
0 = a_{1} + a_{2}
0 = a_{1} + a_{2} + a_{3}

which has no nontrivial solution. But if a_{m} = 0(mod 2) for m = 1, 2, 3 then v_{2}(ϕ_{g}(x̃)) > n + 2, which is impossible. Thus 4U^{n} cannot be in Γ^{2n+1} , and the additive order of u^{m} is indeed 8 for all m. The same process shows that xu^{n}, yu^{n} and xu^{n} + yu^{n} are nonzero in R_{}^{2n+1} for all n.

Products of cyclic 2-groups .

Theorem 1.16.

R^{∗}( /4 ⊗ /2) = \frac{[x, y]}{(4x, 2y, xy^{3} + x^{2}y^{2})}

with |x| = |y| = 1

Proof. Let ρ be a generator of R( /4) and σ the nontrivial representation of /2. Then R^{∗}(G) is generated by c_{1}(ρ) =: x and c_{1}(σ) =: y. By restriction to cyclic subgroups, we see that x has additive order 4 and y, additive order 2. Now consider X = ρ − 1 and Y = σ − 1. Then by expanding (X + 1)^{4} = 1 we get

4X = −(6X^{2} + 4x^{3} + X^{4})

On the other hand, Y^{n} = (−2)^{n−1} Y, so

XY^{3} = 4XY = −X^{4}Y − 4X^{3}Y − 6X^{2}Y = −X^{4}Y − X^{3}Y^{3} + 3X^{2}Y^{2}.

Quotienting by Γ^{5} we get that xy^{3} = x^{2}y^{2}. We now show these are the only relations; the only other possible extra relations (that cannot be ruled out by restrictions to various subgroups) are: x^{n−1} y = 0, x^{n−2} y^{2} = 0 or x^{n−1} y = x^{n−2} y for some n. We use the continuity method to disprove all of these. Let Γ̃^{n} = ⟨X^{n}, Y^{n}, X^{n−1} Y, X^{n−2} Y^{2}⟩, then {Γ̃^{n}}_{n} is an acceptable approximation for {Γ^{n}}_{n}. First suppose that x^{n−2} y = 0 for some n. Then for N arbitrarily large, there exists Z̃ ∈ Γ̃^{n+1} such that X^{n−2} Y = Z̃ + R with R ∈ Γ^{N}; in particular, for any element (i, j) ∈ /4 × /2 we have

v_{2} ( X^{n−2} Y^{2}|_{(i,j)}) = v_{2} ( Z̃|_{(i,j)}) .

Write

Z̃ = a · X^{n+1} + b · X^{n}Y + c · X^{n−1} Y^{2} + d · Y^{n+1} + P · X^{n+2} + Q · X^{n+1} Y + S · X^{n}Y^{2} + T · Y^{n+2}

where a, b, c, d ∈ P, Q, S, T ∈ [X, Y], then evaluate at (2, 1); so that X|_{(2,1)} = Y|_{(2,1)} = −2. Then the 2-valuation of X^{n−1} Y is (−2)^{n} while the 2-valuation of Z̃ is at least n + 1. This shows that x^{n−1} y cannot be zero. The same proof works for x^{n−2} y^{2}. The only possible remaining relation is x^{n−1} y = x^{n−2} y^{2}. Let Z = X^{n−1} Y + X^{n−2} Y^{2}. With notations as above, we have

Z̃|_{(2,0)} = a · (−2)^{n+1} + P · (−2)^{n+2}

while Z|_{(2,0)} = 0. Thus a = 0 (mod 2), so write a = 2a′. We now evaluate at (1, 1):

Z|_{(1,1)} = (i − 1)^{n−2} · 4 + (i − 1)^{n−1} · (−2).

thus v_{2}(Z|_{(1,1)}) = (n + 1)/2. On the other hand:

Z̃|_{(1,1)} = a′· 2 · (i − 1)^{n+1} + b · (i − 1)^{n}(−2) + c · (i − 1)^{n−1} 4 + d · (−2)^{n+1} + R̃

where v_{2}(R̃) ≥ \frac{n+2}{2} . We see that v_{2}(Z̃|_{(1,1)}) ≥ (n + 2)/2, which completes the proof.

Conjecture 1.17. Let G = /2^{n} × /2. Then

R^{∗}(G) = \frac{[x, y]}{(2^{n}x, 2y, xy^{n+1} + x^{2}y^{n})}

2 R^{∗} and PR^{∗} are not Mackey functors

2.1 The graded ring of A_{4}

Let A_{4} be the alternating group on 4 elements, generated by the permutations (12)(34) and (123). There are 4 irreducible complex representations of A_{4}:

• Of dimension 1: the trivial representation , and the representations ρ (resp. ρ̄) that send (123) to e^{2iπ/3} (resp. e^{−2iπ/3} ) and (12)(34) to 1.

• Of dimension 3: the representation θ, which is the quotient of the representation θ̄ acting on ^{4} by permutation of the basis vectors, by the trivial representation. The character χ_{θ} of θ sends 3-cycles to 0 and (12)(34) to −1.

There are the following relations between the representations:

ρ^{2} = ρ̄  (2.1)
ρθ = θ  (2.2)
θ^{2} = + ρ + ρ̄+ 2θ  (2.3)

Additionally λ^{2}(θ) = θ (by a direct calculation of the exterior power) and det(θ) = .

Conjecture 2.1. Let x = c_{1}(ρ) and y = c_{2}(θ), then

R_{}^{∗} (A_{4}) = \frac{[x, y]}{(3x, 12y, 4y + x^{2})}

I do not have a complete proof of this. However, I can show:

Proposition 2.2. Let x = c_{1}(ρ) and y = c_{2}(θ), then

R_{}^{∗} (A_{4}) ⊗ _{2} = _{2}[y]
R_{}^{∗} (A_{4}) ⊗ _{3} = \frac{_{3}[x, y]}{(y + x^{2})}

Proof. Generators. In degree 1, we have x = c_{1}(ρ) = −c_{1}(ρ̄) and 3x = 0. Moreover c_{1}(θ) = c_{1}(detθ) = c_{1}( ) = 0, so x is the only generator of degree 1. In degree 2 we let y = c_{2}(θ). In degree 3:

C_{3}(θ) = γ^{3}(θ − 3) = λ^{3}(θ − 1) = −1 + θ − λ^{2}(θ) + 1 = 0

so there is no additional generator in degree 3. Relations. Apply the total Chern class c_{t} to both sides of (2.3) to obtain:

c_{t}(θ^{2}) = c_{t}(1 + ρ + ρ̄+ 2θ)
= c_{t}(ρ)c_{t}(ρ̄)c_{t}(θ)^{2}
= (1 + xT)(1 − xT)(1 + yT^{2})^{2}
= 1 + (2y − x^{2})T^{2} + (y^{2} − 2yx^{2})T^{4} + zy^{2}T^{6}  (2.4)

while on the left-hand side (looking only at even terms of degree ≤ 6 and keeping in mind that c_{1}(θ) = c_{3}(θ) = 0):

c_{t}(θ^{2}) = c_{t}((σ_{1} + σ_{2} + σ_{3})^{2})
= 1 + 6yT^{2} + 9y^{2}T^{4} + 4y^{3}T^{6}  (2.5)

Putting the previous two equations together yields 4y = −x^{2}. In particular this means that the order of y is a multiple of 3. To show it is not 3, we restrict to H := ( /2) × ( /2). By proposition 1.9 :

R_{}^{∗} ( /2 × /2) = \frac{[t_{1}, t_{2}]}{(2t_{1}, 2t_{2}, t_{1}^{2}t_{2} − t_{1}t_{2}^{2})}

Then Res_{H}(y) = t_{1}^{2} + t_{1}t_{2} + t_{2}^{2}, which has order 2. So the order of y^{i} is a multiple of 2. Finally, restricting x to the subgroup generated by (123) shows that x is not nilpotent, as well as restricting y and yx to the subgroup generated by (12)(34).

