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Let $G$ be a Lie group and consider the space $C_c^\infty(G)$ of compactly supported complex-valued smooth functions on $G$ and $D'(G) = (C_c^\infty(G))'$ the topological dual linear space of $C_c^\infty(G)$.

One can imbue the space with a convolution algebra structure. In this algebra, one can find copies of $U(\mathfrak{g})$, the universal enveloping algebra of the Lie algebra $\mathfrak{g}$ of $G$ ($U(\mathfrak{g})$ is isomorphic to the subalgebra of distributions supported at the identity), as well as the group itself (as the Dirac distributions $\delta_g, g \in G$. Also, you can lift representations of the group $G$ to $D'(G)$ in such a way that it respects all of the above structure (c.f. Cohomological induction and Unitary representations, page 47, by Knapp and Vogan). There is even more structure to this algebra...

My question is: Is there a way of quantizing this algebra in such a way that $ U(\mathfrak{g})$ becomes $U_q(\mathfrak{g})$, the quantized universal enveloping algebra, and keep most of the rest of the structure of $D'(G)$?

I`m looking mostly for literature on the subject, if it exists.

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    $\begingroup$ Apparently, you mean a compact group $G$, because otherwise $D'(G)$ is not closed under convolution. $\endgroup$
    – Sergei Akbarov
    Commented May 3, 2017 at 20:38
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    $\begingroup$ As in @SergeiAkbarov's comment: you'd almost surely want only compactly-supported distributions, since otherwise there're problems with the convolution (e.g., non-associativity). Compactly-supported is enough to capture both the action of the universal enveloping algebra, and the action of the/a maximal compact subgroup. But this doesn't address the question... $\endgroup$
    – paul garrett
    Commented May 3, 2017 at 22:19
  • $\begingroup$ The space ${\mathcal E}'(G)$ of compactly supported distributions is a group algebra, in contrast to the space ${\mathcal D}'(G)$ of all distributions, see link.springer.com/article/10.1007%2FBF02674336 $\endgroup$
    – Sergei Akbarov
    Commented May 4, 2017 at 5:52
  • $\begingroup$ Well, to be honest this compact ~ non-compact thing is still something I'm trying to grasp (I'm just a PhD student struggling) But I found these papers : arxiv.org/pdf/math/0307277.pdf - Topological Hopf algebras, quantum groups, and deformation quantization (which is not exactly the kind of quantization I'm looking for) and projecteuclid.org/euclid.cmp/1104269794 - The hidden group structure of quantum groups: strong duality, rigidity and preferred deformations (which is still deform. quantization but it touches upon the quantization I'm looking for U(g) -> U_q(g). $\endgroup$
    – Henrique Tyrrell
    Commented May 4, 2017 at 12:14
  • $\begingroup$ Henrique, you can just replace ${\mathcal D}'(G)$ by ${\mathcal E}'(G)$ (the dual space to the space ${\mathcal E}(G)=C^\infty(G)$), and this problem will disappear. The proof that ${\mathcal E}'(G)$ is a group algebra is here: link.springer.com/article/10.1023%2FA%3A1020929201133 And the proof that ${\mathcal E}'(G)$ is a Hopf algebra is the same as for ${\mathcal O}'(G)$ and can be seen here: link.springer.com/article/10.1007%2Fs10958-009-9646-1 $\endgroup$
    – Sergei Akbarov
    Commented May 4, 2017 at 15:56

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More a long comment than an answer. Going back to the situation in $\mathbb R^n$, it is not difficult to actually quantize tempered distributions. You want to give a meaning to $$ \iint e^{2π i x\cdot \xi} a(x,\xi)\hat u(\xi) \bar v(x) dx d\xi,\tag{$\ast$} $$ for $u,v\in \mathscr S(\mathbb R^n)$ and $a\in \mathscr S'(\mathbb R^{2n})$. You notice that $ \Omega(u,v)(x,\xi)=e^{2π i x\cdot \xi} \hat u(\xi) \bar v(x) $ defines a function $\Omega(u,v)\in \mathscr S(\mathbb R^{2n})$ and you may thus define $(\ast)$ as the bracket of duality $$ \langle a,\Omega(u,v)\rangle_{\mathscr S'(\mathbb R^{2n}),\mathscr S(\mathbb R^{2n})}=\langle \text{Op}(a) u, v\rangle_{\mathscr S'(\mathbb R^{n}),\mathscr S(\mathbb R^{n})}. $$ Mutatis mutandis, you can do that for the Weyl quantization as well.

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