Mackey functors and functorial extensions

Robert Hartmann Department of Mathematics and Computer Science University of Leicester University Road Leicester LE1 7RH, UK rh91@mcs.le.ac.uk

November 12, 2002

Abstract

Given a finite group X and a normal subgroup G of X, we show that any Mackey functor M for X induces another Mackey functor M̃ for X associated to G. We then consider the question, whether there exists a functorial map M̃ → M. In the case that the order of G and the index of G in X are relatively prime, we give a sufficient condition for the existence of such a map, using canonical induction formulae.

1 Introduction

Let X be a finite group, and let G be a normal subgroup of X. A classical result in the representation theory of finite groups says that, if the order of G and the index of G in X are relatively prime, then any irreducible complex character of G that is stable under conjugation with elements of X, can be extended to a character of X. Moreover, this extension is unique under the additional condition that the determinant of its restriction to a fixed complement of G in X is trivial. However, as Example 3.6 will show, this extension is not functorial, in the following sense.

The rings of complex characters of X and its subgroups form a Mackey functor R for X. That is, every R(Y), for Y being a subgroup of X, is a ℤmodule, and one has conjugation, restriction and induction maps between these modules, satisfying a number of axioms. In Section 2 we show, that the normal subgroup G of X induces another Mackey functor R̃ for X, where the ℤ-module for a subgroup Y of X is given by R̃(Y) = R(Y ∩G)^{Y}, i.e., the virtual characters of Y ∩ G that are stable under Y. Now one can consider maps e : R̃ → R, i.e., families of maps e_{Y} : R̃(Y) → R(Y), one for each subgroup Y of X. If we require additionally that e_{Y} followed by res_{Y ∩G}^{Y} is the identity on R̃(Y) then asking for the existence of such a map is the same as the classical question of extending characters, as described above. But, in this functorial framework, the given answer is not satisfying, since the maps e_{Y} : R̃(Y) → R(Y) do not commute with the restriction maps of the Mackey functors R̃ and R. In Section 3 we show how to construct a different family of maps, that do commute with conjugations and restrictions. We call such a family an r-extension morphism. The essential ingredients in its construction are the Adams operators on the ring of complex characters.

Now, if M is any Mackey functor for X, we still can construct another Mackey functor M̃ for X associated to G, where for a subgroup Y of X we set M̃(Y) = M(Y ∩ G)^{Y}. Also, we still can hope to find an r-extension morphism. However, this question is much harder, and we only give a partial answer to it. First of all, we require the existence of a canonical induction formula for M. Such a formula for the ring of complex characters (and some other representation rings) has been given by Boltje in [1]. We briefly recall the most important facts about this theory in Section 4. Then we show that a canonical induction formula for M induces also one for M̃. Finally, in Section 5 we give sufficient conditions for the existence of r-extension morphisms, using the canonical induction formula. As an application, we prove the existence of an r-extension morphism for the trivial source ring.

Most of the contents of this article are part of the author’s Diplom thesis written at the University of Augsburg under the supervision of Robert Boltje. The author wants to thank Robert Boltje for his advice and many ideas involved in the development of this article.

Notation

For a finite group G, we write H ≤ G to denote that H is a subgroup of G. If H ≤ G and x ∈ G we write ^{x}H for the conjugate subgroup xHx^{−1}. For subgroups H, K ≤ G we write [G/H] respectively [H\G/K] for a set of representatives for the cosets respectively double cosets in G.

2 Mackey functors

Let X be a finite group, and let k be any commutative ring. We use the definition of a Mackey functor due to Green, as given in [1].

2.1 Definition

(a) A k-conjugation functor M = (M, c) for X is a family of k-modules M(Y), for Y ≤ X, together with k-module homomorphisms

c_{x,Y} : M(Y) → M(^{x}Y), for x ∈ X, Y ≤ X,

such that

(C1)  c_{y,Y} = id_{M(Y)} for y ∈ Y, Y ≤ X,

(C2)  c_{x_{1}x_{2},Y} = c_{x_{1},^{x_{2}} Y} ∘ c_{x_{2},Y} for x_{1}, x_{2} ∈ X, Y ≤ X.

(b) A k-restriction functor M = (M, c, r) for X is a k-conjugation functor (M, c) together with k-module homomorphisms

r_{Z}^{Y} : M(Y) → M(Z) for Z ≤ Y ≤ X,

such that

(R1)  r_{Y}^{Y} = id_{M(Y)} for Y ≤ X,
(R2)  r_{W}^{Z} ∘ r_{Z}^{Y} = r_{W}^{Y} for W ≤ Z ≤ Y ≤ X,
(R3)  c_{x,Z} ∘ r_{Z}^{Y} = r_{^{x}Z}^{^{x}Y} ∘ c_{x,Y} for x ∈ X, Z ≤ Y ≤ X.

(c) A k-Mackey functor M = (M, c, r, t) for X is a k-restriction functor (M, c, r) together with k-module homomorphisms

t_{Z}^{Y} : M(Z) → M(Y) for Z ≤ Y ≤ X,

such that

(M1)  t_{Y}^{Y} = id_{M(Y)} for Y ≤ X,
(M2)  t_{Z}^{Y} ∘ t_{W}^{Z} = t_{W}^{Y} for W ≤ Z ≤ Y ≤ X,
(M3)  c_{x,Y} ∘ t_{Z}^{Y} = t_{^{x}Z}^{^{x}Y} ∘ c_{x,Z} for x ∈ X, Z ≤ Y ≤ X,
(M4)  r_{U}^{Y} ∘ t_{Z}^{Y} = ∑_{y∈[U\Y/Z]} t_{U∩^{y}Z}^{U} ∘ r_{U∩^{y}Z}^{^{y}Z} ∘ c_{y,Z},

for Z ≤ Y ≤ X, U ≤ Y. We call the maps c_{x,Y} conjugations, the maps r_{Z}^{Y} restrictions and the maps t_{Z}^{Y} transfer or induction maps.

(d) A morphismus f : M → N of k-conjugation (resp. restriction resp. Mackey) functors M and N is a family of k-module homomorphisms

f_{Y} : M(Y) → N(Y) for Y ≤ X,

that commute with conjugations (resp. conjugations and restrictions resp. conjugations, restrictions and transfers).

Instead of c_{x,Y} (m), often we write just ^{x}m, and sometimes we write m|_{Z} instead of r_{Z}^{Y} (m). A is called a subfunctor of M if A(Y) ⊆ M(Y) is a submodule for all Y ≤ X, and if the structure maps restricted to these submodules have images in the corresponding submodules. In this case we write A ⊆ M. For subgroups Y, Z ≤ X with Z normal in Y, denote by M(Z)^{Y} the set of Y -fixed points in M(Z), i.e.

