Why graded bi-algebras have antipodes

Why graded bi-algebras have antipodes July 7, 2011

Posted by David Speyer in Uncategorized.
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The point of this post is to give a quick proof of a certain fact about bi-algebras. Namely, if $A$ is a graded bialgebra over a field $k$ , with $A_0=k$ , then $A$ 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 $g \mapsto g^{-1}$ . One can write down the map $g \mapsto g^n$ for positive $n$ using just the bi-algebra structure; just take that formula and plug in $n=-1$ . 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 $A$ is a $k$ -vector space with $k$ -linear maps $i: k \to A$ , $m: A \otimes A \to A$ , $\epsilon: A \to k$ and $\Delta: A \to A \otimes A$ such that

(1) Using $m$ as multiplication, and $i(1)$ as the multiplicative unit, $A$ becomes an associative algebra.

(2) The dual of statement (1) holds for $\epsilon$ and $\Delta$ .

(3) $\Delta$ is a map of algebras.

One major example of a bi-algebra is $k[H]$ , where $H$ is any semi-group (with a unit). Here $m$ is the standard multiplication on $k[H]$ , $i(a) = a \cdot e$ where $e$ is the unit; $\epsilon$ is $0$ on the non-unit elements of $H$ and $1$ on $e$ ; and $\Delta(h) = h \otimes h$ for any $h \in H$ .

A bi-algebra is called a Hopf algebra if there is a $k$ -linear map $S: A \to A$ such that $m \circ (S \otimes \mathrm{Id}) \circ \Delta = m \circ (\mathrm{Id} \otimes S) \circ \Delta = i \circ \epsilon$ . (These are all maps from $A \to A$ .) In the case of $k[H]$ , an antipode is precisely a map $s : H \to H$ such that $h s(h) = s(h) h = e$ . I.e. $k[H]$ is a Hopf algebra if and only if the semigroup $H$ has inverses i.e. if and only if $H$ 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 $n$ , define the map $p_n$ from $A \to A$ by composing two maps: First, map $A$ to $A^{\otimes n}$ by using $\Delta$ over and over. Then map $A^{\otimes n} \to A$ by repeatedly using $m$ . Because of the associativity and co-associativity axioms, $p_n$ is well defined. If $A$ is $k[H]$ , then $p_n$ is the $k$ -linear extension of the map $h \mapsto h^n$ from $H$ to itself. (Exercise!)

We would like to define the antipode to be “ $p_{-1}$ “. But what should this mean?

At this point, we introduce the hypothesis that will save us. Suppose that $A$ is graded, meaning that $A = \bigoplus A_i$ , with $i$ , $m$ , $\epsilon$ and $\Delta$ all maps of graded $k$ -vector spaces. So, for example, $m(A_i \otimes A_j)$ is contained in $A_{i+j}$ . Suppose further that $A_0 = k$ .

In this case, $p_n$ must take $A_i$ to $A_i$ . Let’s see what this look like for $i=3$ . First, we repeatedly use $\Delta$ to map $A_3$ to $\bigoplus_{i_1+i_2+\cdots + i_n =3} A_{i_1} \otimes \cdots \otimes A_{i_n}$ . In this direct sum, there are $n$ -terms that look like $A_0^{\otimes i} \otimes A_3 \otimes A_0^{\otimes j}$ ; there are $n(n-1)/2$ terms which look like $A_0^{\otimes i} \otimes A_1 \otimes A_0^{\otimes j} \otimes A_2 \otimes A_0^{k}$ ; there are another $n(n-1)/2$ which look like $A_0^{\otimes i} \otimes A_2 \otimes A_0^{\otimes j} \otimes A_1 \otimes A_0^{k}$ ; and there are $n(n-1)(n-2)/6$ which are made of $A_0$ ‘s and $A_1$ ‘s. Since $A_0$ is just our ground field, we can identify these tensor products with $A_3$ , $A_2 \otimes A_1$ , $A_1 \otimes A_2$ and $A_1 \otimes A_1 \otimes A_1$ respectively.

