The Clifford-cyclotomic group and Euler-Poincar\'e characteristics

TL;DR

This paper investigates the structure of the Clifford-cyclotomic group within unitary matrices over cyclotomic rings, characterizing when it equals the entire group and computing their Euler-Poincaré characteristics.

Contribution

It establishes conditions under which the Clifford-cyclotomic group equals the full unitary group and calculates Euler-Poincaré characteristics of related groups.

Findings

Equality holds for n=8, 12, 16, 24

Index is infinite when groups are not equal

Euler-Poincaré characteristics are computed for several groups

Abstract

For an integer $n\geq 8$ divisible by $4$, let $R_n=\mathbb{Z}[\zeta_n,1/2]$ and let $\operatorname{U}_2(R_n)$ be the group of $2\times 2$ unitary matrices with entries in $R_n$. Set $\operatorname{U}_2^\zeta(R_n)=\{\gamma\in\operatorname{U}_2(R_n)\mid \det\gamma\in\langle\zeta_n\rangle\}$. Let $\mathcal{G}_n\subseteq \operatorname{U}_2^\zeta(R_n)$ be the Clifford-cyclotomic group generated by a Hadamard matrix $H=\frac{1}{2}[\begin{smallmatrix} 1+i & 1+i\\1+i &-1-i\end{smallmatrix}]$ and the gate $T=[\begin{smallmatrix}1 & 0\\0 & \zeta_n\end{smallmatrix}]$. We prove that $\mathcal{G}_n=\operatorname{U}_2^\zeta(R_n)$ if and only if $n=8, 12, 16, 24$ and that $[\operatorname{U}_2^\zeta(R_n):\mathcal{G}_n]=\infty$ if $\operatorname{U}_2^\zeta(R_n)\neq \mathcal{G}_n$. We compute the Euler-Poincar\'{e} characteristics of the groups $\operatorname{SU}_2(R_n)$, $\operatorname{PSU}_2(R_n)$, $\operatorname{PU}_2(R_n)$, $\operatorname{PU}^\zeta_2(R_n)$, and $\operatorname{SO}_3(R_n^+)$.

Equations

$$T_{n}:=\left[\begin{array}[]{cc}1&0\\ 0&\zeta_{n}\end{array}\right]\in\operatorname{U_{2}}({\mathbb{Z}}[\zeta_{n}])\subseteq\operatorname{U_{2}}(R_{n}).$$

$$\mathcal{G}_{n}=\langle\mathcal{C},T_{n}\rangle\qquad\mbox{{(}resp., $\operatorname{S}\!\mathcal{G}_{n}=\mathcal{G}_{n}\cap\operatorname{SU_{2}}(R_{n})${)};}$$

$${\mathcal{G}}_n\le {\mathrm{U}}_2^{\zeta }(R_n)\le {\mathrm{U}}_2(R_n),G(4,n)\le \mathrm{S}{\mathrm{O}}_3(R_n^{+})$$

$$-1\mathrm{mod}d\in ⟨2\mathrm{mod}d⟩\le (\mathbb{Z}/d\mathbb{Z}{)}^{\times }$$

$$\chi (S{\mathcal{G}}_n)=\chi (G(4,n))=\chi (P{\mathcal{G}}_n)=-1/12+1/2n$$

$$H(z)=\frac{1}{2}\left[\begin{array}[]{cc}1+i&z(1+i)\\ \overline{z}(-1+i)&1-i\end{array}\right]$$

$$H:=\frac{1}{2}\left[\begin{array}[]{cc}1+i&1+i\\ 1+i&-1-i\end{array}\right]\in\mathcal{C}.$$

$${\mathcal{H}}_n:=⟨H({\zeta }_n),H({\zeta }_n^2),\dots ,H({\zeta }_n^{n-1}),H({\zeta }_n^n)=H(1)⟩\le S{\mathcal{G}}_n\le \mathrm{S}{\mathrm{U}}_2(R_n).$$

$$1\to \mathrm{S}{\mathrm{U}}_2(R_n)\to {\mathrm{U}}_2^{\zeta }(R_n)\stackrel{\det }{\to }⟨{\zeta }_n⟩\to 1,$$

$$A_1^{\ast }(B)=\left\{\left[\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right]:a^2+b^2+c^2+d^2=1;a,b,c,d\in B\right\}$$

$$1\rightarrow\mbox{$\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}.$}}\raisebox{3.7889pt}{$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}.$}\hskip 4.60007pt$}{}_{2}\rightarrow A_{1}^{*}\xrightarrow{\operatorname{{\rm Ad}}}\operatorname{SO_{3}}\rightarrow 1\,,$$

$$\sigma_{x}=\left[\begin{array}[]{ll}0&1\\ 1&0\end{array}\right],\sigma_{y}=\left[\begin{array}[]{rr}0&-i\\ i&0\end{array}\right],\sigma_{z}=\left[\begin{array}[]{cc}1&0\\ 0&-1\end{array}\right].$$

$$\frac{1}{2}\mathrm{Tr}\left({\left[\begin{array}{cc}z& x-iy\\ x+iy& -z\end{array}\right]}^2\right)=x^2+y^2+z^2,$$

$$\operatorname{{\rm Ad}}\left(\left[\begin{array}[]{cc}a+bi&c+di\\ -c+di&a-bi\end{array}\right]\right)=\left[\begin{array}[]{ccc}a^{2}-b^{2}-c^{2}+d^{2}&2ab+2cd&-2ac+2bd\\ -2ab+2cd&a^{2}-b^{2}+c^{2}-d^{2}&2ad+2bc\\ 2ac+2bd&-2ad+2bc&a^{2}+b^{2}-c^{2}-d^{2}\end{array}\right].$$

