The following is taken (almost) directly from the first draft of my PhD thesis.

## The Quantisation Functor

This functor can be used to motivate the correct notion of (the algebra of functions on) a quantum group. Note that the ‘quantised’ objects that are arrived at via this ‘categorical quantisation’ are nothing but the established definitions so this section should be considered as little more than a motivation. The author feels that introductory texts on quantum groups could include these ideas and that is why they are included here. This quantisation is the translation of statements about a finite group, $G$ into statements about the algebra of functions on $G$, $F(G)$.

This notion of quantisation sits naturally in category theory where two functors — the $\mathbb{C}$ functor and the dual functor — lead towards a satisfactory quantisation.

### The $\mathbb{C}$ Functor

The category of finite sets, $\mathbf{FinSet}$, has the class of all finite sets as objects and functions for morphisms. Also of interest is the category of finite dimensional complex vector spaces, $\mathbf{FinVec}_{\mathbb{C}}$ with linear maps for morphisms. There is a map, the $\mathbb{C}$ map, $\mathbb{C} :\mathbf{FinSet}\rightarrow \mathbf{FinVec}_{\mathbb{C}}$, that associates to each object $X\in\mathbf{FinSet}$, an object $\mathbb{C} X\in\mathbf{FinVec}_{\mathbb{C}}$ — the complex vector space with basis $\{\delta^x:x\in X\}$. This map associates to each morphism $f:X\rightarrow Y$ a morphism $\mathbb{C}f:\mathbb{C} X\rightarrow \mathbb{C} Y$, $\delta^x\mapsto \delta^{f(x)}$ and it isn’t difficult to see that it is a covariant functor.

If $X$ and $Y$ are finite sets then $X\times Y$ is also a finite set. This object is sent to $\mathbb{C}(X\times Y)$ by the $\mathbb{C}$ functor. The following explains how to deal with $\mathbb{C}(X\times Y)$, as well as presenting a number of other useful isomorphisms of vector spaces.

#### Theorem (Tensor Product Isomorphisms)

1. Let $X$ and $Y$ be finite sets. Then, under the isomorphism $\delta_x\otimes\delta_y\leftrightarrow \delta_{(x,y)}$, $\mathbb{C} (X\times Y)\cong \mathbb{C} X\otimes \mathbb{C} Y$.
2. Let $V$ be a finite dimensional complex vector space. Then, under the isomorphism $\lambda\otimes\mathbf{v}\leftrightarrow \lambda\mathbf{v}\leftrightarrow \mathbf{v}\otimes\lambda$, $\mathbb{C} \otimes V\cong V\cong V\otimes \mathbb{C}$.
3. Let $U$ and $V$ be finite dimensional complex vector spaces. Then $(U\otimes V)^*\cong U^*\otimes V^*$.

Proof: Standard results $\bullet$

Therefore, a morphism $f:X\times Y\rightarrow Z$ is sent to the linear map $\mathbb{C} f:\mathbb{C} X\otimes \mathbb{C} Y\rightarrow \mathbb{C} Z$:

$\displaystyle (\mathbb{C}f)(\delta^x\otimes \delta^y)=\delta^{f(x,y)}.$

The dual map, $\mathcal{D}$, is a morphism in the category of finite dimensional vector spaces that sends a vector space to its dual and a linear map $T:U\rightarrow V$ to its transpose:

$\displaystyle \mathcal{D}(T):V^*\rightarrow U^*,\qquad \varphi\mapsto \varphi\circ T.$

It can be shown that for $T:V_1\rightarrow V_1$ and $S:V_2\rightarrow V_3$ that

$\displaystyle (S\circ T)^*=T^*\circ S^*.$

With this result, and the fact that $T^*$ is linear, the dual functor is a contravariant endofunctor.

