Induced module and tensor products.

Question

The following is from the book Algebra Vol. 2 by Cohn.

Let $H$ be a subgroup of $G$ an consider a right $H$-module $U$. From it we can form a right $G$-module $$\tag{1} \text{ind}_H^GU=U^G =U\bigotimes_HkG $$ called the induced $G$-module. To find the representation afforded by $U^G$ we note that the subspace $U\otimes H=\{u\otimes 1|u\in U\}$ of $U^G$ is an $H$-module in a natural way: for any $h\in H$ we have $(u\otimes 1)h=uh\otimes 1$. More generally we can define the subspace $U \otimes H a=\{u \otimes a \mid u \in U\}$ of $(1)$ as an $H^a$-module, where $H^a=a^{-1} H a$, by the rule $$ (u \otimes a) x=u a x a^{-1} \otimes a \quad \text { for } x \in H^a . $$ This is well-defined because $a x a^{-1} \in H$ precisely when $x \in H^a$. When $U \otimes H a$ is considered as right $H^a$-module in this way we shall denote it by $U^a$. Now take a coset representation of $G$ : \begin{equation}\tag{2} G=H t_1 \cup \cdots \cup H t_r \end{equation} With its help we can write $(1)$ as \begin{equation}\tag{3} U^G=U \otimes H t_1 \oplus \cdots \oplus U \otimes H t_r=U^{t_1} \oplus \cdots \oplus U^{t_r} . \end{equation} Each $U^{t_\lambda}$ is an $H^{t_\lambda}$-module and under the action of $G$ these terms are permuted among themselves. Given $x \in G$, we can for each $\lambda=1, \ldots, r$ find a unique $h \in H$ and $\mu, 1 \leqslant \mu \leqslant r$, such that $t_\lambda x=h t_\mu$. Then we have $$ \left(u \otimes t_\lambda\right) x=u h \otimes t_\mu ; $$ this shows how the action of $x$ permutes the terms in $(3)$, as well as acting on each.

My problem: In the beginning he writes that we have $(u\otimes 1)h=uh\otimes 1$, which makes it appear that $h\in H\subset G$ acts on $U\otimes H$ by acting just on $U$. But in the last equation we have $(u\otimes t_\lambda)x=u\otimes (t_\lambda x)=u\otimes ht_\mu=uh\otimes t_\mu$, i.e., $x\in G$ here acts only on the right-hand side of the tensor product while not 'touching' $u\in U$. Why the difference? How should I understand this?

Answer 1

Roughly, one can think about $U \otimes_H kG$ as $U \otimes kG$ with the additional property that you can "pull elements of $H$ across the tensor product." So, elements of $U \otimes_H kG$ are $\color{red}{\text{sums of}}$ elements $u \otimes \alpha \in U \otimes kG$ such that $u \cdot h \otimes \alpha$ = $u \otimes h \cdot \alpha$, where $h \in H$, $u \in U, \alpha \in kG$ and $\cdot$ denotes the actions.

If you are comfortable with cokernels then you can write this as:

$$ U \otimes_H kG = \text{coker}\left( U \otimes kH \otimes kG \xrightarrow{\rho_U \otimes kG \,-\, U \otimes \rho_{kG}} U \otimes kG\right)$$

but if not, ignore this.

In the first case, we have $U \otimes_H kH = U \otimes \{1\}$ (note that we have removed the subscript $H$ from the tensor), why is this?

Well, given any $u \otimes \alpha \in U \otimes_H kH$ we can write this as $\sum_{h\in H} u \otimes c_hh$ for some $c_h \in k$, but now since we are in $U \otimes_H KH$ we can pull the $c_hh$ in each of the summands across the tensor product, giving that $$\sum_{h\in H} u \otimes c_hh = \sum_{h\in H} c_hu \cdot h \otimes 1 $$ inside $U \otimes_H kH$.

Now, given $u \otimes 1 \in U \otimes_H kH$ (and we know now that this describes all elements), how does $h \in H$ act on it?

We have $(u \otimes 1)\cdot h = u \otimes 1 \cdot h = u \otimes h = u \cdot h \otimes 1$ - essentially the $h$ slides straight through onto $U's$ side of the tensor product.

Now, onto $U \otimes_H kG$, generalising the above idea of pulling elements of $H$ across the tensor product, we can see that every element of $U \otimes_H kG$ is of the form $\sum_{i=1}^r u_i \otimes t_i$ for some $u_i \in U$ (the idea here is that you strip elements of $G$ of the $H$ part, leaving only the coset representative.)

