## Tensor Products for Group Actions, Part 1

Suppose $R$ is a ring, $M$ is a right $R$-module and $N$ is a left $R$-module, then we can form the tensor product $M \otimes_R N$ that we all know and love. If we consider that a module is just an abelian group together with an action from a ring, one might ask the question: Can we imitate this construction for group actions?

It turns out that we can; this is what I’ll investigate in this post.

Recall that for a fixed group $G$ a left $G$-set is a set $X$ together with a map $G \times X \to X, (g,x) \mapsto gx$ that satisfies $\forall g,h \in G, x \in X: 1x=x, g(hx)=(gh)x$. An equivalent formulation is that a left $G$-set is a set $X$ together with a group homomorphism $G \to S_X$, where $S_X$ denotes the group of bijections $X \to X$. Yet another way to phrase this definition is that a left $G$-set is a functor from $G$ (regarded as a one-object category) to the category of sets.

There’s an obvious choice of morphisms for left $G$-sets $X$ and $Y$,  namely maps $f: X \to Y$ that satisfy $\forall x \in X, g \in G: f(gx)=gf(x)$. Traditionally called $G$-equivariant maps. If you take the definiton via functors, then these are exactly natural transformations. It follows that the left $G$-sets form a category, which we will denote by $G\textrm{-}\mathbf{Set}$.

In a similar way, one defines right $G$-sets as sets $X$ together with a map $X \times G \to X, (x,g) \mapsto xg$ that satisfies $\forall g,h \in G, x \in X: x1=x, (xg)h=x(gh)$, these can also be thought of as contravariant functors from $G$ to $\mathbf{Set}$ or equivalently as left $G^{op}$-sets, where $G^{op}$ denotes the opposite group of $G$. The category of right $G$-sets will be denoted by $\mathbf{Set}\textrm{-}G$.

If we have two groups $G$ and $H$, then a $(G,H)$-set is a set $X$ that is simultaneously a left $G$-set and a right $H$-set, such that the actions are compatible in the sense that $\forall x \in X, g \in G, h \in H: (gx)h=g(xh)$. We get the category of $(G,H)$-sets $G\textrm{-}\mathbf{Set}\textrm{-}H$ where the morphisms are maps that are both $G$ and $H$-equivariant. Note that if we take $H$ or $G$ to be the trivial group, then $(G,H)$-sets are equivalent to left $G$-sets or right $H$-sets respectively, thus we may always assume to deal with $(G,H)$-sets and will cover all three cases. Also note that if we take both $G$ and $H$ to be the trivial group, then $(G,H)$-sets are just sets.

We are now ready to begin our renarration of the story of tensor products, where groups play the role of rings, sets play the role of abelian groups and left and right $G$-sets play the role of modules.

Since $\textrm{Hom}$-Sets are important in the theory of tensor products for modules, we will study these first.

In the following, $G, H$ and $K$ will denote groups.

Lemma If $X$ is a $(G,H)$-set and $Y$ is a $(G,K)$-set, then $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,Y)$ is a $(H,K)$-set with the actions defined by $\forall h \in H \forall x \in X: (hf)(x) := f(xh)$ and $\forall k \in K, \forall x \in X (fk)(x):=f(x)k$. If $X$ is a $(H,G)$-set and $Y$ is a $(K,G)$-set, then $\textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X,Y)$ is a $(K,H)$-set in the analogous way.

The proof is a routine verification.

As a special case, when $Y$ is just a set and $X$ is a $G$-set, then $\textrm{Hom}_{\mathbf{Set}}(X,Y)$ is a $G$-set, with the action on the opposite side. If $X$ is a $(G,H)$-set, then the two actions we get this way are compatible, so that $\textrm{Hom}_{\mathbf{Set}}(X,Y)$ is a $(H,G)$-set.

We can copy the following definition almost verbatim from the case for modules.

Definition If $X$ is a $(H,G)$-set, $Y$ is a $(G,K)$-set and $Z$ is a $(H,K)$-set, then a map $f: X \times Y \to Z$ is called $G$-balanced and $(H,K)$-equivariant if $\forall x \in X, y \in Y, g \in G: f(xg,y)=f(x,gy)$ and $\forall h \in H, k \in K: f(hx,yk)=hf(x,y)k$. We denote the set of all such maps by $\textrm{Bal}_G^{(H,K)}(X,Y;Z)$. If $H$ and $K$ are the trivial group, we drop them from the notation and just write $\textrm{Bal}_G(X,Y;Z)$.

