# A Simple Introduction to Quantum Groups

In the course of reading some background material for an article by James Worthington on using bialgebraic structures in automata theory, I was led to finally reading up on what a Hopf algebra (sometimes called a “quantum group“) is.

Although it is not strictly related to logic, I’ll write up what I learned here.

My main source for this is sigfpe’s blog post on this. In fact, all I really did was take his post and remove the Haskell from it (and probably add some mistakes). If you can read Haskell, I definitely recommend that post (and his blog in general). I also read the article on quantum groups in the Princeton Companion to Mathematics, which is an amazing book.

We’ll take the ordinary definition of a group and turn it into the definition of a quantum group in two steps.

Step 1: Groorgs.

In the first step, we’ll “symmetrize” the definition of a group to get an object I call a groorg. What does it mean to symmetrize a definition? Well, for one thing, as part of the definition of a group we have a multiplication $\text{mult}\colon G\times G\to G$. Therefore, to make it symmetric, we should also have a comultiplication $\text{comult} \colon G\to G\times G$.

Furthermore, this comultiplication should satisfy laws dual to those satisfied by multiplication. For example, multiplication is associative, which means that for any $a$, $b$, $c\in G$, the two ways of using multiplication to turn the triple $abc$ into a single element are the same. Dually, then, comultiplication should be coassociative, meaning that for any element $a\in G$, the two ways of using comultiplication to turn that single element into a triple should be the same.

OK, so we’ve seen that the dual of multiplication is comultiplication. Another part of the definition of a group is the inverse function $\text{inv}\colon G\to G$, which is no problem, since it can be its own dual. But what about the identity element $e\in G$? This is puzzling until you think of the identity not as an element of $G$, but as a map from a one element set $\{\bullet\}$ to $G$, which, when written it that form, I’ll call $\text{unit}\colon \{\bullet\}\to G$. Then it’s clear that the dual of $\text{unit}$ should be $\text{counit} \colon G\to \{\bullet\}$. The astute reader will note that there is only one such function, so this seems to be a trivial addition, but let’s press on regardless.

We can now begin the definition of a groorg.

Definition of a Groorg, Part 1. A groorg is a set $G$ together with:

1. A map $\text{mult}\colon G\times G\to G$.
2. A map $\text{comult}\colon G\to G\times G$.
3. A map $\text{unit}\colon \{\bullet\}\to G$.
4. A map $\text{counit} \colon G\to \{\bullet\}$.
5. A map $\text{inv}\colon G\to G$.

These must satisfy the following properties:

• Multiplication must be associative and comultiplication must be coassociative.
• $\text{unit}$ should be a unit for multiplication. This means that, given any $a\in G$, if you form an ordered pair with $a$ and $\text{unit}(\bullet)$, then apply $\text{mult}$ to that ordered pair, you get $a$ back.
• $\text{counit}$ should be a counit for comultplication. This means that, given any $a\in G$, if you apply $\text{comult}$ to $a$ to get an ordered pair, then destroy one of the componenents with $\text{counit}$ (i.e., just discard it), then you get $a$ back.

Let’s pause here. If you’ve followed along, you may have noticed that this last condition forces $\text{comult}(a)$ to be $(a,a)$ (which does satisfy the conditions so far). This is obviously quite restrictive, but it has one benefit: it means that we can rewrite the inverse law using comultiplication. To see what I mean, consider the following: If we let $f(a,b) = (\text{inv}(a),b)$ and $g(a,b) = (a,\text{inv}(b))$, then the usual inverse law for a group says that for any $a\in G$, $(\text{mult}\circ f)(a,a) = (\text{mult}\circ g)(a,a) = \text{unit}(\bullet)$. Now, since we know that $\text{comult}$ is forced to take each $a$ to $(a,a)$, we can rewrite the inverse law as: $(\text{mult}\circ f \circ \text{comult})(a) = (\text{mult}\circ g\circ \text{comult})(a) = \text{unit}(\bullet)$.

We can make this completely symmetric by getting $\text{counit}$ into the act:

Definition of a Groorg, Part 2. A groorg is also required to satisfy: $(\text{mult}\circ f \circ \text{comult})(a) = (\text{mult}\circ g\circ \text{comult})(a) = \text{unit}(\text{counit}(a))$ for all $a\in G$ (where $f$ and $g$ are as above.

