Let $G$ be a finite group.

This answer mentions an explicit yet mysteriously preferable isomorphism from the coinduction functor to the induction functor.

This instructive answer observes the element $\frac 1{|G|}\sum_{g\in G}g\in R[G]$ of a group algebra arises from the unique retraction of the counit of the tensor hom adjunction induced by the Hopf algebra counit $R[G]\to R$. Thus the counit of the tensor-hom adjunction has a retraction iff $|G|\in R^\times $.

This answer mentions induction and coinduction should be isomorphic whenever an analogous condition is satisfied for any arrow between finite groups.

I would like some help in uniting these three answers to get a coherent picture. In particular, I would like to see how the isomorphism of the first answer arises canonically, and why it's actually an isomorphism (conceptually).

All I can think of is:

  1. In general if we have an adjoint triple $F\dashv G\dashv H$ there's a canonical composite $FG\Rightarrow 1\Rightarrow HG$ obtained using the counit of the first adjunction and the unit of the second. If we have a retraction of the first counit then we also have a composite $GH\Rightarrow 1\Rightarrow FG$, but I don't know what good that does. Perhaps we should calculate the former composite and see what conditions are required to make it an identity? That seems strange, since even if $F\cong H$ there's no reason to expect the unit and counit to be mutually inverse.

  2. In our case this adjoint triple is induced by an arrow $f=R[G]\overset{R[g]}{\longrightarrow} R[H]$, so we may write $f_!\dashv f^\ast \dashv f_\ast$. In particular, $f_\ast$ is representable in the sense of being an internal hom, so perhaps we should seek to define a natural transformation $H\to F$ using enriched Yoneda. Particularly, in the notation of this answer we have $\mathrm{Nat}(f_\ast,f^!)\cong f^!(f^\ast S)\cong f_!R\otimes _R f^\ast S$, but I know nothing about this creature.

  • $\begingroup$ Exercise 2.2.7 in Martin Lorenz, A Tour of Representation Theory, 12 March 2018 provides a criterion for an algebra homomorphism to induce isomorphic induction and coinduction functors. Perhaps there is something more general beyond the algebra setting (left Kan and right Kan extensions being isomorphic in some way?). $\endgroup$ Jun 17, 2018 at 15:29
  • $\begingroup$ Dear @darijgrinberg, thank you for the reference. I have met that criterion on the nlab. Unfortunately, I don't see the picture clearly enough to recognize some general property of Kan extensions. I am mostly interested to see how to obtain the explicit isomorphism in the first linked answer in a canonical fashion. $\endgroup$
    – Arrow
    Jun 17, 2018 at 20:43
  • 1
    $\begingroup$ My vague understanding was that this has to do with the linearity of the category, since this implies (and in a sense is implied by) the confluence of finite limits and colimits (eg product iff coproduct, pullback iff pushout etc). $\endgroup$ Jun 20, 2018 at 16:01

1 Answer 1


Executive summary: the isomorphism is not particularly mysterious. It arises from the obvious choice of a trace from the group algebra of $G$ to the group algebra of $H$, given by the formula $$\sum_{g \in G} c_g g \mapsto \sum_{h \in H} c_h h,$$ which induces an isomorphism from $RG$ onto $\mathrm{Hom}_{RH}(RG,RH)$ (here $R$ is an arbitrary commutative ring), and hence an isomorphism of the functors from induction (tensor product with $RG$) to co-induction (tensor product with $\mathrm{Hom}_{RH}(RG,RH)$).

Now for the details. First, it may help contextualize things to ask a somewhat more general question: given a homomorphism $R \to S$ of rings, when are the induction $S \otimes_R \bullet$ and co-indunction $\mathrm{Hom}_R(S,\bullet)$ functors isomorphic, and when this is the case, how can one describe the set of all such isomorphisms?