Lemma If $X$ is a $(H,G)$-set, $Y$ is a $(G,K)$-set and $Z$ is any set, then $\textrm{Bal}_G(X,Y;Z)$ is a $(K,H)$-set with the actions defined by $\forall k \in K, x \in X, y \in Y: (kf) (x,y) := f(x,yk)$ and $\forall h \in H, x \in X, y \in Y: (fh) (x,y) :=f(hx,y)$.

This is again a routine verification, similar to the previous lemma.

We now come to the usual “Currying” argument.

Lemma If $X$ is a $(H,G)$-set and $Y$ is a $(G,K)$-set and $Z$ is any set, then there is a natural isomorphism of $(K,H)$-sets $\textrm{Bal}_G(X,Y;Z) \cong \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z))$

Proof We define the “Currying map” $\varphi: \textrm{Bal}_G(X,Y;Z) \to \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z))$ via $f \mapsto \varphi(f)$, where $\varphi(f)(x) = (y \mapsto f(x,y))$. Or more compactly $f \mapsto (x \mapsto (y \mapsto f(x,y)))$
Let us check that for $f \in \textrm{Bal}_G(X,Y;Z)$ the map $x \mapsto \varphi(f)(x)$ is $G$-equivariant, for $g \in G, y \in Y$ $\varphi(f)(x)g$ is defined as $(\varphi(f)(x)g)(y) = \varphi(f)(x)(gy)=f(x,gy)$. Because $f$ is $G$-balanced, this equals $f(xg,y)= \varphi(f)(xg)(y)$,  so $\varphi(f)(xg)=\varphi(f)(x)g$.
Now we check that $\varphi$ is $K$– and $H$-equivariant. Let $k \in K$, then for all $f \in \textrm{Bal}_G(X,Y;Z), x \in X, y \in Y$, we have $\varphi(kf)(x)(y)=kf(x,y)=f(x,yk)$, on the other hand we have $(k \varphi(f)(x))(y)=\varphi(f)(x)(yk)=f(x,yk)$, this shows that $\varphi(kf)=k\varphi(f)$.
Let $h \in H$, then for all $f \in \textrm{Bal}_G(X,Y;Z), x \in X, y \in Y$, we have $\varphi(fh)(x)(y)=f(hx,y)$, on the other hand $(\varphi(f)(x)h)(y)=f(hx,y)$, so $\varphi(fh)=\varphi(f)h$.
So we have shown that $\varphi$ is indeed a well-defined $(K,H)$-equivariant map $\textrm{Bal}_G(X,Y;Z) \to \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z))$.
To see that $\varphi$ is bijective, note that an inverse is given by the “Uncurry”-map $\psi: \textrm{Hom}_{\mathbf{Set}\textrm{-}G}(X, \textrm{Hom}_{\mathbf{Set}}(Y,Z)) \to \textrm{Bal}_G(X,Y;Z), \xi \mapsto \psi(\xi)$, where $\psi(\xi)(x,y)=\xi(x)(y)$.
We omit the verification that $\varphi$ is natural in $X$, $Y$ and $Z$.

We also give the following variants of the Currying isomorphism:

Lemma If $X$ is a $(H,G)$-set, $Y$ is a $(G,K)$-set and $Z$ is a $(H,K)$-set, then we have a natural bijections
$\mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}G}(X,(\mathrm{Hom}_{\mathbf{Set}\textrm{-}K}(Y,Z))$
and $\mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z))$

Proof The proof is very similar to the last one, so we omit some steps.
For the first bijection, let us just show that the map $\varphi: \mathrm{Bal}_G^{(H,K)}(X,Y;Z) \to \mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}G}(X,(\mathrm{Hom}_{\mathbf{Set}\textrm{-}K}(Y,Z)), f \mapsto \varphi(f)$, where $\varphi(f)(x)(y)=f(x,y)$ is well-defined.
For a fixed $x \in X$, we have $\varphi(f)(x)(yk)=f(x,yk)=f(x,y)k=\varphi(f)(x)(y)k$, so $\varphi(f)(x)$ is $K$-equivariant.
Let $h \in H, g \in G$, then $\varphi(f)(hxg)(y)=f(hxg,y)=hf(xg,y)=hf(x,gy)=h\varphi(f)(x)(gy)= (h\varphi(f)g)(y)$.
So we get $\varphi(f)(hxg)=h\varphi(f)(x)g$, thus $\varphi(f)$ is $(H,G)$-equivariant.
For the second bijection, we use the map $\psi: \mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z)), f \mapsto \psi(f)$, where $\psi(f)(y)(x) = f(x,y)$. The verification that this is a well-defined bijection is very similar to the previous computations.

