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Podles and Woronowicz' construct the quantum Lorentz group, by which they mean $SL_q(2,\mathbb{C})$, as a quantum double of the compact quantum group $SU_q(2)$. More precisely, it is a bicrossed product of $K=SU_q(2)$ and its discrete quantum dual $\hat{K}$. So it seems that $SL_q(2,\mathbb{C})$ is not connected as a quantum topological space.

Main Question: Why is the $q$-deformation of the connected group $SL(2,\mathbb{C})$ not connected?


Here are some more precise questions:

  1. Given that $\hat{K}$ is supposed to be a $q$-analogue of the $AN$-component in the $KAN$-decomposition, is there any reasonable sense in which the discrete quantum dual of $SU_q(2)$ tends to the classical $AN$ group in the $q\to1$ limit?

  2. Are there alternative $q$-deformations of $SL(2,\mathbb{C})$ which remain connected?

  3. Is this strange non-connectedness related to the appearance of new "non-smooth" representations of $SL_q(2,\mathbb{C})$ in the work of Podles-Woronowicz, and Pusz?

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Can you say a little more what you mean $SL_q(2,\mathbb{C})$ not being connected? Do you mean more than the fact that there is a discrete quantum group as a factor in the bicrossed product? –  MTS Jan 25 '12 at 5:53
    
I mean precisely than that there is a discrete factor in the bicrossed product. The bicrossed product is supposed to emulate a topological decomposition of a group, so this means that topologically $SL_q(2,\mathbb{C})$ is a product of a compact quantum space and a discrete quantum space. I find this strange, since we are q-deforming a connected Lie group here. I'm looking for some kind of understanding of this phenomenon. –  Bob Yuncken Jan 25 '12 at 9:38
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1 Answer

I think that the assumption that the quantum Lorentz group is not connected is wrong, and this is due to some confusion about "duality", a word used, in this context, with different meanings. This is not special of $SL_2$.

I partly share this confusion and I hope that my message will be corrected by someone with clearer ideas than I have.

The compact group $K$ has a Pontryagin dual $\widehat{K}$. Having fixed on $K$ the standard Poisson-Lie structure, it also has a Poisson dual $K^*$ which is $AN$. the two things are patently not the same.

Now the point is that on one hand the universal enveloping Lie algebra $u(\mathfrak{k})$ may be identified with a quantization of the group $AN$ (this is called quantum duality principle) and on the other hand is related (sorry, here I'm really "handwaving") to $\widehat{K}$ (guess to the convolution algebra). The keypoint is exactly this last relation, which is the right classical analogue of the quantum case.

The relation between $AN$ and $\widehat{K}$ can be considered here a semiclassical effect.

EDIT: Say $K$ is a compact Lie group and $R(K)$ its Lie algebra of representative functions. Then $R(K)$ is isomorphic to the group algebra of its Pontryagin dual $\mathbb C[\widehat{K}]$ (this can be seen as a version of Fourier transform).

You may then say that the universal enveloping algebra $U(\mathfrak{k})$ is dual to $\mathbb C[\widehat{K}]$.

On the other hand $U(\mathfrak k)$ is a quantization of the algebra of functions on the dual of the Lie algebra, $\mathfrak k^\ast$, with respect to the linear Lie-Poisson bracket.

So the same object is related on one side to the discrete group $\widehat{K}$ and on the other side to the Poisson manifold $\mathfrak k^\ast$. Here there is no quantum group appearing, since we are assuming the trivial Poisson-Lie bracket on $K$.

In the paper cited above we are assuming the standard Poisson-Lie structure on $K=SU(2)$. This implies that the Poisson manifold $\mathfrak{su}(2)^*$ has to be replaced by $AN$ with the dual Poisson-Lie group structure and that $U(\mathfrak k)$ gets replaced by $U_q(\mathfrak k)$.

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Thanks, though I can't say I really understand yet. One thing I can say is that in the construction of $SL_q(2,\mathbb{C})$, the algebra defining the $\hat{K}$ component is a direct sum of finite dimensional matrix algebras, so it definitely has discrete spectrum. That strikes me as not connected in any interpretation. What this has to do with Poisson duality I couldn't say. Perhaps that does explain the relation between $\hat{K}$ and $AN$, but I'll need to chew on that. –  Bob Yuncken Jan 25 '12 at 14:28
    
I edited my answer adding some explanations. –  Nicola Ciccoli Jan 25 '12 at 18:00
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