Shuffle Hopf algebra: how to prove its properties in a slick way?

Question

Let $k$ be a commutative ring with $1$, and let $V$ be a $k$-module. Let $TV$ be the $k$-module $\bigoplus\limits_{n\in\mathbb N}V^{\otimes n}$, where all tensor products are over $k$.

We define a $k$-linear map $\mathrm{shf}:\left(TV\right)\otimes\left(TV\right)\to TV$ by

$\mathrm{shf}\left(\left(a_1\otimes a_2\otimes ...\otimes a_i\right)\otimes\left(a_{i+1}\otimes a_{i+2}\otimes ...\otimes a_n\right)\right)$

$= \sum\limits_{\sigma\in\mathrm{Sh}\left(i,n-i\right)} a_{\sigma^{-1}\left(1\right)} \otimes a_{\sigma^{-1}\left(2\right)} \otimes ... \otimes a_{\sigma^{-1}\left(n\right)}$

for every $n\in \mathbb N$ and $a_1,a_2,...,a_n\in V$. Here, $\mathrm{Sh}\left(i,n-i\right)$ denotes the set of all $\left(i,n-i\right)$-shuffles, i. e. of all permutations $\sigma\in S_n$ satisfying $\sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right)$ and $\sigma\left(i+1\right) < \sigma\left(i+2\right) < ... < \sigma\left(n\right)$.

We define a $k$-linear map $\eta:k\to TV$ by $\eta\left(1\right)=1\in k=V^{\otimes 0}\subseteq TV$.

We define a $k$-linear map $\Delta:TV\to \left(TV\right)\otimes\left(TV\right)$ by

$\Delta\left(a_1\otimes a_2\otimes ...\otimes a_n\right) = \sum\limits_{i=0}^n \left(a_1\otimes a_2\otimes ...\otimes a_i\right)\otimes\left(a_{i+1}\otimes a_{i+2}\otimes ...\otimes a_n\right)$

for every $n\in \mathbb N$ and $a_1,a_2,...,a_n\in V$.

We define a $k$-linear map $\varepsilon:TV\to k$ by

$\varepsilon\left(x\right)=x$ for every $x\in V^{\otimes 0}=k$, and

$\varepsilon\left(x\right)=0$ for every $x\in V^{\otimes n}$ for every $n\geq 1$.

Then the claim is:

1. The $k$-module $TV$ becomes a Hopf algebra with multiplication $\mathrm{shf}$, unit map $\eta$, comultiplication $\Delta$ and counit $\varepsilon$. It even becomes a graded Hopf algebra with $n$-th graded component $V^{\otimes n}$.

2. The antipode $S$ of this Hopf algebra satisfies

$S\left(v_1\otimes v_2\otimes ...\otimes v_n\right) = \left(-1\right)^n v_n\otimes v_{n-1}\otimes ...\otimes v_1$

for every $n\in \mathbb N$ and any $v_1,v_2,...,v_n\in V$.

I call this Hopf algebra the shuffle Hopf algebra, although I am not sure whether this is the standard notion. What I know is that the algebra part of it is called the shuffle algebra (note that it is commutative), while the coalgebra part of it is called the tensor coalgebra or deconcatenation coalgebra.

Question: Is there a slick, or at least a not-too-long proof (I'm speaking of <10 pages in detail) for the statements 1 and 2? The best I can come up with is this here:

For 1, we WLOG assume that $V$ is a finite free $k$-module (because all we have to prove are some identities involving finitely many elements of $V$; now we can see these elements as images of a map from a finite free $k$-module $W$, and by functoriality it is thus enough to prove these identities in $W$). Then, we have $V^{\ast\ast}\cong V$, and we notice that the graded dual of our above graded Hopf algebra (we don't know that it is a graded Hopf algebra yet, but at least it has the right signature) is the tensor Hopf algebra of $V^{\ast}$, for which Hopf algebraicity is much easier to show. (Note that this only works with the graded dual, not with the standard dual, because $TW$ is free but not finite free.)