$$A_1^{\ast }(R^{+})=\mathrm{S}{\mathrm{U}}_2(R)\twoheadrightarrow \mathrm{PS}{\mathrm{U}}_2(R)\hookrightarrow \mathrm{S}{\mathrm{O}}_3(R^{+}).$$

$$\pi (g)=\mathrm{Ad}(\frac{1}{\sqrt{\det g}}g)$$

$${\mathrm{U}}_2(R)\twoheadrightarrow \mathrm{P}{\mathrm{U}}_2(R)\stackrel{\overline{\pi }}{\hookrightarrow }\mathrm{S}{\mathrm{O}}_3(R^{+}).$$

$$1\to ⟨\pm 1⟩\to A_1^{\ast }(\mathbb{R})\stackrel{\mathrm{Ad}}{\to }\mathrm{S}{\mathrm{O}}_3(\mathbb{R})\to 1$$

$$1\rightarrow\mbox{$\raisebox{-2.54025pt}{$l$}\mu\raisebox{-4.2194pt}{\scalebox{2.0}{$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}.$}}\raisebox{3.7889pt}{$\color[rgb]{1,1,1}\definecolor[named]{pgfstrokecolor}{rgb}{1,1,1}\pgfsys@color@gray@stroke{1}\pgfsys@color@gray@fill{1}.$}\hskip 4.60007pt$}{}_{2}(R^{+})=\langle\pm 1\rangle\rightarrow A_{1}^{*}(R^{+})\xrightarrow{\operatorname{{\rm Ad}}}\operatorname{SO_{3}}(R^{+})\,.$$

$$T_{q}=\left[\begin{array}[]{ccc}\frac{\sqrt{21}+3}{8}&\frac{1}{4}&\frac{-\sqrt{21}+3}{8}\\[3.25201pt] \frac{-\sqrt{21}+3}{8}&\frac{\sqrt{21}+3}{8}&\frac{1}{4}\\[3.25201pt] \frac{1}{4}&\frac{-\sqrt{21}+3}{8}&\frac{\sqrt{21}+3}{8}\end{array}\right]\in\operatorname{SO_{3}}(R^{+}).$$

$$M_{q}=\left[\begin{array}[]{cc}\frac{4+\sqrt{21}+i}{4}&\frac{1+i}{4}\\[3.25201pt] \frac{-1+i}{4}&\frac{4+\sqrt{21}-i}{4}\end{array}\right];$$

$$(1,-1,-1),(-1,1,-1),(-1,-1,1)$$

$${\varphi }_1(M)$$

$${\varphi }_2(M)$$

$${\varphi }_3(M)$$

$${\varphi }_4(M)$$

$${\theta }_{12}(M):={\theta }_{21}(M)$$

$${\theta }_{13}(M):={\theta }_{31}(M)$$

$${\theta }_{14}(M):={\theta }_{41}(M)$$

$${\theta }_{34}(M):={\theta }_{43}(M)$$

$${\theta }_{24}(M):={\theta }_{42}(M)$$

$${\theta }_{23}(M):={\theta }_{32}(M)$$

$${\mathrm{Sel}}_2^{+}(\mathsf{R}):=\{x\in F_{+}^{\times }\mid {\mathrm{val}}_{\mathfrak{p}}x\equiv 0(\mathrm{mod}2)\text{ for every finite prime p of R}\}/(F^{\times }{)}^2.$$

$${\mathsf{R}}^{\times }\stackrel{2}{\to }{\mathsf{R}}_{+}^{\times }\to {\mathrm{Sel}}_2^{+}(\mathsf{R})\to \mathrm{Cl}(\mathsf{R})\stackrel{2}{\to }\mathrm{Cl}(\mathsf{R}).$$

Full text

1. Introduction

Let $\operatorname{U_{2}}=\{g\in\operatorname{GL_{2}}(\mathbb{C})\mid gg^{\dagger}=1\}$ be the group of $2\times 2$ unitary matrices stabilizing the standard hermitian form on $\mathbb{C}^{2}$ with $\dagger$ denoting conjugate-transpose. Let $\operatorname{U_{2}^{\zeta}}$ and $\operatorname{SU_{2}}$ be its subgroups of matrices whose determinants are roots of unity or 1 respectively. For a subring $R\subseteq\mathbb{C}$, write $\operatorname{U_{2}}(R):=\operatorname{U_{2}}\cap\operatorname{GL_{2}}(R)$ for the subgroup of $\operatorname{U_{2}}$ whose matrix entries lie in $R$; similarly $\operatorname{U_{2}^{\zeta}}(R):=\operatorname{U_{2}^{\zeta}}\cap\operatorname{GL_{2}}(R)$, and $\operatorname{SU_{2}}(R):=\operatorname{SU_{2}}\cap\operatorname{\operatorname{SL}_{2}}(R)$. Let $\operatorname{SO_{3}}=\{g\in\operatorname{SL_{3}}({\mathbb{R}})\mid gg^{t}=1\}$. For a subring $R^{+}\subseteq{\mathbb{R}}$, write $\operatorname{SO_{3}}(R^{+})$ for the subgroup of $\operatorname{SO_{3}}$ whose entries lie in $R^{+}$.

