Are the Drinfeld compact quantum groups simply connected ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:14:58Z http://mathoverflow.net/feeds/question/84694 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84694/are-the-drinfeld-compact-quantum-groups-simply-connected Are the Drinfeld compact quantum groups simply connected ? Amin 2012-01-01T19:22:01Z 2012-01-03T21:17:51Z <p>To fix notations : let G be simply connected simple compact group, and $U_q(\mathcal{G})$ the Drinfeld-Jimbo universal algebra quantization of its complexified algebra defined as usual, with q not root of unity. Recall that it's a *-algebra, and I denote $G_q$ the unital $C^*$-algebra dual to it as usual (thus completion of matrix coefficient of representations...).<br> My question is very naive and loose, so all suggestions are welcome. Is there a general notion of fundamental group (e.g. using like étale), such that $G_q$ has trivial fundamental group?</p> http://mathoverflow.net/questions/84694/are-the-drinfeld-compact-quantum-groups-simply-connected/84702#84702 Answer by unknown (google) for Are the Drinfeld compact quantum groups simply connected ? unknown (google) 2012-01-01T22:13:56Z 2012-01-01T22:37:04Z <p>The group $G$ being simply connected is not a notion that one can detect from the usual universal enveloping algebra $U(\mathfrak g)$, so one shouldn't expect there to be a notion for quantum groups $U_q(\mathfrak g)$ corresponding to $G$ being simply connected. You should think of $U_q(\mathfrak g)$ as just giving the local structure around the identity, and so global notions like being simply connected don't make sense (for exactly the same reason that $G$ and $G/N$ have the same Lie algebra for any finite normal subgroup $N\lhd G$).</p> <p>An "intrinsic" perspective is as follows. Let <code>$\hat{\mathcal O}_{G,e}$</code> be the complete local ring of $G$ at the identity $e\in G$. Then <code>$U(\mathfrak g)$</code> is a certain dual of <code>$\hat{\mathcal O}_{G,e}$</code> (it's the direct limit of <code>$(\hat{\mathcal O}_{G,e}/\mathfrak m_{G,e}^n)^\ast$</code>). Thus $U(\mathfrak g)$ only sees the local group structure in a formal neighborhood of $e\in G$.</p> <p>There may indeed, however, be a natural notion of being simply connected if you look at the models of quantum groups like $\mathbb Z[q,q^{-1}]\langle a,b,c,d\rangle$ modulo the relations $ab=q^{-1}ba$, $ac=q^{-1}ca$, $cd=q^{-1}dc$, $bd=q^{-1}db$, $bc=cb$, $ad-da=(q^{-1}-q)bc$, and $ad-q^{-1}bc=1$ (corresponding to the classical group $\operatorname{SL}(2)$).</p> http://mathoverflow.net/questions/84694/are-the-drinfeld-compact-quantum-groups-simply-connected/84837#84837 Answer by MTS for Are the Drinfeld compact quantum groups simply connected ? MTS 2012-01-03T21:17:51Z 2012-01-03T21:17:51Z <p>This is a negative answer to your question, or at least a partially negative one. I don't know if you've thought about things in this way, but there is a naive way to formulate the idea of the fundamental group of a space in terms of its $C^*$-algebra of continuous functions. The problem is that, at least according to my quick calculations, the noncommutative space associated to the quantized function algebra $\mathcal{O}_q(SU_2)$ does not have trivial fundamental group. This may mean that this notion of fundamental group doesn't make much sense for noncommutative spaces, or that quantum phenomena arise which just make these quantum groups not simply connected, or something else. My suspicion is that the first is true, my reasoning being that if there was a good notion of fundamental group for noncommutative spaces, we would have heard of it already. Perhaps somebody can chime in on that?