M(Z)^{Y} = {m ∈ M(Z) | c_{y,Z}(m) = m for all y ∈ Y }.

2.2 Theorem Let (M, c, r, t) be a k-Mackey functor for X, and let G be a normal subgroup of X. Then

M̃(Y) := M(Y ∩ G)^{Y} , for Y ≤ X,
c̃_{x,Y} := c_{x,Y ∩G} for x ∈ X, Y ≤ X,
r_{Z}^{Y} := r_{Z∩G}^{Y ∩G} for Z ≤ Y ≤ X,

t̃_{Z}^{Y} := ∑_{y∈[Y/Z(Y ∩G)]} c_{y,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} for Z ≤ Y ≤ X,

defines a k-Mackey functor (M̃, c̃, r̃, t̃) for X.

Proof First of all, note that the structure maps are well defined:

c̃_{x,Y} (m) = c_{x,Y ∩G}(m) ∈ M(^{x}(Y ∩ G)) = M(^{x}Y ∩ G)

for all m ∈ M̃(Y), and if ^{x}y = xyx^{−1} ∈ ^{x}Y then

c_{xyx^{−1},^{x}Y ∩G} ∘ c̃_{x,Y} (m) = c_{xyx^{−1}x,Y ∩G}(m) = c_{x,Y ∩G} ∘ c_{y,Y ∩G}(m) = c̃_{x,Y} (m),

so that c̃_{x,Y} (m) ∈ M(^{x}Y ∩ G)^{^{x}Y} = M̃(^{x}Y). Similarly, r_{Z}^{Y} maps M̃(Y) to M̃(Z), and for the transfer maps one has

c_{y_{0},Y ∩G} ∘ t̃_{Z}^{Y} = ∑_{y∈[Y/Z(Y ∩G)]} c_{y_{0}y,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} =

= ∑_{y′∈[Y/Z(Y ∩G)]} c_{y′,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} = t̃_{Z}^{Y} ,

since also the elements y′ = y_{0}y, y ∈ [Y/Z(Y ∩ G)], constitute a set of representatives for the cosets. It remains to show the validity of the conditions from Definition 2.1. Axioms (C1,2), (R1,2,3) and (M1) follow immediately from the corresponding conditions for the Mackey functor M. We will show (M2-4): (M2): Let W ≤ Z ≤ Y ≤ X. Then

t̃_{Z}^{Y} ∘ t̃_{W}^{Z} = ∑_{y∈[Y/Z(Y ∩G)]} c_{y,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} ∘ (  ∑_{z∈[Z/W (Z∩G])} c_{z,Z∩G} ∘ t_{W ∩G}^{Z∩G}) =

= ∑_{y∈[Y/Z(Y ∩G)]} ∑_{z∈[Z/W (Z∩G])} c_{yz,Y ∩G} ∘ t_{W ∩G}^{Y ∩G}.

On the other hand,

t̃_{W}^{Y} = ∑_{x∈[Y/W (Y ∩G])} c_{x,Y ∩G} ∘ t_{W ∩G}^{Y ∩G},

so that it remains to show that if [Y/Z(Y ∩ G)] is a fixed set of representatives for the cosets in Y/Z ( Y ∩ G), and [Z/W (Z∩G)] for the cosets in Z/W (Z∩G), then the set

{yz | y ∈ [Y/Z(Y ∩ G)], z ∈ [Z/W (Z ∩ G)]}

forms a set of representatives for the cosets in Y/W (Y ∩ G). Hence let y ∈ Y be arbitrary, then there exists y′ ∈ [Y/Z(Y ∩ G)] such that y ∈ y′Z(Y ∩ G), so that we find z ∈ Z such that y ∈ y′z(Y ∩ G). Now, there exists z′ ∈ [Z/W (Z ∩ G)] with z ∈ z′(Z ∩ G)W, and we obtain

y ∈ y′z′(Z ∩ G)W (Y ∩ G) = y′z′W (Y ∩ G).

Conversely, if yz(Y ∩ G)W = y′z′(Y ∩ G)W with y, y′ ∈ [Y/Z(Y ∩ G)] and z, z′ ∈ [Z/W (Z ∩ G)], then there exist g ∈ Y ∩ G and w ∈ W satisfying yzgw = y′z′, hence

y^{−1}y′ = zgwz′^{−1} ∈ Z(Y ∩ G)W Z = Z(Y ∩ G),

and it follows that y = y′ and z′ = zgw, so that

z′z^{−1} = gw ∈ Z ∩ W (Y ∩ G).

Using Dedekind identity, we conclude

z′z^{−1} ∈ Z ∩ Y ∩ GW = Z ∩ GW = (Z ∩ G)W,

hence z = z′, which proves (M2). (M3): Let Z ≤ Y ≤ X and x ∈ X. Then

t̃_{^{x}Z}^{^{x}Y} ∘ c̃_{x,Z} = ∑_{^{x}y∈[^{x}Y/(^{x}Y ∩G)(^{x}Z)]} c_{^{x}y,^{x}Y ∩G} ∘ t_{^{x}Z∩G}^{^{x}Y ∩G} ∘ c_{x,Z∩G} =

= ∑_{y∈[Y/(Y ∩G)Z]} c_{xyx^{−1},^{x}Y ∩G} ∘ c_{x,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} =

= c_{x,Y ∩G} ( ∑_{y∈[Y/(Y ∩G)Z]} c_{y,Y ∩G} ∘ t_{Z∩G}^{Y ∩G}) = c̃_{x,Y} ∘ t̃_{Z}^{Y} .

Finally (M4): Let U, Z ≤ Y ≤ X. We will proceed in three steps. First we consider the case Z = Z(Y ∩ G). Then Y ∩ G = Z ∩ G and

t̃_{Z}^{Y} = ∑_{x∈[Y/Z(Y ∩G)]} c_{x,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} = ∑_{x∈[Y/Z]} c_{x,Y ∩G}.