Using the co-unit axiom repeatedly, one shows that all of the $n(n-1)(n-2)/6$ different maps $A_3 \to A_1 \otimes A_1 \otimes A_1$ which we get in this way are the same map. And similarly for the maps $A_3 \to A_2 \otimes A_1$ , $A_3 \to A_1 \otimes A_2$ and $A_3 \to A_3$ .
Let’s call these maps $q_{111}$ , $q_{21}$ , $q_{12}$ and $q_{3}$ .

Going to $A^{\otimes n}$ is just the first half of computing $p_n$ . The second half is mapping $A^{\otimes n}$ back to $A$ by repeated uses of $m$ . As before, if we want to understand the terms that end up in $A_3$ , we need to understand maps $A_{i_1} \otimes \cdots \otimes A_{i_n} \to A_3$ , with $\sum i_k =3$ . And, as before, this comes down to just understanding 4 different maps: $r_{111} : A_1 \otimes A_1 \otimes A_1 \to A_3$ , and similarly defined $r_{21}$ , $r_{12}$ and $r_3$ .

Putting it all together,

$\displaystyle{p_n = n r_3 \circ q_3 + \frac{n(n-1)}{2} r_{21} \circ q_{21}}$

$\displaystyle{+ \frac{n(n-1)}{2} r_{12} \circ q_{12} + \frac{n(n-1)(n-2)}{6} r_{111} \circ q_{111} .}$

This suggests an obvious definition for $p_{-1}$ from $A_3$ to $A_3$ : just plug in $n=-1$ above to get
$\displaystyle{p_{-1} = (-1) r_3 \circ q_3 + r_{21} \circ q_{21}+ r_{12} \circ q_{12} - r_{111} \circ q_{111} .}$

Here is the key fact:

For every $i$ , the map $p_n$ from $A_i \to A_i$ is a sum of a finite number (specifically $2^{i-1}$ ) of linear maps which are independent of $n$ , with coefficients which are polynomials in $n$ .

So we can define $p_{a}$ for any integer $a$ . Now, when $j$ and $k >0$ , it is easy to see that $m \circ (p_j \otimes p_k) \circ \Delta = p_{j+k}$ . This generalizes the equation $(h^j)(h^k) = h^{j+k}$ . Writing out a proof directly from the Hopf algebra axioms is a good exercise. For any $A_i$ , the fact that $m \circ (p_j \otimes p_k) \circ \Delta = p_{j+k}$ and $p_{j+k}$ give the same map $A_i \to A_i$ is a polynomial identity. So it is also true for negative integers. In particular, $m \circ (p_{-1} \otimes p_1) \circ \Delta$ must be $p_0$ , which is $i \circ \epsilon$ . In other words, $p_{-1}$ is an antipode.

The same method of proof shows another, related result: Let $A$ be a bialgebra, and let $J$ be the kernel of $\epsilon$ . Set $\widehat{A} = \lim_{\leftarrow} A/J^k$ , the $J$ -adic completion of $A$ . Then $\widehat{A}$ is a Hopf algebra. Proof sketch: Show that $p_n$ descends to a map $A/J^k \to A/J^k$ , and that, for each $k$ , this map is a polynomial in $n$ . Exercise: When $A$ is commutative, what is the algebraic geometry meaning of this statement?

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Comments»

1. Kevin Walker - July 8, 2011: “We’ll start out trying to build an antipode in an arbitrary Hopf algebra”. Did you mean “arbitrary bi-algebra”?
2. David Speyer - July 9, 2011: Fixed, thanks.
3. Theo - July 13, 2011: At one of the problem sessions at the workshop, Alex Chirvasitu told me a generalization of the above result:

Theorem (probably Sweedler?): Let $C$ be a coalgebra and $A$ an algebra (both over a field $k$, for convenience; (co)unital and (co)associative, but not necessarily (co)commutative). Recall that $\operatorname{Hom}_k(C,A)$ of linear maps is a (noncommutative, unital) $k$-algebra with the convolution product. Recall also that the _coradical_ $C_0$ of $C$ is the direct sum of its simple subcoalgebras; it is a subcoalgebra of $C$. Given a linear map $f: C \to A$, let $f_0$ denote its restriction to $C_0$. Suppose that $f_0$ is invertible for the convolution product on $\operatorname{Hom}_k(C_0,A)$. Then $f$ is convolution-invertible.