H = /2 × /2 is a normal, abelian, 2-Sylow of G. The image of G under the restriction map

Res_{H}^{G} : R_{}^{∗} (A_{4}) −→ R_{}^{∗} ( /2 × /2)

is generated by powers of Res_{H}(y) = t_{1}^{2} + t_{1}t_{2} + t_{2}^{2}, since Res_{H}(x) = 0. On the other hand, G acts on R_{}^{∗} (H) by cyclic permutations of the elements t_{1}, t_{2}, t_{1} + t_{2}. The element z = t_{1}^{3} + t_{2}^{3} + t_{1}^{2}t_{2} is invariant under this action. But z is not a combination of powers of t_{1}^{2} + t_{1}t_{2} + t_{2}^{2} since it has odd degree, and thus does not belong to the image of the restriction map. Therefore:

Im(Res_{H}^{G}) ⊊ R_{}^{∗} (H)^{N_{G}(H)}

which means that R^{∗} is not a Mackey functor.

2.2 The permutation ring PR^{∗}(A_{4})

There are 5 conjugacy classes of subgroups of A_{4}, which correspond to isomorphism classes of subgroups: 1, C_{2}, C_{3}, C_{2} × C_{2}, A_{4}. The permutation representations associated to each of these classes are:

• [A_{4}/A_{4}] =

• [A_{4}/C_{2} × C_{2}] = + ρ + \overline{ρ}

• [A_{4}/C_{3}] = + θ

• [A_{4}/C_{2}] = + ρ + \overline{ρ} + θ

• [A_{4}] = + ρ + \overline{ρ} + 3θ

So PR(A_{4}) is generated by , ρ + \overline{ρ} and θ, and (with notation as above) the graded permutation ring PR^{∗}(A_{4}) is generated by z = c_{2}(θ) and t = c_{2}(ρ + \overline{ρ}). As we did above, we can apply the total Chern class c_{t} to both sides of the equation:

θ^{2} = 1 + ρ + ρ̄+ 2θ

and obtain that 4z = t. The natural map PR^{∗}(A_{4}) → R_{}^{∗} (A_{4}) induced by the inclusion PR(A_{4}) ↪ R_{}(A_{4}) maps z to y, and thus the order of z in PR^{∗} is a multiple of 2 and 3 (it is 12 if conjecture 2.1 is true), and z is not nilpotent. We have proven:

Proposition 2.3. Let y = c_{2}(θ), then

PR^{∗}(A_{4}) ⊗ _{2} = _{2}[y]
PR^{∗}(A_{4}) ⊗ _{3} = _{3}[y]

The permutation ring PR^{∗}(C_{2} × C_{2}) is isomorphic to R_{}^{∗} (C_{2} × C_{2}): there are 5 conjugacy classes of subgroups of C_{2} × C_{2}, which are the two trivial ones and three subgroups isomorphic to C_{2}; The permutation representations corresponding to these groups are ρ_{(1,0)} + ρ_{(0,1)}, ρ_{(1,1)} + ρ_{(0,1)} and ρ_{(1,0)} + ρ_{(1,1)} respectively. This means that the ring PR^{∗} is generated by t_{1} + t_{2}, t_{1} + (t_{1} + t_{2}) = t_{2} and t_{2} + (t_{1} + t_{2}) = t_{1}. In other words,

PR^{∗}(C_{2} × C_{2}) = \frac{[t_{1}, t_{2}]}{(2t_{1}, 2t_{2}, t_{1}^{2}t_{2} − t_{1}t_{2}^{2})} = R_{}^{∗} (C_{2} × C_{2})

so Swan’s lemma also fails for PR^{∗}, as is shown by restriction of y.

3 The saturated ring R

3.1 Theory

3.1.1 Definitions and first properties

Definition. On the λ-ring R_{}(G), define the saturated filtration {F^{n}}_{n} as follows:

F^{n}(G) = ∑_{H≤G} Ind_{H}^{G}(Γ^{n}(H))

This means that F^{n}(G) is generated by elements of the form:

x = Ind_{H}^{G}(γ^{i_{1}}(ρ_{1}) ⋯γ^{i_{m}}(ρ_{m})), i_{1} + ⋯+ i_{m} ≥ n

with each ρ_{l} an irreducible representation of H. Let w(x) = i_{1} + ⋯+ i_{m} be the weight of x.

Lemma 3.1. (i) Induction and restriction of representations preserve the filtration F.

(ii) F^{i}(G) · F^{j}(G) ⊆ F^{i+j} (G).

(iii) F^{0}(G) = R(G), F^{1}(G) = I, where I is the augmentation ideal.

Proof. (i) Obviously induction preserves the filtration. For restriction, let x ∈ F^{i}(G). It is enough to prove that if x = Ind_{H}^{G}(y) with y ∈ Γ^{i}(H) then Res_{K}(x) ∈ F^{i}(K) for all k ≤ G. For this, we use the double coset formula: let S be a set of (H, K)-double coset representatives of G. For s ∈ S, let

H_{s} = sHs^{−1} ∩ K ≤ K

and let

y^{s}(g) = y(s^{−1} gs), for all g ∈ H_{s}.

Then y^{s} is a representation of H_{s} and

Res_{K}Ind_{H}^{G}(y) = ∑_{s∈K\G/H} Ind_{H_{s}}^{K}(y^{s})

Note that each y^{s} is in Γ^{i}(H_{s}), since exterior powers commute with conjugation (so the λ-operations commute with the operation y ↦ y^{s}). Thus Res_{K}(x) ∈ F^{i}(K).

(ii) It is sufficient to prove that if x̃ = Ind_{H}^{G}(x) and ỹ = Ind_{K}^{G}(y) with x ∈ Γ^{i}(H) and y ∈ Γ^{j}(K) then x̃· ỹ ∈ F^{i+j} (G). We proceed by induction on the order of G. Suppose H < G is a proper subgroup (otherwise there is nothing to prove). By the projection formula:

x̃ỹ= Ind_{H}^{G}(x) Ind_{K}^{G}(y) = Ind_{H}^{G}(x Res_{H} Ind_{K}^{G}(y)).

Since restriction preserves the filtration, Res_{H} Ind_{K}^{G}(y) ∈ F^{j}(H), so:

x · Res_{H} Ind_{K}^{G}(y) ∈ F^{i}(H)F^{j}(H) ⊆ F^{i+j} (H).

where the inclusion is true by induction. In conclusion:

Ind_{H}^{G}(x · Res_{H} Ind_{K}^{G}(y)) ∈ F^{i+j} (G).

(iii) The fact that F^{0}(G) = R(G) comes from the fact that Γ^{0}(G) = R(G). For F^{1}(G), simply observe that, since ε(Ind_{H}^{G}(ρ)) = [G : H]ε(ρ):

F^{1}(G) = ∑_{H≤G} Ind_{H}^{G}(ker(ε|_{H})) = ker(ε) = I.

Definition. The graded ring

R_{}(G) = ⨁_{i≥0} F^{i}(G)/F^{i+1} (G).

is called the saturated (graded representation) ring associated to G. In the sequel, we drop the subscript whenever is clear from the context.

Theorem 3.2. R^{∗} is a Mackey functor, with

Ind_{K}^{H} : R^{∗}(K) −→ R^{∗}(H)
Res_{K}^{H} : R^{∗}(H) −→ R^{∗}(K)
c_{g} : R^{∗}(H) −→ R^{∗}(^{g}H)

Proof. The only nontrivial statement is that conjugation of representations gives rise to a well-defined homomorphism c_{g} : R^{∗}(H) → R^{∗}(H^{g}). The rest of the axioms follow from basic representation theory. So, consider an element of F^{i}(H) of the form Ind_{K}^{H}(C_{i_{1}}(ρ_{1}) ⋯C_{i_{n}}(ρ_{n})) for some representations ρ_{l} ∈ R(K) and K ≤ H. Note that since exterior powers commute with conjugation, one has c_{g}(C_{i}(ρ)) = C_{i}(c_{g}(ρ)) (in other words, c_{g} is compatible with the Γ-filtration). Then

c_{g} Ind_{K}^{H}(C_{i_{1}}(ρ_{1}) ⋯C_{i_{n}}(ρ_{n})) = Ind_{^{g}K}^{^{g}H} c_{g}(C_{i_{1}}(ρ_{1}) ⋯C_{i_{n}}(ρ_{n})) = Ind_{^{g}K}^{^{g}H}(C_{i_{1}}(c_{g}(ρ_{1})) ⋯C_{i_{n}}(c_{g}(ρ_{n})))

and Ind_{^{g}K}^{^{g}H}(C_{i_{1}}(c_{g}(ρ_{1})) ⋯C_{i_{n}}(c_{g}(ρ_{n}))) ∈ F^{i}(^{g}H) as required.

