This is part of an ongoing series on quantum groups and knot invariants, the first few entries of which were posted on Tumblr last year. This particular post doesn’t require any familiarity with those earlier installments.
The goal of this post is to introduce (and provide some motivation for why we should care about) a class of algebraic structures called Hopf algebras. We assume familiarity with the basic concepts of groups, rings, vector spaces and fields, but not much beyond that. All rings are assumed to be unital (and all ring homomorphisms preserve the identity), but we do not require the rings to be commutative.
Algebras over a field
We begin by defining (associative) algebras.
Morally speaking, an algebra over a field (algebras can be defined more generally over any commutative ring but we will not need this more general definition for this series) is a vector space that also has the structure of a ring, in such a way that these two structures are “as compatible as possible”. A vector space has notions of “addition” and “scalar multiplication”, while a ring has its own notions of “addition” and “multiplication”. For an algebra we want those two notions of addition to be the same, and the two notions of multiplication to be as similar as they can be given that, for a general ring, multiplication is not necessarily commutative.
Even more compactly, we can remember the slogan: an associative algebra is a monoid in the category of vector spaces.
Here we began with a vector space and then specified some additional linear maps between vector spaces that (we claim) gave our vector space a suitable ring-structure. There is also an equivalent definition going the other way: we begin with a ring and impose a condition in terms of morphisms between rings that give this ring the structure of a vector space.
Before giving this alternative definition and proving it is equivalent, we had better check that the algebra have defined really does have the structure of a ring.
ProofWe define multiplication of elements by . We define the identity element in by where is the identity element of our base field .
To show that is a ring, we need to show that is an abelian group under addition (something we get for free from the fact that is a vector space), that it is a monoid under multiplication (which we get from the three equations we gave in Definition 1) and that multiplication distributes over addition (which we get from the fact we specified that was a linear map). There is nothing else to prove.
Now we have:
ProofWe start with the “only if” direction. Suppose that is an algebra over .
By Lemma 2 we know that is a ring. We claim that the required morphism from to the center of is precisely the linear map . By definition we have so this map preserves multiplicative identities. Moreoever, since is a -linear map whose domain is itself it is in fact a ring homomorphism: since for any we must have .
Since for any we have we must also have It follows that .
For the “if” direction, suppose is a ring and that is a ring homomorphism with . To show that is a vector space over we have to show that it is an abelian group under addition (which it must be because it is a ring) and define a suitable notion of scalar multiplication.
We set . This operation is commutative, as required, precisely because is ine center of . It also has the required distributivty properties because is a ring morphism. This completes the proof.
As well as the definition of associative algebras as particularly nice rings, or as a “monoids in the category of vector spaces”, a third perspective is possible. This is a somewhat more concrete view which can be useful for doing calculations. We think of algebras as not merely abstract vector spaces but as vector spaces with some fixed basis, and define multiplication directly in terms of the result of multiplying two basis elements together. We are free to choose the basis however we want, so we can fix one of the basis elements to be the multiplicative identity element .
In this notation, the associativity axiom and the fact that should act as a unit with respect to multiplication can both be captured as a relationship between structure constants.
ProofSince the algebra is associative we must have Expanding both sides of this equation in terms of the structure constants we eventually obtain from which the claimed identity follows at once.
For the unit, we have that . Expanding both sides ofthis equation in terms of the structure constants gives and again the claimed identity follows.
Before giving a few examples of algebras (see below), we note one way of constructing new algebras from existing algebras. First we fix some notation: for any two vector spaces and let be the unique linear map which maps each basis element to . Now:
ProofThe maps and are clearly linear (they are the composition or tensor product of linear maps). That they satisfy the required axioms can be checked by a direct calculation, which we will omit here.
We end this section by defining an algebra homomorphism: a linear map between two algebras that preserves the algebra stucture. In the diagram notion we have been using, this is:
Some examples
You are probably already familiar with many examples of associative algebras. You might not be familiar with all of these, but we will return to many of them later.