For 2, we prove that $v_1\otimes v_2\otimes ...\otimes v_n\mapsto \left(-1\right)^n v_n\otimes v_{n-1}\otimes ...\otimes v_1$ is indeed a $\ast$-inverse of $\mathrm{id}$ by checking the appropriate equalities combinatorially (i. e., showing that positive and negative terms cancel out).

These things are ultimately not really difficult, but extremely annoying to write up. Somehow it seems to me that there are simpler proofs, but I am unable to find any proof of this at all in literature (except of the "obviously" kind of proof).

One reason why I am thinking that there are simpler proofs is that the similar statements for the tensor Hopf algebra (this is another Hopf algebra with underlying $k$-module $TV$; it has the same counit and unit map as the shuffle Hopf algebra, but the multiplication is the standard tensor algebra multiplication, and the comultiplication is the so-called shuffle comultiplication) are significantly easier to prove. In particular, 2 holds verbatim for the tensor Hopf algebra, but the proof is almost trivial (since $v_1\otimes v_2\otimes ...\otimes v_n$ equals $v_1\cdot v_2\cdot ...\cdot v_n$ in the tensor Hopf algebra).

What would Grothendieck do? Is there a good functorial interpretation, i. e., is the algebraic group induced by the shuffle Hopf algebra (since it is commutative) of any significance?

Answer 1

Hi Darij,

A geometric way of thinking about the shuffle algebra

A geometric way of seeing $T(V)$ with the shuffle product is by considering functions on the loop space of $V^*$ (i.e. the space of continuous maps from $S^1$ to $V^*$) that are given by iterated integrals. You'll see that the product of two iterated integrals is precisely the shuffle product. Moreover, this way you can see the coproduct coming from the concatenation of loops and the antipode from reversing orientation (if I remember well).

But here you have to be in a situation when $V^{**}\cong V$.

Let me be a bit more precise. For convenience I will work with $T(V^*)$ equipped with the shuffle product $\star$, the deconcatenation coproduct $\Delta$, and $S$ as you defined it.

Now let me consider the algebra $\mathcal A$ of functionals on $L(V^*)=C_*^0(S^1,V^*)$ (the subscript $*$ means that I ask that $0$ is sent to $0$). There is an algebra monomorphism $T(V^*)\to \mathcal{A}$ given as follows: $$ \xi_1\otimes\cdots\otimes \xi_n\mapsto \left(\gamma\mapsto \int_{0<t_1<\cdots<t_n<1}\xi_1(\gamma(t_1))\dots\xi_n(\gamma(t_n))dt_1\dots dt_n\right) $$

Now observe that the composition of loops and the orientation reversing define algebra morphisms $\Delta_A:\mathcal A\to\mathcal A\otimes\mathcal A$ and $S_A:\mathcal A\to\mathcal A$.

The point is that $\Delta_A$ and $S_A$ do not really satisfy the axioms you want (e.g. coassociativity of $\Delta_A$) BUT their restriction onto the image of $T(V^*)$ does (you have to use an avatar of Stokes' Theorem to see this - or, shortly: iterated integrals are not sensitive to reparametrization), and actually coincide with $\Delta$ and $S$ (very simple computation).

---

Below are a few algebraic considerations (independant from the above answer).

The main point concerning the antipode is that

any connected filtered bialgebra is a filtered Hopf algebra, the antipode being defined as $S(x)=\sum_{k\geq0}(\eta\circ\epsilon-id)^{*k}(x)$

Here $*$ denotes the convolution product. In your example the bialgebra you consider is actually graded so the result applies. You can find the above claim (and its proof) in these lecture notes (I think you are going to like them) by Dominique Manchon (Corollary II.3.2).

The group algebra of a free group

The degree completion of $T(V)$ is the structure ring of the pro-unipotent completion of a free group.

I hope this can help.

Answer 2

Suppose $V$ is a free $k$-module with base $X = \{x_i,i\in I\}$.