Note that R^{∗} is actually a Green functor, that is, a Mackey functor with an R-algebra structure compatible with restriction and satisfying the projection formula.

Since Γ^{n} ⊆ F^{n} for all n ≥ 0, there is a natural map of graded rings:

η : R^{∗}(G) −→ R^{∗}(G)

induced by the identity.

Definition. We say that R_{}^{∗} (G) is saturated if the natural map η is an isomorphism. A group G is saturated (over ) if R_{}^{∗} (G) is saturated. For H ≤ G, if the induction i_{∗} : R(H) → R(G) is compatible with the filtration (Γ^{n}), then H is Γ-compatible with G.

Note that by definition, the saturated ring R^{∗}(G) is generated by Chern classes of irreducible characters G, as well as classes of the form Ind_{H}^{G}(c_{i}(ρ)) with ρ a virtual character of H ≤ G. Denote

d_{i}(ρ) := c_{i}^{H}(ρ) := Ind_{H}^{G}(c_{i}(ρ))

Below are some conditions under which G is saturated:

Lemma 3.3. If the restriction maps i^{∗} : R(G) → R(H) are surjective for all H ≤ G, then G is saturated.

Proof. Let ρ be a virtual representation of G, with ρ in Γ^{M}(G) say. Then:

i_{∗}(ρ) = i_{∗}i^{∗}(σ) = i_{∗}( )ρ ∈ Γ^{n}(G).

So all virtual representations in F^{n}(G) (which are induced from subgroups of G) are also in the Γ^{n}(G), and thus R^{∗}(G) = R^{∗}(G).

Note however that the converse isn’t true: in the case of D_{p}, restriction of representations to C_{p} isn’t surjective, but the induction still preserves the filtration.

Proposition 3.4. The filtrations (F^{n})_{n} and (Γ^{n})_{n} induce the same topology on R(G).

Proof. Let U ⊆ R(G) be open for the F-topology. Then by definition, for any x ∈ U there is an integer N such that x + F^{N} ⊂ U. Since Γ^{N} ⊆ F^{N}, we also have x + Γ^{N} ⊆ U. Thus U is also open for the Γ-topology. To prove that a set U open in the Γ-topology is also open in the F-topology, we need to show that for each N, there is an M such that F^{M} ⊆ Γ^{N}. We use [Ati61, Th. 6.1]: let H ≤ G, and recall that R(H) can be viewed as an R(G)-module via the restriction homomorphism. Then the I(H)-adic topology is equal to the I(G)-adic topology. By proposition 1.3, these topologies are also equal to the Γ-topology on H. In particular, for some k,m we have

Γ^{N}(H) ⊃ I(G)^{k} · R(H) ⊃ Γ^{m}(H)

Pick k, l large enough that we also have I(G)^{k} ⊂ Γ^{N}(G). Then

Ind_{H}^{G}(Γ^{m}(H)) ⊂ Ind_{H}^{G}(I(G)^{k} · R(H)) ⊂ I(G)^{k} ⊂ Γ^{N}(G)

Now let

M = max_{H≤G} {min {m | Ind_{H}^{G}(Γ^{m}(H)) ⊂ Γ^{N}(G)}} ,

then F^{M}(G) = ∑ _{H≤G} Ind_{H}^{G}(Γ^{M}(H)) ⊂ Γ^{N}(G).

Corollary 3.5. If the natural map η : R(G) → R^{∗}(G) is surjective, then it is an isomorphism and the filtrations (F^{n}) and (Γ^{n}) are equal.

Proof. If η is surjective, then R^{∗}(G) is generated by Chern classes of elements of R(G). Let P_{w} denote a polynomial of total weight w, then any x ∈ F^{n}(G) can be written as

x = P_{n}(C_{i_{1}}(ρ_{1}), ⋯, C_{i_{k}}(ρ_{k})) + y_{n+1}

= P_{n}(C_{i_{1}}(ρ_{1}), ⋯, C_{i_{k}}(ρ_{k})) + P_{n+1} (C_{i_{1}}(ρ_{1}), ⋯, C_{i_{k}}(ρ_{k})) + y_{n+2}

= ∑_{l=1}^{m} P_{n+l} (C_{i_{1}}(ρ_{1}), ⋯, C_{i_{k}}(ρ_{k})) + y_{m+1}

where the ρ_{j}’s are irreducible representations of G and y_{t} ∈ F^{t}. So x is in Γ^{n}(G) + F^{m}(G) for all m, that is, x is in the topological closure \overline{Γ^{n}(G)} of Γ^{n}(G). But Γ^{n}(G) is a closed set in R(G), thus x ∈ Γ^{n}(G).

Finally, we can use transitivity of induction and restriction: suppose σ ∈ Γ^{n}(H) with H ≤ K ≤ G, and R^{∗}(K) is saturated. Then Ind_{H}^{K}(σ) ∈ Γ^{n}(K), thus we do not need to restrict representations to H, but only to K.

3.1.2 Stable elements, Swan’s lemma

Consider a sugroup H ≤ G.

Definition. An element x ∈ R^{∗}(H) is called stable if

Res_{^{g}H∩H}^{^{g}H} (c_{g}(x)) = Res_{^{g}H∩H}^{H} (x)

for all g ∈ G.

We want to prove:

Theorem 3.6. Let G ≥ H ≥ Syl_{p}(G) where Syl_{p}(G) is a p-Sylow of G and let R^{∗}(G)_{(p)} denote the p-primary component of R^{∗}(G). Then:

Res_{H}^{G} : R^{∗}(G)_{(p)} −→ R^{∗}(H)_{(p)}.

is injective, and its image consists of the stable elements in R^{∗}(H)_{(p)} Proof. Step 1. Res_{H}^{G} is injective on p-primary components. Let x ∈ F^{n}(G), then Ind_{H}^{G} Res_{H}^{G}(x) = [G/H] (mod F^{n+1} (G)). But

[G/H] = [G/H] − ( [G/H] − ε( [G/H])) = ε( [G/H]) = [G : H] (mod F^{1}(G))

so that

Ind_{H}^{G} Res_{H}^{G}(x) = [G : H]x (mod F^{n+1} (G))

which is injective on p-primary components since H contains a p-Sylow of G. Thus Res_{H}^{G} is injective on p-primary components, as needed. Step 2. If x ∈ Im(Res_{H}^{G}) then x is stable. Write x = Res_{H}^{G}(y). Then, since y is a character of G we have c_{g}(y) = y for all g ∈ G and:

Res_{^{g}H∩H}^{^{g}H} (c_{g}(x)) = Res_{^{g}H∩H}^{^{g}H} (c_{g} Res_{H}^{G}(y)) = Res_{^{g}H∩H}^{^{g}H} (Res_{^{g}H}^{G} c_{g}(y)) = Res_{^{g}H∩H}^{^{g}H} (Res_{^{g}H}^{G}(y))
= Res_{^{g}H∩H}^{G} (x) = Res_{^{g}H∩H}^{H} Res_{H}^{G}(y)
= Res_{^{g}H∩H}^{H} (x)

Step 3. If x ∈ R^{∗}(H)_{(p)} is stable then x ∈ Res_{H}^{G}(R^{∗}(G)_{(p)}). First observe that if x is stable, then by the double coset formula:

Res_{H}^{G} Ind_{H}^{G}(x) = ∑_{g∈[H\G/H]} Ind_{^{g}H∩H}^{H} (x^{g}) = ∑_{g∈[H\G/H]} Ind_{^{g}H∩H}^{H} Res_{^{g}H∩H}^{^{g}H} c_{g}(x)

= ∑_{g∈[H\G/H]} Ind_{^{g}H∩H}^{H} Res_{^{g}H∩H}^{H} (x) = ∑_{g∈[H\G/H]} [H/^{g}H ∩ H]x.

Plugging x = into this equality, we get:

Res_{H}^{G}( [G/H]) = Res_{H}^{G} Ind_{H}^{G}( ) = ∑_{g∈[H\G/H]} [H/^{g}H ∩ H]

so that whenever x is stable, we have Res_{H}^{G} Ind_{H}^{G}(x) = Res_{H}^{G}( [G/H])x = [G : H]x. Now, say |Syl_{p}(G)| = p^{i}, and choose an integer q so that q[G : H] = 1 (mod p^{i}). Then

Res_{H}^{G} Ind_{H}^{G}(qx) = q[G : H]x = x (mod p^{i})

which implies that x ∈ Res_{H}^{G}(R^{∗}(G)_{(p)}) whenever x ∈ R^{∗}(H)_{(p)} is stable.