Call the composition of these two functors by the quantisation functor:

$\displaystyle \mathcal{Q}:\textbf{FinSet}\rightarrow \textbf{FinVec}_{\mathbb{C}},\qquad \mathcal{Q}=\mathcal{D}\circ\mathbb{C}.$

It will be seen that the image of a group under this functor is the algebra of functions on the group. This gives us a routine to quantise groups and related objects: apply the $\mathcal{Q}$ functor to objects, morphism and commutative diagrams in the category of finite sets to get quantised objects, morphisms and commutative diagrams in the category of finite dimensional vector spaces.

It will be seen that the image of a finite group under this functor, $\mathcal{Q}(G)=F(G)$, has the structure of aHopf-algebra: whose axioms are found simply by quantising the group axioms on $G$. That is $F(G)$ satisfies:

$\displaystyle\mathcal{Q}(\text{group axioms})\sim\{\mathcal{Q}(\text{associativity}),\mathcal{Q}(\text{identity}),\mathcal{Q}(\text{inverses})\}.$

There are, however, vector spaces together with morphisms that also satisfy these axioms but aren’t the algebra of functions on any group — because the multiplication is no longer commutative. These are the algebras of functions on quantum groups:

”Algebras of functions” on quantum groups are algebras, $F(\mathbb{G})$, that satisfy the quantisations of the group axioms, except letting go of commutativity in the algebra means that $\mathbb{G}$ is a virtual object.

## Algebra of Functions on a Group and the Group Ring

As a group is a ‘space’, it is natural to study the algebra of functions on it. Let $G$ be a finite group and let $F(G)$ be the set of complex-valued functions on $G$. There is a natural $\mathrm{C}^*$-algebra structure on $F(G)$ defined by:

$(f+g)(x)=f(x)+g(x)\qquad(f,\, g\in F(G))$
$(\lambda f)(x)=\lambda \,f(x)\qquad (\lambda\in \mathbb{C})$
$M(f\otimes g)(x)=f(x)g(x)\qquad$
$f^*(x)=\overline{f(x)}\qquad$
$\|f\|:=\max_{x\in G}|f(x)| \qquad$

The unit is the indicator function $\mathbf{1}_{G}$. As in the previous discussion, there are relations that will always hold ‘up’ in $F(G)$ as quantised versions of the relations ‘down’ in $G$. The quantisation functor is used to see exactly what these relations look like in $F(G)$. Note that $F(G)$ is referred to as the algebra of functions on $G$ and is a commutative $\mathrm{C}^*$-algebra.

Also associated to a finite group is another canonical algebra: the group ring. For $G$ a finite group, let $\mathbb{C} G$ be a complex vector space with basis elements $\{\delta^g\,:\,g\in G\}$. The scalar multiplication and vector addition are, for a $\nu=\sum_t \alpha_t\delta^t$ and $\mu=\sum_t\beta_t\delta^t$, the natural

$\displaystyle\lambda \nu=\sum_{t\in G}(\lambda\alpha_t)\delta^t\text{ and }$

$\displaystyle\nu+\mu=\sum_{t\in G}(\alpha_t+\beta_t)\delta^t.$

and the multiplication is given by:

$\displaystyle\nabla(\delta^s\otimes\delta^t)=\delta^{st}.$

The vector space $\mathbb{C} G$ together with the multiplication $\nabla$ is a complex associative algebra called the group ring of $G$. Take an element $\nu$ of $\mathbb{C} G$:

$\displaystyle\nu=\sum_{g\in G}\alpha_g\delta^g.$

If the elements of $\mathbb{C} G$ are considered complex-valued functions on $G$ via the embedding $s\hookrightarrow\delta_s$, $\nu(\delta_s)\hookleftarrow\nu(s)=\alpha_s$, a quick calculation shows that this multiplication $\nabla$ is nothing but the convolution:

$\displaystyle\nabla(\nu\otimes \mu)(s)=(\nu\star \mu)(s)=\sum_{t\in G}\nu(st^{-1})\mu(t).$

The unit is $\delta^e=:\mathbf{1}_{\widehat{G}}$. There is also an involution:

$\displaystyle\nu^*=\sum_{t\in G}\overline{\alpha_t}\delta^{t^{-1}},$

turning $\mathbb{C} G$ into a *-algebra. Note that $\mathbb{C} G$ is commutative if and only if $G$ is abelian. Considering $\mathbb{C} G$ as a Hilbert space with an orthonormal basis $\{\delta^g:g\in G\}$, $\mathbb{C} G$ acts on $\mathbb{C} G$ by left multiplication so $\mathbb{C} G$ can be seen as an algebra of linear operators on the Hilbert space $\mathbb{C} G$ and thus a $\mathrm{C}^*$-algebra with the operator norm.

