## tensor products for group actions, part 2

In this previous post, tensor products of $G$-sets were introduced and some basic properties were proved, this post is a continuation, so I’ll asume that you’re familiar with the contents of that post.

In this post, unless specified otherwise, $G,H,K,I$ will denote groups. Groups will sometimes be freely identified with the corresponding one-object category. Left $G$-sets will sometimes just be referred to as $G$-sets.

After some lemmas, we will begin this post by some easy consequences of the Hom-Tensor adjunction, which is the main result of the previous post.

Lemma The category $G\textrm{-}\mathbf{Set}$ is complete and cocomplete and the limits and colimits look like in the category of sets (i.e. the forgetful functor $G\textrm{-}\mathbf{Set} \to \mathbf{Set}$ is continuous and cocontinuous) with the “obvious” actions from $G$. Similarly for $\mathbf{Set}\textrm{-}G$.

Proof This is not too difficult to prove directly (you can reduce the existence of (co)limits to (co)products and (co)equalizers by general nonsense), but it also follows directly from the fact that $G\textrm{-}\mathbf{Set}$ is the functor category $[G, \mathbf{Set}]$. The reason is that if $\mathcal{C}$ and $\mathcal{D}$ are categories and $\mathcal{D}$ is (co)complete (and $\mathcal{C}$ is small to avoid any set-theoretic trouble), then the functor category $[\mathcal{C},\mathcal{D}]$ is also (co)complete and the (co)limits may be computed “pointwise”. In the case of $[G, \mathbf{Set}]$, $G$ has only one object, so the (co)limits look like they do in $\mathbf{Set}$.

Lemma If $X$ is a $(G,H)$-set, $Y$ is a $(H,K)$-set and $Z$ is a $(K,I)$-set, then we have a natural isomorphism of $(G,I)$-sets
$(X \otimes_H Y) \otimes_K Z \cong X \otimes_H (Y \otimes_K Z)$

Proof The proof is the same as the proof for modules, mutatis mutandis. Use the universal property of tensor products a lot to get a well-defined maps $(X \otimes_H Y) \otimes_K Z \to X \otimes_H (Y \otimes_K Z) , (x \otimes y) \otimes z \mapsto x \otimes (y \otimes z)$ and  $X \otimes_H (Y \otimes_K Z) \to (X \otimes_H Y) \otimes_K Z, x\otimes (y\otimes z) \mapsto (x \otimes y) \otimes z$.

Lemma If $H \leq G$ is a subgroup, and we regard $G$ as a $(G,H)$-set via left and right multiplication, then $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G,-): G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is naturally isomorphic to the restriction functor $\mathrm{res}_H^G: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ (this functor takes any $G$-set, which we may think of as a group homomorphism or a functor and restricts it to the subgroup/subcategory given by $H$.)

Proof Define a (natural) map $\varphi: \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G,X) \to \mathrm{Res}_H^G(X)$ via $\varphi(f)=f(1)$. This is $H$-equivariant, because $\varphi(hf)(1)=f(1h)=f(h)=hf(1)=h\varphi(f)$. On the other hand, given $x \in \mathrm{Res}_H^G(X)$ (which is just $X$ as a set), we can define $f \in \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G,X)$ via $f(g)=gx$. This defines an inverse for $\varphi$.

Corollary The restriction functor $\mathrm{Res}_H^G$ has a left adjoint $G \otimes_H - := \mathrm{Ind}_H^G$.

The notation $\mathrm{Ind}$ is chosen because we can think of this functor as an analog to the induced representation from linear representation theory, where we think of group actions as non-linear represenations. (Similar to the induced representation, one can give an explicit description of $\mathrm{Ind}_H^G$ after choosing coset representatives for $G/H$ etc.)
In linear represenation theory, the adjunction between restriction and induction is called Frobenius reciprocity, so if we wish to give our results fancy names (as mathematicians like to do) we can call this corollary “non-linear Frobenius reciprocity”.

If we take $H$ to be the trivial subgroup, we obtain a corollary of the corollary:

Corollary The forgetful functor $G\textrm{-}\mathbf{Set}\to \mathbf{Set}$ has a left adjoint, the “free $G$-set functor”.

Proof If $H$ is the trivial group, then $H$-sets are the same as sets and the restriction functor $G\textrm{-}\mathbf{Set}\to H\textrm{-}\mathbf{Set}$ is the same as the forgetful functor. Since $G \otimes_H$ commutes with coproducts and $H$ is a one-point set, we can also describe this more explicitly: for a set $X$, we have $G \otimes_H X \cong G \otimes_H \coprod_{x \in X} H \cong \coprod_{x \in X} G \otimes_H H \cong \coprod_{x \in X} G := G^{(X)}$

We can also use the Hom-Tensor adjunction to get a description of some tensor products. Let $1$ denote a one-point set (simultanously the trivial group), considered as a $(1,G)$-set with (necessarily) trivial actions.