We now come to the main definition of this post.

Definition Let $X$ be a right $G$-set, $Y$ be a left $G$-set, then a tensor product $X \otimes_G Y$ is a set together with a $G$-balanced map $\psi_{\otimes}: X \times Y \to X \otimes_G Y$ that satisfies the following universal property: for any set $Z$ and $G$-balanced map $f: X \times Y \to Z$, there is a unique map $\overline{f}: X \otimes_G Y \to Z$ such that $\overline{f} \circ \psi_{\otimes} = f$.

Lemma In the situation of the previous definition, the tensor product $(\psi_{\otimes}, X \otimes_G Y)$ exists and if $(\psi'_{\otimes}, (X \otimes_G Y)')$ is a second tensor product, then there exists a unique bijection $g: X \otimes_G Y \to (X \otimes_G Y)'$ such that $\psi'_{\otimes}=\psi_{\otimes} \circ f$.

Proof We first prove uniqueness. Suppose $(\psi_{\otimes}, X \otimes_G Y)$ and $(\psi'_{\otimes}, (X \otimes_G Y)')$, then because of the universal property of $X \otimes_G Y$, we get a unique map $f: X \otimes_G Y \to (X \otimes_G Y)'$ such that $\psi'_{\otimes}=f \circ \psi_{\otimes}$ and due to universal property of $(X \otimes_G Y)'$ we get a unique map $g:(X \otimes_G Y)' \to X \otimes_G Y$ such that $\psi_{\otimes}=g \circ \psi'_{\otimes}$, then $g \circ f$ satisfies $g \circ f \circ \psi_{\otimes} = g \circ \psi'_{\otimes} = \psi_{\otimes}$, but by the universal property of $X \otimes_G Y$, there is a unique map $h: X \otimes_G Y \to X \otimes_G Y$ that satisfies $h \circ \psi_{\otimes} = \psi_{\otimes}$. We have just computed that $g$ satisfies this, but $\textrm{id}_{X \otimes_G Y}$ does, too. Thus $g \circ f = \textrm{id}_{X \otimes_G Y}$. The proof that $f \circ g = \textrm{id}_{(X \otimes_G Y)'}$ follows analogously.

Now we show existence. Consider the equivalence relation $\sim$ on $X \times Y$ generated by $\forall g \in G, x \in X, y \in Y: (xg,y) \sim (x,gy)$. We set $X \otimes_G Y := (X \times Y)/\sim$ and let $\psi_{\otimes}: X \times Y \to X \otimes_G Y$ be the quotient map. By construction, $\psi_{\otimes}$ is $G$-balanced.
We denote the image $\psi_{\otimes}(x,y)$ by $x \otimes y$. We have the relation $gx \otimes y = x \otimes gy$. (Note that unlike in the case of modules, $\psi_{\otimes}$ is surjective. So every element is an “elementary tensor”.)
Suppose $Z$ is a set and $f:X \times Y \to Z$ is $G$-balanced, then the map $\overline{f}: X \otimes_G Y \to Z, x \otimes y \mapsto f(x,y)$ is well-defined and it is the unique map $X \otimes_G Y \to Z$ that satisfies $f = \overline{f} \circ \psi_{\otimes}$. (This follows just from the universal property of a quotient set.)
In the case of modules, when we have a $(R,S)$-bimodule $M$ and a $(S,T)$-bimodule $N$, then $M \otimes_S N$ is a $(R,T)$-bimodule. We know show the analogous result for $G$-sets.

Lemma If $X$ is a $(H,G)$-set and $Y$ is a  $(G,K)$-set, then $X \otimes_G Y$ is a $(H,K)$-set.

Proof Let $h \in H$, then the map $X \times Y \to X \otimes_G Y, (x,y) \mapsto hx \otimes y$ is $G$-balanced, because $X$ is a $(H,G)$-set, so this map descends to a well-defined map $X \otimes_G Y \to X \otimes_G Y, (x,y) \mapsto (hx,y)$, which shows that the map $H \times (X \otimes_G Y) \to X \otimes_G Y, (h, x \otimes y) \mapsto hx \otimes y$ is well-defined. It’s clear that this makes $X \otimes_G Y$ into a left $H$-set. In the same manner, we get a right action of $K$ on $X \otimes_G Y$ given by $(x \otimes y)k =x \otimes yk$. These actions are compatible, since $(h(x \otimes y))k = hx \otimes yk = h((x \otimes y)k)$.