Note that $\mathbb{C} G$ can be identified with the algebraic dual of $F(G)$ via

$\displaystyle\delta^g(\delta_s)=\delta_{g,s},$

and as $G$ is finite dimensional:

$\displaystyle F(G)^*=\mathbb{C} G\text{ and }\mathbb{C} G^*=F(G).$

## Quantising Finite Groups

In this section the category theory approach is taken to the category of finite groups. A group is an object in $\mathbf{FinSet}$ together with morphisms $m:G\times G\rightarrow G$, $e:\{\bullet\}\rightarrow G$ and $^{-1}:G\rightarrow G$ that satisfy:

$\displaystyle m\circ(I_G\circ m)=m\circ (m\times I_G)$
$\displaystyle m\circ (e\times I_G)=I_G=m\circ(I_G\times e)$
$\displaystyle m\circ(I_G\times{}^{-1})\circ \Delta_G= e\circ \varepsilon_G=m\circ({}^{-1}\times I_G)\circ \Delta_G.$

The second commutative diagram invokes the isomorphism $\{\bullet\}\times G\cong G\cong G\times\{\bullet\}$ while the third uses the maps $\Delta_G:G\rightarrow G\times G$, $g\mapsto (g,g)$ and $\varepsilon_G:G\rightarrow \{\bullet\}$.

Now apply the covariant $\mathbb{C}$ functor to $G$, the three morphisms and these three commutative diagrams. Firstly the image of $G$ is $\mathbb{C} G$. The image of the group multiplication is the linear multiplication $\nabla:\mathbb{C} G\otimes \mathbb{C} G\rightarrow \mathbb{C} G$:

$\displaystyle (\mathbb{C} m)(\delta^s\otimes \delta^t)=\delta^{m(s,t)}=\delta^{st}=\nabla(\delta^s\otimes \delta^t).$

Note that $\mathbb{C}\{\bullet\}\cong \mathbb{C}$ and so $(\mathbb{C} e):\mathbb{C}\rightarrow \mathbb{C} G$:

$\displaystyle(\mathbb{C} e)(1)\cong (\mathbb{C} e)(\delta^{\bullet})=\delta^{e(\bullet)}=\delta^e.$

Note that $\delta^e$ is the unit of $\mathbb{C} G$ and so denote by $\eta_{\mathbb{C} G}:=\mathbb{C} e$ the unit map. The image of ${}^{-1}$ is the linear map $\text{inv}:\mathbb{C} G\rightarrow \mathbb{C} G$, $\delta^g\mapsto \delta^{g^{-1}}$. Note

$\displaystyle (\mathbb{C} \Delta_G)(\delta^g)=\delta^{\Delta_G g}=\delta^{(g,g)}\cong \delta^g\otimes\delta^g,$

and denote $\mathbb{C}\Delta_{G}=:\Delta_{\mathbb{C} G}$. Finally

$\displaystyle(\mathbb{C} \varepsilon_G)(\delta^g)=\delta^{\varepsilon_G(g)}=\delta^{\bullet}\cong 1,$

and denote $\mathbb{C} \varepsilon_G=:\varepsilon_{\mathbb{C} G}$.