</p> <p>Anyway, let me tell you what I've been thinking about. The fundamental group of a space $X$ is defined as homotopy classes of based continuous maps $S^1 \to X$. Continuous maps $S^1 \to X$ are in bijection with unital $<em>$-homomorphisms $C(X) \to C(S^1)$, at least for $X$ compact. So for a compact noncommutative space with corresponding unital <code>$C^*$</code>-algebra $A$, loops should be defined as morphisms (of unital $C^</em>$-algebras) $A \to C(S^1)$.</p> <p>I believe that the correct notion of homotopy of morphisms is the following: two morphisms $\Phi_0, \Phi_1 : A \to C(S^1)$ are homotopic if there is a morphism $\Psi : A \to C([0,1]) \otimes C(S^1)$ such that $\mathrm{ev}_0 \circ \Psi = \Phi_0$ and $\mathrm{ev}_1 \circ \Psi = \Phi_1$. This is spelled out in Chapter 1 of Elements of Noncommutative Geometry, by Varilly, Gracia-Bondia, and Figuera.</p> <p>To say that a loop takes basepoint to basepoint is a little trickier for noncommutative spaces. However, for the quantum groups this is ok because there is still one classical point, namely the identity element, which corresponds to the counit $\varepsilon : \mathcal{O}_q(G) \to \mathbb{C}$. For a loop/morphism $\Phi : \mathcal{O}_q(G) \to C(S^1)$, then, we can say that $\Phi$ preserves basepoints if $\Phi(x)(1) = \varepsilon(x)$ for all $x \in \mathcal{O}_q(G)$. Or in more fancy terms we could say that $\mathrm{ev}_1 \circ \Phi = \varepsilon$.</p> <p>Finally, we have to define the correct analogue of the constant loop. Again this may be difficult in general, but for the quantum groups we have the counit, so the constant loop is the morphism $x \mapsto \varepsilon(x)1$ of $\mathcal{O}_q(G)$ to $C(S^1)$.</p> <p>So this gives us most of the elements to define the fundamental group of a noncommutative space with associated $C^*$-algebra $A$: we take based morphisms $A \to C(S^1)$ modulo homotopy, and say that a loop is trivial if it is homotopic (through based morphisms) to the constant loop. I haven't said how you're supposed to compose loops, but this at least defines the fundamental group as a set.</p> <hr> <p>Now the bad news: for $\mathcal{O}_q(SU_2)$, this doesn't give the right answer. For the sake of completeness, $\mathcal{O}_q(SU_2)$ (for $0 &lt; q &lt; 1$, say) is the unital $C^*$-algebra generated by elements $a,b$ with relations <code>$$a^*a + b^*b =1, \quad aa^* + q^2 bb^* = 1,$$</code></p> <p><code>$$bb^* = b^*b, \quad ab = qba, \quad ab^* = qb^*a.$$</code> The first two relations show that if $\Phi : \mathcal{O}_q(SU_2) \to A$ is a morphism to a commutative $C^*$-algebra $A$, then $(1-q^2)\Phi(b^*b) = 0$, so $\Phi(b) = 0$. (Thus we really need $q \neq \pm 1$ for this to work.)</p> <p>Now $C(S^1)$ is the universal $C^*$-algebra generated by a single unitary element $u$. This is a quantum subgroup of $\mathcal{O}_q(SU_2)$ via the map $\pi : \mathcal{O}_q(SU_2) \to C(S^1)$ defined by $$\pi(a) = u, \quad \pi(b) = 0.$$ Combining this with the preceding statement, any morphism $\Phi : \mathcal{O}_q(SU_2) \to C(S^1)$ must factor through the quotient map $\pi$, so homotopy classes of morphisms $\mathcal{O}_q(SU_2) \to C(S^1)$ are in bijection with homotopy classes of morphisms $C(S^1) \to C(S^1)$. These clases correspond to homotopy classes of maps $S^1 \to S^1$, so this means that the fundamental group is $\mathbb{Z}$, rather than the trivial group.</p> <p>Again, my guess is that this notion of fundamental group is just too rigid to make sense for all noncommutative spaces. One needs at least one classical point in order for this to make sense, which excludes any simple algebras, such as noncommutative tori. The question is interesting, though, and I would be glad to know if anybody else has thought about other notions of a fundamental group for noncommutative spaces.</p>