Abstract
We show that the first \({\ell }^2\)-Betti number of the duals of the free unitary quantum groups is one, and that all \({\ell }^2\)-Betti numbers vanish for the duals of the quantum automorphism groups of full matrix algebras.
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Acknowledgements
The authors thank Julien Bichon for pointing out the reference [10], which provides a reference for the result shown in the “Appendix”.
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Communicated by Thomas Schick.
D. K. gratefully acknowledges the financial support from the Villum foundation (Grant 7423). Sven Raum’s research leading to these results has received funding from the People Programme (Marie Curie Actions) of the European Union’s Seventh Framework Programme (FP7/2007-2013) under REA Grant Agreement No. (622322).
Appendix
Appendix
In this section we provide a proof of the fact that \({{\text {Pol}}}(O_2^+)\) is a domain; i.e. that it has no non-trivial zero-divisors. This fact can be deduced from [10, Chapter I.1] using the well known identification of \({{\text {Pol}}}(O_2^+)\) with \({{\text {Pol}}}(SU_{-1}(2))\) (cf. [2, Proposition 5 & 6]) and the fact that the underlying rings of \({{\text {Pol}}}(SU_{-1}(2))\) and of \({{\text {Pol}}}(SL_{-1}(2))\) are isomorphic. For the benefit of the reader, we give a short proof in operator algebraic terminology, only using the identification \({{\text {Pol}}}(O_2^+) \cong {{\text {Pol}}}(SU_{-1}(2))\).
We denote by \(\alpha \) and \(\gamma \) the canonical generators of \({{\text {Pol}}}(SU_{-1}(2))\), and recall [44] that the defining relations are
Note that this implies that \(\alpha ^*\alpha =\alpha \alpha ^*\) and that \(\gamma \gamma ^*\) is a central element. In the following we use the convention that \(\alpha ^{i}=(\alpha ^*)^{-i}\) for \(i<0\) and \(x^0=1\) for \(x\ne 0\). By [44] the set
constitutes a linear basis for \({{\text {Pol}}}(SU_{-1}(2))\). For a non-zero element \(x{=}\sum \lambda _{ijk}\alpha ^i \gamma ^{j}\gamma ^{*k} \in {{\text {Pol}}}(SU_{-1}(2))\) we define its degrees with respect to the basis:
Proposition 3.1
The ring \({{\text {Pol}}}(SU_{-1}(2))\) is a domain.
Proof
We first prove the following claim:
Claim 1
For \(i,j\in {\mathbb Z}\) there exists a polynomial \(p_{i,j}\in {\mathbb C}[X,Y]\) such that \(\alpha ^i\alpha ^j= \alpha ^{i+j}p_{i,j}(\gamma ,\gamma ^*)\)
Proof of Claim 1
When i and j have the same sign this is clear — the constant polynomial 1 does the job. If \(i\geqslant 0\), \(j<0\) and \(i>|j|\) then
and, similarly, if \(i\leqslant -j\) we get
The remaining case (\(i<0\) and \(j\geqslant 0\)) follows by symmetry. \(\square \)
Claim 2
The element \(\alpha ^i\) is not a left zero-divisor for any \(i\in {\mathbb Z}\).
Proof of Claim 2
Since \(\alpha ^*\alpha =\alpha \alpha ^*\), it suffices to prove that \(\alpha \alpha ^*\) (and hence none of its powers) is a left zero-divisor. To this end, assume that \(x\in {{\text {Pol}}}(SU_{-1}(2))\) satisfies \(\alpha \alpha ^* x=0\). Since \(\alpha \alpha ^*=1-\gamma \gamma ^*\), this means \(x=\gamma \gamma ^*x\) and by expanding x as \(x=\sum _{i,j,k}\lambda _{ijk}\alpha ^i\gamma ^j\gamma ^{*k}\) and using that \(\gamma \gamma ^*\) is central this translates into
If these terms were non-zero, we could apply \(\deg _{\gamma , \gamma ^*}(-)\) on both sides to obtain a contradiction. Hence \(x=0\). \(\square \)
We now turn to the actual proof of the fact that \({{\text {Pol}}}(SU_{-1}(2))\) is a domain. Let \(x=\sum _{ijk}\lambda _{ijk} \alpha ^i \gamma ^j\gamma ^{*k}\) and \(y=\sum _{lmn}\mu _{lmn}\alpha ^l\gamma ^m\gamma ^{*n} \) in \({{\text {Pol}}}(SU_{-1}(2))\) be non-zero elements and denote their \(\alpha \)-degrees by \(i_0\) and \(l_0\), respectively. Assuming that \(xy=0\), we have
and hence
Applying Claim 1, we see that the last term has the form \(\alpha ^{i_0+l_0}p(\gamma ,\gamma ^*)\) for some \(p\in {\mathbb C}[X,Y]\) and hence its \(\alpha \)-degree is \(i_0+l_0\) if \(p(\gamma ^*,\gamma )\ne 0\). Similarly, we get that
and hence the equality (5) can only happen if \(p(\gamma ,\gamma ^*)=0\). We therefore have
and, by Claim 2, this implies that
However, since \({\mathscr {B}}\) is a linear basis for \({{\text {Pol}}}(SU_{-1}(2))\), the map \({\mathbb C}[X,Y]\ni p\mapsto p(\gamma ,\gamma ^*)\in {{\text {Pol}}}(SU_{-1}(2))\) is injective, and since \({\mathbb C}[X,Y]\) is a domain one of the factors in the last product needs to be zero, which contradicts the fact that \(i_0\) and \(l_0\) are chosen such that there exist \(j,k,l,m\in {\mathbb N}_0\) with \(\lambda _{i_0jk}\ne 0\) and \(\mu _{l_0,n,m}\ne 0\). \(\square \)
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Kyed, D., Raum, S. On the \({\ell }^2\)-Betti numbers of universal quantum groups. Math. Ann. 369, 957–975 (2017). https://doi.org/10.1007/s00208-017-1531-5
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DOI: https://doi.org/10.1007/s00208-017-1531-5