The prototypical example is the algebra of polynomials in one variable over , denoted . The unit is the constant polynomial and multliplication is given in the natural way. More generally the ring of polynomials in multiple (commuting) variables is also an algebra, as is the ring of polynomials in multiple non-commuting variables . These algebras all are infinite dimensional (as vector spaces).
The complex numbers are a two dimensional algebra over the real numbers . More generally any field extension is an algebra over its base field. And trivially any field is an algebra over itself.
If is a (finite) group, then let be the vector space whose basis is the elements of . This vector space becomes an algebra if we define the product of two basis elements to be their product in , and extend this -linearly to all elements.
If G is a group, then let be the vector space whose basis is the set of all linear maps from .
The Hecke algebra and Temperley-Lieb algebras defined in an earlier post in this series are associative algebras in this sense. Recall in particular that the Temperley-Lieb algebra is the algebra with generators subject to the relations where .
For any positive integer , the ring of -valued matrices is an algebra. This example shows in particular that while the image of under must be contained in the center of an algebra, it need not be equal to it.
Many other matrix algebras are of interest. We will be particularly interested in , the universal enveloping algebra of the Lie algebra . In particular, is the (infinite dimensional!) algebra with generators and subject to the relations
If is a quiver (i.e. a directed multigraph) then the path algebra is defined to be the vector space with basis all (directed) paths in (including the paths of zero length: the vertices). Multiplication of two paths is given by concatenation (if the start and ends of the two paths agree) or is defined to be zero (if they do not).
If is any vector space over then let where in particular , and let We give this new (infinite-dimensional) vector space the structure of an associative algebra – the tensor algebra – by using the natural isomoprhism and extending it linearly to all of . That is, given and where we define their product to be
Representations of groups and algebras
Let be a (not necessarily finite) group. By a representation of we mean a map . the set of invertible linear transformations acting on a (finte dimensional) vector space . Because linear transformations on a finite dimensional vecotr space are equivalent to matrices, the map extends to a map from .
It is slightly more fashionable (useful?) to talk about modules rather than representations. A module is to an algebra (or any ring) as a vector space is to a field. More formally
There is an equivalent notion of right module, defined in the obvious way. Note that any algebra has two ‘trivial’ modules: the zero module (on which every element of acts like zero) and the algebra itself (here is called the regular representation of ).
To help motivate the importance of modules, we recall some ring theory.
The singleton set and the full ring are both ideals of . We can order the set of ideals of by inclusion to give them the structure of a lattice, where is the common least element and is the common greatest element. If is an ideal of we say that is maximal if, for any other ideal , implies that either or .
An ideal in a ring is roughly analogous to the idea of a normal subgroup in a group. One difference is that, in general, an ideal is not itself a ring. However, given a ring and a two-sided ideal we can construct a new ring, called the quotient ring, in much the same way we use a normal subgroup to construct a quotient group.
Note that there is an implicit claim here that addition and multiplication are well-defined (i.e. that they are independent of the choice of representative of each equivalence class). Once this claim is checked it is straightforward to see that must be a ring.
If you know a little group theory, the following result should not surprise you:
It is clear from the definition that (left) ideals of an algebra are (left) -modules. Further, the quotient ring is also an -module: . This then is one motivation to study modules of an algebra.
A morphism between modules is a linear map that respects the action of the algebra on these modules. In terms of diagrams:
A invertible module morphism whose inverse is also a module morphism is called an isomorphism. Two modules are said to be isomorphic if there exists an isomorphism between them. The category of -modules is the category whose objects are the modules of and whose arrows are the module morphisms between .
Isomorphic modules are “essentially the same”. The associated representations are not necessarily identical but are equivalent as matrices: we have for every .
If and are ideals with then, as -modules, there is obviously a map . We say in this case that is a submodule of . Indeed, recall that the notion of category generalizes the notion of partially ordered set. The lattice of ideals of is therefore contained within the category of -modules: morphisms between modules generalize the notion of inclusions between ideals.
Clearly, just as the zero ideal is contained within every ideal, every submodule contains the trivial zero module as a submodule. The modules with no submodules except for themselves and this trivial module are called simple. Simple modules turn out to be very important to the theory of modules (as usual, a reader familar with enough group theory to have heard of the Jordan-H/“{o}lder theorem should not be surprised by this).
