Suppose is a ring, is a right -module and is a left -module, then we can form the tensor product 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 a left -set is a set together with a map that satisfies . An equivalent formulation is that a left -set is a set together with a group homomorphism , where denotes the group of bijections . Yet another way to phrase this definition is that a left -set is a functor from (regarded as a one-object category) to the category of sets.
There’s an obvious choice of morphisms for left -sets and , namely maps that satisfy . Traditionally called -equivariant maps. If you take the definiton via functors, then these are exactly natural transformations. It follows that the left -sets form a category, which we will denote by .
In a similar way, one defines right -sets as sets together with a map that satisfies , these can also be thought of as contravariant functors from to or equivalently as left -sets, where denotes the opposite group of . The category of right -sets will be denoted by .
If we have two groups and , then a -set is a set that is simultaneously a left -set and a right -set, such that the actions are compatible in the sense that . We get the category of -sets where the morphisms are maps that are both and -equivariant. Note that if we take or to be the trivial group, then -sets are equivalent to left -sets or right -sets respectively, thus we may always assume to deal with -sets and will cover all three cases. Also note that if we take both and to be the trivial group, then -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 -sets play the role of modules.
Since -Sets are important in the theory of tensor products for modules, we will study these first.
In the following, and will denote groups.
Lemma If is a -set and is a -set, then is a -set with the actions defined by and . If is a -set and is a -set, then is a -set in the analogous way.
The proof is a routine verification.
As a special case, when is just a set and is a -set, then is a -set, with the action on the opposite side. If is a -set, then the two actions we get this way are compatible, so that is a -set.
We can copy the following definition almost verbatim from the case for modules.
Definition If is a -set, is a -set and is a -set, then a map is called -balanced and -equivariant if and . We denote the set of all such maps by . If and are the trivial group, we drop them from the notation and just write .
Lemma If is a -set, is a -set and is any set, then is a -set with the actions defined by and .
This is again a routine verification, similar to the previous lemma.
We now come to the usual “Currying” argument.
Lemma If is a -set and is a -set and is any set, then there is a natural isomorphism of -sets
Proof We define the “Currying map” via , where . Or more compactly
Let us check that for the map is -equivariant, for is defined as . Because is -balanced, this equals , so .
Now we check that is – and -equivariant. Let , then for all , we have , on the other hand we have , this shows that .
Let , then for all , we have , on the other hand , so .
So we have shown that is indeed a well-defined -equivariant map .
To see that is bijective, note that an inverse is given by the “Uncurry”-map , where .
We omit the verification that is natural in , and .
We also give the following variants of the Currying isomorphism:
Lemma If is a -set, is a -set and is a -set, then we have a natural bijections
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 , where is well-defined.
For a fixed , we have , so is -equivariant.
Let , then .
So we get , thus is -equivariant.
For the second bijection, we use the map , where . 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 be a right -set, be a left -set, then a tensor product is a set together with a -balanced map that satisfies the following universal property: for any set and -balanced map , there is a unique map such that .
Lemma In the situation of the previous definition, the tensor product exists and if is a second tensor product, then there exists a unique bijection such that .
Proof We first prove uniqueness. Suppose and , then because of the universal property of , we get a unique map such that and due to universal property of we get a unique map such that , then satisfies , but by the universal property of , there is a unique map that satisfies . We have just computed that satisfies this, but does, too. Thus . The proof that follows analogously.
Now we show existence. Consider the equivalence relation on generated by . We set and let be the quotient map. By construction, is -balanced.
We denote the image by . We have the relation . (Note that unlike in the case of modules, is surjective. So every element is an “elementary tensor”.)
Suppose is a set and is -balanced, then the map is well-defined and it is the unique map that satisfies . (This follows just from the universal property of a quotient set.)
In the case of modules, when we have a -bimodule and a -bimodule , then is a -bimodule. We know show the analogous result for -sets.
Lemma If is a -set and is a -set, then is a -set.
Proof Let , then the map is -balanced, because is a -set, so this map descends to a well-defined map , which shows that the map is well-defined. It’s clear that this makes into a left -set. In the same manner, we get a right action of on given by . These actions are compatible, since .
One of the basic properties of the tensor product of modules is that it defines a functors between module categories, this works also for -sets.
Lemma/Definition The tensor product can be made into a bifunctor
Proof Suppose and are -sets, and is a -equivariant map, and are -sets and is a -equivariant map. Consider the map . This is -balanced, because .
So we get a well-defined map , which we will denote by . is -equivariant because ist left -equivariant and is right -equivariant and the -set structure on a tensor product is defined by acting with on the left factor and with on the right factor.
If is another -set and is another -set and are morphisms of the respective type, then if we consider the map both and are maps which satisfy , thus by the uniqueness part of the universal property, they must be equal. That 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 -set , the functor , is left-adjoint to the -functor .
Proof The universal property of the tensor product can be reformulated in the form that the map is a bijection. It’s not difficult to check that this map is natural in and and that if is a -set, then it restricts to a bijection . If we compose this bijection with the Currying bijection , we get a natural bijection
This concludes this first post on tensor products of -sets, I will investigate more properties of this construction in future posts. Feel free to leave comments below.