There are two relatively obvious conditions that $S$ must satisfy if this is the case. First, since $S \otimes_R \bullet$ is right-exact, if there is such an isomorphism then $\mathrm{Hom}_R(S,\bullet)$ must be right-exact as well, implying that $S$ is projective as a left $R$-module. Second, since $S \otimes_R \bullet$ commutes with arbitrary direct sums, so must $\mathrm{Hom}_R(S,\bullet)$, implying that $S$ is finitely generated as a left $R$-module.

Suppose now that $S$ is projective and finitely generated as a left $R$-module. Then, just as for vector spaces over a field, the map $\mathrm{Hom}_R(S,R) \otimes_R M \to \mathrm{Hom}_R(S,M)$ given by $\phi \otimes m \mapsto [s \mapsto \phi(s) m]$ is a functorial isomorphism: this is true if $S=R$, hence for finitely generated free modules, hence for their summands, the finitely generated projective $R$-modules.

Summing up: an isomorphism from induction to co-induction can exist only if $S$ is finitely generated and projective as an $R$-module, and in this case it is the same thing as an isomorphism of $(S,R)$-bimodules $$S \longrightarrow \mathrm{Hom}_R(S,R).$$ In fact, there is an isomorphism $$\mathrm{Hom}_{R,R}(S,R) \to \mathrm{Hom}_{S,R}(S, \mathrm{Hom}_R(S,R))$$ given by sending a "trace" $t: S \to R$ to the map $s \mapsto [s' \mapsto t(s' s)].$ So what we are looking for is a trace from $S$ to $R$ (that is, a morphism of $R,R$-bimodules) such that the corresponding map from $S$ to $\mathrm{Hom}_R(S,R)$ is an isomorphism (necessarily of $S,R$-bimodules).

Now we specialize all this to group algebras. Thus let $H \leq G$ be a subgroup of the finite group $G$, and let $R$ be a commutative ring (we could more generally work with a map from one group $H$ to another $G$, but there is a simple obstruction: we need $RG$ to be a projective $RH$-module, which fails e.g. if the map is trivial and the order of $H$ is not invertible in $R$). Thus $RG$ is a finitely generated free, and hence projective, $RH$ module. An obvious candidate for a trace from $RG$ to $RH$ is the projection map $$\sum_{g \in G} c_g g \mapsto \sum_{h \in H} c_h h.$$ One checks that this induces an isomorphism of $RG$ onto $\mathrm{Hom}_{RH}(RG,RH)$, mapping $g \in G$ to the function $$g' \mapsto \begin{cases} g' g \quad \hbox{if $g' g \in H$, and} \\ 0 \quad \hbox{ else.} \end{cases}$$ and hence an isomorphism from induction onto co-induction. Given an $RH$-module $M$ this map may be written explicitly as follows: it is given on $g \otimes m \in RG \otimes M$ by $$g \otimes m \mapsto \left[ g' \mapsto \begin{cases} g' g m \quad \hbox{if $g' g \in H$, and} \\ 0 \quad \hbox{else.} \end{cases} \right]$$ In the case in which the order of $H$ is invertible in $R$, the inverse of this map is recorded in the first answer you linked to in your question.

This answer is long enough as it is, but I would like to mention that although the unit-counit compositions you mention in your question are not necessary for understanding the induction-coinduction isomorphism, they are useful for analyzing relative projectivity of modules over group algebras, and in particular for the theory of vertices and sources and the Green correspondence, as indicated in the next paragraph.

The composition of the unit for the co-induction and the co-unit for the induction produces an endomorphism of the identity functor on $RG$-mod, or in other words a central element of $RG$. This central element can be computed explicitly: it is simply the index $[G:H]$. As a consequence, if this index is invertible in $R$ then every $RG$-module is a summand of its induction-restriction to $H$, and in particular is relatively $H$-projective. So in case $R$ is a field of characteristic $p$, every $RG$-module is relatively $H$-projective provided $H$ contains a Sylow $p$-subgroup of $G$.


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