Now, how do we act on this by any $x \in G$? Let's take an element $u \otimes t_i$ as an example. Here, we must find a $h\in H$ and $t_j$ such that $t_i \cdot x = h \cdot t_j$ (I'll leave it up to you to convince yourself that this is always be done). Then we have

$$ (u \otimes t_i)\cdot x = u \otimes t_i \cdot x = u \otimes h \cdot t_j = u \cdot h \otimes t_j$$

Observe that in writing $t_i \cdot x$ as $h \cdot t_j$ we now have the ability to pull the $h$ across the tensor product and act on $u$ with it instead.

In conclusion: You always start by acting on the right, but you can only pull elements of $H$ across the tensor.

Answer 2

The main point here is not really about group algebras: if you are happy with rings which might not be commutative, so the notions of left and right modules over them are different, then the point is that, given a ring $R$, there is a notion of tensor product for modules over $R$, but you need to take a right $R$-module $M$ and a left $R$-module $N$. Their tensor product is then a pair $(M\otimes_R N, t)$ where $M\otimes_R N$ is an abelian group and $t\colon M\times N \to M\otimes_R N$ is an "$R$-bilinear map", that is, it has the following properties:

1. the map $t\colon M\times N \to M\otimes_R N$ is "$R$-bilinear" in that $$ \begin{split} t(m_1+m_2,n) &= t(m_1,n)+t(m_2,n)\\ t(m,n_1+n_2) &= t(m,n_1)+t(m,n_2)\\ t(m.r,n) &= t(m,r.n). \end{split} $$ and moreover
2. any $R$-bilinear map $\theta\colon M\times N\to A$ taking values in an abelian group $A$ factors through $t$ in the sense that it induces a unique homomorphism of abelian groups $\tilde{\theta} \colon M\otimes_R N \to A$ such that $\theta = t\circ \tilde{\theta}$. In other words, if let $R\text{-Bil}(M,N;A)$ denote the $R$-biinear maps from $M\times N$ taking values in $A$, then $t$ induces an isomorphism $$ \tag{$\dagger$} R\text{-Bil}(M,N;A) \cong \text{Hom}_{\mathbb Z}(M\otimes_R N,A). $$

One way to construct $M\otimes_R N$ is as follows: take the free abelian group on the set $M\times N$ and quotient by the relation subgroup generated by $$ \begin{split} {\large\{} (m_1+m_2, n)&-(m_1, n) - (m_2,n), (m,n_1+n_2)- (m,n_1) - (m,n_2), \\ & \hspace{5mm} (m.r,n)-(m,r.n)\mid m,m_1,m_2 \in M, n,n_1,n_2 \in N{\large\}}. \end{split} $$

If we write $m\otimes n$ for the image of $(m,n)$ in the quotient $M\otimes_R N$, then the map $t$ is given by $t(m,n) = m\otimes n$, and the fact that this construction yields an abelian group with the required property follows from the isomorphism theorem.

The bijection $\dagger$ can be re-expressed as an adjunction using two easy facts:

(i) There is an obvious bijection: $$ \text{Map}(X\times Y,Z) \cong \text{Map}(X,\text{Maps}(Y,Z)) $$ (where here $\text{Map}(X,Y)$ denotes all functions from $X$ to $Y$). Indeed if $\theta\colon X\times Y \to Z$, then $x\mapsto \theta(x,-)$ is a function from $X$ to $\text{Map}(Y,Z)$, and similarly if $\eta\colon X\to \text{Map}(Y,Z)$, then $(x,y)\mapsto \eta(x)(y)$ is a map from $X\times Y$ to $Z$.

(ii) If $N$ is a left $R$-module, then the space of homomorphisms of abelian groups $\text{Hom}_{\mathbb Z}(N,A)$ is a right $R$-module, where if $\phi\colon N \to A$, then $(\phi.r)(n) = \phi(r.n)$ for all $r \in R, n \in N$. Similarly if $M$ is a right $R$-module, $\text{Hom}_{\mathbb Z}(M,A)$ is a left $R$-module where $(r.\psi)(m) = \psi(m.r)$ for all $m \in M, r\in R$.