Corollary 3.7. If H ⊴ G is a normal subgroup such that H ⊇ Syl_{p}(G), then

R^{∗}(G)_{(p)} ≅ Im(Res_{H}^{G}) = R^{∗}(H)_{(p)}^{G/H}

Proof. If H is normal, the stability condition becomes c_{g}(x) = x, that is, x is invariant by the action of G/H.

Corollary 3.8. If H := Syl_{p}(G) is abelian, then

Res_{H}^{G} : R^{∗}(G)_{(p)} −→ R^{∗}(H)^{N_{G}(H)}

is an isomorphism.

Proof. By theorem 3.6, Res_{H}^{G} is an isomorphism on stable elements in R^{∗}(H)_{(p)}. Moreover, R^{∗}(H) is p^{i}- torsion for some i, so that Res_{H}^{G} is in fact surjective on all stable elements of R^{∗}(H). Obviously any stable element is in particular invariant under the action of N_{G}(H), so we only need to show that all elements of R^{∗}(H)^{N_{G}(H)} are stable.

Let g ∈ G and let C be the centralizer of ^{g}H ∩ H in G. Since H is abelian, H, ^{g}H ⊆ C, and H and ^{g}H are p-Sylows of C. Thus for some t ∈ C we have^{tg}H = H. Then

Res_{^{g}H∩H}^{H} c_{tg}(x) = Res_{^{g}H∩H}^{H} c_{t}c_{g}(x)
= c_{t}( Res_{(^{g}H∩H)^{t}}^{H^{t}} c_{g}(x))
= Res_{^{g}H∩H}^{H^{t}} c_{g}(x) since t ∈ C
= Res_{^{g}H∩H}^{^{g}H} c_{g}(x) since H^{t} = ^{g}H

Now if x ∈ R^{∗}(H)^{N_{G}(H)} then c_{tg}(x) = x thus:

Res_{^{g}H∩H}^{^{g}H} c_{g}(x) = Res_{^{g}H∩H}^{H} (x)

and x is stable.

3.2 Examples

3.2.1 Groups that are saturated

Proposition 3.9. Abelian groups are saturated over .

Proof. Let G be an abelian group and define Ĝ := Hom(G, ^{∗}). Then any group homomorphism (of abelian groups) ϕ : G → H induces a map ϕ̂ : Ĥ → Ĝ, which is injective if, and only if, ϕ is surjective. Additionally, there is a natural isomorphism between G and its double dual ˆ Ĝ given by associating to g the evaluation in g. Now if H ≤ G, then the injection H → G induces a map ϕ̂ : Ĝ → Ĥ, and also a map ˆ ϕ̂ : ˆ Ĥ → ˆ Ĝ. The latter map is injective, which means by the above that ϕ̂ is surjective. Thus the characters of H all come from restrictions of characters of G, and G is saturated.

Proposition 3.10. The group Q_{8} is saturated.

Proof. The quaternion group contains one subgroup isomorphic to C_{2}, which is generated by −1, and three subgroups isomorphic to C_{4}, which all contain C_{2} and are all conjugate, generated respectively by i, j and k. Since all these groups are saturated, we only need to check that the maximal saturated subgroup H = ⟨k⟩ ≅ C_{4} is Γ-compatible with Q_{8}, which we do manually. Recall that

R^{∗}(Q_{8}) = \frac{[x_{1}, x_{2}, y]}{(2x_{i}, 8y, x_{i}^{2}, x_{1}x_{2} − 4y)} , R^{∗}(C_{4}) = \frac{[t]}{(4t)}

where, in particular, y = c_{2}(Δ), with Δ the irreducible representation of Q_{8} of degree 2, and t = c_{1}(ρ) with ρ the usual generator of R(C_{4}). Note that i_{∗}(C_{1}(ρ)) ∈ Γ^{1}(Q_{8}) = I_{Q_{8}}. Moreover, the representation Δ restricts on C_{4} to ρ + ρ^{−1} ,and so

Res_{C_{4}}^{Q_{8}}(y) = c_{2}(ρ + ρ^{^{−} 1}) = −c_{1}(ρ)^{2} = −t^{2}.

therefore C_{1}(ρ)^{2} = Res(−Y), and so

i_{∗}(C_{1}(ρ)^{2}) = i_{∗}(Res(Y)) = [Q_{8}/C_{4}].Y ∈ Γ^{2}(Q_{8}).

Thus:

i_{∗}(C_{1}(ρ)^{n}) = i_{∗}(C_{1}(ρ)^{2m+l} ) = i_{∗}(C_{1}(ρ))i_{∗}(Y^{m}) ∈ Γ^{l} · Γ^{2m} for l = 0, 1

which means that C_{4} is Γ-compatible with Q_{8}, and therefore Q_{8} is saturated.

Proposition 3.11. Let p be an odd prime, then the dihedral group D_{p} of order 2p is saturated. Proof. Recall that

R^{∗}(G) = \frac{[x, y]}{(2x, py, xy)}.

Since D_{p} = C_{p} ⋊ C_{2} and C_{p}, C_{2} are abelian, these are the maximal saturated subgroups of D_{p}. The signature ε of D_{p} restricts on C_{2} to the representation ρ, which generates R(C_{2}). Thus C_{2} is Γ-compatible with D_{p}, and we only need to look at C_{p}. Since Res(Y) = −C_{1}(ρ)^{2} the same argument as in the proof of proposition 3.10 applies.

Another idea is to quantify the obstruction to i_{∗} being a ring homomorphism. Lemma 3.12. Suppose H is normal in G and y ∈ R(H) is stable by G, that is:

y(στσ^{−1} ) = y(τ)

for any τ ∈ H, σ ∈ G. Then for any x ∈ R(H)

i_{∗}(x)i_{∗}(y) = i_{∗}(xy)i_{∗}(1)

Proof. If y is stable by G then the induced character i_{∗}(y) is i_{∗}(1)y′, where the values of y′are those of y on H and 0 on G \ H. In other words we have Res(y′) = y. Thus:

i_{∗}(x)i_{∗}(y) = i_{∗}(x)i_{∗}(1)y′= i_{∗}(x Res(y′))i_{∗}(1) = i_{∗}(xy)i_{∗}(1).

3.2.2 The saturated ring of A_{4} Proposition 3.13.

R^{∗}(A_{4}) = \frac{[x, y, z]}{(3x, 2y, 2z, y^{3} − z^{2})} ≅ \frac{[x, y′, z]}{(3x, 6y′, 2z, 2y′− x^{2}, x^{6} + y^{′3}− z^{2})}

with x = c_{1}(ρ), y′= c_{2}(θ), z = d_{3}(ρ_{(1,0)} − ρ_{(0,1)}) where ρ_{(i,j)} are representations of C_{2} × C_{2}. Proof. The group A_{4} has the following p-Sylow structure:

(i) For p = 2, we have Syl_{2}(A_{4}) ≅ C_{2} × C_{2}, generated by (12)(34) and (13)(24). The group C_{2} × C_{2} is normal in A_{4}.

(ii) For p = 3, we have Syl_{3}(A_{4}) ≅ C_{3}, generated by say (123); it is self-normalizing. 3-primary part. Since C_{3} is self-normalizing and abelian, the action of its normalizer is trivial. Thus

R^{∗}(A_{4})_{(3)} ≅ R^{∗}(C_{3}) = \frac{[t]}{(3t)}

2-primary part. Elements of A_{4} act on C_{2} × C_{2} by circular permutations of the three nontrivial elements. Since C_{2} × C_{2} is abelian, it is saturated so:

R^{∗}(C_{2} × C_{2}) ≅ \frac{[t_{1}, t_{2}]}{(2t_{1}, 2t_{2}, t_{1}^{2}t_{2} − t_{1}t_{2}^{2})}

the stable elements are those invariant under circular permutations of t_{1}, t_{2}, t_{1} + t_{2}. There are no such elements in degree 1. For n ≥ 2, we have R^{n}(C_{2} × C_{2}) = ⟨t_{1}^{n}, t_{2}^{n}, t_{1}^{n−1} t_{2}⟩, and the action of A_{4} is described by the permutation matrix:

( \begin{matrix}0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{matrix})

whose (one-dimensional) invariant subspace is t_{1}^{n} + t_{2}^{n} + t_{1}^{n−1} t_{2}. Let y = t_{1}^{2} + t_{1}t_{2} + t_{2}^{2} and z = t_{1}^{3} + t_{1}^{2}t_{2} + t_{2}^{3}, then

R^{∗}(A_{4})_{(2)} ≅ R^{∗}(C_{2} × C_{2})^{A_{4}} ≅ \frac{[y, z]}{(2y, 2z, y^{3} − z^{2})} .