The image of the commutative diagrams above are therefore given by:

$\displaystyle \nabla\circ(I_{\mathbb{C} G}\circ \nabla)=\nabla\circ (\nabla\otimes I_{\mathbb{C} G})$
$\displaystyle \nabla\circ (\eta_{\mathbb{C} G}\otimes I_{\mathbb{C} G})=I_{\mathbb{C} G}=\nabla\circ(I_{\mathbb{C} G}\otimes \eta_{\mathbb{C} G})$
$\displaystyle \nabla\circ(I_{\mathbb{C} G}\otimes\text{inv})\circ \Delta_{\mathbb{C} G}= \eta_{\mathbb{C} G}\circ \varepsilon_{\mathbb{C} G}=\nabla\circ(\text{inv}\otimes I_{\mathbb{C} G})\circ \Delta_{\mathbb{C} G}.$

Indeed, the first two commutative diagrams here show that $\mathbb{C} G$ together with $\nabla$ and $\eta_{\mathbb{C} G}$ is an algebra.

To fully quantise the group, the contravariant dual functor must be applied to $\mathbb{C} G$, the morphisms and the commutative diagrams. First note that $(\mathbb{C} G)^*=F(G)$. The multiplication $\nabla:\mathbb{C} G\otimes \mathbb{C} G\rightarrow \mathbb{C} G$ has a dual:

$\displaystyle\nabla^*:(\mathbb{C} G)^*=F(G)\rightarrow (\mathbb{C} G\otimes\mathbb{C} G)^*\cong (\mathbb{C} G)^*\otimes(\mathbb{C} G)^*=F(G)\otimes F(G)\cong F(G\times G),$

where the last isomorphism can be seen as a consequence of $\mathbb{C}(X\times Y)^*=F(X\times Y)$. Embed the group $G$ in the group ring $\mathbb{C} G$ via $g\hookrightarrow\delta^g$ and consider for $f\in F(G)$:

$\displaystyle\nabla^*f(g,h)\hookrightarrow\nabla^*f(\delta^g\otimes\delta^h)=f\circ \nabla(\delta^g\otimes\delta^h)=f(\delta^{gh})=f(gh).$

This map $\nabla^*=:\Delta:F(G)\rightarrow F(G)\otimes F(G)$, $\Delta f(g,h)=f(gh)$, is the comultiplication on $F(G)$. Note that $\Delta(\delta_{g})$ is the indicator function on $m^{-1}(g)$ so that after the identification $F(G\times G)\cong F(G)\otimes F(G)$,

$\displaystyle\Delta(\delta_{g})=\sum_{t\in G}\delta_{gt^{-1}}\otimes\delta_{t}.$

Now consider the unit map $\eta_{\mathbb{C} G}:\mathbb{C}\rightarrow \mathbb{C} G\cong \mathbb{C} \otimes\mathbb{C} G$, $\delta^{\bullet}\cong 1\mapsto \delta^e\cong 1\otimes \delta^e$. The dual of this map is

$\displaystyle\eta_{\mathbb{C} G}^*:(\mathbb{C}\otimes \mathbb{C} G)^*\cong \mathbb{C}^*\otimes \mathbb{C} G^*\cong \mathbb{C}\otimes F(G)\cong F(G)\rightarrow \mathbb{C}^*\cong \mathbb{C}.$

Consider an element $f\in F(G)$:

$\displaystyle\eta_{\mathbb{C} G}^*(f)(\delta^{\bullet})$

$\displaystyle =f\circ \eta_{\mathbb{C} G}(\delta^{\bullet})=f(\delta^e)\hookleftarrow f(e),$

so that $\eta_{\mathbb{C} G}^*(f)=f(e)$. This map $\eta_{\mathbb{C} G}^*=:\varepsilon$ is called the counit.

The inverse map $\text{inv}:\mathbb{C} G\rightarrow \mathbb{C} G$ has a dual $\text{inv}^*:F(G)\rightarrow F(G)$ which (via the embedding) is given by:

$\displaystyle\text{inv}^*(f)(g)\hookrightarrow \text{inv}^*(f)(\delta^g)=f\circ\text{inv}(\delta^g)=f(\delta^{g^{-1}})=f(g^{-1}).$

This map $\text{inv}^*=:S$ is called the antipode.

