The point of this post is to give a quick proof of a certain fact about bi-algebras. Namely, if is a graded bialgebra over a field , with , then is a Hopf algebra.
This statement came up at the cluster algebras work shop in Oregon a few weeks ago, and most people seemed to feel it was mysterious. But, in fact, the concept of the proof is very simple. When your Hopf algebra is a group algebra, then the antipode is the map . One can write down the map for positive using just the bi-algebra structure; just take that formula and plug in . Of course, the details are a little messier than that; hence this post.
I’ll give the necessary definitions below the fold, but this is written for people who are already happy with the definitions of bi-algebras and Hopf algebras.
A bi-algebra is a -vector space with -linear maps , , and such that
(1) Using as multiplication, and as the multiplicative unit, becomes an associative algebra.
(2) The dual of statement (1) holds for and .
(3) is a map of algebras.
One major example of a bi-algebra is , where is any semi-group (with a unit). Here is the standard multiplication on , where is the unit; is on the non-unit elements of and on ; and for any .
A bi-algebra is called a Hopf algebra if there is a -linear map such that . (These are all maps from .) In the case of , an antipode is precisely a map such that . I.e. is a Hopf algebra if and only if the semigroup has inverses i.e. if and only if is a group. In general, an antipode should be thought of as a generalized inverse.
We’ll start out trying to build an antipode in an arbitrary bi-algebra. Of course, we’ll fail: Not all semi-groups are groups. But our failure will suggest a method that often succeeds.
For any positive integer , define the map from by composing two maps: First, map to by using over and over. Then map by repeatedly using . Because of the associativity and co-associativity axioms, is well defined. If is , then is the -linear extension of the map from to itself. (Exercise!)
We would like to define the antipode to be ““. But what should this mean?
At this point, we introduce the hypothesis that will save us. Suppose that is graded, meaning that , with , , and all maps of graded -vector spaces. So, for example, is contained in . Suppose further that .
In this case, must take to . Let’s see what this look like for . First, we repeatedly use to map to . In this direct sum, there are -terms that look like ; there are terms which look like ; there are another which look like ; and there are which are made of ‘s and ‘s. Since is just our ground field, we can identify these tensor products with , , and respectively.
Using the co-unit axiom repeatedly, one shows that all of the different maps which we get in this way are the same map. And similarly for the maps , and .
Let’s call these maps , , and .
Going to is just the first half of computing . The second half is mapping back to by repeated uses of . As before, if we want to understand the terms that end up in , we need to understand maps , with . And, as before, this comes down to just understanding 4 different maps: , and similarly defined , and .
Putting it all together,
This suggests an obvious definition for from to : just plug in above to get
Here is the key fact:
For every , the map from is a sum of a finite number (specifically ) of linear maps which are independent of , with coefficients which are polynomials in .
So we can define for any integer . Now, when and , it is easy to see that . This generalizes the equation . Writing out a proof directly from the Hopf algebra axioms is a good exercise. For any , the fact that and give the same map is a polynomial identity. So it is also true for negative integers. In particular, must be , which is . In other words, is an antipode.
The same method of proof shows another, related result: Let be a bialgebra, and let be the kernel of . Set , the -adic completion of . Then is a Hopf algebra. Proof sketch: Show that descends to a map , and that, for each , this map is a polynomial in . Exercise: When is commutative, what is the algebraic geometry meaning of this statement?