## From the Poincaré group to Heisenberg doubles

There’s a nice geometric way to understand the Heisenberg double of a Hopf algebra, using what one might call its “defining representation(s).” In fact, it’s based on the nice geometric way to understand any semidirect product of groups, so I’ll start with that.

First, consider the Poincaré group, the group of symmetries of Minkowski spacetime.  Once we pick an origin of Minkowski spacetime, making it into a vector space $\mathbb{R}^{3,1}$, the Poincaré group becomes a semidirect product

$\mathrm{ISO}(3,1)\cong\mathbb{R}^{3,1} \rtimes \mathrm{SO}(3,1)$

and the action on $\mathbb{R}^{3,1}$ can be written

$(v,g)\cdot x = v + g x$

In fact, demanding that this be a group action is enough to determine the multiplication in the Poincaré group.  So, this is one way to think about the meaning of the multiplication law in the semidirect product.

In fact, there’s nothing so special about Minkowski spacetime in this construction.  More generally, suppose I’ve got a vector space $V$ and a group $G$ of symmetries of $V$.   Then $V$ acts on itself by translations, and we want to form a group that consists of these translations as well as elements of $G$.  It should act on $V$ by this formula:

$(v,g)\cdot x = v + g x$

Just demanding that this give a group action is enough to determine the multiplication in this group, which we call $V \rtimes G$.   I won’t bother writing the formula down, but you can if you like.

In fact, there’s nothing so special about $V$ being a vector space in this construction.  All I really need is an abelian group $H$ with a group $G$ of symmetries.  This gives us a group $H\rtimes G$, whose underlying set is $H \times G$, and whose multiplication is determined by demanding that

$(h,g) \cdot h' = h + gx$

is an action.

In fact, there’s nothing so special about $H$ being abelian.  Suppose I’ve got a group $H$ with a group $G$ of symmetries.  This gives us a group $H\rtimes G$, built on the set $H \times G$, and with multiplication  determined by demanding that

$(h,g)\cdot x = h (g x)$

give an action on $H$.  Here $gx$ denotes the action of $g\in G$ on $x\in H$, and $h(gx)$ is the product of $h$ and $gx$.

For example, if $H$ is a group and $G=\mathrm{Aut}(H)$ is the group of all automorphisms of $H$, then the group $H\rtimes \mathrm{Aut}(H)$ is called the holomorph of $H$.

What I’m doing here is defining $H \rtimes G$ as a concrete group: it’s not just some abstract group as might be defined in an algebra textbook, but rather a specific group of transformations of something, in this case transformations of $H$.  And, if you like Klein Geometry, then whenever you see a concrete group, you start wondering what kind of geometric structure gets preserved by that group.

So: what’s the geometric meaning of the concrete group $H \rtimes G$?  This really involves thinking of $H$ in two different ways: as a group and as a right torsor of itself.  The action of $G$ preserves the group structure by assumption: it acts by group automorphisms.  On the other hand, the action of $H$ by left multiplication is by automorphisms of $H$ as a right $H$ space.  Thus,  $H \rtimes G$ preserves a kind of geometry on $H$ that combines the group and torsor structures.  We can think of these as a generalization of the “rotations” and “translations” in the Poincaré group.

But I promised to talk about the Heisenberg double of a Hopf algebra.

In fact, there’s nothing so special about groups in the above construction.  Suppose $H$ is a Hopf algebra, or even just an algebra, and there’s some other Hopf algebra $G$ that acts on $H$ as algebra automorphisms.  In Hopf algebraists’ lingo, we say $H$ is a “$G$ module algebra”.  In categorists’ lingo, we say $H$ is an algebra in the category of $G$ modules.

Besides the Hopf algebra action, $H$ also acts on itself by left multiplication.  This doesn’t preserve the algebra structure, but it does preserve the coalgebra structure: $H$ is an $H$ module coalgebra.

So, just like in the group case, we can form the semidirect product, sometimes also called a “smash product” in the Hopf algebra setting, $H \rtimes G$, and again the multiplication law in this is completely determined by its action on $H$. We think of this as a “concrete quantum group” acting as two different kinds of “quantum symmetries” on $H$—a “point-fixing” one preserving the algebra structure and a “translational” one preserving the coalgebra structure.

The Heisenberg double is a particularly beautiful example of this.   Any Hopf algebra $H$ is an $H^*$ module algebra, where $H^*$ is the Hopf algebra dual to $H$.  The action of $H^*$ on $H$ is the “left coregular action” $\rightharpoonup$ defined as the dual of right multiplication:

$(h\rightharpoonup \alpha)(k) = \alpha(kh)$

for all $h,k\in H$ and all $\alpha \in H^*$.

One could use different conventions for defining the Heisenberg double, of course, but not as many as you might think.  Here’s an amazing fact:

$H \rtimes H^* = H \ltimes H^*$

So, while I often see $\rtimes$ and $\ltimes$ confused, this is the one case where you don’t need to remember the difference.

But wait a minute—what’s that “equals” sign up there.   I can hear my category theorist friends snickering.  Surely, they say, I must mean $H \rtimes H^*$ is isomorphic to $H \rtimes H^*$.

But no.  I mean equals.

I defined $H \rtimes H^*$ as the algebra structure on $H \otimes H^*$ determined by its action on $H$, its “defining representation.”   But every natural construction with Hopf algebras has a dual.  I could have instead defined an algebra $H \ltimes H^*$ as the algebra structure on $H \otimes H^*$ determined by its action on $H^*$.  Namely, $H^*$ acts on itself by right multiplication, and $H$ acts on $H^*$ by the right coregular action.  These are just the duals of the two left actions used to define $H \rtimes H^*$.

That’s really all I wanted to say here.  But just in case you want the actual formulas for the Heisenberg double and its defining representations, here they are in Sweedler notation:

Define a map $\rhd \colon (H \otimes H^*) \otimes H \to H$ by

$(h\otimes \alpha) \rhd a := h(\alpha \rightharpoonup a)$

where $\rightharpoonup$ denotes the left regular action of $H^*$ on $H$.  Then a quick calculation shows this is a representation iff we define the multiplication in $H \otimes H$ by

$(h' \otimes \alpha')\cdot (h \otimes \alpha) = h'(\alpha'_{(1)} \rightharpoonup h) \otimes \alpha'_{(2)} \alpha$

This defines the algebra that I called $H \rtimes H^*$, and this is exactly the usual formula for a semidirect product of Hopf algebras.  $H$ is its “defining representation.”

On the other hand, the dual of this defining representation is a right representation on $H^*$ given by

$\beta \lhd (h\otimes \alpha) := (\beta \leftharpoonup h) \alpha$

Since everything in Hopf algebra theory has an equivalent dual statement, we could just as well have started with this action and defined the algebra structure on $H \otimes H^*$ by demanding that this give a representation.  A quick calculation show this is a representation iff we define the multiplication in $H \otimes H$ by

$(h' \otimes \alpha')\cdot (h \otimes \alpha) = h'h_{(1)} \otimes (\alpha'\leftharpoonup h_{(2)}) \alpha$.

This is the usual formula for $H\ltimes H^*$.  But this must be equal to the other formula for the multiplication, so we can define multiplication in the Heisenberg double by either one:

$(h' \otimes \alpha')\cdot (h \otimes \alpha) = h'(\alpha'_{(1)} \rightharpoonup h) \otimes \alpha'_{(2)} \alpha= h'h_{(1)} \otimes (\alpha'\leftharpoonup h_{(2)}) \alpha$

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