Moreover, for y ∈ Y we have

t̃_{U∩^{y}Z}^{U} = ∑_{u∈[U/(U∩^{y}Z)(U∩G)]} c_{u,U∩G} ∘ t_{U∩^{y}Z∩G}^{U∩G} = ∑_{u∈[U/U∩^{y}Z]} c_{u,U∩^{y}Z},

since U ∩ G = U ∩^{y}(Y ∩ G) = U ∩^{y}(Z ∩ G) ≤ U ∩^{y}Z, and hence U ∩ G = ^{y}Z. Therefore we have

r_{U}^{Y} ∘ t̃_{Z}^{Y} = r_{U∩G}^{Y ∩G} ( ∑_{y∈[Y/Z]} c_{y,Y ∩G}) = ∑_{y∈[Y/Z]} r_{U∩G}^{Y ∩G} ∘ c_{y,Y ∩G}

and

∑_{x∈[U\Y/Z]} t̃_{U∩^{x}Z}^{U} ∘ r_{U∩^{x}Z}^{^{x}Z} ∘ c̃_{x,Z} = ∑_{x∈[U\Y/Z]} ∑_{u∈[U/U∩^{x}Z]} c_{u,U∩G} ∘ r_{U∩G}^{^{x}Z∩G} ∘ c_{x,Z∩G} =

= ∑_{x∈[U\Y/Z]} ∑_{u∈[U/U∩^{x}Z]} r_{U∩G}^{Y ∩G} ∘ c_{ux,Y ∩G}.

Hence we need to show that the set

{ux | x ∈ [U\Y/Z], u ∈ [U/U ∩^{x}Z]}

is a system of representatives for Y/Z. So, for arbitrary y ∈ Y, choose first x ∈ [U\Y/Z] with y ∈ UxZ, then let u ∈ U and z ∈ Z be such that y = uxz. Now pick u′ ∈ [U/U ∩^{x}Z] such that u ∈ u′(U ∩^{x}Z), and let ũ ∈ U and z̃ ∈ Z be such that u = u′ũ and ũ = ^{x}z̃. Then we have

y = u′ũxz = u′xz̃x^{−1}xz = u′xz̃z ∈ u′xZ.

Conversely, if x, x̃ ∈ [U\Y/Z], u ∈ [U/U ∩^{x}Z] and ũ ∈ [U/U ∩ x̃Z] are such that uxZ = ũx̃Z, then there exists z ∈ Z such that ux = ũx̃z, hence

x̃ = u^{−1}uxz^{−1} ∈ UxZ,

which implies that x = x̃, and furthermore

u = ũxzx^{−1} ∈ ũ(^{x}Z ∩ U),

since ^{x}z = u^{−1}u ∈ ^{x}Z ∩ U. This shows u = ũ, and the first case in proved. In a second step, we consider the special case Y = Z(Y ∩ G). Here we have

t̃_{Z}^{Y} = t_{Z∩G}^{Y ∩G},

so that

(1)  r_{U}^{Y} ∘ t̃_{Z}^{Y} = r_{U∩G}^{Y ∩G} ∘ t_{Z∩G}^{Y ∩G} =

= ∑_{y∈[U∩G\Y ∩G/Z∩G]} t_{U∩^{y}Z∩G}^{U∩G} ∘ r_{U∩^{y}Z∩G}^{^{y}Z∩G} ∘ c_{y,Z∩G}.

On the other hand,

(2)  ∑_{x∈[U\Y/Z]} t̃_{U∩^{x}Z}^{U} ∘ r_{U∩^{x}Z}^{^{x}Z} ∘ c̃_{x,Z} =

= ∑_{x∈[U\Y/Z]} ∑_{u∈[U/(U∩G)(U∩^{x}Z)]} c_{u,U∩G} ∘ t_{U∩^{x}Z∩G}^{U∩G} ∘ r_{U∩^{x}Z∩G}^{^{x}Z∩G} ∘ c_{x,Z∩G} =

= ∑_{x∈[U\Y/Z]} ∑_{u∈[U/(U∩G)(U∩^{x}Z)]} t_{U∩^{ux}Z∩G}^{U∩G} ∘ r_{U∩^{ux}Z∩G}^{^{ux}Z∩G} ∘ c_{ux,Z∩G},

and we have to show that expressions (1) and (2) coincide. Let R denote a set of representatives for the double cosets in U\Y/Z, and for any x ∈ R, let R_{x} denote a set of representatives for the cosets in U/(U ∩ G)(U ∩^{x}Z). For every x ∈ R and u ∈ R_{x} we have ux ∈ Y = (Y ∩ G)Z, so we can write

ux = g_{ux}z_{ux},

where g_{ux} ∈ Y ∩ G and z_{ux} ∈ Z. Now inspection of expression (2) shows that we can replace every occuring ux by uxz_{ux}^{−1} without changing it. Thus, it is now enough to show that the set

R′ = {g_{ux} | x ∈ R, u ∈ R_{x}}

is a system of representatives for the double cosets in U ∩ G\Y ∩ G/Z ∩ G. Now, given any g ∈ Y ∩ G, we choose x ∈ R with g ∈ UxZ, and u ∈ U, z ∈ Z such that g = uxz. Next, pick u′ ∈ R_{x} such that

u ∈ u′(U ∩ G)(U ∩^{x}Z) = (U ∩ G)u′(U ∩^{x}Z),

and let u = u_{1}u′u_{2}, where u_{1} ∈ U ∩ G and u_{2} = xz_{2}x^{−1} ∈ U ∩ ^{x}Z for some z_{2} ∈ Z. Then we have

g = u_{1}u′u_{2}xz = u_{1}u′xz_{2}x^{−1}xz = u_{1}u′xz_{2}z.

Now, as described above, we write u′x = g̃z̃, where g̃ ∈ R′ and z̃ ∈ Z. Then g = u_{1}g̃z̃z_{2}z and therefore g^{−1}u_{1}^{−1}g = z̃z_{2}z ∈ Z ∩ G. This yields

g = u_{1}g̃(z̃z_{2}z) ∈ (U ∩ G)g̃(Z ∩ G)

with g̃ ∈ R′. Conversely, suppose we have x_{1}, x_{2} ∈ R′ with x_{2} = ux_{1}z for some u ∈ U ∩ G and z ∈ Z ∩ G. Then there are elements y_{1}, y_{2} ∈ R, u_{1} ∈ R_{y_{1}}, u_{2} ∈ R_{y_{2}} and z_{1}, z_{2} ∈ Z such that x_{1} = u_{1}y_{1}z_{1} and x_{2} = u_{2}y_{2}z_{2}. It follows that

y_{2} = u_{2}^{−1}x_{2}z_{2}^{−1} = u_{2}^{−1}uu_{1}y_{1}z_{1}zz_{2}^{−1} ∈ Uy_{1}Z,

hence y_{1} = y_{2}. Furthermore,

u_{2} = x_{2}z_{2}^{−1}y_{2}^{−1} = ux_{1}zz_{2}^{−1}y_{1}^{−1} = uu_{1}(y_{1}z_{1}zz_{2}^{−1}y_{1}^{−1}),

thus u_{2} ∈ (U ∩ G)u_{1}(^{y_{1}}Z ∩ U) = u_{1}(U ∩ G)(^{y_{1}}Z ∩ U) and we conclude u_{1} = u_{2} which proves the second case. Finally, in the third step we show the general case. Write

r_{U}^{Y} ∘ t̃_{Z}^{Y} = (r_{U}^{Y} ∘ t̃_{Z(Y ∩G)}^{Y}) ∘ t̃_{Z}^{Z(Y ∩G)} =

(applying the result from the first step)