It is not true that every simple module can be identified with an ideal. However, something close to this is true:
ProofWe only sketch a proof here. The key is a version of the ismorphism theorem for modules which we state but will not prove: if is a module morphism then is a submodule of and is a submodule of .
In particular, is itself an -module (left or right) and any (left) ideal is a (left) module. If is an ideal of such that is simple, then must be maximal: otherwise there exists an ideal with and is a submodule of .
Now suppose is a (non-zero) simple (left) module and choose some . Define a module morphism by . This map is clearly not identically zero, so – as must be a submodule of – it can only be equal to itself. is then isomorphic to and therefore must be a maximal ideal.
So ideals are not merely an arbitrary special case of modules: the connection between modules and ideals goes in both directions.
Having motivated the idea that modules are important for understanding algebras, we turn briefly to the question of constructing modules.
Given any algebra , we get an -module for free: itself is a module. This is obvious if we compare the diagrams above to the diagrams that appeared after Definition 1. In fact, we get infinitely many modules: the direct sum of any two modules is again a module, so , , … and so on are all -modules. Modules of this form are called free modules.
For groups, the situation is even better. If is a module for a group algebra , then so is . The action of on is given by for all . We also have a “trivial” one-dimensional module, the underlying field . Here the action of on is given by for all . Furthermore, if is a representation, then there is a second representation – called the dual representation or sometimes the contragredient representation – denoted and defined by , where denotes the conjugate transpose of the matrix .
For more general algebras, none of these constructions necessarily works. Indeed most algebras do not have any notion of ‘inverses’ of their basis elements.
However, if there is a map and is a -module, then we can always extend to an -module by defining an action . In particular, if is a quotient algebra of then every -representation can be extended in this way to an -representation. On the other hand, if is merely a subalgebra of then there is no obvious way to turn a representation of into a representation of (though we can, of course, turn every representation of the larger algebra into a represenation of the smaller algebra ).
Some examples
Let be the algebra of upper-triangular matrices and let be the (left) ideal consisting of matrices of the form Then is a maximal ideal, and is (isomorphic to) the simple one-dimensional module on which acts like multiplication by the leading diagonal element .
Let be the Temperley-Lieb algebra on generators , and let be the (two-sided) ideal of with basis every non-trivial word in these generators. Then is a maximal ideal and is (isomorphic to) the simple-one dimensional module on which each generator acts like zero.
If is the cyclic group then there is a (non-trivial) one-dimensional representation given by . The dual of this representation is given by
If is the cyclic group then the regular representation is map from to ) given by The extenstion to the regular representation for should be obvious.
The universal enveloping algebra has a representation given by
More generally, the same universal enveloping algebra has a representation given by
The Temperley-Lieb algebra has a representation given by where . Since the Temperley-Lieb algebra is a quotient of the Hecke algebra, this map in turn defines a representation of the Hecke algebra.
Coalgebras
One advantage of defining algebras in terms of commutative diagrams in some category is that, as good category theorists, we can then ask what happens if we reverse the direction of all the arrows. As you might have guessed, we obtain the definition of a new algebraic structure called a coalgebra.
Coalgebras are not just a curiosity, but arise naturally in combinatorics. If we think about multiplication (in an algebra) as a way of combining two elements to obtain a third, then comultiplication – the linear map – can be thought of as a way of decomposing an element into all the different possible pairs that could have been combined to obtain it. We will see this in the examples later.
It is worth unpacking the definition of a little bit. If then the fact is a linear map means that where the exact choice of and is not unique. We can fix a basis for , and express as a sum in terms of this basis, but if haven’t done this there are lots of possible different choices of and .