Combining (i) and (ii) yields isomorphisms of abelian groups $$ \begin{split} \text{Hom}_{\mathbb Z}(M\otimes N,A) \cong R\text{-Bil}(M,N;A) \cong \text{Hom}_{R}(M,\text{Hom}_{\mathbb Z}(N,A))\\ \text{Hom}_{\mathbb Z}(M\otimes N,A) \cong R\text{-Bil}(M,N;A) \cong \text{Hom}_{R}(N,\text{Hom}_{\mathbb Z}(M,A)) \end{split} $$ which is usually referred to as the "tensor-Hom" adjunction.

In general $M\otimes_R N$ does not have more structure than that of an abelian group unless $M$ or $N$ has some other additional structure, but where they do, the tensor product will inherit any such structure which is compatible with the $R$-module structure. In particular, if either $M$ or $N$ is a bimodule, say $M$ is an $(S,R)$-module (i.e. a left $S$-module and a right $R$-module, then $M\otimes_R N$ will inherit a left $S$-module structure, as one can check, for example, using the explicit construction of $M\otimes_R N$.

One natural situation where this arises is when $R$ is a subring of a larger ring $S$. Then we may view $S$ as an $(S,R)$-bimodule, and (taking $M=S$ if you like) form from an $R$-module $N$ the $S$-module $S\otimes_R N$.This process is usually referred to as "extension of scalars". For example, if we take $R= \mathbb Z$ and $S=\mathbb Q$, then if $A$ is an abelian group, $\mathbb Q\otimes_{\mathbb Z} A$ is a $\mathbb Q$-vector space. Since the elements of $\mathbb Z$ are invertible in $\mathbb Q$, the extension of scalars $\mathbb Q\otimes_{\mathbb Z} A$ annihilates any torsion in $A$, and thus if $A$ is finitely generated, $\dim(\mathbb Q\otimes_{\mathbb Z} A)$ is what is the rank of $A$.

There is a natural adjunction for extension of scalars: if $W$ is an $R$-module and $V$ is an $S$-module, then by the second of our tensor-Hom adjunctions, in its enhanced version since $S$ is a bimodule, gives $$ \begin{split} \text{Hom}_S(S\otimes_R W, V) \cong \text{Hom}_R(W,\text{Hom}_S(S,V)). \end{split} $$ But now $\text{Hom}_S(S,V) \cong V$ since if $\phi\colon S\to V$ is $S$-linear, then $\phi(s)=\phi(s.1) = s\phi(1)$, so that $\phi\mapsto \phi(1)$ gives an isomorphism from $\text{Hom}_S(S,V)$ to $V$. Thus the right-hand side of the above isomorphism can be identified with $\text{Hom}(W,V)$. But we are not done, because we still need to pay attention to the $R$-module structures. But since we viewed $S$ as a right $R$-module, the map $\phi\mapsto \phi(1)$ satisfies $$ (\phi.r)(1)= \phi(1.r) = \phi(r.1)=r.\phi(1), $$ hence the $R$-module structure on $\text{Hom}_S(S,V)$ is the restriction of the $S$-module structure to $R$. Thus if we write $\text{Res}^S_R(V)$ for the $R$-module $V$ obtained by restricting the action of $S$ to $R$, we obtain $$ \text{Hom}_S(S\otimes_R W,V) \cong \text{Hom}_R(W,\text{Res}^S_R(V)), $$

The functor of (co)induction for representations of a finite group $G$ with subgroup $H$ is then the case where we take $R=\mathsf kH$ and $S= \mathsf kG$. If $U$ is a $\mathsf kH$-module, then $$ \text{co-Ind}_{H}^{G}(U) := \mathsf kG\otimes_{\mathsf kH} U. $$

The adjunction we got for extension of scalars in general is, in this case, Frobenius reciprocity:

$$ \begin{split} \text{Hom}_{\mathsf kG}(\text{co-Ind}_{H}^{G}(W),V) = \text{Hom}_{\mathsf kH}(W,\text{Res}_{H}^{G}(V)) \end{split} $$

Moreover, describing the structure of $\mathsf kG\otimes_{\mathsf kH} W$ in the way the OP is essentially using Lagrange's theorem to see that $\mathsf kG$ is a free $\mathsf kH$ module of rank $|G|/|H|$, and a section of the quotient map $q\colon G \to G/H$ gives a choice of a $\mathsf kH$-basis which you can use to make the action of $G$ explicit.

P.S. Sorry, looking back at the question, it seems Cohn uses right $\mathsf kH$- and $\mathsf kG$-modules, but I can't bring myself to do that.