Throughout this paper $n=2^{s}d$ is a positive integer with $d$ odd. Unless explicitly stated otherwise, we assume $s\geq 2$. Let $\zeta_{n}:=e^{2\pi i/n}$, $K_{n}:={\mathbb{Q}}(\zeta_{n})$, and $R_{n}={\mathbb{Z}}[\zeta_{n},1/2]$. Set $F_{n}=K_{n}^{+}:={\mathbb{Q}}(\zeta_{n}+\overline{\zeta_{n}})$ and $R_{n}^{+}={\mathbb{Z}}[\zeta_{n}+\overline{\zeta}_{n},1/2]$. Then $i\in R_{n}$ and $R_{n}=R_{n}^{+}\oplus R_{n}^{+}i$ since $1/2\in R_{n}^{+}$. The Clifford group $\mathcal{C}$ can be defined as $\mathcal{C}=\operatorname{U_{2}}(R_{4})$ [fgkm, Section 2.1]. Set

[TABLE]

Define the Clifford-cyclotomic group [fgkm, Section 2.2] (resp., special Clifford-cyclotomic group) by

[TABLE]

we have $\mathcal{G}_{n}\subseteq\operatorname{U_{2}^{\zeta}}(R_{n})$. In general, $\operatorname{U_{2}^{\zeta}}(R_{n})\subsetneq\operatorname{U_{2}}(R_{n})$. For a subgroup $H\leq\operatorname{U_{2}}(R_{n})$, denote by $\operatorname{P\!}H$ the image of $H$ in $\operatorname{PU_{2}}(R_{n})$. The Adjoint map $\operatorname{{\rm Ad}}:\operatorname{SU_{2}}(R_{n})\rightarrow\operatorname{SO_{3}}(R_{n}^{+})$ induces maps $\pi:\operatorname{U_{2}}(R_{n})\rightarrow\operatorname{SO_{3}}(R_{n}^{+})$ and $\overline{\pi}:\operatorname{PU_{2}}(R_{n})\rightarrow\operatorname{SO_{3}}(R_{n}^{+})$; see Section 5.

Let $G(r,s)$ be the subgroup of $\operatorname{SO_{3}}({\mathbb{R}})$ generated by rotations of order $r$ and order $s$ about chosen perpendicular axes. For an appropriate choice of axes one has $G(4,n)\subseteq\operatorname{SO_{3}}(R_{n}^{+})$. In Theorem 5.1 we show that $\pi(\mathcal{G}_{n})=G(4,n)$. The subgroup structures

[TABLE]

play a large role in exact synthesis for quantum gates in single-qubit quantum computation. The following results are known:

Theorem 1.1**.**

1. (a)

$G(4,8)=\operatorname{SO_{3}}(R_{8}^{+})$* and $\mathcal{G}_{8}=\operatorname{U_{2}}(R_{8})$ [s2, fgkm], $G(4,12)=\operatorname{SO_{3}}(R_{12}^{+})$ and $\mathcal{G}_{12}=\operatorname{U_{2}}(R_{12})$ [s2, brs], $G(4,16)=\operatorname{SO_{3}}(R_{16}^{+})$ and $\mathcal{G}_{16}=\operatorname{U_{2}}(R_{16})$ [s2], $G(4,24)=\operatorname{SO_{3}}(R_{24}^{+})$ and $\mathcal{G}_{24}=\operatorname{U_{2}}(R_{24})$ [fgkm].* 2. (b)

For an integer $n$ we have $\operatorname{U_{2}^{\zeta}}(R_{n})=\operatorname{U_{2}}(R_{n})$ if and only if

[TABLE]

[fgkm, Theorem 5.3]**. 3. (c)

Let $S_{4}$ be the symmetric group on $4$ letters and $D_{m}$ be the dihedral group of order $2m$. We have $G(4,n)\cong S_{4}\ast_{D_{4}}D_{n}$ [rs]. 4. (d)

If $n=2^{s}$, $s\geq 5$, then $G(4,n)$ is of infinite index in $\operatorname{SO_{3}}(R_{n}^{+})$ [s2].

Serre [s2] introduced Euler-Poincaré characteristics to the study of $G(4,n)$ and $\mathcal{G}_{n}$, as well as observing that $\operatorname{SO_{3}}(R_{n}^{+})$ for $n=2^{s}$ acts on a tree by looking at it over ${\mathbb{Q}}_{2}$. Theorem 1.1(d) follows from computing the Euler-Poincaré characteristic $\chi$ of $G(4,n)$ and $\operatorname{SO_{3}}(R_{n}^{+})$ for $n=2^{s}\geq 8$:

Theorem 1.2**.**

(Serre [s2])* Suppose $n=2^{s}\geq 8$.*

1. (a)

$\chi(G(4,n))=-1/12+1/2n$. 2. (b)

$\chi(\operatorname{SO_{3}}(R_{n}^{+}))=-2^{-2^{s-2}}\zeta_{F_{n}}(-1)$.

In this paper we prove the following theorem, settling affirmatively a conjecture of Sarnak [sa, p. $15^{\rm IV}$]:

Theorem 1.3**.**

Suppose $4|n$ with $n\geq 8$.

1. (a)

We have $\mathcal{G}_{n}=\operatorname{U_{2}^{\zeta}}(R_{n})$ if and only if $n=8,12,16,24$. 2. (b)

We have $\operatorname{S}\!\mathcal{G}_{n}=\operatorname{SU_{2}}(R_{n})$ if and only if $n=8,12,16,24$. 3. (c)

We have $G(4,n)=\operatorname{SO_{3}}(R_{n}^{+})$ if and only if $n=8,12,16,24$.

In all cases above where there is not equality, the index is infinite.