When the exact choice of choices of and is not important, we can use Sweedler notation. We write the sum as
Here and are not specific elements but merely placeholders representing an arbitrary summand of the whole sum. With this notation, the coassociativty of – the fact that – can be described as
The relationship between and – the fact that – can be expressed as
Given any vector space recall that we can define the dual space by
In particular, an algebra and a coalgebra both have dual spaces. From the duality of their definitions, we might hope that the dual space of an algebra can be given a coalgebra structure in some natural way (and vice versa). This turns out to be almost, but not quite, true:
ProofWe define multiplication of two elements by for all , where . This multiplication is associative precisely because is coassociative. For the identity element, recall that is a linear map and hence . Now for any linear map we must have and similarly
Suppose is a finite dimensional vector space, so that . If is a basis for , then has dual basis , where the linear functions are defined by
ProofHere the key result we need is that, for a finite dimensional vector space , we have . This means that if is a basis for and is a basis for then is a basis for . We define maps and by, for ,
To show that is coassociative requires a little calculation.
First we note that and that
We can rewrite the second of these equations as
Comparing coefficients, we want to show that
Now we make use of the fact that and . We have
We also have
But in fact we have as this is exactly the relationship between structure constants that tells us that is associative. So the desired equality follows at once and is coassociative, as claimed.
For the counit relationship we want to show that . That is, we want to show that
Indeed, uing the structure constants, we have and . It is not hard to see that the claimed identity follows from the fact that .
Note the lack of symmetry here: in general, the dual space of an arbitrary infinite-dimensional algebra is not a coalgebra. This is a sign that algebras and coalgebras differ in some substantial ways. More generally, we have the so-called Fundamental Theorom of Coalebgras:
ProofLet . Let . Then where we are free to chose both and to be linearly independent, and we will do so.
Let be the subspace of spanned by . We want to show that is a of – in other words, that .
Note that by coassociativty we have . This means that
But we have chosen is a linearly independent set. So this previous equation tells us that for all .
It follows then that and as is a linearly independent set it follows that for all .
Symmetrically, from the fact that , we can show that for all .
Hence it follows that where for this last equality we use the general fact that if and are vector spaces with then . This completes the proof.
No similar result holds for algebras. For example, consider the algebra of polynomials in one variable. This is an infinite dimensional algebra generated by the element . But this then means that cannot be contained in any finite dimensional subalgebra.
Although we will not define them here, there are dual concepts of all the algebra conceptswe have described so far. There are coideals, cokernels, comodules and so on.
Some examples
The matrix coalgebra has elements the set of matrices. This vector space has basis where is the matrix whose entries are all zero except for a 1 in position . A coalgebra structure on this vector space is given on basis elements by where is the Kronecker matrix product and is the Kronecker delta, that is
If is a (finite) group then and for all . Elemebts of other coalgebras that have this second property are correspondingly called grouplike elements.
If is a partially ordered set, let . The incidence coalgebra is the -vector space with basis and comultiplication and counit given by where is the Kronecker delta defined above.
There is a coalgebra structure on , given by – for any – and . Elements of other coalgebras that have this second property are called primitive.
The ring of polynomials can be given a coalgebra structure. We have and .
Similarly, for a quiver we can define a path coalgebra. The counit of any non-trivial path is zero while the comultiplication of a path is the sum of all pairs of paths whose concatenation gives that path. In fact, the previous example is just a special case of this for the quiver with one vertex and one edge labelled .
We previously defined an algebra structure on the infinite dimensional space . This space can also be given the structure of a coalgebra. Let be the set of all permutations with the property that and . We call this the set of riffle shuffles. For we now define We define the counit by and otherwise Note that the elements are primitive in the sense defined earlier: .
Bialgebras and Hopf algebras
Similarly to how we motivated an algebra as being both a vector space and a module in a compatible way, a bialgebra is both an algebra and a coalgebra in some compatible way.
In order to define the right notion of compatibility, we return to the notion of an algebra homomorphism. First, recall that if is an algebra then so is . Something very similar holds for coaelgbras.
Equally, there is a dual notion of coalgebra homomorphism::
Motivated by the definition of an algebra morphism and a coalgebra morphism, we now have the right notion of compatibility:
You can think of the diagrams as either telling us that and are algebra morphisms or equivalently that and are coalgebra morphisms. To move from one perspective to another, we simply rotate the diagrams in the obvious way.
Recall that the reason we were led to discuss coalgebras was the representation theory of algebras. In particular, we are trying to find algebras which share some of the nice represention theory properties of group algebras.
