On the ${\ell }^2$-Betti numbers of universal quantum groups

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.

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

$$\begin{aligned} \begin{array}{lr} \alpha ^*\alpha + \gamma ^*\gamma = 1&{} \qquad \qquad \alpha \gamma + \gamma \alpha = 0\\ \alpha \alpha ^* + \gamma \gamma ^*=1&{} \qquad \qquad \alpha \gamma ^*+ \gamma ^*\alpha = 0\\ \gamma \gamma ^*-\gamma ^*\gamma = 0&{} \qquad \qquad \\ \end{array} \end{aligned}$$

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

$$\begin{aligned} {\mathscr {B}}:=\{\alpha ^i \gamma ^{j}\gamma ^{*k} \mid i\in {\mathbb Z},j,k\in {\mathbb N}_0\} \end{aligned}$$

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:

$$\begin{aligned} \deg _\alpha (x)&:={{\text {max}}}\{i\in {\mathbb Z}\mid \exists j,k\in {\mathbb N}_0: \lambda _{ijk}\ne 0 \};\\ \deg _{\gamma ,\gamma ^*}(x)&:={{\text {max}}}\{p\in {\mathbb N}\mid \exists i\in {\mathbb Z}, k,l\in {\mathbb N}_0 : \lambda _{ijk}\ne 0 \ \text {and} \ p=j+k\}. \end{aligned}$$

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

$$\begin{aligned} \alpha ^i \alpha ^j=\underbrace{\alpha \cdots \alpha }_{i}\underbrace{\alpha ^*\cdots \alpha ^*}_{-j}=\alpha ^{i+j} (\alpha \alpha ^*)^{-j}=\alpha ^{i+j}(1-\gamma \gamma ^*)^{-j}, \end{aligned}$$

and, similarly, if $i\leqslant -j$ we get

$$\begin{aligned} \alpha ^i \alpha ^j=\underbrace{\alpha \cdots \alpha }_{i}\underbrace{\alpha ^*\cdots \alpha ^*}_{-j}=\alpha ^{*(-i-j)} (\alpha \alpha ^*)^{i}=\alpha ^{i+j}(1-\gamma \gamma ^*)^{i}. \end{aligned}$$

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

$$\begin{aligned} \sum _{i,j,k}\lambda _{ijk}\alpha ^i\gamma ^j\gamma ^{*k}=\sum _{i,j,k}\lambda _{ijk}\alpha ^i\gamma ^{j+1}\gamma ^{*(k+1)}. \end{aligned}$$

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

$$\begin{aligned} 0=\sum _{\begin{array}{c} i,j,k\\ l,m,n \end{array}}\lambda _{ijk} \mu _{lmn} \alpha ^i \gamma ^j\gamma ^{*k}\alpha ^l\gamma ^m\gamma ^{*n} \end{aligned}$$

and hence

$$\begin{aligned}&\sum _{\begin{array}{c} i \leqslant i_0, l \leqslant l_0 \\ i + l < i_0 + l_0 \\ j,k,m,n \end{array}} \lambda _{ijk} \mu _{lmn} \alpha ^i \gamma ^j\gamma ^{*k}\alpha ^l\gamma ^m\gamma ^{*n} \nonumber \\&\quad = -\sum _{\begin{array}{c} j,k\\ m,n \end{array}} \lambda _{i_0jk}\mu _{l_0mn} \alpha ^{i_0} \gamma ^j\gamma ^{*k}\alpha ^{l_0}\gamma ^m\gamma ^{*n} \nonumber \\&\quad = -\sum _{\begin{array}{c} j,k\\ m,n \end{array}} \lambda _{i_0jk} \mu _{l_0mn}(-1)^{(j+k)|l_0|} \alpha ^{i_0}\alpha ^{l_0} \gamma ^j\gamma ^{*k}\gamma ^m\gamma ^{*n}\nonumber \\&\quad = -\sum _{\begin{array}{c} j,k\\ m,n \end{array}} \lambda _{i_0jk} \mu _{l_0mn}(-1)^{(j+k)|l_0|} \alpha ^{i_0}\alpha ^{l_0} \gamma ^{j+m}\gamma ^{*(k+n)} \text {.} \end{aligned}$$

(5)

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

$$\begin{aligned} \deg _\alpha \left( \sum _{\begin{array}{c} i<i_0,l<l_0\\ j,k,m,n \end{array}}\lambda _{ijk} \mu _{lmn} \alpha ^i \gamma ^j\gamma ^{*k}\alpha ^l\gamma ^m\gamma ^{*n} \right) <i_0+l_0 \end{aligned}$$

and hence the equality (5) can only happen if $p(\gamma ,\gamma ^*)=0$. We therefore have

$$\begin{aligned} 0&=\sum _{\begin{array}{c} j,k\\ m,n \end{array}}\mu _{l_0mn}\lambda _{i_0jk}(-1)^{(j+k)|l_0|} \alpha ^{i_0}\alpha ^{l_0} \gamma ^{j+m}\gamma ^{*(k+n)}\\&= \alpha ^{i_0}\alpha ^{l_0}\left( \sum _{\begin{array}{c} j,k\\ m,n \end{array}}\mu _{l_0mn}\lambda _{i_0jk}(-1)^{(j+k)|l_0|}\gamma ^{j+m}\gamma ^{*(k+n)}\right) \end{aligned}$$

and, by Claim 2, this implies that

$$\begin{aligned} 0&=\sum _{\begin{array}{c} j,k\\ m,n \end{array}}\mu _{l_0mn}\lambda _{i_0jk}(-1)^{(j+k)|l_0|} \gamma ^{j+m}\gamma ^{*(k+n)} \\&=\left( \sum _{j,k} \lambda _{i_0,j,k} (-1)^{(j+k)|l_0|}\gamma ^j\gamma ^{*k} \right) \left( \sum _{m,n} \mu _{l_0mn} \gamma ^m\gamma ^{*n}\right) \end{aligned}$$

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 $\mu _{l_0,n,m}\ne 0$. $\square$