Define y′= y − x^{2} to obtain the presentation

R^{∗}(A_{4}) = \frac{[x, y′, z]}{(3x, 6y′, 2z, 2y′− x^{2}, x^{6} + y^{′3}− z^{2})} .

Explicit generators. We determine the images of the classes c_{1}(ρ), c_{2}(θ) under the natural map η. For this, we use that the maps Res and η commute.

• The class c_{1}(ρ) restricts to the degree one generator t of R^{∗}(C_{3}) and to 0 on C_{2}× C_{2}. Thus its 3-primary part must be equal to x and its 2-primary part to 0, and [c_{1}(ρ)] = x.

• The class c_{2}(θ) restricts to −t^{2} on C_{3} and to t_{1}^{2} + t_{1}t_{2} + t_{2}^{2} on C_{2} × C_{2}, thus its 3-torsion part is −x^{2} and its 2-primary part is y, so [c_{2}(θ)] = y − x^{2} = y′.

• For z: write i_{∗} = Ind_{C_{2}×C_{2}}^{A_{4}} , i^{∗} = Res_{C_{2}×C_{2}}^{A_{4}} , and consider:

i^{∗}(z) = t_{1}^{3} + t_{1}^{2}t_{2} + t_{2}^{3}
i_{∗}i^{∗}(z) = i_{∗}(t_{1}^{3} + t_{1}^{2}t_{2} + t_{2}^{3})
3z = z = i_{∗}(t_{1}^{3} + t_{1}^{2}t_{2} + t_{2}^{3})

and note that t_{1}(t_{1}^{2} + t_{1}t_{2} + t_{2}^{2}) = t_{1}^{3}. Then

i_{∗}(t_{1}^{3}) = i_{∗}(t_{1}(t_{1}^{2} + t_{1}t_{2} + t_{2}^{2})) = i_{∗}(t_{1} · i^{∗}(c_{2}(θ)))
= i_{∗}(t_{1})c_{2}(θ)

Now 2t_{1} = 0 so 2i_{∗}(t_{1}) = 0, but R^{1}(G) is 3-torsion, thus i_{∗}(t_{1}) = 0. This means that i_{∗}(t_{1}^{3}) = 0 and

z = i_{∗}(t_{1}^{2}t_{2} + t_{2}^{3})

a direct computation then shows that c_{3}(ρ_{(1,0)} − ρ_{(0,1)}) = t_{1}^{2}t_{2} + t_{2}^{3}, so that z = i_{∗}(ρ_{(1,0)} − ρ_{(0,1)}) = d_{3}(ρ_{(1,0)} − ρ_{(0,1)}).

3.2.3 The saturated ring of PSL(2, p)

Let G = PSL(2, p) be the projective special linear group over _{p}, where p is an odd prime such that p ≡ 3, 5(mod 8). The order of G is |G| = \frac{p(p+1)(p−1)}{2} Sylow structure and stable elements. For each prime l dividing the order of G, let H_{l} = Syl_{l}(G) and N_{l} = N_{G}(H_{l}). There are 4 possible cases:

(i) l = p. Then H_{p} ≅ C_{p} is generated by the matrix ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) . The normalizer of H_{p} is the group:

N_{p} = \{ ( \begin{matrix}a & b \\ 0 & a^{−1} \end{matrix} ) ∈ PSL(2, p)\} ,

with action

( \begin{matrix}a & b \\ 0 & a^{−1} \end{matrix} ) · ( \begin{matrix}1 & n \\ 0 & 1 \end{matrix}) · ( \begin{matrix}a^{−1} & −b \\ 0 & a \end{matrix} ) = ( \begin{matrix}1 & a^{2}n \\ 0 & 1 \end{matrix}) ,

inducing ρ ↦ ρ^{a^{2}} for a generator ρ of R(C_{p}). On R^{∗}(C_{p}) ≅ \frac{[x]}{(px)} , this induces x ↦ a^{2}x. The subring

generated by x^{\frac{p−1}{2}} is stable by this action, and conversely if a is an element of multiplicative order (p − 1), then a monomial x^{m} being stable by the action x ↦ a^{2}x implies that m is a multiple of \frac{p−1}{2} . Thus

R^{∗}(H_{p})^{N_{p}} ≅ \frac{[t]}{(pt)} , |t| = \frac{p − 1}{2} .  (3.1)

(ii) l is an odd prime dividing p − 1. Then H_{l} ≅ C_{l^{i}} for some integer i, generated by ( \begin{matrix}n & 0 \\ 0 & n^{−1} \end{matrix} ) for some

n of order l^{i} in _{p}^{×} . A straightforward computation gives that N_{l} := N^{G}(H_{l}) is generated by diagonal

matrices (which commute with the elements of H_{l}) together with the matrix ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}) which sends

an element h ∈ H_{l} to its inverse. The induced action on the representation ring is ρ ↦ ρ^{−1} , which translates as x ↦ −x in the graded ring. Thus

R^{∗}(H_{l})^{N_{l}} ≅ \frac{[y]}{(l^{i}y)} , |y| = 2.  (3.2)

(iii) l = r is an odd prime dividing p + 1. We prove that H_{r} is cyclic. Note that the r-Sylow of G is isomorphic to that of G′ := PSL(2, p^{2}) since the index of G in G′ is coprime to r. Let α ∈ _{p^{2}}^{×} have

multiplicative order r^{i}. The matrix A′= ( \begin{matrix}α & 0 \\ 0 & α^{−1} \end{matrix} ) generates a cyclic group H′_{r}of order r^{i} in G′, which

is thus an r-Sylow subgroup. Note that α ∉ _{p}^{×} , however any matrix of G similar to A generates an

isomorphic group in G. One can take for example A = ( \begin{matrix}0 & −1 \\ 1 & α + α^{−1} \end{matrix} ) , the companion matrix to the

minimal polynomial of α. The normalizer N′_{r}of C′_{r^{i}} in G′is a dihedral group of order p^{2} − 1, generated by all diagonal matrices

together with the matrix ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}) which sends A to its inverse. The change of basis sending A to

A′ allows us to view N_{r} as a subgroup of N′_{r}, and thus the elements of N_{r} act either trivially or by inversion on H_{r}. It remains to show that there exists a matrix S such that S^{−1} AS = A^{−1} . Let a = α + α^{−1} . By a direct

calculation, one shows that any matrix of the form ( \begin{matrix}−x & y \\ ax + y & x \end{matrix}) in GL(2, p) satisfies this property,

thus S ∈ PSL(2, p) exists if and only if there is a pair (x, y) ∈ _{p}^{2} such that −x^{2} − axy − y^{2} = 1. This equation is equivalent to X^{2} + 1 = bY^{2}, with X = x + \frac{a^{2}}{4}y, Y = y and b = \frac{a^{2}}{4} − 1. There are (p + 1)/2 squares in _{p} (including 0), so there are (p + 1)/2 elements of the form X^{2} + 1, and, if b ≠ 0 then there are also (p + 1)/2 elements of the form bY^{2}. Thus whenever b ≠ 0, the set of elements of the form X^{2} + 1 and the set of elements of the form bY^{2} have nontrivial intersection, and there is a solution to x^{2}axy + y^{2} = −1. Now, b = 0 if and only if a^{2} = 4, that is, a = ±2(mod p). But then α is a solution of t^{2} ± 2t + 1, that is, α = α^{−1} has multiplicative order 2, in contradiction with our assumption. Thus b is always nonzero, which completes the proof. We have:

R^{∗}(H_{r})^{N_{r}} ≅ \frac{[y]}{(r^{i}y)} , |y| = 2.  (3.3)

(iv) l = 2. Since p ≡ 3, 5( mod 8), the 2-Sylow subgroup of G has order 4. There are two cases:

• if p ≡ 5( mod 8), let a satisfy a^{2} ≡ −1(mod p). Then

H_{2} = ⟨ h_{1} := ( \begin{matrix}a & 0 \\ 0 & −a \end{matrix} ) , h_{2} := ( \begin{matrix}0 & a \\ a & 0 \end{matrix}) ⟩ .