= ∑_{x∈[U\Y/Z(Y ∩G)]} t̃_{U∩^{x}(Z(Y ∩G))}^{U} ∘ r_{U∩^{x}(Z(Y ∩G))}^{^{x}(Z(Y ∩G))} ∘ c̃_{x,Z(Y ∩G)} ∘ t̃_{Z}^{Z(Y ∩G)} =

= ∑_{x∈[U\Y/Z(Y ∩G)]} t̃_{U∩^{x}Z(Y ∩G)}^{U} ∘ r_{U∩^{x}Z(Y ∩G)}^{^{x}Z(Y ∩G)} ∘ t̃_{^{x}Z}^{^{x}Z(Y ∩G)} ∘ c̃_{x,Z} =

(applying the result from the second step)

= ∑_{x∈[U\Y/Z(Y ∩G)]} ∑_{y∈[U∩^{x}Z(Y ∩G)\^{x}Z(Y ∩G)/^{x}Z]} t̃_{U∩^{yx}Z}^{U} ∘ r_{U∩^{yx}Z}^{^{yz}Z} ∘ c̃_{yz,Z}

and it remains to show that, if R denotes a set of representatives for the double cosets in U\Y/Z(Y ∩ G), and if for every x ∈ R, R_{x} denotes a set of representatives for the double cosets in U ∩ ^{x}Z(Y ∩ G)\^{x}Z(Y ∩ G)/^{x}Z, then the set

R′ = {yx | x ∈ R, y ∈ R_{x}}

is a set of representatives for U\Y/Z. For any given y_{0} ∈ Y, choose x ∈ R such that

y_{0} ∈ Ux(Y ∩ G)Z = U(Y ∩ G)xZ.

Then there exist u ∈ U, g ∈ Y ∩ G and z ∈ Z such that y_{0} = ugxz. Now pick y ∈ R_{x} such that g ∈ (U ∩^{x}Z(Y ∩G))y(^{x}Z), then we can write g = u′y(xz′x^{−1}) for some u′ ∈ U ∩^{x}Z(Y ∩ G) and z′ ∈ Z, and we obtain

y_{0} = uu′yxz′x^{−1}xz = uu′yxz′z ∈ UyxZ.

Conversely, if x_{1}, x_{2} ∈ R, y_{1} ∈ R_{x_{1}} and y_{2} ∈ R_{x_{2}} are such that y_{2}x_{2} = uy_{1}x_{1}z for some u ∈ U and z ∈ Z, then we can write y_{2} = g(^{x_{2}}z_{2}), where g ∈ Y ∩ G and z_{2} ∈ Z. We obtain

x_{2} = (gx_{2}z_{2}x_{2}^{−1})^{−1}uy_{1}x_{1}z = x_{2}z_{2}^{−1}x_{2}^{−1}g^{−1}uy_{1}x_{1}z,

hence x_{2} = g^{−1}uy_{1}x_{1}zz_{2}^{−1}, so that

x_{2} ∈ (Y ∩G)U(^{x_{1}}Z(Y ∩G))x_{1}Z = U(Y ∩G)x_{1}Z(Y ∩G)x_{1}^{−1}x_{1}Z = Ux_{1}(Y ∩G)Z,

which implies that x_{1} = x_{2}. Moreover,

y_{2} = uy_{1}x_{1}zx_{1}^{−1} = uy_{1}(^{x_{1}}z),

and

u = y_{2}(^{x_{1}}z)^{−1}y_{1}^{−1} ∈ (^{x_{1}}Z(Y ∩ G)) ∩ U,

hence

y_{2} ∈ (^{x_{1}}Z(Y ∩ G) ∩ U)y_{1}(^{x_{1}}Z),

hence y_{2} = y_{1}, which completes the proof. □

3 Extension morphisms

Let X be a finite group, G a normal subgroup of X, and let (M, c, r, t) denote a k-Mackey functor for X. Define (M̃, c̃, r̃, t̃) as in Theorem 2.2.

3.1 Lemma The maps

d : M → M̃, d_{Y} = r_{Y ∩G}^{Y} for Y ≤ X

define a morphism of Mackey functors.

Proof For Z ≤ Y ≤ X and x ∈ X one has

c̃_{x,Y} ∘ d_{Y} = c_{x,Y ∩G} ∘ r_{Y ∩G}^{Y} = r_{^{x}Y ∩G}^{^{x}Y} ∘ c_{x,Y} = d_{^{x}Y} ∘ c_{x,Y} ,
r_{Z}^{Y} ∘ d_{Y} = r_{Z∩G}^{Y ∩G} ∘ r_{Y ∩G}^{Y} = r_{Z∩G}^{Y} = r_{Z∩G}^{Z} ∘ r_{Z}^{Y} = d_{Z} ∘ r_{Z}^{Y}

and

t̃_{Z}^{Y} ∘ d_{Z} = ∑_{y∈[Y/(Y ∩G)Z]} c_{y,Y ∩G} ∘ t_{Z∩G}^{Y ∩G} ∘ r_{Z∩G}^{Z} =

= ∑_{y∈[Y ∩G\Y/Z]} t_{^{y}Z∩(Y ∩G)}^{Y ∩G} ∘ r_{^{y}Z∩(Y ∩G)}^{^{y}Z} ∘ c_{y,Z} = r_{Y ∩G}^{Y} ∘ t_{Z}^{Y} = d_{Y} ∘ t_{Z}^{Y} .

□

3.1 Definition For Y ≤ X and m ∈ M̃(Y) = M(Y ∩ G)^{Y} we call m̂ ∈ M(Y) an extension of m to M(Y) if d_{Y} (m̂) = m. A morphism e : M̃ → M of conjugation functors is called an extension morphism, if for all Y ≤ X

d_{Y} ∘ e_{Y} = id_{M̃(Y)}.

If e is even a morphism of restriction functors, we call e an r-extension morphism.