And we finally have a requirement which says that the two ways of computing $(ab, ab)$ by using comultiplication and multiplication are equal.

Definition of a Groorg, Part 3. A groorg is also required to satisfy: $\text{comult}(\text{mult}(a,b)) = \text{mult}(\text{comult}(a),\text{comult}(b))$, where the second $\text{mult}$ is the natural multiplication on $G\times G$.

This concludes the definition of a groorg.

But, what use is it? As we’ve observed, in every groorg, we must have that $\text{comult}$ sends $a$ to $(a,a)$, so that the groorg just reduces to an ordinary group. Furthermore, every group becomes a groorg by defining comult in that way (and by defining counit in the only possible way).

What we’ve gained is that we now have a definition of a group which is equivalent to the old one and which is symmetric, which will lend itself well to our next step.

How can we turn this concept of a group into one of a “quantum” group? If you’re like me, the only thing you know about quantum mechanics is that you often hear the word “superposition” used in conjunction with it. That’s not much, but it turns out to be enough in this case.

Instead of having the composition $\text{mult}(g,h)$ of two group elements be another group element, let’s have it be a superposition of group elements. It turns out that what this should mean is a linear combination of group elements. So, let $\mathbb{C}[G]$ be the $\mathbb{C}$-vector space generated by taking $G$ as a formal set of basis vectors. Instead of requiring that $\text{mult}$ be a map from $G\times G$ to $G$, we will let it be a map from $G\times G$ to $\mathbb{C}[G]$. So we have the following:

Provisional Definition of a Quantum Group. A quantum group is a set $G$ together with

1. A map $\text{mult}\colon G\times G \to \mathbb{C}[G]$.
2. A map $\text{comult}\colon G\to \mathbb{C}[G\times G] = \mathbb{C}[G]\otimes\mathbb{C}[G]$.
3. A map $\text{unit}\colon \{\bullet\}\to \mathbb{C}[G]$.
4. A map $\text{counit}\colon G\to \mathbb{C}[\{\bullet\}] = \mathbb{C}$
5. A map $\text{inv}\colon G\to \mathbb{C}[G]$.

satisfying …

Before we can think about what properties these functions should satisfy, we have to settle a question: We know how to multiply two group elements to get a superposition of group elements, but how should we multiply two superpositions of group elements? For example, what should the product of $g_1 + \pi g_2$ and $ig_3 + 2 g_4$ be? (Note that I am using the same symbol to stand for the group element and the formal basis vector corresponding to it.)

The quickest way to define the multiplication of superpositions is to notice that, since $\{g_1\otimes g_2\mid g_1,g_2\in G\}$ is a basis for $\mathbb{C}[G]\otimes \mathbb{C}[G]$, the map $\text{mult}\colon G\times G\to \mathbb{C}[G]$ extends to a linear map $\text{mult}\colon \mathbb{C}[G]\otimes \mathbb{C}[G]\to \mathbb{C}[G]$, which we can use to multiply superpositions. In the above example, $(g_1 + \pi g_2)\otimes (ig_3 + 2 g_4)$ is equal to $ig_1\otimes g_3 + 2g_1\otimes g_4 + i\pi g_2\otimes g_3 + 2\pi g_2\otimes g_4$, so the product of the two superpositions would be $i\text{mult}(g_1,g_3) + 2\text{mult}(g_1,g_4) + i\pi\text{mult}(g_2,g_3) + 2\pi\text{mult}(g_2,g_4)$.

Now, our symmetric definition of a group above translates exactly, and we no longer need to mention the basis explicitly:

Definition of a Quantum Group (or Hopf Algebra). A quantum group is a $\mathbb{C}$-vector space $V$ together with:

1. A linear map $\text{mult}\colon V\otimes V\to V$
2. A linear map $\text{comult}\colon V\to V\otimes V$
3. A linear map $\text{unit} \colon \mathbb{C}\to V$
4. A linear map $\text{counit} \colon V\to \mathbb{C}$
5. A linear map $\text{inv}\colon V\to V$

satisfying the analogues of the laws given in the definition of a groorg.

Some Combinatorial Examples

I believe there are many examples of the usefulness of this concept in physics. However because I don’t know any physics, I won’t give them.

Here are two combinatorial examples from sigfpe’s blog post:

Example 1: A Quantum Group on Finite Strings.