I claim that N_{2} ≅ A_{4}. First, we have C_{G}(h_{1}) ∩ N_{G}(H_{2}) = {Id}, as a direct calculation shows, and similarly for h_{2} and h_{1}h_{2} =: h_{3}. Therefore, if N ∈ N_{2} acts nontrivially on H_{2}, it must permute

all 3 nontrivial elements. If T = ( \begin{matrix}x & −ax \\ x & ax \end{matrix} ) , with x^{2} = \frac{1}{2a}, then Th_{1}T^{−1} = h_{2} and Th_{2}T^{−1} = h_{3}.

Both 2 and a are nonresidues mod p since p ≡ 5(mod 8) and if a were a residue, then PSL(2, p) would contain an element of order 4, contradicting H_{2} ≅ C_{2} × C_{2}. Thus there is an x satisfying x^{2} = 1/2a. Moreover T is unique up to multiplication by an element of C_{G}(H_{2}) = H_{2}, which shows that N_{2} = ⟨T, H_{2}⟩≅ A_{4}.

• if p ≡ 3(mod p), let b satisfy b^{2} ≡ −2(mod p). Then

H_{2} = ⟨ ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , ( \begin{matrix}b & 1 \\ 1 & −b \end{matrix} ) ⟩

Again, we have N_{2} ≅ A_{4} acting by cyclic permutations, generated by H_{2} together with the ma-

trix T = ( −\frac{b+2}{2}  \frac{b−2}{2}^{\frac{1}{b}  \frac{1}{b}} ) .

In both cases the normalizer acts as cyclic permutations on the nontrivial elements of H_{2}, and thus:

R^{∗}(H_{2})^{N_{2}} ≅ \frac{[y, z]}{(2y, 2z, y^{3} − z^{2})} , |y| = 2, |z| = 3.  (3.4)

Putting all of this together, we get:

Theorem 3.14. Let G = PSL(2, p) be the projective special linear group over _{p}, where p is an odd prime such that p ≡ 3, 5(mod 8). Write:

|G| = 4 · p · l_{1}^{i_{1}} ⋯l_{n}^{i_{n}} · r_{1}^{j_{1}} ⋯r_{m}^{j_{m}}, with l_{k}|(p − 1), r_{k}|(p + 1).

Then:

R^{∗}(G) ≅ \frac{[x_{1}, ⋯, x_{n}, y_{1}, ⋯y_{m}, z, t, u]}{(l_{k}^{i_{k}}x_{k}, r_{k}^{j_{k}}y_{k}, 2z, 2t, pu, z^{3} − t^{2})}  (3.5)

with |x_{k}| = |y_{k}| = |z| = 2, |t| = 3, |u| = (p − 1)/2.

4 Completed rings

Let R̂(G) be the completion of R(G) with respect to the I-adic topology, with I the augmentation ideal. Since this topology coincides with the F-topology and the Γ-topology, the generators of the graded rings R^{∗}(G) and R^{∗}(G) are topological generators for R̂(G). Remark. The restriction and induction maps naturally extend to R̂(G) and satisfy the same restricitoncorestriction formulas as in the not complete case.

4.1 p-groups

Notation. Fix a prime p and assume G is a p-group of exponent e. Then the characters of G all have values in [μ], with μ a primitive p^{e}-th root of unity. We want to look at the p-adic completion of [μ]. The p-adic valuation on [μ] is well-defined: any two embeddings (μ) → \overline{}_{p} are Galois-conjugated and thus have the same image, so we can pick some primitive p^{e}-th root of unity in \overline{}_{p} and call it μ again. Then the image of (μ) in \overline{}_{p} is _{p}(μ). There is a unique valuation on _{p}(μ) extending that on _{p}, and since [μ] is the ring of integers of (μ), there is a unique p-adic valuation on [μ] extending that on . The completed ring _{p}[μ] is the ring of integers of _{p}(μ). Let K = _{p}[μ], and let A = _{p}[μ]. Back to representation rings: we consider the evaluation morphisms R(G) → [μ], which we see as having values in A. This makes R(G) into a subring of the ring of functions F(G, A).

Proposition 4.1. The topological closure of R(G) in F(G, A) is R(G) ⊗ _{p}.

Proof. The elements of a basis for R(G) (that is, the irreducible characters of G) are still linearly independent in F(G, K), thus they are linearly independent over A and over _{p}. Thus R(G) ⊗ _{p} is a subring of F(G, A). In fact, since _{p} is closed in A, so is R(G) ⊗ _{p} in F(G, A); thus R(G) ⊗ _{p} is the closure of R(G) in F(G, A).

The evaluation morphisms ϕ : R(G) → A are continuous with respect to the I-adic topology on R(G) and the p-adic topology on A, so each ϕ can be extended to ϕ̂: R̂(G) → A. Define

i : R̂(G) → F(G, A)

as i(χ) = (g ↦ ϕ̂_{g}(χ)).

Proposition 4.2. The map i is injective.

Proof. We want to show that if ϕ̂_{g}(χ) = 0 for all g ∈ G then χ = 0, or in other words, if (x_{n})_{n} is a sequence of characters in R(G) such that (x_{n}) converges and lim x_{n}(g) = 0 for all g ∈ G, then x_{n} converges to the zero character.

• First assume G is cyclic of order q = p^{r} and fix a generator γ of G. Recall that:

R(G) ≅ \frac{[ρ]}{(ρ^{q} − 1)}

where ρ(γ) = μ, and Γ^{M} = I^{M} = (1 − ρ)^{M}. Note that R(G) is also generated as a ring by X := ρ_{1}.

Let (x_{n})_{n} ∈ R(G)^{}be a sequence of characters and suppose that (x_{n}(g))_{n} tends to zero p-adically, for all g ∈ G. First note that since (x_{n}(1)) tends to zero, we must have x_{n} ∈ I for n large enough. Fix M ∈ and suppose that there is an N such that x_{n} ∈ I^{M} for all n > N; I claim that for n large enough, x_{n} ∈ I^{M+1} = X^{M+1} . Recall that

R^{∗}(G) ≅ \frac{[\overline{X}]}{(q\overline{X})}

where \overline{X} denotes the class of X in R^{∗}(G). In particular, we have Γ^{M}/Γ^{M+1} = /p^{r} . Write:

x_{n} = (a_{n,0} + a_{n,1}X + ⋯+ a_{n,m}X^{m}) X^{M} = a_{n,0}X^{M} + f(X)X^{M+1}

for some polynomial f. Then for all g ∈ G,

v_{p}(ϕ_{g}(x_{n})) ≥ min {v_{p} ( ϕ_{g} ( a_{n,0}X^{M}) ) , v_{p} ( ϕ_{g} ( f(X)X^{M+1} ) ) } .

Now, we know that, for n large enough and for all g ∈ G, v_{p}(ϕ_{g}(x_{n})) ≥ v_{p}(ϕ_{g}(X^{M})) + r, which is possible only if a_{n,0} ≡ 0 (mod p^{r}). Since Γ^{M}/Γ^{M+1} ≅ /p^{n} , this implies that x_{n} ∈ Γ^{M+1} , which proves the claim.

• Consider the map:

ψ : R̂(G) → ⨁ R̂(C)

where the direct sum is over all cyclic subgroups of G, and ψ is induced by restriction of characters. If i(χ) = 0 for some χ ∈ R̂(G), in particular i(Res_{C}^{G}(χ)) = 0 for all cyclic subgroups C ≤ G. Since i is injective on cyclic groups, this implies that ψ(χ) = 0. To see that ψ is injective, let χ ∈ R̂(G) satisfy ψ(χ) = 0, that is, Res_{C}^{G}(χ) = 0 for all cyclic subgroups C. A theorem of Artin ([Ser77, 13.1, Th. 30]) states that every rational character of G is a linear combination with rational coefficients of permutation characters defined by cyclic subgroups of G. Thus we can write:

= ∑_{C≤G cycl.} λ_{C} [G/C],

with each λ_{C} ∈ . Let m ∈ such that m · λ_{C} ∈ for all C. Now, for any cyclic subgroups C ≤ G, we have Res_{C}^{G}(χ) = 0, and in particular λ_{C} · Res_{C}^{G}(χ) = 0. Thus

m · χ = m · · χ = ∑_{C≤G cyc.} m · λ_{C} [G/C] · χ = ∑_{C≤G cycl.} m · λ_{C}Ind_{C}^{G}(Res_{C}^{G} χ) = 0

and thus m · χ = 0 in R̂(G). Since I-adic completion is exact (see [AM69, Th. 10.12]), there is no nontrivial torsion in R̂(G), and thus χ = 0.