These are the most important dualisations of maps but there are two more namely $\Delta_{\mathbb{C} G}$ and $\varepsilon_{\mathbb{C} G}$. Note that $\Delta_{\mathbb{C} G}(\delta^g)=\delta^g\otimes\delta^g$ maps from $\mathbb{C} G$ to $\mathbb{C} G\otimes\mathbb{C} G$ so that

$\displaystyle\Delta_{\mathbb{C} G}^*:(\mathbb{C} G\otimes \mathbb{C} G)^*\cong F(G)\otimes F(G)\rightarrow F(G).$

Let $f,\,g\in F(G)$ and $\delta^s\hookleftarrow s\in G$:

$\displaystyle\Delta^*_{\mathbb{C} G}(f\otimes g)(\delta^s)=(f\otimes g)\Delta_{\mathbb{C} G}(\delta^s)=(f\otimes g)(\delta^s\otimes \delta^s)=f(\delta^s)g(\delta^s)\hookleftarrow f(s)g(s),$

so that $\Delta_{\mathbb{C} G}^*=:M$ is just the pointwise multiplication on $F(G)$. Finally consider the map $\varepsilon_{\mathbb{C} G}:\mathbb{C} G\rightarrow \mathbb{C}$, $\delta^g\mapsto 1$. Its dual $\varepsilon_{\mathbb{C} G}^*:\mathbb{C}\rightarrow F(G)$ is the unit map of $F(G)$ ($\mathbf{1}_G=\sum_{t}\delta_t$ is the unit of the algebra $F(G)$) as can be seen by taking any $\delta^s\hookleftarrow s\in G$:

$\displaystyle\varepsilon_{\mathbb{C} G}^*(\lambda)(\delta^s)=\lambda\circ\varepsilon_{\mathbb{C} G}(\delta_s)=\lambda\cdot 1=\lambda,$

that is $\varepsilon_{\mathbb{C} G}(\lambda)=\lambda \cdot \mathbf{1}_G$, i.e. $\varepsilon_{\mathbb{C} G}^*=\eta_{F(G)}$.

Now that the morphisms have been identified:

$\displaystyle \mathcal{Q}(m)=\Delta\qquad;\qquad \delta_g\mapsto \mathbf{1}_{m^{-1}(g)}$
$\displaystyle \mathcal{Q}(e)=\varepsilon \qquad;\qquad \delta_g\mapsto \delta_{g,e}$
$\displaystyle \mathcal{Q}({}^{-1})=S\qquad;\qquad \delta_g\mapsto \delta_{g^{-1}}$
$\displaystyle \mathcal{Q}(\Delta_G)=M\qquad;\qquad f\otimes g\mapsto fg$
$\displaystyle \mathcal{Q}(\varepsilon_G)=\eta_{F(G)}\qquad;\qquad \lambda\mapsto \lambda \cdot \mathbf{1}_G,$

applying the dual functor to the last set of commutative diagrams gives coassociativity, the counital property and the antipodal property:

$\displaystyle (\Delta\otimes I_{F(G)})\circ \Delta=(I_{F(G)}\otimes \Delta )\circ \Delta$
$\displaystyle (\varepsilon\otimes I_{F(G)})\circ \Delta\cong I_{F(G)}\cong (I_{F(G)}\otimes \varepsilon)\circ \Delta$
$\displaystyle M\circ (S\otimes I_{F(G)})\circ \Delta=\eta_{F(G)}\circ \varepsilon=M\circ (I_{F(G)}\otimes S)\circ \Delta$

The first two commutative diagrams here show that $F(G)$ together with $\Delta$ and $\varepsilon$ is a coalgebra.

Now take an object $H$ in the category of finite vector spaces with morphisms that satisfy these ‘quantised’ axioms. Such an object is a finite Hopf-algebra and if it is non-commutative:

$\displaystyle M_{H}(a\otimes b)\neq M_H(b\otimes a),$

then it can be considered (if we make a few extra assumptions) the algebra of functions on a quantum group.

The preceding quantisation gives, more or less, the correct definition of a finite quantum group. To be more precise, we need to say quite a little more.