Lemma For a $G$-set $X$, $1 \otimes_G X$ is naturally isomorphic to the set of orbits $X/G$ and both are left adjoint to the functor $\mathbf{Set} \to G\textrm{-}\mathbf{Set}$ which endows every set with a trivial $G$-action.

Proof Let $Y$ be a set and $X$ be a $G$-set. Denote $Y^{triv}$ the $G$-set with $Y$ as its set and a trivial action. If we have any $f \in \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,Y^{triv})$, then $f$ must be constant on the orbits, since $f(gx)=gf(x)=f(x)$, so $f$ descends to a map of sets $X/G \to Y$. Conversely, if we have any map $h: X/G \to Y$, then we can define a $G$-equivariant map $f:X \to Y^{triv}$ by setting $f(x)=h([x])$, where $[x]$ denotes the orbit of $x$. These maps are mutually inverse natural bijection which shows that “set of orbits”-functor is left adjoint to $Y \mapsto Y^{triv}$. On the other hand, we can identify $Y^{triv}$ with $\mathrm{Hom}_{\mathbf{Set}}(1,Y)$ (where the $G$ action is induced from the trivial right $G$-action on $1$), so the left adjoint must be given by $X \mapsto 1 \otimes_G X$. Since adjoints are unique (by a Yoneda argument), we have a natural bijection $1 \otimes_G X \cong X/G$

The set of orbits $X/G$ carries some information about the $G$-set, but we can do a more careful construction which also includes $X/G$ in a natural way as part of the information.

Definition If $X$ is a $G$-set, then the action groupoid $X//G$ is the category with $\mathrm{Obj}(X//G) := X$ and $\mathrm{Hom}_{X//G}(x,y):= \{g \in G \mid gx=y\}$. Composition is given by $\mathrm{Hom}_{X//G}(y,z) \times\mathrm{Hom}_{X//G}(x,y) \to \mathrm{Hom}_{X//G}(x,z), (h,g) \mapsto hg$.

The fact that this is called a groupoid is not important here, one can think of that as just a name (it means that every morphism in $X//G$ is an isomorphism).
The set of isomorphism classes of $X//G$ correspond to the orbits $X/G$. For $x \in X//G$, the endomorphisms $\mathrm{End}_{X//G}$ is the stabilizer group $G_x$. The following lemma shows how to reconstruct a $G$-set $X$ from $X//G$, assuming that we know how all the Hom-sets lie inside $G$.

Lemma (“reconstruction lemma”) If $X$ is a $G$-set, then we define the functor $G: (X//G)^{op} \to G\textrm{-}\mathbf{Set}$ with $G(x)=G$ for all $x \in X//G$ and for $g \in \mathrm{Hom}_{(X//G)}(x,y)$, we define the map $G(g): G(y) \to G(x)$ via $a \mapsto ag$. Then we have $\varinjlim\limits_{x \in (X//G)^{op}}G(x) \cong X$

Proof For $x \in X//G$, define a map $G(x)=G \to X$ via $g \mapsto gx$. This defines a cocone over $G(.)$, so we get an induced map $\varphi: \varinjlim\limits_{x \in (X//G)^{op}}G(x) \to X$.  $\varinjlim\limits_{x \in (X//G)^{op}}G(x)$ can be described explicitly as $\coprod_{x \in (X//G)^{op}} G(x)/\sim$, where the equivalence relation $\sim$ is generated by $ga \in G(x) \sim a \in G(gx)$. To see that $\varphi$ is surjective, note that $x \in X$ is the image of $1 \in G(x)$. To see that $\varphi$ is injective, suppose $g \in G(x)$ and $h \in G(y)$ are sent to the same element, i.e. $gx=hy$, then we have $(h^{-1}g)x=y$, so that we may assume $h=1$. Then $gx=y$ implies that $1 \in G(y)=G(gx) \sim g \in G(x)$, so the two elements which map to the same element are already equal in $\varinjlim\limits_{x \in (X//G)^{op}}G(x)$.

The previous lemma can be thought of as a generalization of the orbit-stabilizer theorem.  (The proof has strong similarities as well.) For illustration, let us derive the usual orbit-stabilizer theorem from it.

Lemma Let $X$ be a $G$-set, then we have an isomorphism of $G$-sets $G/G_x \cong Gx$, where $Gx$ is the orbit of $x$ (with the restricted action) and $G/G_x$ is the coset space of the stabilizer subgroup with left multiplication as the action.