Let $\Sigma$ be an alphabet, and let $\Sigma^*$ be the set of all finite strings with characters from $\Sigma$. We may put a quantum group structure on $\mathbb{C}[\Sigma^*]$ as follows:

• We let $\text{mult}(\sigma,\tau) = \sigma\tau$, the concatenation of $\sigma$ and $\tau$.
• We let $\text{unit}(1) = \varepsilon$, the empty string.
• We let $\text{counit}(\varepsilon) = 1$ and $\text{counit}(\sigma) = 0$ where $\sigma \ne \varepsilon$.
• We let $\text{inv}(\sigma) = (-1)^{|\sigma|}\bar{\sigma}$, where $|\sigma|$ is the length of $\sigma$ and $\bar{\sigma}$ is the reverse of $\sigma$.
• We let $\text{comult}(\sigma)$ be the sum of $\sigma_0\otimes \sigma_1$ where $\sigma_0$ and $\sigma_1$ can be shuffled together to give $\sigma$. This means that the characters in $\sigma_0$ occur in $\sigma$ in the same order, and when you remove them you get $\sigma_1$.

For an example of the comultiplication, if $a$ and $b\in \Sigma$, then $\text{comult}(ab) = \varepsilon \otimes ab + a \otimes b + b \otimes a + ab\otimes \varepsilon$.
We can verify the inverse law in this case: If we apply $\text{inv}$ on the right to $\text{comult}(ab)$, we get $\varepsilon\otimes ba - a \otimes b - b \otimes a + ab\otimes \varepsilon$. Applying $\text{mult}$ to this, we get $ba -ab -ba + ab = 0$.

Example 2. Another Quantum Group on Finite Strings.
This is also a quantum group on $\mathbb{C}[\Sigma^*]$.

• We let $\text{mult}(\sigma,\tau)$ be the sum of all possible ways of shuffling $\sigma$ and $\tau$ together.
• We let $\text{unit}(1) = \varepsilon$, the empty string.
• We let $\text{counit}(\varepsilon) = 1$ and $\text{counit}(\sigma) = 0$ where $\sigma \ne \varepsilon$.
• We let $\text{inv}(\sigma) = (-1)^{|\sigma|}\bar{\sigma}$, where $|\sigma|$ is the length of $\sigma$ and $\bar{\sigma}$ is the reverse of $\sigma$.
• We let $\text{comult}(\sigma)$ be the sum of all $\sigma_0\otimes \sigma_1$ such that $\sigma = \sigma_0\sigma_1$.

Again, we can verify the inverse law in a specific case: We have that $\text{comult}(ab) = \varepsilon\otimes ab + a\otimes b + ab\otimes \varepsilon$. If we apply $\text{inv}$ on the right, we get $\varepsilon\otimes ab - a\otimes b + ab\otimes \varepsilon$. Now applying $\text{mult}$, we get $ab -ab -ba + ba = 0$.

Example 3: A Quantum Group on Finite Binary Trees.
This example is from here. We think of a finite binary tree as a finite tree where each node has either zero or two children, and where we distinguish between left and right. This picture from the above paper shows how, if you select a leaf of a finite binary tree, you may divide the tree into the tree to the left of the leaf and the tree to the right of the leaf:

If you select a multiset of $q$ leaves, you may similarly divide your tree into $q$ trees.

We may now define a quantum group on $\mathbb{C}[\mathcal{T}]$, where $\mathcal{T}$ is the set of finite binary trees as follows:

• $\text{mult}(T_1,T_2)$ is the sum of all trees generated as follows: Suppose that $T_2$ has $q$ leaves. Divide $T_1$ into $q$ trees as above, and stick them onto the leaves of $T_2$. There are many graphical examples of this here.
• $\text{unit}(1)$ is the tree with a single node and no leaves.
• $\text{comult}(T)$ is the sum of all $T_1\otimes T_2$, where $T_1$ and $T_2$ are trees that $T$ can be divided into. There are many graphical examples of this here.
• $\text{counit}(T)$ is 1 if $T$ is the tree with only one node, and zero otherwise.
• For the definition of $\text{inv}$, I refer you to the above paper. However, there are many graphical examples of the antipode here.

## 2 thoughts on “A Simple Introduction to Quantum Groups”

1. quantummoxie says:

In case anyone is interested, I have a brief outline of how groups relate to visual symmetry on my blog that might also help the uninitiated understand groups in general.