We prove Theorem 1.3 by computing Euler-Poincaré characteristics with $4|n$, $n\geq 8$, generalizing Theorem 1.2. We prove that

[TABLE]

in Theorem 6.3. Then in Theorem 6.6 we compute $\chi$ of $\operatorname{SU_{2}}(R_{n})$, $\operatorname{PSU_{2}}(R_{n})$, $\operatorname{PU_{2}}(R_{n})$, $\operatorname{PU_{2}^{\zeta}}(R_{n})$, and $\operatorname{SO_{3}}(R_{n}^{+})$. We gain a foothold on these Euler-Poincaré characteristics by considering the group scheme $\operatorname{SU_{2}}({\mathbb{Z}}[1/2])$ over ${\mathbb{Z}}[1/2]$, denoted $A_{1}^{\ast}$. We have $A_{1}^{\ast}({\mathbb{R}})=\operatorname{SU_{2}}(\mathbb{C})$ and $A_{1}^{\ast}(R_{n}^{+})=\operatorname{SU_{2}}(R_{n})$. The results of Serre [s1] (which depend on theorems of Harder) apply to compute $\chi(\operatorname{SU_{2}}(R_{n}))$ because because $A_{1}^{\ast}$ is simply connected and simple. We then deduce $\chi$ of the other groups from this using properites of Euler-Poincaré characteristics. The relationship between $\chi(\operatorname{PU_{2}}(R_{n}))$ and $\chi(\operatorname{SO_{3}}(R_{n}^{+}))$ is particularly interesting – it involves embedding $\operatorname{PU_{2}}(R_{n})$ in $\operatorname{SO_{3}}(R_{n}^{+})$ via the Adjoint representation with attendant invariant $\overline{c}(R_{n})$ defined in Definition 4.10(a).

2. The special Clifford-cyclotomic group

For a complex number $z$ of absolute value $1$, define the unitary matrix

[TABLE]

of determinant $1$. In particular $H(1)\in\mathcal{C}$. Following [fgkm, (2)], we take our Hadamard matrix to be

[TABLE]

We have $H=T_{4}^{-1}H(1)$ with $T_{n}\in\operatorname{U_{2}}(R_{n})$ as in (1) and $T_{n}^{-j}H(1)T_{n}^{j}=H(\zeta_{n}^{j})\in\operatorname{SU_{2}}(R_{n})$ for integers $j$ if $4|n$. With $4|n$, set

[TABLE]

Proposition 2.1**.**

Assuming $4|n$, we have

1. (a)

$[\mathcal{G}_{n}:\operatorname{S}\!\mathcal{G}_{n}]=n$, 2. (b)

$\mathcal{G}_{n}=\langle H,T_{n}\rangle=\langle H(1),T_{n}\rangle$, 3. (c)

$\operatorname{S}\!\mathcal{G}_{n}=\mathscr{H}_{n}$.

Proof.

(a) follows from the exact sequence

[TABLE]

since the roots of unity in $R_{n}$ are $\langle\zeta_{n}\rangle$ as $n$ is even.

(b) is shown in [fgkm, Section 2.2].

For (c), let $w$ be a word in $H(1)$ and $T_{n}$ of determinant $1$ with $k$ occurences of $H(1)$. We proceed by induction on $k$. If $k=0$, then the word must be $1$. If $k=1$, the word must be $T_{n}^{-j}H(1)T_{n}^{j}=H(\zeta_{n}^{j})$ for some $0\leq j\leq n$. Suppose inductively that every word in $H(1)$ and $T_{n}$ of determinant $1$ with at most $k_{0}$ occurences of $H(1)$ is in $\mathscr{H}_{n}$, and let $w$ be a word in which $H(1)$ appears $k_{0}+1$ times. Choose $a$ with $0\leq a<n$ such that $w$ begins with $T_{n}^{a}H(1)$. Then $H(\zeta_{n}^{-a})^{-1}w\in\operatorname{S}\!\mathcal{G}_{n}$ has at most $k_{0}$ occurences of $H(1)$, and so is in $\mathscr{H}_{n}$ by assumption. Hence $w\in\mathscr{H}_{n}$ and $\operatorname{S}\!\mathcal{G}_{n}=\mathscr{H}_{n}$. ∎

Theorem 2.2**.**

Assume $4|n$. Then $\operatorname{U_{2}^{\zeta}}(R_{n})=\mathcal{G}_{n}$ if and only if $\operatorname{SU_{2}}(R_{n})=\operatorname{S}\!\mathcal{G}_{n}$.

Proof.

First, suppose that $\operatorname{SU_{2}}(R_{n})=\mathscr{H}_{n}$ and let $\alpha\in\operatorname{U_{2}^{\zeta}}(R_{n})$. Let $\det\alpha=\zeta_{n}^{j}$, where $0\leq j<n$. Then $\alpha=T^{j}\alpha^{\prime}$, where $\det\alpha^{\prime}=1$ and so $\alpha^{\prime}\in\operatorname{SU_{2}}(R_{n})$. Since the generators of $\operatorname{SU_{2}}(R_{n})$ belong to $\langle H,T_{n}\rangle$, it follows that $\alpha$ does too.

In the other direction, suppose that $\operatorname{U_{2}^{\zeta}}(R_{n})=\mathcal{G}_{n}$. Then $\operatorname{SU_{2}}(R_{n})=\operatorname{S}\!\mathcal{G}_{n}$ trivially by definition. ∎

3. $\operatorname{SU_{2}}(R_{n})$ and

$\operatorname{SO_{3}}(R_{n}^{+})$

Definition 3.1**.**

1. (a)

Throughout this paper $R^{+}$ is the ring of $S$-integers in a totally real number field $F$, where $S$ contains the archimedean places and all places above $2$. We put $R=R^{+}[i]$ and $K=F(i)$. Both $R^{+}$ and $R$ are Dedekind domains. 2. (b)

Define $A_{1}^{*}$ to be the group scheme over ${\mathbb{Z}}[1/2]$ with

[TABLE]

for any ${\mathbb{Z}}[1/2]$-algebra $B$ with group operation defined by matrix multiplication. In particular, $A_{1}^{\ast}(B)=\operatorname{SU_{2}}(B[i])$. For example, $A_{1}^{*}({\mathbb{R}})=\operatorname{SU_{2}}(\mathbb{C})$.