I like to take as a definition of the shuffle algebra $Sh(X)$ that it is the topological dual of the Hopf algebra $k\langle\langle X\rangle\rangle$ of non commutative series in the variables $x_i$ (i.e. the completed tensor algebra $\widehat{T(V^*)}$ wrt the augmentation ideal). This last algebra is easy to understand.

Then one can derive the formula for the shuffle product and we find the same as yours. And your points 1 and 2 are obvious by duality.

Note that in this point of vue, instead of looking about a morphism of algebras $\varphi: Sh(X) \to A$ we just look at the corresponding generating series $\Phi = \sum_m \varphi(m) m^*$ ($m$ a basis). The fact that "$\varphi$ is a morphism of algebras" translates as "$\Phi$ is diagonal in $A\langle\langle X\rangle\rangle$" (i.e. $\Delta \Phi = \Phi\otimes \Phi$ and $\varepsilon(\Phi)= 1$).

Answer 3

I have come across a nice way to think about this Hopf algebra. Let $v_1\otimes\ldots\otimes v_n$ be a monomial in $TV$. Rather than thinking of it as just a word, think of it as a path of $n$ steps in $V$. Geometrically (if $V$ is defined over $\mathbb{R}$) it could be a piecewise linear path $\mu:[0,n]\rightarrow V$, where $\mu(k) = v_1+\ldots+v_k$. Ofcourse it's not really a geometric thing, it's a sequence of points, so we don't need anything to be over $\mathbb{R}$.

The coproduct comes from splitting up the path at the integer values (including 0 and n). The coassociativity is obvious, both sides of the equation split a path into three. The counit sends any non-trivial path to 0, the trivial path to 1.

The product is not the concatenation of paths of course, let $\mu$ and $\nu$ be paths of length $n$ and $m$ respectively. Their direct product $\mu\times\nu$ has domain $[0,n]\times[0,m]$ and codomain $V\times V$, not the tensor product. So how do we get a path from this direct product? Well we take paths in $[0,n]\times[0,m]$ where each step is a positive unit step in either the horizontal or vertical direction. The image of such a step is of the form $(v,0)$ or $(0,v)$, forget the zeros and we have a new path of the same type as we started with. But this required choice, we chose a path in $[0,n]\times[0,m]$, what possible choices were there? Well the set of such paths is easily seen as the set of shuffles! And so the product is given by summing all the possible paths.

The associativity of the product is now easily seen as the product of three paths involves a summation over all possible paths in not a square but now a cube. The unit is the trivial path.

So how about the bialgebra relation? Well this now has a simple combinatorial description. Take two paths $\mu$ and $\nu$, on one side of the bialgebra compatibility we have to take their product and then their coproduct. Their product is indexed by paths through a square, taking the coproduct splits such a path at a point, so taking the product then the coproduct gives a sum over paths in a square with a chosen point. Reindex: a sum over all integer points of the square with a path from $(0,0)$ to that point, followed by a path from the point to $(n,m)$ the other corner of the square.

Now onto the other side of the equation, take the coproducts of each of path, this is indexed by an integer point in $[0,n]$ and an integer point in $[0,m]$, which taken together is an integer point in the square. Now take the products of the paths, left side with left side and right side with right side. We get precise what we hoped for, all possible paths from $(0,0)$ to the point in the square, followed by a path from the point to $(n,m)$.

So we have a bialgebra, and the antipode just reverses the paths. Now I should stress that all I have done is to make the combinatorics apparent as paths in a square, if you want a quick proof, just do the combinatorics.

Answer 4

I believe the right way to consider this algebra is to view it as the free zinbiel algebra. A zinbiel algebra has a single operation o which must satisfy

(x o y) o z = x o (y o z + z o y)

The zinbiel operation o in your algebra is the sum over all of the (p,q) shuffles which in the notation of the question have

${\sigma^{-1}(1)} < {\sigma^{-1}(i+1)}$

So the commutative product defined in the question is a.b = a o b + b o a. The answer to your question now comes from a result which will state that the free zinbiel algebra viewed as a commutative algebra is itself free, then you just need to check that your shuffle coproduct is defined on the generators. This will show that it is a bialgebra.