3.2 Proposition Assume that |G| ∈ k^{×}. Then

e_{Y} : M̃(Y) → M(Y), m ↦ \frac{1}{[Y : Y ∩ G]}t_{Y ∩G}^{Y}(m)

defines an r-extension morphism.

Proof Let Z ≤ Y ≤ X and let m ∈ M̃(Y). Then

d_{Y} ∘ e_{Y} (m) = r_{Y ∩G}^{Y} (\frac{1}{[Y : Y ∩ G]}t_{Y ∩G}^{Y}(m)) =

= \frac{1}{[Y : Y ∩ G]} ∑_{y∈[Y ∩G\Y/Y ∩G]} t_{^{y}(Y ∩G)∩(Y ∩G)}^{Y ∩G} ∘ r_{^{y}(Y ∩G)∩(Y ∩G)}^{^{y}(Y ∩G)}(^{y}m) =

= \frac{1}{[Y : Y ∩ G]} ∑_{y∈[Y/Y ∩G]} m = m.

Clearly e commutes with conjugations, and

r_{Z}^{Y} ∘ e_{Y} (m) = \frac{1}{[Y : Y ∩ G]}r_{Z}^{Y} ∘ t_{Y ∩G}^{Y}(m) =

= \frac{1}{[Y : Y ∩ G]} ∑_{y∈[Z\Y/Y ∩G]} t_{Y ∩G∩Z}^{Z} ∘ r_{Y ∩G∩Z}^{Y ∩G}(^{y}m) =

= \frac{1}{[Y : Y ∩ G]} ∑_{y∈[Y/Z(Y ∩G)]} t_{Z∩G}^{Z} ∘ r_{Z∩G}^{Y ∩G}(m) =

= \frac{[Y : Z(Y ∩ G)]}{[Y : Y ∩ G]}t_{Z∩G}^{Z} ∘ r_{Z∩G}^{Y ∩G}(m) =

= \frac{1}{[Z : Z ∩ G]}t_{Z∩G}^{Z} ∘ r_{Z∩G}^{Y ∩G}(m) = e_{Z} ∘ r_{Z}^{Y} (m).

□

3.3 Remark If k = ℤ then we can extend scalars to ℚ, i.e., consider the ℚ-Mackey functors ℚM (by that we mean ℚ ⊗_{ℤ} M), and similarly ℚM̃. We can apply Proposition 3.2, but in general the image of M̃ under e will not be in M but only in ℚM.

Now we consider the Mackey functor R = (R, c, res, ind) that assigns to each Y ≤ X the ring R(Y) = R_{ℂ}(Y) of complex characters of Y. Moreover, assume that (|G|, [X : G]) = 1, so that G has a complement S in X. In this situation, there is a well known result about extensions of characters:

3.4 Theorem ([3], 13.3) For any irreducible character χ of G that is stable under X, there is a unique irreducible character χ̂ of X such that χ̂|_{G} = χ and det(χ̂|_{S}) = 1. Any other extension of χ is then of the form χ̂φ for some φ ∈ Ŝ.

3.5 Corollary If χ is an irreducible character of G and Stab_{X}(χ) = Y < X then χ has an extension χ̂ to an irreducible character of Y, and ind_{Y}^{X}(χ̂) is an extension of ∑_{x∈[X/Y]} ^{x}χ to a character of X. This induces an extension morphism

e_{1} : R̃ → R.

However, e_{1} does not commute with restrictions, as the following example shows:

3.6 Example Let G be the quaternion group of order 8, and let X be a semidirect product of G and a cyclic group S of order 3, such that S acts non-trivially on G. Let χ be the irreducible character of G of degree 2, then χ ∈ R̃(X) and

e_{1} ∘ r̃es_{S}^{X}(χ) ≠ res_{S}^{X} ∘ e_{1}(χ).

Let Ψ_{X}^{n} : R(X) → R(X) denote the n-th Adams operator, for any integer n, i.e.

Ψ_{X}^{n}(χ)(x) = χ(x^{n})

for any χ ∈ R(X) and x ∈ X.

3.7 Proposition There is a unique r-extension morphism

ext : R̃ → R

with the property that for all cyclic subgroups C of X and all φ ∈ R̃(C) one has ext(φ) = φ · 1. Moreover, for any Y ≤ X, the map ext^{Y} : R̃(Y) → R(Y) is given by

(3)  ext^{Y} (χ) = Ψ_{Y}^{n} (χ_{1}),

where χ_{1} ∈ R(Y) is any extension of χ ∈ R̃(Y), and n is an integer such that

n ≡ 1 mod |G| and n ≡ 0 mod [X : G].

Proof Let π denote the set of prime divisors of |G|, then every x ∈ X can be written as x = x_{π} · x_{π′} = x_{π′} · x_{π} with x_{π} ∈ G and x_{π′} being an element of π′-order.

Now let Y ≤ X. If χ_{1} and χ_{2} are two extensions of χ ∈ R̃(Y), and if y ∈ Y then

Ψ_{Y}^{n} (χ_{1})(y) = χ_{1}(y^{n}) = χ_{1}(y_{π}) = χ(y_{π}) = χ_{2}(y_{π}) = Ψ_{Y}^{n} (χ_{2})(y).

Hence, ext is well defined by the equation (3), and clearly ext commutes with conjugations. Moreover, for Z ≤ Y ≤ X, let χ_{1} ∈ R(Y) be any extension of χ ∈ R̃(Y), and let ψ_{1} ∈ R(Z) be any extension of ψ = χ|_{Z∩G}, then one has for any z ∈ Z that

res_{Z}^{Y} (ext^{Y} (χ))(z) = ext^{Y} (χ)(z) = Ψ^{n}(χ_{1})(z) = χ_{1}(z^{n}) = χ_{1}(z_{π}) =

= χ(z_{π}) = χ|_{Z∩G}(z_{π}) = ψ(z_{π}) = ψ_{1}(z^{n}) = Ψ^{n}(ψ_{1})(z) = ext^{Z}(r̃es_{Z}^{Y} (χ))(z), hence ext is indeed an r-extension morphism. If C is a cyclic subgroup of X, φ ∈ R̃(C) and y ∈ C then

ext^{C}(φ)(y) = φ(y_{π}) = (φ · 1)(y_{π} · y_{π′}) = (φ · 1)(y).