Proof We may replace $X$ with $Gx$ so that we have a transitive action. Then the previous lemma gives us an isomorphism $X \cong \varinjlim\limits_{x \in (X//G)^{op}}G(x)$.
Consider the one-object category $(G_x)$. This can be identified with a full subcategory of $X//G$ corresponding to the object $x$. Because we have a transitive action, all objects in $X//G$ are isomorphic (isomorphism classes correspond to orbits), so that the inclusion functor $G_x \to X//G$ is also essentially surjective, so it is a category equivalence.
We may thus replace the colimit by the colimit $\varinjlim\limits_{x \in (G_x)^{op}}G(x)$. As $(G_x)^{op}$ has just one object, this colimit is a colimit over a bunch of parallel morphism $G(x) \to G(x)$, so it is the simultanous coequalizer of these morphisms. We know how to compute coequalizers in $G\textrm{-}\mathbf{Set}$: the same way that we compute coequalizers in $\mathbf{Set}$. So we have the families of maps $\cdot g: G(x)=G \to G, a \mapsto ag$, where $g$ varies over $G_x$. The coequalizer is the quotient $G/\sim$, where $\sim$ is generated by $a \sim ag$ for each $a \in G$ and $g \in G_x$. But this is exactly the equivalence relation that defines $G/G_x$.

There is another case where the colimit takes a simple form after replacing $X//G$ with an equivalent category.

Lemma A $G$-set $X$ is free in the sense that it is in the essential image of the “free $G$-set functor” $Y \mapsto G \otimes_{1} Y$ or equivalently it is a coproduct of copies of $G$ with the standard action iff the action of $G$ on $X$ is free in the sense that $\forall x \in X \forall g \in G: (gx=x \Rightarrow g=1)$.

Proof It’s clear that if we have a disjoint union $X= \coprod_{i \in I} G$, then no element of $G$ other than $1$ can fix an element in $X$. For the other direction, suppose that we have the condition $\forall x \in X \forall g \in G: (gx=x \Rightarrow g=1)$. This implies that the morphism sets in the action groupoid are really small: Suppose $g,h \in \mathrm{Hom}(x,y)$ such that $gx=y=hx$, which implies that $h^{-1}gx=x$, so $h^{-1}g=1$ by assumption, thus $h=g$. This means that for any pair of objects in $X//G$, there is at most one morphism between them. So if we consider the set $X/G$ as a discrete category (i.e. the only morphisms are the identities), then if we take a representative for each orbit $X/G$, this defines an inclusion of categories $X/G \to X//G$. As elements in $X/G$ represent isomorphism classes in $X//G$, this inclusion is always essentially surjective. By our computations of the Hom-sets, it is also fully faithful if the action of $G$ on $X$ is free. So if we apply the “reconstruction lemma” we get $X \cong \varinjlim\limits_{x \in (X//G)^{op}}G(x) \cong \varinjlim\limits_{x \in X/G} G$. But a colimit over a discrete category is just a coproduct, so this is isomorphic to $G^{(X/G)}$ which shows that $X$ is free.

After some further lemmas, we will come to the main result of this post, which is also an application of the reconstruction lemma.

In the previous post, I described $G$-sets in different ways, among them as functors $G \to \mathbf{Set}$, but I didn’t do the same for $(G,H)$-sets. The following lemma remedies this deficiency.

Lemma $(G,H)$-sets may be identified with left $G \times H^{op}$-sets or with functors $G \to \mathbf{Set}\textrm{-}H$ or with functors $H^{op} \to G\textrm{-}\mathbf{Set}$. In other words, we have equivalences of categories $G\textrm{-}\mathbf{Set}\textrm{-}H \cong G\times H^{op}\textrm{-}\mathbf{Set} \cong [G,\mathbf{Set}\textrm{-}H] \cong [H^{op},G\textrm{-}\mathbf{Set}]$.

The proof of this lemma is a lot of rewriting of definitions, not more difficult than proving the corresponding statements for one-sided $G$-sets.

This lemma has a useful consequence, which one could also verify by hand:

Observation If $F: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is a functor and $X$ is a $(G,K)$-set, then $F(X)$ is a $(H,K)$-set in a “natural” way.

Proof Think of $X$ as functor $X:K^{op} \to G\textrm{-}\mathbf{Set}$, composing with $F$, gives us a functor $F(X): K^{op} \to H\textrm{-}\mathbf{Set}$, which we may also think of as a $(H,K)$-set.
More explicitly, the action of $K$ on $F(X)$ can be described as follows: for $k \in K$, the right-multiplication-map $X \to X, x \mapsto xk$ is left $G$-equivariant, so it induces a left $H$-equivariant map $F(X) \to F(X)$, we can define the action of $k$ on $F(X)$ via this map.

