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This is an attempt to salvage the wreckage of my hope to motivate (finite) quantum groups a lá this question.

Let $\{S_i\}_{i=0}^n$ be a family of finite sets and let $\varphi$ be a map

$$\varphi:S_1\times S_2\times \cdots \times S_n\rightarrow S_0.$$

This map may be extended to a multilinear map

$$\varphi_1:\mathbb{C}S_1\times \cdots\times \mathbb{C}S_n\rightarrow \mathbb{C}S_0.$$

Here $\mathbb{C}S_i$ is the complex vector space with basis $\{\delta^s:s\in S_i\}$.

This map may be further extended to a linear map:

$$\tilde{\varphi_1}:\mathbb{C}S_1\otimes \cdots\otimes \mathbb{C}S_n\rightarrow \mathbb{C}S_0.$$

Finally we can take the transpose of this map to get a map:

$$\tilde{\varphi_1}^*:\mathbb{C}S_0^*\rightarrow \mathbb{C}S_1^*\otimes \cdots\otimes \mathbb{C}S_n^*.$$

I am looking for a word to describe this quantisation where a map on cartesian copies of $S_i$ is turned into a map into tensor copies of $\mathbb{C}S_i^*$. The answer should be "$\underline{\qquad}$ Quantisation".

I had hoped to call it categorical or functorial quantisation but I know now this isn't appropriate. I might be happy enough to just call it "Tensoring Quantisation" but would love a better word.

Context:

One way to generalise classical finite group concepts to the quantum setting is to use the above regime.

For example, applying this regime to multiplication, inclusion of the unit and inverses yields comultiplication, the counit and the antipode.

Translating the axioms of associativity, identity and inverses gives coassociativity, the counitary property and the antipodal property.

Similarly, applying this regime to group actions $X\times G\rightarrow X$ yields corepresentations $V\rightarrow V\otimes F(G)$. The translations of the axioms of actions gives the axioms of corepresentations.

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  • $\begingroup$ Just out of curiosity, why do you want to call this "quantisation"? $\endgroup$ – Matthias Klupsch Aug 16 '16 at 14:13
  • $\begingroup$ @MatthiasKlupsch because I am using it to motivate the definition of a quantum group; i.e. the axioms of a Hopf algebra are 'translations' of the group axioms under this 'regime'. I am trying to say that quantum groups are the quantisations of groups. $\endgroup$ – JP McCarthy Aug 16 '16 at 14:26
  • $\begingroup$ I see. I was asking because as it was possible that you had some obscure physics interpretation in mind. Note also that the notions of quantum groups and Hopf algebras are not always used synonymously. Saying that quantum groups are quantisations of groups is problematic, since Hopf algebras are a generalization of group algebras (or their duals, if you prefer that viewpoint). $\endgroup$ – Matthias Klupsch Aug 17 '16 at 6:44
  • $\begingroup$ Maybe you are looking for the notion of Hopf monoid of a symmetric monoidal category? Groups are the Hopf monoids in the category of sets w.r.t. cartesian product and Hopf algebras are the Hopf monoids in the category of vector spaces w.r.t. tensor product. Your ("quantisation") functor is monoidal so it maps Hopf monoids to Hopf monoids (actually to "Hopf comonoids", since it is contravariant, but the axioms for Hopf monoids are self-dual). $\endgroup$ – Matthias Klupsch Aug 17 '16 at 6:46
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Matthias Klupsch Aug 17 '16 at 13:02
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With help of Matthias's comments (both on this question and on the previous one), I have come to see that my problems emanate from the way I am doing the category theory. The below answers both this question and the previous.

Rather than talking about the category of finite groups (and group homomorphisms), we are better off starting in the category of finite sets where the morphisms are functions.

A finite group $G$ is an object in this category together with three morphisms $m:G\times G\rightarrow G$, $^{-1}:G\rightarrow G$ and $e:\{0\}\rightarrow G$ that satisfy three commutative diagrams expressing associativity, identity and inverses.

There is a functor from the category of finite sets to the category of finite dimensional vector spaces that sends the object $G$ to the object $\mathbb{C}G$ --- the complex vector space with basis $\{\delta^g:g\in G\}$. For this to be a morphism, the images of $m$, $^{-1}$ and $e$ under this $\mathbb{C}$ functor must be morphisms in the category of finite dimensional vector spaces i.e. they must be linear maps.