Finally, to prove uniqueness, assume that e_{0} is another morphism satisfying the stated conditions. Then, for any Y ≤ X, any χ ∈ R̃(Y) and any y ∈ Y one has

e_{0}^{Y} (χ)(y) = r_{<y>}^{Y}(e_{0}^{Y} (χ))(y) = e_{0}^{<y>}(r̃es_{<y>}^{Y}(χ))(y) =
(r̃es_{<y>}^{Y}(χ) · 1)(y) = ext^{<y>}(r̃es_{<y>}^{Y}(χ))(y) = ext^{Y} (χ)(y),

hence e_{0} = ext. □

4 Canonical induction formulae

We will use the theory of canonical induction formulae, as described in [1]. We will briefly recall the construction, but refer the reader to [1] for details.

4.1 Let (M, c, r) be a k-restriction functor for the finite group X. For every Y ≤ X we define

S_{M}(Y) := ⨁_{Z≤X} M(Z),

and

M_{+}(Y) := S_{M}(Y)/I(kY) · S_{M}(Y),

where I(kY) denotes the augmentation ideal of kY, i.e., the kernel of the algebra homomorphism ϵ : kY → k mapping ∑ α_{y}y to ∑ α_{y}, where y ∈ Y and α_{y} ∈ k. We write

[Z, m]_{Y} := m + I(kY) · S_{M}(Y)

for the elements in M_{+}(Y), where Z ≤ Y and m ∈ M(Z). Then M_{+} is a Mackey functor with the following structure maps:

c_{+}_{x,Y} : M_{+}(Y) → M_{+}(^{x}Y), [U, m]_{Y} ↦ [^{x}U,^{x}m]_{^{x}Y}

r_{+}_{Z}^{Y} : M_{+}(Y) → M_{+}(Z), [U, m]_{Y} ↦ ∑_{y∈[Z\Y/U]}[Z ∩^{y}U, r_{Z∩^{y}U}^{^{y}U}(^{y}m)]_{Z}

t_{+}_{Z}^{Y} : M_{+}(Z) → M_{+}(Y), [V, m]_{Z} ↦ [V, m]_{Y} .

From now on, let X be a finite group, and let G be a normal subgroup of X. Moreover, let (M, c, r, t) denote a ℤ-Mackey functor for X, and A ⊆ M a restriction subfunctor. Define M̃ and Ã as in Theorem 2.2. We will make the following 4.2 Assumptions

(i) A(Y) is a free abelian group, for every Y ≤ X. (ii) A has a stable basis (B(Y))_{Y ≤X}, such that for all Z ≤ Y ≤ X and b ∈ B(Y) the element r_{Z}^{Y} (b) is a linear combination of the basis elements in B(Z) with non-negative coefficients.

4.3 Clearly (Ã, c̃, r̃) is a restriction subfunctor of M̃. If M and A satisfy 4.2, then so do M̃ and Ã, since Ã(Y), Y ≤ X, has the stable basis

B̃(Y) = { ∑_{y∈Y/Stab_{Y} (φ)}^{y}b | b ∈ B(Y ∩ G)} .

4.4 For Y ≤ X define

M_{Y} := {(Z, b) | Z ≤ Y, b ∈ B(Z)},

then Y acts on M_{Y} by ^{y}(Z, b) = (^{y}Z,^{y}b). Moreover,

(Z_{1}, b_{1}) =_{Y} (Z_{2}, b_{2}) :⟺ (Z_{1}, b_{1}) = ^{y}(Z_{2}, b_{2}) for some y ∈ Y.

defines an equivalence relation on M_{Y}. By [Z, b]_{Y} we denote the equivalence class of (Z, b), and by M_{Y} /Y the set of equivalence classes. Note that A_{+}(Y) is then the free abelian group with basis M_{Y} /Y. Similarly, Ã_{+}(Y) is the free abelian group with basis M̃_{Y} /Y, which is the set of Y -conjugacy classes of pairs (Z, b′) where Z ≤ Y and b′ ∈ B̃(Z). We denote the basis elements in Ã_{+} again by [Z, b′]_{Y}.

For b ∈ B(Z) and W ≤ Z ≤ Y , write r_{W}^{Z} (b) as a linear combination of the basis elements in B(W), and call the (non-negative) coefficient m_{(W,c)}^{(Z,b)} at c ∈ B(W) the multiplicity of (W, c) in (Z, b). Then

(W, c) ≤ (Z, b) :⟺ W ≤ Z and m_{(W,c)}^{(Z,b)} > 0.

defines a partial ordering on M_{Y}.

4.5 We define the induction morphism

b = b^{M} : M_{+} → M

by

b_{Y} : M_{+}(Y) → M(Y), [U, m]_{Y} ↦ t_{U}^{Y} (m).

Then b is a morphism of Mackey functors (see [1], 3.1). A canonical induction formula a : M → A_{+} is a morphism of restriction functors such that b ∘ a = id_{M}.

4.6 Let p : M → A be a morphism of conjugation functors. Define

a_{Y} = a_{Y}^{(M,A,p)} : ℚM(Y) → ℚA_{+}(Y)

a_{Y} (m) = \frac{1}{|Y |} ∑_{V <U<Y} |V |μ(V, U)[L, r_{V}^{U}(p_{U}(r_{U}^{Y} (m)]_{Y} ,

where μ denotes the Möbius function on the poset of subgroups of X. Then a is a morphism of restriction functors (see [1],6.1).

4.7 Proposition (Boltje, [1], 6.4) The morphism a defined in 4.6 is a canonical induction formula if and only if for all Y ≤ X and m ∈ M(Y) one has p_{Y} (m) − m ∈ ∑_{Z<Y} t_{Z}^{Y} (M(Z)).

However, even if this condition is satisfied, the definition of the map a contains denominators, so the image of M(Y) under a_{Y} will in general not be contained in A_{+}(Y). We need an extra integrality condition:

4.8 Theorem (Boltje, [1],9.3) Let Y ≤ X, and assume that for a set π of primes the following condition holds:

For all m ∈ M(Y) and all T ◁ U ≤ Y such that U/T is a cyclic

(∗)_{π} π -group, and all b ∈ B(T) which are fixed under U, the coefficients of the two elements p_{T} (r_{T}^{Y} (m)) and r_{T}^{U}(p_{U}(r_{U}^{Y} (m))) in A(T) with respect to the basis B(T) coincide.

Then

|Y |_{π′} · a_{Y}^{(M,A,p)} ∈ A_{+}(Y).

In particular, if condition (∗)_{π} is satisfied for all Y ≤ X and the set π of all primes, then a defines an integral canonical induction formula.

Next we will show how, given a canonical induction formula for (M, A, p), we can construct a formula for (M̃, Ã, p̃), where

p̃: M̃ → Ã, p̃_{Y} = p_{Y ∩G} for Y ≤ X.