This implies the above result because in a (nonnegatively) graded bialgebra, the coradical of the whole coalgebra is the coradical of its degree-$0$ part.

The proof is not much harder than what’s given above. The idea is that $C$ is the ind-limit of its coradical filtration, and you can invert $f$ as a power series.

But the underlying _reason_, I think, is interesting and geometric. The point is that coalgebras are a fairly basic form of “noncommutative spaces”. They do not allow for interesting topology, but any _set_ gives a coalgebra of linear combinations of the points in the set, and you can also easily add nilpotent “fuzz” to the points in the set, and stay in the world of cocommutative coalgebras.

Then the Theorem boils down to the following idea. The coradical of a coalgebra is essentially the “closed points” in a “noncommutative space”, and the structure theory of coalgebras assures that every coalgebra consists entirely of “infinitesimal neighborhoods of closed points”. A linear map $C \to A$ is something like an “$A$-valued function on $\operatorname{cospec}(C)$”. Then the Theorem just says that to tell whether a function is invertible in an infinitesimal neighborhood of a point, it’s enough to know that it’s invertible at the point, but it says this in the “noncommutative” world.
4. Theo - July 13, 2011: (Also, I’ve clearly been on MathOverflow too long — I just expect MathJax and Markdown to work everywhere. Apologies, then, for the TeX-heavy comment above; I think WordPress must require different conventions.)
5. darij - August 19, 2011: Very instructive proof, even if the standard recursive construction is much easier. Here is a way to rewrite it in a way Hopf algebraists would immediately recognize the trick:

Consider the convolution algebra $L\left(A,A\right)$ of all linear maps $A\to A$ . The unity of this algebra is the map $i\circ \epsilon$ . Now, define a map $t\in L\left(A,A\right)$ by $t=\mathrm{id}-i\circ\epsilon$ . Then, $t$ maps $A_0=k$ to $0$ , and thus is “locally nilpotent”; more precisely, every $i\geq 0$ satisfies $t^{\ast i+1}\mid_{A_i} = 0$ (where $^{\ast i+1}$ means “ $\left(i+1\right)$ -th power in the convolution algebra”). Hence, for every fixed $i$ , the map $\mathrm{id}^{\ast n}\mid_{A_i} = \left(i\circ\epsilon + t\right)^{\ast n}\mid A_i$ depends polynomially on $n$ (because in any algebra, if $t$ is a nilpotent element, then $\left(1+t\right)^n$ depends polynomially on $n$ ). Since this map is exactly your $p_n$ , restricted to $A_i$ , this explains your key fact.

By the way, a few flaws (I can’t stop pointing them out):
- There is an unprocessed LaTeX formula ($s : H \to H$) in the text.
- In the “key fact”, you probably don’t really mean to say $2^{n-1}$ – the number should be independent on $n$ for the proof to work. Here is, by the way, the explicit version of the key fact:

$p_n = \sum\limits_{k=0}^{i} \sum\limits_{i_1,i_2,...,i_k>0;\ i_1+i_2+...+i_k=i} \binom{n}{k} r_{i_1,i_2,...,i_k} \circ q_{i_1,i_2,...,i_k}$ ,

unless I am mistaken.
- I think you also have to require that $\epsilon$ is an algebra map in the definition of a bi-algebra.
6. David Speyer - October 4, 2011: Typos corrected. Thanks, darij! I thought that $\epsilon$ being a map of algebras follows from what I’ve already stated, but I could be wrong.

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