One further such nice property of group algebras comes from the fact that each element of the basis is invertible.
Not every bialgebra can be given the structure of a Hopf algebra, but when this is possible there is always only one way to do it. This is equivalent to the fact that if a monoid has an identity element (ie if it is a group) then that identity element must be unique (and the proof of both results is similar).
ProofLet . Then and we can choose such that they form a linearly independent set. By the diagram above (and the similar diagram defining an antipode) we have which implies that, because the are linearly independent, we must have for all .
Since and are linear maps, it follows that and since this is true for an arbitrary it follows that .
Finally, we can put everything together and start talking about the representation theory of Hopf algebras. We only state a few initial definitions and results here but we will return to this topic in the next part of this series.
ProofIn fact in this case all three isomorphisms , and are simply the identity map, so we can omit them here.
Suppose that and are two -modules. Then the vector space naturally has the structure of a algebra: for any , and we have
Now we can use the fact that is an algebra morphism to give the structure of a -module. For any we have We can call this new -module . That this operation satisfies the required associativty axioms follows at once from the fact that is coassociative.
For the unit object we use the fact that is a algebra homomorphism. This makes a -module and the axioms relating to the other structure maps of ensure this module has the required properties.
ProofClearly is a monoidal category because is a bialgebra. What we have to show is that this category is rigid.
For any algebra , if is a (right) -module then the dual is a left -module. If , and , then we define a left action of on by This operation actually extends to a duality between the category of left modules and the category of right modules.
For a Hopf algebra, the existence of the antipode gives us a second way of translating between left and right modules. IF is a left -module, then let be the right module with the action of on defined by
Combining these two notions of duality gives us the duals we want. If is any finite dimensional left -module, then is also a (finite dimensional) left -module. That this module has the required properties is left as an exercise.
Note that if is a finite dimensional -module with basis and is the corresponding dual basis of then the maps and – called and respectively – are given explicitly by and
We will consider the category of modules of a Hopf algebra in more detail in later instalments in this series.
Some examples
As stated in the motivation, the antipode for any group algebra is the map that sends to for every . Indeed, is actually a bialgebra for any monoid . An antipode for this bialgebra – making it a Hopf algebra – exists if and only if every element of is invertible: that is, if is in fact a group.
We have given algebra and coalgebra structures for the vector space with basis the set of all matrices. These two structures are in fact compatible, making the vector space a bialgebra. However it can be shown that this bialgebra has no antipode.
We also saw previously that the tensor space can be given both an algebra and a coalgebra structure. These structures are also compatible, making a bialgebra. Furthermore, an antipode for this bialgebra is given by and meaning that is actually a Hopf algebra. Moreover, if is any bialgebra with primitive elements, then there is a morphism from which maps each of the he canonical basis elements to a distinct primitive element of . This means that, if is a -module for any bialgebra , it is also a module for some tensor space algebra (using the construction given earlier).
For the antipode is defined by , and . It is straightforward to check that this linear map has the required properties. For example,
Staying with the algebra , we can now give an explicit example of how the coproduct lets us build up new representations. Since for each of the primitive elements , we can define a new representation by where on the last line denotes the Kronecker product. Direct calculation then gives us that The reader can check that this is indeed a valid representation of the algebra. In fact, for further calculation gives us that and so the new representation is equivalent to the direct sum of two of the other representations we introduced previously. This is the first hint of a much more general result called the Clebsch-Gordan decomposition formula which we might return to later.
An algebra is called commutative if, for every , we have Similarly a coalgebra is cocommutative if, for every , we have Most of the bialgebras we have seen so far are either commutative or cocommutative or both. For example, a group algebra is commutative exactly when the underlying group is abelian, but is cocommutative no matter the group. Sweedler’s Hopf algebra is the four dimensional algebra with generators subject to the relations The coproduct is given by the counit is given by and the antipode is given by Sweedler constructed this algebra in 1969 as an example of a Hopf algebra that was neither commutative nor cocommutative. We will see many more examples of such Hopf algebras in the next post in this series.