One of the basic properties of the tensor product of modules is that it defines a functors between module categories, this works also for $G$-sets.

Lemma/Definition The tensor product $\otimes_G$ can be made into a bifunctor $H\textrm{-}\mathbf{Set}\textrm{-}G \times G\textrm{-}\mathbf{Set}\textrm{-}K \to H\textrm{-}\mathbf{Set}\textrm{-}K$

Proof Suppose $X$ and $X'$ are $(H,G)$-sets, and $\alpha: X \to X'$ is a $(H,G)$-equivariant map, $Y$ and $Y'$ are $(G,K)$-sets and $\beta:Y \to Y'$ is a $(G,K)$-equivariant map. Consider the map $f: X \times Y \to X' \otimes_G Y', (x,y) \mapsto \alpha(x) \otimes \beta(y)$. This is $G$-balanced, because $f(xg,y)=\alpha(xg) \otimes \beta(y) = \alpha(x)g \otimes \beta(y) = \alpha(x) \otimes g \beta(y) = \alpha(x) \otimes \beta(gy) = f(x,gy)$.
So we get a well-defined map $X \otimes_G Y \to X' \otimes_G Y', x \otimes y \mapsto \alpha(x) \otimes \beta(y)$, which we will denote by $\alpha \otimes \beta$. $\alpha \otimes \beta$ is $(H,K)$-equivariant because $\alpha$ ist left $H$-equivariant and $\beta$ is right $K$-equivariant and the $(H,K)$-set structure on a tensor product is defined by acting with $H$ on the left factor and with $K$ on the right factor.
If $X''$ is another $(H,G)$-set and $Y''$ is another $(G,K)$-set and $\alpha': X' \to X'', \beta': Y' \to Y''$ are morphisms of the respective type, then if we consider the map $f: X \times Y \to X'' \otimes Y'', (x,y) \mapsto \alpha'(\alpha(x)) \otimes \beta'(\beta(y))$ both $(\alpha' \otimes \beta') \circ (\alpha \otimes \beta)$ and $(\alpha' \circ \alpha) \otimes (\beta' \circ \beta)$ are maps $h:X \otimes_G Y \to X'' \otimes_G Y''$ which satisfy $h \circ \psi_{\otimes} = f$, thus by the uniqueness part of the universal property, they must be equal. That $\mathrm{id} \otimes \mathrm{id} = \mathrm{id}$ is clear from the definition.

Much of the utility of tensor products of modules lies in the Hom-Tensor-adjunction, which also has an analog for group actions. There’s not much left to do to prove it.

Theorem For a $(H,G)$-set $X$, the functor $G\textrm{-}\mathbf{Set}\textrm{-}K \to H\textrm{-}\mathbf{Set}\textrm{-}K$, $Y \mapsto X \otimes_G Y, (f: Y \to Y') \mapsto (\mathrm{id}_X \otimes f: (X \otimes_G Y \to X \otimes_G Y'))$ is left-adjoint to the $\mathrm{Hom}$-functor $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,-)$.

Proof The universal property of the tensor product $(\psi_\otimes, X \otimes_G Y)$ can be reformulated in the form that the map $\mathrm{Hom}_{\mathbf{Set}}(X \otimes_G Y,Z) \to \mathrm{Bal}_G(X,Y;Z), f \mapsto f \circ \psi_\otimes$ is a bijection. It’s not difficult to check that this map is natural in $Y$ and $Z$ and that if $Z$ is a $(H,K)$-set, then it restricts to a bijection  $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}K}(X \otimes_G Y,Z) \cong \mathrm{Bal}_G^{(H,K)}(X,Y;Z)$. If we compose this bijection with the Currying bijection $\mathrm{Bal}_G^{(H,K)}(X,Y;Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,(\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z))$, we get a natural bijection
$\mathrm{Hom}_{H\textrm{-}\mathbf{Set}\textrm{-}K}(X \otimes_G Y,Z) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}\textrm{-}K}(Y,(\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(X,Z))$

This concludes this first post on tensor products of $G$-sets, I will investigate more properties of this construction in future posts. Feel free to leave comments below.