By $\operatorname{SO_{3}}$ we mean the group of $3\times 3$ matrices of determinant $1$ that stabilize the standard inner product on ${\mathbb{R}}^{3}$. It is defined as a group scheme over ${\mathbb{Z}}$ by $\det(g)=1$ and $gg^{t}=1$. There is an exact sequence of group schemes

[TABLE]

given by $\operatorname{SU_{2}}$ acting by conjugation on the three-dimensional real vector space $V$ of trace-[math] $2\times 2$ hermitian ($m^{\dagger}=m$) matrices in the Pauli basis

[TABLE]

In terms of hermitian matrices the standard form is $\langle A,B\rangle=\frac{1}{2}\operatorname{Tr}(AB)$:

[TABLE]

which is obviously preserved under conjugation by $\operatorname{SU_{2}}$. This is the Adjoint action of $\operatorname{SU_{2}}$ on its Lie algebra $iV$ of trace-[math] skew-hermitian matrices in disguise. Explicitly, we have [na, Appendix A]

[TABLE]

The map $\operatorname{{\rm Ad}}$ factors as

[TABLE]

The adjoint action $\operatorname{{\rm Ad}}$ given in (9) extends to a group homomorphism $\pi:\operatorname{U_{2}}(R)\rightarrow\operatorname{SO_{3}}(R^{+})$ via conjugation on the trace-[math] $2\times 2$ hermitian matrices in the Pauli basis. We have

[TABLE]

for an arbitrary choice of $\sqrt{\det g}$. The map $\pi$ in turn factors as

[TABLE]

We view $\operatorname{PSU_{2}}(R)$ as a subgroup of $\operatorname{SO_{3}}(R^{+})$ via (10) and we view $\operatorname{PU_{2}}(R)$ as a subgroup of $\operatorname{SO_{3}}(R^{+})$ via (11) with $\operatorname{PSU_{2}}(R)\leq\operatorname{PU_{2}}(R)\leq\operatorname{SO_{3}}(R^{+})$.

Remark 3.2**.**

In Section 4 we will define a map $\phi$ from $\operatorname{SO_{3}}(R^{+})$ into a finite elementary abelian $2$-group (the Selmer group $\operatorname{Sel}_{2}^{+}(R^{+})$) with kernel $\operatorname{PSU_{2}}(R)$. From this it follows that $\operatorname{PSU_{2}}(R)$ and $\operatorname{PU_{2}}(R)$ are normal subgroups of $\operatorname{SO_{3}}(R^{+})$, cf. Corollary 4.9.

4. $\operatorname{SO_{3}}(R^{+})/\operatorname{PSU_{2}}(R)$ and $\operatorname{SO_{3}}(R^{+})/\operatorname{PU_{2}}(R)$

The short exact sequence (8) remains short exact on ${\mathbb{R}}$-points

[TABLE]

with $A_{1}^{*}({\mathbb{R}})=\operatorname{SU_{2}}(\mathbb{C})$, but in general for $R^{+}$ we only have

[TABLE]

In our situation $A_{1}^{*}(R^{+})$ does not surject onto $\operatorname{SO_{3}}(R^{+})$. In particular the map $\operatorname{{\rm Ad}}$ factors as (see the next section) and the map from $\operatorname{SU_{2}}(R)$ to $\operatorname{PU_{2}}(R)$ is not surjective. Indeed, for us $R$ is a localization of an order in a number field, so the group of roots of unity of $R$ is finite, generated by some root of unity $\zeta$. Then $\zeta$ is not a square in $R$, so $\left[\begin{smallmatrix}\zeta&0\\ 0&1\end{smallmatrix}\right]$ is an element of $\operatorname{PU_{2}}(R)$ whose determinant is not a square. Therefore it cannot be the image of any element of $\operatorname{SU_{2}}(R)$. Since the map $\operatorname{PU_{2}}(R)\to\operatorname{SO_{3}}(R^{+})$ is injective, this implies that $\operatorname{SU_{2}}(R)\to\operatorname{SO_{3}}(R^{+})$ is not surjective either, proving the following proposition.

Proposition 4.1**.**

Let $R^{+}$ be the $S$-integers in a toally real field $F$, where $S$ contains the archimedean primes and all primes above $2$, and let $R=R^{+}[i]$. Then the group $\operatorname{SO_{3}}(R^{+})/\operatorname{PSU_{2}}(R)$ is nontrivial.

Even the map $\operatorname{PU_{2}}(R)\to\operatorname{SO_{3}}(R^{+})$ may not be surjective.

Example 4.2**.**

*The map $\operatorname{PU_{2}}({\mathbb{Z}}[\sqrt{21},i,1/2])\hookrightarrow\operatorname{SO_{3}}({\mathbb{Z}}[\sqrt{21},1/2])$ is not surjective.

*Let $R^{+}={\mathbb{Z}}[\sqrt{21},1/2]$ and $R=R^{+}[i]$. Let $u=\frac{5+\sqrt{21}}{2}\in(R^{+})^{\times}$ which is totally positive and not the norm of a unit in $R$. (One checks that $R^{\times}$ is generated by $u,i,1+i$ and hence that $u$ is not a norm from $R^{\times}$.) Choose $q\in\left(\frac{-1,-1}{R^{+}}\right)$ of norm $u$, such as $\frac{4+\sqrt{21}+i+j+k}{4}$. The homomorphism from the unit Hamilton quaternions over $R^{+}$ to $\operatorname{SO_{3}}$ takes $q$ to

[TABLE]

The $2\times 2$ matrix $M$ corresponding to $q$ is

[TABLE]

it has the property that $MM^{\dagger}=u\,{\rm Id}_{2\times 2}$. The element of $\operatorname{PU_{2}}(\mathbb{C})$ mapping to $T_{q}$ is obtained by dividing $M_{q}$ by an element of $\mathbb{C}$ of norm $u$. However, lifting this element to an element of $\operatorname{PU_{2}}(R)$ would require finding an element of $R$ of norm $u$, which does not exist. Hence $T_{q}\in\operatorname{SO_{3}}(R^{+})$ is not the image of any element of $\operatorname{PU_{2}}(R)$. **