If we define: $$\begin{align} \varphi:\mathbb{C}G\times\mathbb{C}G\rightarrow \mathbb{C}G,\qquad(\delta^g,\delta^h)\mapsto \delta^{gh}, \end{align}$$ and extend bilinearly we get a bilinear map rather than a linear map. However we know there is a linear map $\nabla:\mathbb{C}G\otimes \mathbb{C}G\rightarrow \mathbb{C}G$, namely $\delta^g\otimes \delta^h\mapsto \delta^{gh}$ and we send the morphism $m$ to the morphism $\nabla$ using the $\mathbb{C}$ functor so that $\mathbb{C}m=\nabla$. In a similar way any morphisms on cartesian copies of objects are sent to morphisms on tensor products; e.g.

$$I_G\times m:G\times G\times G\rightarrow G\times G$$ is sent to $$\mathbb{C}(I_G\times m)=I_{\mathbb{C}G}\otimes \nabla:\mathbb{C}G\otimes\mathbb{C}G\otimes \mathbb{C}G\rightarrow \mathbb{C}G\otimes\mathbb{C}G.$$

The finite set morphism $^{-1}$ is defined in the obvious way and we write $\mathbb{C}(^{-1})=\operatorname{inv}$. Also $\mathbb{C}e:\mathbb{C}\{0\}\cong\mathbb{C}\rightarrow \mathbb{C}G$, $1\mapsto \delta^e$ is denoted by $\eta$.

Now the image of a commutative diagram under a functor is another commutative diagram and what we have if we look at the three commutative diagrams expressing the group axioms is:

\begin{eqnarray} \nabla\circ(\nabla\otimes I_{\mathbb{C} G})=\nabla\circ(I_{\mathbb{C} G}\otimes\nabla) \\ \nabla\circ(\eta\otimes I_{\mathbb{C} G})\cong I_{\mathbb{C} G}\cong \nabla\circ(I_{\mathbb{C} G}\otimes\eta) \\ \nabla\circ(\operatorname{inv}\otimes I_{\mathbb{C} G})\circ \Delta_{\mathbb{C} G}=\eta\circ\varepsilon_{\mathbb{C} G}=\nabla\circ(I_{\mathbb{C} G}\otimes\operatorname{inv})\circ\Delta_{\mathbb{C} G}. \end{eqnarray}

Here $\Delta_{\mathbb{C}G}:\mathbb{C}G\rightarrow \mathbb{C}G\otimes \mathbb{C}G$, $\delta^g\mapsto \delta^g\otimes \delta^g$ is the diagonal map and $\varepsilon_{\mathbb{C}G}:\mathbb{C}G\rightarrow \mathbb{C}$, $\delta^g\mapsto 1$.

Now apply the dual map $\mathcal{D}$ to $\mathbb{C}G$. The dual map is an contravariant endofunctor of the category of finite dimensional vector spaces and sends a vector space to it's dual and a linear map $T:U\rightarrow V$ to its transpose ($\varphi\in V^*$):

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

Once again this map sends linear maps to linear maps and so is a functor and the image of the commutative diagrams expressing $\mathbb{C}(\text{group axioms})$ is a commutative diagram expressing the three Hopf algebra axioms.

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

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

Note $$\mathcal{Q}(G)=F(G),\,\mathcal{Q}(m)=\Delta, \mathcal{Q}(^{-1})=S\text{ and }\mathcal{Q}(e)=\varepsilon.$$

Also the image of a commutative diagram under a functor is a commutative diagram so we have:

$$\begin{align} \mathcal{Q}(\text{associativity})&=\text{coassociativity} \\ \mathcal{Q}(\text{identity})&=\text{counitary property} \\ \mathcal{Q}(\text{inverses})&=\text{antipodal property} \end{align}$$

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

$$\Delta^*_{H}(a\otimes b)=\Delta^*_H(b\otimes a),$$

then it can be considered (if we make a few extra assumptions) the algebra of functions on a quantum group and I am simply going to call this regime by Quantisation via Category Theory.

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