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I am reading a paper "2-vector spaces and groupoid" by Jeffrey Morton and I need a help to understand the following.

Let $X$ and $Y$ be finite groupoids. Let $[X, \mathbb{Vect}]$ be a functor category from $X$ to $\mathbb{Vect}$. Suppose we have a functor $f: Y \rightarrow X$. The author defines a pushforward $f_{*} :[Y, \mathbb{Vect}] \rightarrow [X, \mathbb{Vect}]$ as follows.

For each object $x \in X$, the comma category $(f \downarrow x)$ has object which are objects $y \in Y$ with maps $f(y) \rightarrow x$ in $X$, and morphisms which are morphisms $a: y \rightarrow y'$ whose images make the evident triangle in $X$ commute.

Then the author defines for each $F\in [Y, \mathbb{Vect}]$, $f_{*}(F)(x):=colim F(f \downarrow x)$.

He also shows that $f_{*}(F)(y)=\bigoplus_{f(x)\cong y} \mathbb{C}[Aut(y)]\otimes_{\mathbb{C}[Aut(x)]}F(x)$ and fimilarly calculate $f^*f_*$ and $f_*f^*$.

Later he says that this description accords with the susal description of these functors in the left adjunction. Then he mentions that the right adjoint is given as

$f_*F(x)= \bigoplus_{[y], f(y)\cong x}hom_{\mathbb{C}[Aut(x)]}(\mathbb{C}[Aut(y), F(y)).$

Then we says:

The Nalayama isomorphism gives the duality between the two descriptions of $f_*$, in terms of $hom_{\mathbb{C}[Aut(x)]}$ and $\otimes_{\mathbb{C}[Aut(x)]}$ by means of the exterior trace map.

I understood the calculation of $f_*F(x)$ using a colimit defined above. But I don't know what it means by saying left or right adjoint. Also I don't know how to get the formula for the right adjoint. I even don't know what is Nakayama isomorphism. I searched for it but I couldn't find a good resource.

So I would like to know what is going on here. Especially, I'd like to know;

  1. Why is the first construction using a colimit called "left adjoint"?

  2. What is the right adjoint and how is it defined and how to calculate it to get the formula above.

  3. What is Nakayama isomorphism?

I have never studied these things so I don't know where to look up. I also want good references.

I appreciate any help. Thank you in advance.

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@QiaochuYuan Thanks for the comment. I'm not familiar with the representation theory (of finite groups). –  Primo Oct 31 '12 at 1:31
    
There is a Nakayama lemma in ordinary commutative algebra, but I don't see how this is related. As for the double description – when you have a pullback $f^*$ defined by $f^* G (y) = F(f(y))$, then, assuming you have enough limits and colimits, $f^*$ has left and right adjoints $f_!$ (a.k.a. left Kan extension) and $f_*$ (a.k.a. right Kan extension). But usually these are very different kinds of functors. –  Zhen Lin Oct 31 '12 at 8:10

1 Answer 1

up vote 5 down vote accepted

You should learn something about the representation theory of finite groups. It suffices to understand the case that $X, Y$ have one object (note that the definition is as a direct sum of things that happen at one object); call those objects $x, y$ and let $G = \text{Aut}(x), H = \text{Aut}(y)$. Then $f$ may be regarded as a morphism $f : H \to G$. The functor categories $[X, \text{Vect}]$ and $[Y, \text{Vect}]$ are precisely the categories $\text{Rep}(G), \text{Rep}(H)$ of representations of $G$ and $H$ respectively. The pullback functor

$$f^{\ast} : \text{Rep}(G) \to \text{Rep}(H)$$

is known in representation theory as restriction and commonly denoted $\text{Res}_H^G$. Restriction happens to have two equivalent descriptions, and it will be cleaner for me to describe these by working in slightly greater generality. Namely, let $f : R \to S$ be a homomorphism of rings. $f$ induces a pullback functor

$$\text{Res}_R^S : S\text{-Mod} \to R\text{-Mod}$$

from the category of left $S$-modules to the category of left $R$-modules. (Recall that the category of left $R$-modules is the functor category $[R, \text{Ab}]$). When $R = \mathbb{C}[H], S = \mathbb{C}[G]$ are group algebras we recover the picture above.

Restriction can be described in two ways. The first description is (writing $_RM_S$ for an $(R, S)$-bimodule $M$)

$$_S M \mapsto \text{Hom}_S(_S S_R, _SM)$$

where $_S S_R$ is $S$ regarded as a left $S$-module in the obvious way and as a right $R$-module via the map $f$ above, and

$$_S M \mapsto _R S_S \otimes_S M$$

where $_R S_S$ is $S$ regarded as a right $S$-module in the obvious way and as a left $R$-module via the map $f$ above. Via the tensor-hom adjunction, this shows that restriction has both a left and a right adjoint. The left adjoint is called extension of scalars in module theory and induction in group theory and is given by

$$_R N \mapsto _S S_R \otimes_R N.$$

In group theory it is usually denoted $\text{Ind}_H^G$.

I don't know if the right adjoint has a standard name in module theory. In group theory I think it is sometimes called coinduction (so maybe in module theory it should be called coextension?) and is given by

$$_R N \mapsto \text{Hom}_R(_R S_S, _R N).$$

For an arbitrary map $f : R \to S$ of rings, there is no reason to expect these two functors to be naturally isomorphic. However, if $f : \mathbb{C}[H] \to \mathbb{C}[G]$ is a map on group algebras induced by a group homomorphism, then induction and coinduction are naturally isomorphic (which is why you won't see textbooks on the representation theory of finite groups talk about coinduction). The explicit isomorphism between them is what Morton calls the Nakayama isomorphism; it is an isomorphism

$$\text{Nak} : \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V) \to \mathbb{C}[G] \otimes_{\mathbb{C}[H]} V$$

(I am dropping the bimodule subscripts for ease of notation) given by

$$\text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V) \ni \phi \mapsto \frac{1}{|H|} \sum_{g \in G} g^{-1} \otimes \phi(g) \in \mathbb{C}[G] \otimes_{\mathbb{C}[H]} V.$$

From a representation-theoretic perspective, the existence of this isomorphism merely reflects that $\mathbb{C}[G]$, as a representation of $H$, is self-dual, and this in turn follows from the fact that it has real character. But the Nakayama isomorphism is a preferred isomorphism for reasons I do not completely understand.

share|improve this answer
    
Thank you very much. I got an idea what's going on. Could you give me a reference of the fact that coinduction is given by Hom? –  Primo Nov 1 '12 at 3:41
    
@Primo: it follows from the tensor-hom adjunction applied to the second description of restriction. –  Qiaochu Yuan Nov 1 '12 at 3:43

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