In this section we will prove that $\operatorname{SO_{3}}(R^{+})/\operatorname{PSU_{2}}(R)$ and $\operatorname{SO_{3}}(R^{+})/\operatorname{PU_{2}}(R)$ are finite abelian $2$-groups, with $\operatorname{SO_{3}}(R^{+})/\operatorname{PSU_{2}}(R)$ nontrivial by Proposition 4.1. Denote by $F_{+}$ the totally positive elements of $F$. For any subset $S\subseteq F$, denote by $S_{+}\subseteq F_{+}$ the totally positive elements of $S$.

Definition 4.3**.**

Let $M_{x},M_{y},M_{z}\in\operatorname{SO_{3}}(R^{+})$ be the diagonal matrices with entries

[TABLE]

respectively.

Define that following $R^{+}$-valued functions for $M\in\operatorname{SO_{3}}(R^{+})$:

[TABLE]

and

[TABLE]

Definition 4.4**.**

Let $\sf{R}$ be the $S$-integers in a totally real number field $F$. Define the Selmer group

[TABLE]

We denote by $\operatorname{Cl}({\sf R})$ the class group of $\sf{R}$.

It is not difficult to compute $\operatorname{Sel}^{+}_{2}({\sf R})$ in examples using the following elementary proposition.

Proposition 4.5**.**

There is an exact sequence of abelian groups

[TABLE]

In particular, let $r=[F:{\mathbb{Q}}]$ and let $s$ be the number of finite primes in $S$. Then the kernel of the signature map ${\sf R}^{\times}\to({\mathbb{Z}}/2{\mathbb{Z}})^{r}$ is precisely ${\sf R}^{\times}_{+}$, while ${\sf R}^{\times}/({\sf R}^{\times})^{2}\cong({\mathbb{Z}}/2{\mathbb{Z}})^{r+s}$. Thus if the image of the signature map is isomorphic to $({\mathbb{Z}}/2{\mathbb{Z}})^{v}$, then ${\sf R}^{\times}_{+}/({\sf R}^{\times})^{2}\cong({\mathbb{Z}}/2{\mathbb{Z}})^{r+s-v}$.

This makes it straightforward to compute the following examples:

Proposition 4.6**.**

Let $R_{n}$, $R_{n}^{+}$ be as in the introduction.

1. (a)

Suppose $n=2^{s}$, $n\geq 8$. Then $\operatorname{Sel}_{2}^{+}(R_{n}^{+})\cong{\mathbb{Z}}/2{\mathbb{Z}}$. 2. (b)

Suppose $n=3\cdot 2^{s}$, $4|n$. Then $\operatorname{Sel}_{2}^{+}(R_{n}^{+})\cong{\mathbb{Z}}/2{\mathbb{Z}}$.

Proof.

Let $\mathcal{O}_{n}:={\mathbb{Z}}[\zeta_{n}+\overline{\zeta}_{n}]$, the ring of integers in $F_{n}:={\mathbb{Q}}(\zeta_{n})^{+}$.

(a): Let $n=2^{s}$, $n\geq 8$. Then $\mathcal{O}_{n}$ has odd class number by [w, Theorem 10.4(b)] and so $R_{n}^{+}=\mathcal{O}_{n}[1/2]$ has odd class number. Every totally positive unit in $\mathcal{O}_{n}$ is a square by Weber’s Theorem [we] and there is one prime $\mathfrak{p}$ in $F_{n}$ above $2$. Hence $(R_{n}^{+})_{+}/[(R_{n}^{+})^{\times}]^{2}\cong{\mathbb{Z}}/2{\mathbb{Z}}\cong\operatorname{Sel}_{2}^{+}(R_{n}^{+})$.

(b): Let $n=3\cdot 2^{s}$, $s\geq 3$. Then $\mathcal{O}_{n}$ has odd class number by applying [w, Theorem 10.4] to $F_{n}/{\mathbb{Q}}(\sqrt{3})$. We have $(R_{n}^{+})_{+}^{\times}/[(R_{n}^{+})^{\times}]^{2}\cong{\mathbb{Z}}/2{\mathbb{Z}}$ by [IJKLZ1, Theorem 3.13(b)].

∎

The functions of Definition 4.3 satisfy the following properties.

Lemma 4.7**.**

For $M\in\operatorname{SO_{3}}(R^{+})$ and

[TABLE]

we have:

1. (a)

$\phi_{i}(\operatorname{{\rm Ad}}(A))=a_{i}^{2}$, $1\leq i\leq 4$. 2. (b)

$\theta_{ij}(\operatorname{{\rm Ad}}(A))=a_{i}a_{j}$, $1\leq i,j\leq 4$, $i\neq j$. 3. (c)

$\phi_{1}(M)+\phi_{2}(M)+\phi_{3}(M)+\phi_{4}(M)=1$. 4. (d)

$\phi_{i}(M)\phi_{j}(M)=\theta_{ij}(M)^{2}$, $1\leq i,j\leq 4$, $i\neq j$, and $\theta_{\pi(1)\pi(2)}(M)\theta_{\pi(3)\pi(4)}(M)$ does not depend on the choice of $\pi\in\mathcal{S}_{4}$. 5. (e)

There exists a well a unique well-defined function $\phi:\operatorname{SO_{3}}(R^{+})\to F^{\times}/(F^{\times})^{2}$ which agrees with each $\phi_{i}$, $1\leq i\leq 4$, whenever the latter is nonzero. 6. (f)

The image of $\phi$ lies in $\operatorname{Sel}^{+}_{2}(R^{+})\subseteq F^{\times}/(F^{\times})^{2}$. In other words $\phi(M)$ has even valuation at all primes of $R^{+}$ and is totally positive. 7. (g)

For $i\in\{x,y,z\}$ we have $\phi(M)=\phi(MM_{i})=\phi(M_{i}M)$.

