# 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 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 $\varphi$.

Via the Hom-Tensor-Adjunction this implies

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

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

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 $X \cong \varinjlim\limits_{x \in (X//G)^{op}}F(x)$.
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 $\varinjlim\limits_{x \in (G_x)^{op}}F(x)$. As $(G_x)^{op}$ has just one object, this colimit is a colimit over a bunch of parallel morphism $F(x) \to F(x)$, 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 $\cdot g: F(x)=G \to G, a \mapsto ag$, 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 $\coprod_{x \in X/G} G$ 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 $W: G\textrm{-}\mathbf{Set} \to H\textrm{-}\mathbf{Set}$ is naturally equivalent to $X \otimes_G$ for a $(H,G)$-set $X$. One can explicitly choose $X = W(G)$ (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 $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G) \otimes_G X, Y) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$
Using the reconstruction lemma, we get $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(X,\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(\varinjlim\limits_{x \in (X//G)^{op}}F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$
For every $x \in (X//G)^{op}$, $F(x)=G$, so $\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$ 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 $F(gx) \to F(x), a \mapsto ag$, this induces a map $\varphi_g: \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \to \mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(gx),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y))$ 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 $\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)$, this left multiplication translates to right multiplication on $W(G)$. Because of the construction of the right $G$-action on $W(G)$, this right multiplication is the map that is induced from right multiplication $G \to G$. We may summarize this computation by stating that $\varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{G\textrm{-}\mathbf{Set}}(F(x),\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(G),Y)) \cong \varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(F(x)),Y)$
Using the assumption that $W$ preserves colimits, we get $\varprojlim\limits_{x \in (X//G)^{op}}\mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(F(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(\varinjlim\limits_{x \in (X//G)^{op}}W(F(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(\varinjlim\limits_{x \in (X//G)^{op}} F(x)),Y) \cong \mathrm{Hom}_{H\textrm{-}\mathbf{Set}}(W(X),Y)$ where we used the reconstruction lemma again in the last step.
We conclude $W(X) \cong W(G) \otimes_G X$ 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 $Y$, every 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.