As in 4.6, we define ã : M̃ → Ã_{+} by

(4)  ã_{Y} (m) = \frac{1}{|Y |} ∑_{V <U<Y} |V |μ(V, U)[L, r_{V}^{U}(p̃_{U}(r_{U}^{Y} (m)]_{Y} ,

for Y ≤ X and m ∈ M̃(Y).

4.9 Theorem Let (M, c, r, t) be a ℤ-Mackey functor for X, A ⊆ M a subfunctor satisfying the assumptions in 4.2, and let p : M → A be a morphism of conjugation functors, such that condition (∗)_{π} from Theorem 4.8 is satisfied for the set π of all primes. Moreover, let G be a normal subgroup of X such that the order of G and the index in X are relatively prime. Define M̃ as in Theorem 2.2. Then (∗)_{π} holds also for the Mackey functor M̃, its subfunctor Ã and the morphism p̃, and the morphism ã, defined by equation (4), is an integral canonical induction formula.

Proof Clearly p̃ is a morphism of conjugation functors, hence defining ã by (4.6) certainly yields a morphism of restriction functors ℚM̃ → ℚÃ_{+}. Now, according to Proposition 4.7, for ã to be a canonical induction formula, we need to show that for any Y ≤ X and m ∈ ℚM̃(Y), we have

m − p̃_{Y} (m) ∈ ∑_{Z<Y} t̃_{Z}^{Y} (ℚM̃(Z)).

Since a is a canonical induction formula, again by 4.7, we know that

m − p̃_{Y} (m) = m − p_{Y ∩G}(m) = ∑_{H<Y ∩G} t_{H}^{Y ∩G}(m_{H})

for some elements m_{H} ∈ ℚM(H), thus

∑_{y∈[Y/Y ∩G]}^{y}(m − p̃_{Y} (m)) = ∑_{y∈[Y/Y ∩G]} ∑_{H<Y ∩G}^{y}t_{H}^{Y ∩G}(m_{H}) =

= ∑_{H<Y ∩G} ∑_{y∈[Y/(Y ∩G)H]}^{y}t_{H∩G}^{Y ∩G}(m_{H}) = ∑_{H<Y ∩G} t̃_{H}^{Y}(m_{H}).

Since m is stable under Y, and since p̃ is a morphism of conjugation functors, also m − p̃_{Y} (m) is stable under Y, so we obtain

[Y : Y ∩ G](m − p̃_{Y} (m)) ∈ ∑_{H<Y ∩G} t̃_{H}^{Y}(ℚM̃(H))

and dividing by [Y : Y ∩G] shows that ã is indeed a canonical induction formula.

To show integrality, we have to verify condition (∗)_{π} from Theorem 4.8 for the set π of all primes. So let T ◁ U ≤ Y such that U/T is cyclic, let b ∈ B̃(T)^{U}, and let m ∈ M̃(Y) be arbitrary. We need to show that the coefficients of the two elements p̃_{T} (r_{T}^{Y} (m)) and r_{T}^{U}(p̃_{U}(r_{U}^{Y} (m))) in Ã(T) at the basis element b coincide. We can write

(5)  b = ∑_{t∈T/S} ^{t}b_{1}

for some b_{1} ∈ B(T ∩ G), where S = Stab_{T} (b_{1}).

First we claim that b_{1} is stable under U ∩ G. Set V = Stab_{U∩G}(b_{1}), then clearly the element

b′ = ∑_{u∈[U∩G/V]}^{u}b_{1}

is stable under U ∩ G, and so is b by assumption. Comparing the definition of b′ and the description of b in (5), both are linear combinations of basis elements in B(T ∩ G), so the summands in b′ must all appear in b, and b must be a sum of conjugates of b′. However, the sum for b has [T : S] terms, and the sum for b′ has [U ∩ G : V] terms, and these numbers are relatively prime, by hypothesis. Thus V = U ∩ G.

Now,

p̃_{T} (r_{T}^{Y} (m)) = p_{T ∩G}(r_{T ∩G}^{Y ∩G}(m)

and

r_{T}^{U}(p̃_{U}(r_{U}^{Y} (m))) = r_{T ∩G}^{U∩G}(p_{U∩G}(r_{U∩G}^{Y ∩G}(m))).

Moreover, T ∩ G ◁ U ∩ G and U ∩ G/T ∩ G is cyclic, hence the coefficients of the two elements at b_{1} ∈ B(T ∩ G)^{U∩G} coincide, since condition (∗)_{π} holds for p. However, these are the same coefficients as at b, so (∗)_{π} holds also for p̃. □

4.10 Remark We just mention that there is an explicit formula for a without denominators, but refer the interested reader to [1], formula (9.5a). Then it is easy to write down a similar formula for ã.

5 Extensions for Mackey functors

Throughout this section, we assume that X is a finite group and G a normal subgroup of X such that (|G|, [X : G]) = 1. Let (M, c, r, t) be a ℤ-Mackey functor for X, A ⊆ M a subfunctor and p : M → A a morphism of conjugation functors. Let M̃ and Ã be defined as in Theorem 2.2. Furthermore, we assume that 4.2 is satisfied, and that a = a^{(M,A,p)} is an integral canonical induction formula.

Our goal is to generalize the result of Proposition 3.7, so we want to give sufficient conditions for the existence of an r-extension morphism

ext : M̃ → M.

Our first result allows us to reduce the question to the subfunctor A:

5.1 Proposition If there exists a morphism ext_{∗} : Ã → M of restriction functors such that

d_{Y} (ext_{∗}^{Y} (a)) = a

for all a ∈ Ã(Y) and all Y ≤ X, then

ext = b ∘ ext_{+} ∘ ã

defines an r-extension morphism M̃ → M, where

ext_{+}^{Y}([Z, a]_{Y}) := [Z, ext_{∗}^{Z}(a)]_{Y}

for [Z, a]_{Y} ∈ Ã_{+}(Y) and Y ≤ X.

Proof Clearly everything commutes with conjugations, and for Z, U ≤ Y ≤ X, and a ∈ Ã(Z) one has

ext_{+}^{U}(r̃_{+}_{U}^{Y} ([Z, a]_{Y})) = ext_{+}^{U} ( ∑_{x∈[U\Y/Z]}[U ∩^{x}Z, r_{U∩^{x}Z}^{^{x}Z}(^{x}a)]_{U} ) =

= ∑_{x∈[U\Y/Z]}[U ∩^{x}Z, ext_{∗}^{U∩^{x}Z}(r_{U∩^{x}Z}^{^{x}Z}(^{x}a))]_{U} =

= ∑_{x∈[U\Y/Z]}[U ∩^{x}Z, r_{U∩^{x}Z}^{^{x}Z}(^{x}ext_{∗}^{Z}(a))]_{U} =

= r_{+}_{U}^{Y} ([Z, ext_{∗}^{Z}(a)]_{Y} = r_{+}_{U}^{Y} (ext_{+}^{X}([Z, a]_{Y})).