Proof.

(c) follows immediately by summing the definitions of the $\phi_{i}$’s. (a) and (b) follow immediately by plugging in the definition of $\operatorname{{\rm Ad}}$ (9).

(d) is not as trivial but can be derived from the defining equations of $\operatorname{SO_{3}}$ by a simple Gröbner Basis calculation.

To see (e) observe that by (c) at least one of the $\phi_{i}(M)$’s is always nonzero and by (d) all the nonzero $\phi_{i}(M)$’s always agree once one mods out by squares.

(f) follows since by (c) at each prime of $R^{+}$ at least one of the $\phi_{i}(M)$’s must have valuation [math]. The total positivity follows from the definitions of $\phi_{i}$ and the fact that for $M\in\operatorname{SO_{3}}({\mathbb{R}})$ we always have $\operatorname{Tr}(M)\geq-1$.

Finally, (g) holds, because the sets $\{\phi_{j}(M)\}$, $\{\phi_{j}(MM_{i})\}$, $\{\phi_{j}(M_{i}M)\}$ for $1\leq j\leq 4$ are visibly equal. ∎

Theorem 4.8**.**

The map $\phi:\operatorname{SO_{3}}(R^{+})\to\operatorname{Sel}^{+}_{2}(R^{+})$ is group homomorphism and

[TABLE]

is an exact sequence.

Proof.

In view of Lemma 4.7(g), we may assume that $\phi_{1}(MN)\neq 0$. It can be checked using a simple Gröbner basis calculation that for $M,N\in\operatorname{SO_{3}}(R^{+})$ and $1\leq i\leq 4$, the equation

[TABLE]

follows from the defining equations of $\operatorname{SO_{3}}$, where $\{i,j,k,\ell\}=\{1,2,3,4\}$ and the sign is $-1$ when $1$ appears in the subscript and $1$ otherwise. Hence $\phi(M)\phi(N)=\phi(MN)$ as long as $\phi_{i}(M)$, $\phi_{i}(N)$ are both nonzero for the same $i$.

If three of the $\phi_{i}(M)$ are [math], then $M\in\{I_{3},M_{x},M_{y},M_{z}\}$ and it is simple to check that $\phi(MN)=\phi(M)\phi(N)$; similarly if three of the $\phi_{i}(N)$ are [math]. Otherwise, there is no problem unless two are [math] for $M$ and the other two are [math] for $N$. Suppose that $\phi_{1}(M)=\phi_{2}(M)=\phi_{3}(N)=\phi_{4}(N)$ (the other cases are similar). Then we have

[TABLE]

where $a^{2}+b^{2}=c^{2}+d^{2}=1$. In this case it is easy to check that $\phi_{1}(MN)=0$, contradicting our choice of $N$ (and it is also easy to check that $\phi(MN)=\phi(M)\phi(N)$). Thus, $\phi$ is a group homomorphism.

That $\phi\circ\operatorname{{\rm Ad}}=1$ follows from 4.7(a). Now suppose $M\in\ker\phi$, hence the $\phi_{i}(M)$ are all squares in $R^{+}$. Let $a_{i}=\sqrt{\phi_{i}(M)}$ with signs chosen so that $a_{i}a_{j}=\theta_{ij}(M)$; we can do this by 4.7(b). Now it is again straightforward to check that the equations

[TABLE]

follow from the defining equations of $\operatorname{SO_{3}}$. ∎

Proof.

It can be checked using a simple Gröbner Basis calculation that for $M,N\in\operatorname{SO_{3}}(R^{+})$ the equation

[TABLE]

follows from the defining equations of $\operatorname{SO_{3}}$. Hence, $\phi(M)\phi(N)=\phi(MN)$ as long as all three $\phi_{1}(M)$, $\phi_{1}(M)$, $\phi_{1}(MN)$ are nonzero. In general, one has give expressions analogous to (14) for $\phi_{i}(M)$, $\phi_{j}(M)$, $\phi_{k}(M)$ with $1\leq i,j,k\leq 4$. These additional cases are similar. Thus, $\phi$ is a group homomorphism.

That $\phi\circ\operatorname{{\rm Ad}}=1$ follows from 4.7(a). Now suppose $M\in\ker\phi$, hence the $\phi_{i}(M)$ are all squares in $R^{+}$. Let $a_{i}=\sqrt{\phi_{i}(M)}$ with signs chosen so that $a_{i}a_{j}=\theta_{ij}(M)$, we can do this by 4.7(b). Now it is again straightforward to check that the equations

[TABLE]

follow from defining equations of $\operatorname{SO_{3}}$. ∎

Corollary 4.9**.**

The subgroups $\operatorname{PSU_{2}}(R)$ and $\operatorname{PU_{2}}(R)$ of $\operatorname{SO_{3}}(R^{+})$ are normal.

Definition 4.10**.**

1. (a)

Set $C(R)=\operatorname{SO_{3}}(R^{+})/\operatorname{PSU_{2}}(R)$, $\overline{C}(R)=\operatorname{SO_{3}}(R^{+})/\operatorname{PU_{2}}(R)$, $c(R)=\#C(R)$, and $\overline{c}(R)=\#\overline{C}(R)$. Hence $c(R)$, $\overline{c}(R)$ are powers of 2 with $c(R)\neq 1$. We have

[TABLE]

Example 4.2 shows that $\overline{c}({\mathbb{Z}}[\sqrt{21},i,1/2])\neq 1$. 2. (b)

Let $r(n)$ be the number of primes in $K_{n}:={\mathbb{Q}}(\zeta_{n})$ above $2$ and $r_{+}(n)$ be the number of primes in $F_{n}:={\mathbb{Q}}(\zeta_{n})^{+}$ above $2$.