Thus R̂(G) can be viewed as a subring of F(G, A). Since R(G) = ⊕ I, we have R̂(G) = ⊕ Î, with Î = lim_{n≥1} I^{n}/I^{n+1} . Since I^{n}/I^{n+1} is finite for all n ≥ 1, the limit Îis compact. It contains I as a dense subset, so it is the closure I, that is, Î= I ⊗ _{p}. Thus R̂(G) is a subring of _{p} ⊕ Î= R(G) ⊗ _{p}. In summary, the generators of R^{∗}(G) or R^{∗}(G) give us generators of the completed ring R̂(G), itself a dense subring of the completion R(G) ⊗ _{p} of R(G) in F(G, A). So generators of the graded rings are (topological) generators of R(G) ⊗ _{p}. Remark. When G is a p-group, there is no prime-to-p torsion in R^{∗}(G). Thus the kernel of the map:

Γ^{n}(G) ⊗ _{p} −→ \frac{Γ^{n}(G)}{Γ^{n+1} (G)} ⊗ _{p}

is exactly Γ^{n+1} (G) ⊗ _{p}. This implies

R^{∗}(G) ⊗ _{p} ≅ \frac{Γ^{n}(G) ⊗ _{p}}{Γ^{n+1} (G) ⊗ _{p}} .

Or in other words, tensoring with _{p} commutes with taking graded rings. This is also true for saturated rings.

4.2 General finite groups

Let G be any finite group, and fix a prime p. Let H_{p} = Syl_{p}(G) and R̂_{p}(G) = R̂(G) ⊗ _{p}. The restriction and induction maps are both continuous, so they extend to maps

Res : R̂_{p}(G) ⇆ R̂_{p}(H_{p}) : Ind

Proposition 4.3. The map Res : R̂_{p}(G) −→ R̂_{p}(H_{p}) is injective.

Proof. The restriction-corestriction formula is still valid on R̂_{p}(G):

Ind_{H}^{G} Res_{H_{p}}^{G}(x) = [G/H_{p}]x

I claim that [G/H_{p}] ∈ R̂_{p}(G)^{×} . Indeed:

[G/H_{p}] = ε( [G/H_{p}]) + ( [G/H_{p}] − ε( [G/H_{p}])) = ε( [G/H_{p}])(1 − u)

where u ∈ I. Note that (ε( [G/H_{p}]), p) = 1, so ε( [G/H_{p}]) is invertible in R̂_{p}(G). Moreover, since u ∈ I, the sum 1 + u + u^{2} + ⋯converges and (1 − u) is invertible with inverse

(1 − u)^{−1} = 1 + u + u^{2} + u^{3} + · · ·.

Thus [G/H_{p}] ∈ R̂_{p}(G)^{×} . This means that Ind_{H_{p}}^{G} Res_{H_{p}}^{G} is injective, and thus Res is.

Note that R̂_{p}(−) is again a Mackey functor (check!), so that:

R̂_{p}(G) ≅ (R̂_{p}(H_{p}))^{N_{G}(H_{p})}

5 Saturated rings as Tambara functors

5.1 Representation rings and tensor induction

First the actual definition (from Pierre’s notes). Fix a group G and a field . We view a G-set X as a category with an object for each point and arrows between objects x, y for each g ∈ G such that g · x = y. A vector bundle is defined as a functor between X and the category of -vector spaces and linear maps. Let K_{G}(X) be the semigroup of isomorphism classes of vector bundles over X, under direct sum.

Lemma 5.1. Let X be a transitive G-set with a distinguished point x ∈ X, and let H = Stab(x). Then there is an isomorphism (depending on x) between K_{G}(X) and the semiring of representations R(H).

Proof. Let W be a representation of H and consider the induced representaiton V = Ind_{H}^{G} = [G] ⊗_{}[H]W. Define a vector bundle on X as follows: for each y ∈ X, write y = g · x and let (V)_{y} = g · W = g ⊗ W ⊂ V. This depends only on y and the action of g takes V_{x} to V_{gx}, so this is a vector bundle. Conversely if V is a vector bundle on X, define W = V_{x}. This is a well-defined H-module (since it is stable by H), so W is a representation. Those two constructions are mutually inverse.

Remark. The isomorsphism above depends on x; choosing the basepoint y = g · x as a basepoint instead, one obtains the isomorphic representation of gHg^{−1} which is given by precomposing the action og H on V_{x} by conjugation with g.

We can now define our restrictions and transfers:

Definition. Let f : X → Y be a map of G-sets. Then we define:

• The restriction f^{∗} : K_{G}(Y) → K_{G}(X), f^{∗}(V)_{x} = V_{f^{∗}(x)}

• The induction (or transfer) f_{∗} : K_{G}(X) → K_{G}(Y), f_{∗}(V)_{y} = ⊕V_{f^{−1} (y)}

• The norm (or tensor induction) f : K_{G}(X) → K_{G}(Y), f♯(V)_{y} = ⊗V_{f^{−1} (y)} Theorem 5.2. The function K_{G}(−) with these three maps is a Tambara functor.

Proof. check!

In [Tam93], Tambara gives a formula for the norm of a sum. This is part of the proof that semi-Tambara functors (then called semi-TNR functors) are actual Tambara functors, that is, the norm extends from semirings to rings. Let f : X → Y be a map of G-sets. Let

V = {(y, C)|y ∈ Y, C ⊂ f^{−1} (y)}

then the pullback X ×_{Y}V can be written as the disjoint union of two sets:

• U = {(x, C)|x ∈ X, C ⊂ f^{−1} (y), x ∈ C}

• U′= {(x, C)|x ∈ X, C ⊂ f^{−1} (y), x ∉ C}. Then we have two commutative diagrams of G-sets:

\xymatrix{U \ar[r]^{r} \ar[d]_{t} & X \ar[d]^{f} & U′ \ar[r]^{r′} \ar[d]_{t′} & X \ar[d]^{f} \\ V \ar[r]_{s} & Y & V \ar[r]_{s} & Y}

where r, r′, s are the projections and t : U → V sends (x, C) to (f(x), C) (similarly for t′). Proposition 5.3 ([Tam93, Prop. 4.1]). Let X and Y be G-sets and S a semi-TNR functor. Let a, b ∈ S(X).

f_{♯}(a + b) = s_{∗} ( t_{♯}r^{∗}(a) · t′_{♯}r^{′∗}(b))  (5.1)

We now apply this formula to representation rings and the norm map N_{H}^{G}, for H a normal subgroup of G of prime index. We have X = G/H, Y = G/G = {∗} and f is the constant map. Moreover, V is just the set of subsets of G/H. Let O(V) be a complete set of representatives for the G-orbits in V, and pick a set S ⊂ G of coset representatives of G/H. Then, for a representation ρ of H and a subset C = {g_{1}, ⋯, g_{i}} of G/H with g_{j} ∈ S, write

ρ^{⊗C} = ρ^{g_{1}} ⊗ ρ^{g_{2}} ⊗ ⋯⊗ ρ^{g_{i}}

where ρ^{g} is the conjugate of ρ by g. Note that this is still a representation of H. Proposition 5.4. Let H be a subgroup of G and let τ, σ be representations of H.