Thus ext is a morphism of restriction functors, and

d_{Y} ∘ ext^{Y} = r_{Y ∩G}^{Y} ∘ b_{Y} ∘ ext_{+}^{Y} ∘ ã_{Y} = b_{Y ∩G} ∘ ext_{+}^{Y ∩G} ∘ ã_{Y ∩G} ∘ r_{Y ∩G}^{Y} =

= b_{Y ∩G} ∘ ã_{Y ∩G} = b_{Y ∩G} ∘ a_{Y ∩G} = id_{M(Y ∩G)}.
□

We will have to assume that elements of the subfunctor Ã of M̃ have extensions, more precisely, the need the following condition

5.2 For every Y ≤ X and every basis element b ∈ B(Y ∩ G) there exists a (distinguished) extension b_{1} ∈ A(Stab_{Y} (b)), such that for any x ∈ X, the element ^{x}b_{1} coincides with the distinguished extension of the basis element ^{x}b in B(^{x}Y ∩ G).

Additionally, we need an analog to the Adams operators, that is:

5.3 There is a morphism Ψ : M → M of restriction functors, such that d_{Y} (Ψ(a)) = a for all Y ≤ X and a ∈ Ã(Y), and such that we have Ψ(a_{1}) = Ψ(a_{2}) for any two extensions a_{1}, a_{2} ∈ M(Y) of a.

5.4 Theorem Suppose that conditions 5.2 and 5.3 are satisfied. Then there exists an rextension morphism ext = ext^{M,A} : M̃ → M.

Proof Let Z ≤ Y ≤ X. Every basis element in B̃(Z) is of the form b′ = ∑_{z∈Z/S} ^{z}b, where b ∈ B(Z ∩ G) and S = Stab_{Z}(b). Let b_{1} ∈ A(S) be the distinguished extension of b to S, then b̂ = t_{S}^{Z}(b_{1}) is an extension of b′ to Y. Now define

ext_{+}^{Y} : Ã_{+}(Y) → M_{+}(Y)

ext_{+}^{Y}([Z, b′]_{Y}) = [Z, Ψ(b̂)]_{Y} .

Then clearly ext_{+} commutes with conjugations, and

r_{+}_{U}^{Y} (ext_{+}^{X}([Z, b′]_{Y})) = r_{+}_{U}^{Y} ([Z, Ψ(b̂)]_{Y}) =

= ∑_{x∈[U\Y/Z]}[^{x}Z ∩ U, r_{^{x}Z∩U}^{^{x}Z}^{x}Ψ(b̂)]_{U} =

= ∑_{x∈[U\Y/Z]}[^{x}Z ∩ U, Ψ(r_{^{x}Z∩U}^{^{x}Z}(^{x}b̂))]_{U}

But r_{^{x}Z∩U}^{^{x}Z}(^{x}b̂) is an extension of r_{^{x}Z∩U∩G}^{^{x}Z}(^{x}b̂), hence the above equals

∑_{x∈[U\Y/Z]} ext_{+}^{U}([^{x}Z ∩ U, r_{^{x}Z∩U∩G}^{^{x}Z}(^{x}b̂)]_{U}) =

= ext_{+}^{U}( ∑_{x∈[U\Y/Z]}[^{x}Z ∩ U, r_{^{x}Z∩U}^{^{x}Z}(d_{^{x}Z}(^{x}b̂)]_{U}) =

= ext_{+}^{U}( ∑_{x∈[U\Y/Z]}[^{x}Z ∩ U, r_{^{x}Z∩U}^{^{x}Z}(^{x}b′)]_{U}) =

= ext_{+}^{U}(r̃_{+}_{U}^{Y} ([Z, b′]_{Y}))

Therefore ext := b ∘ ext_{+} ∘ a is an r-extension morphism by Proposition 5.1. □

5.5 Remark If one sets M = R, the character ring over the complex numbers with the usual conjugation, restriction and induction maps, one can easily give another proof of Proposition 3.7, using the canonical induction formula from [1]. Here the subfunctor A = R^{ab}, where for Y ≤ X, we denote by R^{ab}(Y) the free abelian group with basis Ŷ = Hom(Y, ℂ^{×}) the group of linear characters of Y.

As an application, we will show that there is an r-extension morphism for the trivial source ring. Let F be a field of prime characteristic p, and let X be a finite group. Recall that an F X-module V is called a trivial source module if V is a direct summand of a permutation F X-module. The Grothendieck group of the category of trivial source F X-modules is denoted by T_{F} (X), and it forms a ring under the tensor product. A canonical induction formula for T_{F} has been given by Boltje in [2]. Here the subfunctor A is again R^{ab}, where R^{ab}(Y) is the free abelian group with basis Ŷ (F) = Hom(Y, F^{×}), the group of linear characters from Y to F^{×}. Here we identify ψ ∈ Ŷ (F) with the isomorphism class of the rank one F Y -module F_{ψ}, on which y ∈ Y acts by multiplication with ψ(y).

Now condition 5.2 just requieres that linear characters can be extended from Y ∩ G to Y in a distinguished way. In fact, we can extend them such that the restriction of the extension to a complement of Y ∩ G in Y is trivial. This obviously satisfies the conditions in 5.2.

For the morphism Ψ in 5.3 we will use the Adams operators for the trivial source ring as defined by Boltje in [2]. We again choose n to be an integer such that

n ≡ 1 mod |G| and n ≡ 0 mod [X : G].

By [2], Proposition 3.3, we have

Ψ_{Y}^{n} ([F_{ψ}]) = [F_{ψ^{n}}],

hence condition 5.3 is clearly satisfied. We obtain

5.6 Proposition There exists an r-extension morphism ext : T̃_{F} → T_{F}.

References

[1] R. Boltje: A general theory of canonical induction formulae., J.Algebra 206 (1998), no.1,293-343.

[2] R. Boltje, Linear source modules and trivial source modules, Proc. Sympos. Pure Math. 63 (1998), 7-30.

[3] I. Isaacs: “Character Theory of Finite Groups,” Academic Press, New York-London, 1976.