The following lemma is an analog of the classical Eilenberg-Watts theorem from homological algebra which describes colimit-preserving functors $R\textrm{-}\mathbf{Set} \to S\textrm{-}\mathbf{Set}$ as tensor products with a $(S,R)$-bimodule.

Thereom (Eilenberg-Watts theorem for group actions) Every colimit-preserving functor $F: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is naturally equivalent to $X \otimes_G$ for a $(H,G)$-set $X$. One can explicitly choose $X = F(G)$ (with the $(H,G)$-set structure from the previous observation, as $G$ is a $(G,G)$-set.)

Proof Let $X$ be a $G$-set and $Y$ be a $H$-set, then we have a natural bijection $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G) \otimes_G X, Y) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y))$
Using the reconstruction lemma, we get $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y)) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(\varinjlim\limits_{x \in (X//G)^{op}}G(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y)) \cong \varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y))$
For every $x \in (X//G)^{op}$, $G(x)=G$, so $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y)) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y))$ via the map $f \mapsto f(1)$. We need to consider how this identification behaves under the morphisms involved in the colimit. For $g \in G$, we have the map $G(gx) \to G(x), a \mapsto ag$, this induces a map $\varphi_g: \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y)) \to \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G(gx),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y))$ given by $\varphi_g(f)(h)=f(hg)$. If we make the indentification described above by evaluating both sides at $1$, we get $\varphi_g(f)(1)=f(1g)=gf(1)$. Using the definition of the $G$-action on the Hom-set $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y)$, this left multiplication translates to right multiplication on $F(G)$. Because of the construction of the right $G$-action on $F(G)$, this right multiplication is the map that is induced from right multiplication $G \to G$. We may summarize this computation by stating that $\varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(G(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G),Y)) \cong \varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G(x)),Y)$
Using the assumption that $F$ preserves colimits, we get $\varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(G(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(\varinjlim\limits_{x \in (X//G)^{op}}F(G(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(\varinjlim\limits_{x \in (X//G)^{op}} G(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(F(X),Y)$ where we used the reconstruction lemma again in the last step.
We conclude $F(X) \cong F(G) \otimes_G X$ by the Yoneda lemma.

This theorem (like the classical Eilenberg-Watts-theorem) is remarkable not only because it gives a concrete description of every colimit-preserving functor between certain categories, but also because it shows that such functors are completely determined by the image of one object $G$ and how it acts on the endomorphisms of that object (which are precisely the right-multiplications.)

It’s natural to ask at this point when two functors of the form $X \otimes_G$ and $Y \otimes_G$ for $(H,G)$-sets are naturally isomorphic. It’s not difficult to see that it is sufficient that $X$ and $Y$ are isomorphic as $(H,G)$-sets. The following lemma shows that this is also necessary, among other things.

Lemma For $(H,G)$-sets $X$ and $Y$, ever natural transformation $\eta:X \otimes_G \to Y \otimes_G$ is induced by a unique $(H,G)$-equivariant map $f: X \to Y$

Proof Assume we have a natural transformation $\eta_A: X \otimes_G A \to Y \otimes_G A$, then we have in particular a left $H$-equivariant map $\eta_G: X \otimes_G G \to Y \otimes_G G$. We have $X \otimes_G G \cong X$ and $Y \otimes_G G \cong Y$, so this gives us a $H$-equivariant map $X \to Y$ which I call $f$. Clearly $f$ is uniquely determined by this construction. For a fixed $g \in G$, right multiplication by $g$ defines a left $G$-equivariant map $G \to G$. Under the isomorphism $X \otimes_G G \cong X$ these maps describe the right $G$ action on $X$. Naturality with respect to these maps implies that $f$ is right $G$-equivariant.

This lemma allows a reformulation of the previous theorem.

Theorem (Eilenberg-Watts theorem for group actions, alternative version)
The following bicategories are equivalent:
– The bicategory where the objects are groups, 1-morphisms between two groups $G, H$ are $(G,H)$-sets $X$, where the composition of 1-morphisms is given by taking tensor products and 2-morphisms between two $(G,H)$-sets are given by $(G,H)$-equivariant maps.
– The 2-subcategory of the 2-category of categories $\mathbf{Cat}$ where the objects are all the categories $G\textrm{-}\mathbf{Set}$ for groups $G$, 1-morphisms are colimit-preserving functors $G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ and 2-morphisms are natural transformations between such functors.

This concludes my second blog post. If you want, please share or leave comments below.

## tensor products for group actions

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 $\forall g \in G, x \in X, y \in Y: (gx,y) \sim (x,gy)$. 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.