An analogous result is that the free associative algebra is a free Lie algebra with the associated Lie bracket.

One way to prove this result would be to decompose the Zinbiel operad as a left module for the commutative operad. But I imagine that the result is already in the literature somewhere. I guess that there are other names for zinbiel algebras, perhaps shuffle algebras or something similar.

They do occur naturally, for instance if you want to decompose the direct product of two simplices (which isn't a simplex) into simplices in a natural way; see p278 of Allen Hatcher's Algebraic Topology.

Answer 5

Well, assume again that $V$ is a free $k$-module with base $X=x_i,i∈I$. One has to avoid the fact that $k\langle\langle X\rangle\rangle$ is NOT a Hopf algebra, be it with shuffle or concatenation except when $X=\emptyset$ because you have to take Sweedler's dual and cannot consider complete dual. A striking (but limited to free - finite or infinitely generated - $V$) proof of the first statement goes as follows

a) The Hopf algebra $k\langle X\rangle$ with concatenation as product and co-shuffle as coproduct is graded - in finite dimensions - over $\mathbb{N}^{(I)}$.

b) Then, the shuffle Hopf algebra is exactly the graded dual of it with the pairing given by
$$ \langle x_{i_1}\otimes\ldots\otimes x_{i_n}\mid x_{j_1}\otimes\ldots\otimes x_{j_n}\rangle=\delta_{i_1,j_1}\ldots \delta_{i_n,j_n} $$ and 0 if $n\not=m$.

c) (For statement 2.) the antipode is just $S^*=S$ (the adjoint of $S$).

Answer 6

For 1. If you are willing to accept facts about the bar construction, consider $A = \Sigma^{-1} V\oplus k$ and define an algebra structure on $A$ by insisting that the product on $\Sigma^{-1} V$ is zero. This makes $A$ into a commutative algebra. Now, we have $BA$ the bar construction on $A$, and $BA$ is the same as your $TV$. A map $C \to BA$ of graded coalgebras is completely determined by the projection $C\to BA \to A$ of degree $-1$. Given a map $f: C\to A$ of degree $-1$, it extends to a coalgebra map $C\to BA$ provided that it satisfies $0 = m(f\otimes f)\Delta$ where $m$ is the product on $A$, and $\Delta$ is the coproduct on $C$. Now, check that if $A$ is a commutative algebra, then the map $BA\otimes BA \to A$ given by $[a]\otimes 1 \mapsto a$, and $1\otimes [a] \mapsto a$, and zero on all other tensor factors, satisfies the condition. Then, check that the induced map $BA\otimes BA \to BA$ is the shuffle product. Similarly, check that it is associative by projecting to $A$.

Answer 7

Oh, well, there is maybe some way out (I learned from Martin Bordmann at some point) It does not avoid all compuptations but works slightly more transparent than just brute force...

The main point is that the deconcatenation $\Delta$ makes the tensor algebra $TV$ the cofree coalgebra in a certain subcategory of coalgebras (with a unique group-like element $1$ and nilpotent augmentation ideal). Then the idea is to \emph{define} the shuffle multiplication as a coalgebra morphism $\mu\colon TV \times TV \longrightarrow TV$ which we only need to specify on cogenerators where one sets $\mu\colon \xi\otimes\eta \mapsto \mathrm{pr}_V(\xi)\epsilon(\eta) + \epsilon(\xi)\mathrm{pr}_V(\eta)$ for $\xi, \eta \in TV$ with $\mathrm{pr}_V$ being the projection onto $V$.

Being a coalgebra morphism, the associativity of $\mu$ can then again be checked on cogenerators only, which is pretty easy since $\mu \circ (\mathrm{id} \otimes \mu)$ as well as $\mu \circ (\mu \otimes \mathrm{id})$ are still both coalg morphism and the relevant coalgs are cofree.

The advantage of this point of view is that one can elaborate on it a bit to also include the symmetrized versions as well as the graded versions of it. Beside that one learns something important about the cofreeness (at least in this restricted category).

OOPS: YBL's answer just popped in: I think this is essentially the same idea.