N_{H}^{G}(τ + σ) = ∑_{C∈O(V)} Ind_{StabC}^{G}( τ^{⊗C} σ^{⊗(X\C)} )  (5.2)

This formula does not depend on the set of orbit representatives O(V). Proof. We apply Tambara’s formula:

N_{H}^{G}(τ + σ) = s_{∗} ( t_{♯}r^{∗}(τ) · t′_{♯}r^{′∗}(σ))

The map r sends a pair (x, C) to x, thus r^{∗}(τ)_{(x,C)} = (τ)_{x} = τ^{x}. Similarly r^{′∗}(σ)_{(x,C)} = σ^{x}. Now for any C ∈ V = P(X),

(t_{♯}r^{∗}(τ))_{C} = ⨂_{x∈C} r^{∗}(τ)_{(x,C)} = ⨂_{x∈C} τ^{x}

and ( t′_{♯}r^{′∗}(σ)) _{C} = ⨂ _{x∉C} σ^{x}. Let

ρ_{C}^{x} = \{ \begin{matrix}τ^{x} & if x ∈ C \\ σ^{x} & if x ∉ C \end{matrix}

then ( t_{♯}r^{∗}(τ) · t′_{♯}r^{′∗}(σ))_{C} = ⨂ _{x∈X} ρ_{C}^{x}. Let ρ_{C} = ⨂ _{x∈X} ρ_{C}^{x}, then ρ is a representation of StabC, and ρ_{g·C} = (ρ_{C})^{g}, the conjugate representation.

s_{∗} ( t_{♯}r^{∗}(τ) · t′_{♯}r^{′∗}(σ))_{y} = ⨁_{C∈P(X)} ρ_{C} = ⨁_{C∈O(V)} ( ⨁_{D∈G·C}ρ_{D})

= ⨁_{C∈O(V)} ( ⨁_{g∈G/StabC} (ρ_{g.C})) = ⨁_{C∈O(V)} ( ⨁_{g∈G/StabC} (ρ_{C})^{g})

= ∑_{C∈O(V)} Ind_{StabC}^{G}( τ^{⊗C} σ^{⊗(X\C)} )

As promised, this formula does not depend on the set of orbit representatives O(V): picking another representative boils down to conjugating τ^{⊗C} σ^{⊗(X\C)} by some element g ∈ G, which does not change the induced representation.

Corollary 5.5. Let H be a normal subgroup of G of prime index p and let a, b ∈ R(H). Then

N_{H}^{G}(a + b) = N_{H}^{G}(a) + N_{H}^{G}(b) + ∑_{C∈O(V)} Ind_{H}^{G}(a^{⊗C} b^{⊗(X\C)} ).

Moreover:

N_{H}^{G}(−a) = \{ \begin{matrix}−N_{H}^{G}(a) + Ind_{H}^{G}(a^{2}) & if p = 2 \\ −N_{H}^{G}(a) & otherwise \end{matrix}

Proof. This is seen by applying the addition formula to a and −a and noticing that, in the odd prime case, the middle terms cancel.

5.2 Steenrod operations

Theorem 5.6. Let p be a prime, and let ρ ∈ R(G)^{+} be an actual representation of G. Then

N_{G}^{G×C_{p}} (ρ) = (ψ_{1}(ρ)^{p}) ⊗ + ∑_{i=1}^{p−1} \binom{p}{i + 1} ( \frac{1}{p}ψ_{1}(ρ)^{p} − \frac{1}{p}ψ_{p}(ρ)) ⊗ (σ − 1)^{i}

where σ is a choice of generator for R(C_{p}) and the ψ_{k} are the Adams operations.

Lemma 5.7. With notation as above

N_{G}^{G×C_{p}} (ρ) = ( \frac{1}{p}ψ_{1}(ρ)^{p} + \frac{p − 1}{p} ψ_{p}(ρ)) ⊗ + ∑_{i=1}^{p−1} ( \frac{1}{p}ψ_{1}(ρ)^{p} − \frac{1}{p}ψ_{p}(ρ)) ⊗ σ^{i}.

Proof. Note that

R(G × C_{p}) ≅ R(G) ⊗ R(C_{p}) ≅ ⨁_{i=0}^{p−1} R(G) ⊗ σ^{i}.

Recall that N_{H}^{G}(ρ)(g × γ) acts diagonally as ρ(g), with a cyclic permutation corresponding to γ. Fix g ∈ G and let B = {e_{i}}_{i∈I} be a basis of eigenvectors for ρ(g), with ρ(g) · e_{i} = λ_{i}e_{i}. Let B^{⊗p} = {e_{i_{1}} ⊗ ⋯⊗ e_{i_{p}}|e_{j} ∈ B} , a basis for V^{⊗p} . Pick a generator γ of C_{p}, then γ acts on V^{⊗p} by cyclic permutation of the factors; as such V^{⊗p} can be decomposed into isotypical components, on which γ acts as multiplication by ω^{k} for k = 0, ⋯, p − 1, with ω a primitive p-th root of unity. Let E_{k} be the subspace on which γ is multiplication by ω^{k}. Then N_{G}^{G×C_{p}} (ρ) = ⨁ (ρ ⊗ ⋯⊗ ρ)|_{E_{k}} ⊗ σ^{k}, and

E_{0} = ⟨ ∑_{i=0}^{p−1} σ(γ)^{i} · u | u ∈ B^{⊗p} ⟩

E_{k} = ⟨ ∑_{i=0}^{p−1} ω^{−ik} σ(γ)^{i} · u | u ∈ B^{⊗p} ⟩

Let u = e_{i_{1}} ⊗ ⋯⊗ e_{i_{p}} ∈ B^{⊗p} and w = ∑ σ(γ)^{i} · u ∈ E_{0}. Then ρ(g) · w = λ_{i_{1}} ⋯λ_{i_{p}}w, thus

Trρ(g)|_{E_{0}} = ∑_{(i_{1},···,i_{p})∈I^{p}} λ_{i_{1}} ⋯λ_{i_{p}} = \frac{1}{p} ( ∑_{i∈I} λ_{i}) ^{p} + \frac{p − 1}{p} ∑_{i∈I} λ_{i}^{p}.

So (ρ ⊗ ⋯⊗ ρ)|_{E_{0}} = \frac{1}{p}ρ^{p} + \frac{p−1}{p} ψ_{p}(ρ). Now, since all the E_{k} are isomorphic, we can specialize to E_{1}:

Trρ(g)|_{E_{1}} = ∑_{(i_{1},···,i_{p})∈I\Δ} λ_{i_{1}} ⋯λ_{i_{p}} = \frac{1}{p} ( ∑_{i∈I} λ_{i}) ^{p} − \frac{1}{p} ∑_{i∈I} λ_{i}^{p}

where Δ is the diagonal of I^{p}. Thus we have (ρ ⊗ ⋯⊗ ρ)|_{E_{0}} = \frac{1}{p}ρ^{p} − \frac{1}{p}ψ_{p}(ρ). This proves the lemma.

proof of theorem 5.6.This follows from the Lemma by a change of basis, keeping in mind that

∑_{i=k}^{n} \binom{i}{k} = \binom{n + 1}{k + 1} .

6 Table of results

The variables x_{i} denote degree one generators, y_{i} for degree two, etc.

Group  Order  R^{∗}(G)  R^{∗}(G) (if unsaturated)
Cyclic C_{N}  N  \frac{[x]}{(Nx)}  sat.
Elementary C_{p}^{k}  p^{k}  \frac{[x_{1},···,x_{k}]}{(px_{i},x_{i}^{p}x_{j}−x_{i}x_{j}^{p})}  sat.
Quaternion Q_{8}  8  \frac{[x_{1},x_{2},y]}{(2x_{i},8y,x_{i}^{2},x_{1}x_{2}−4y)}  sat.
Dihedral D_{p} (p odd)  2p  \frac{[x,y]}{(2x,py,xy)}  sat.
Alternating A_{4}  12  \frac{[x,y,z]}{(3x,2y,2z,y^{2}−z^{3})}  \frac{[x,y]}{(3x,12y,4y+x^{2})} (partial proof)
Alternating A_{5} (wrong!)  60  \frac{[x_{1},x_{2},y,z]}{(3x_{1},5x_{2},2z,2y,y^{3}−z^{3})}  probably unsat.

References

[AM69] M. F. Atiyah and I. G. Macdonald. Introduction to commutative algebra. Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1969.

[Ati61] M. F. Atiyah. Characters and cohomology of finite groups. Inst. Hautes Études Sci. Publ. Math., (9):23–64, 1961.

[GM14] Pierre Guillot and Ján Mináč. Milnor K-theory and the graded representation ring. Journal of K-theory, 2014.

[Ser77] Jean-Pierre Serre. Linear representations of finite groups. Springer-Verlag, 1977. Translated from the second French edition by Leonard L. Scott, Graduate Texts in Mathematics, Vol. 42.

[Tam93] D. Tambara. On multiplicative transfer. Comm. Algebra, 21(4):1393–1420, 1993.

[Web] Peter Webb. A guide to mackey functors. Online notes.