We now state a result from [IJKLZ2] which we will need:

Proposition 4.11**.**

([IJKLZ2, Proposition 2.3])*

Suppose $n\geq 8$ and $4|n$ with $r(n)$, $r_{+}(n)$ as in Definition 4.10(b).*

1. (a)

$\operatorname{PU_{2}}(R_{n})/\operatorname{PSU_{2}}(R_{n})\cong({\mathbb{Z}}/2{\mathbb{Z}})^{1+r(n)-r_{+}(n)}$. 2. (b)

$\operatorname{PU_{2}}(R_{n})/\operatorname{PU_{2}^{\zeta}}(R_{n})\cong({\mathbb{Z}}/2{\mathbb{Z}})^{r(n)-r_{+}(n)}$. 3. (c)

$\operatorname{PU_{2}^{\zeta}}(R_{n})/\operatorname{PSU_{2}}(R_{n})\cong{\mathbb{Z}}/2{\mathbb{Z}}$.

Combining Proposition 4.11(a) with (15) then gives the following:

Proposition 4.12**.**

For $R_{n}={\mathbb{Z}}[\zeta_{n},1/2]$, $4|n$, $n\geq 8$, we have $c(R_{n})=2^{1+r(n)-r_{+}(n)}\overline{c}(R_{n})$.

We can compute $c(R)$ and $\overline{c}(R)$ in some important examples with $R=R_{n}:={\mathbb{Z}}[\zeta_{n},1/2]$.

Theorem 4.13**.**

1. (a)

Suppose $n=2^{s}$, $n\geq 8$. Then $c(R_{n})=2$ and $\overline{c}(R_{n})=1$. 2. (b)

Suppose $n=3\cdot 2^{s}$, $s\geq 2$. Then $c(R_{n})=2$ and $\overline{c}(R_{n})=1$.

Proof.

Suppose $n=2^{s}$ or $n=3\cdot 2^{s}$, $n\geq 8$. Then there is one prime in $K_{n}={\mathbb{Q}}(\zeta_{n})$ above $2$ and $r(n)=r_{+}(n)=1$. Hence by Proposition 4.12, $c(R_{n})=2\overline{c}(R_{n})$. But by Proposition 4.5, $\operatorname{Sel}_{2}^{+}(R_{n}^{+})\cong{\mathbb{Z}}/2{\mathbb{Z}}$. Hence $c(R_{n})\leq 2$ and we therefore must have $c(R_{n})=2$ and $\overline{c}(R_{n})=1$.

∎

5. Amalgamated products and the Clifford-cyclotomic group

Set $\overline{\mathcal{G}}_{n}=\pi(\mathcal{G}_{n})\subseteq\operatorname{SO_{3}}(R_{n}^{+})$ and $\overline{\operatorname{S}\!\mathcal{G}}_{n}=\operatorname{{\rm Ad}}(\operatorname{S}\!\mathcal{G}_{n})\subseteq\operatorname{SO_{3}}(R_{n}^{+})$.

Theorem 5.1**.**

Assume $4|n$, $n\geq 8$. We have

[TABLE]

Proof.

By Proposition 2.1(b) and Theorem 1.1(c) it suffices to show that $\pi(H_{n}T^{2m})$ and $\pi(T_{n})$ are rotations of order 4 and $n$ about orthogonal axes. Now

[TABLE]

which has determinant $i=\zeta_{8}^{2}$. Define

[TABLE]

We calculate using (9):

[TABLE]

which is a rotation around the $y$-axis by $\pi/2$, while

[TABLE]

[TABLE]

is a rotation by $2\pi/n$ about the $z$-axis. ∎

The finite subgroups of $\operatorname{SU_{2}}(\mathbb{C})$ are well-known; see [v, Théorème I.3.7]. Let $D_{n}$ be the dihedral group of order $2n$. Denote by $E_{48}$ the tetrahedral group, i.e., the degree-$2$ central extension of $S_{4}$, and by $Q_{4n}$ the quaternion group of order $4n$ (called dicyclique in [v]). We have $Q_{4n}/\langle\pm 1\rangle\cong D_{n}$.

Corollary 5.2**.**

Let $\mathcal{H}$ be the pullback of $\overline{\mathcal{G}}_{n}$ under the surjective map $\operatorname{SU_{2}}(\mathbb{C})\stackrel{{\scriptstyle\operatorname{{\rm Ad}}}}{{\rightarrow}}\operatorname{SO_{3}}({\mathbb{R}})$. Then $\operatorname{S}\!\mathcal{G}_{n}\subseteq\mathcal{H}$ with $[\mathcal{H}:\operatorname{S}\!\mathcal{G}_{n}]=2$ and

[TABLE]

Proof.

We have $\operatorname{P\!}\mathcal{G}_{n}/\operatorname{P\!}\operatorname{S}\!\mathcal{G}_{n}\cong\mu_{n}/\mu_{n}^{2}\cong{\mathbb{Z}}/2{\mathbb{Z}}$ since $n$ is even. But

[TABLE]

and $\operatorname{P\!}\operatorname{S}\!\mathcal{G}_{n}\cong\operatorname{{\rm Ad}}(\operatorname{S}\!\mathcal{G}_{n})=\overline{\operatorname{S}\!\mathcal{G}}_{n}$. Hence $\operatorname{{\rm Ad}}^{-1}(\overline{\mathcal{G}}_{n}):=\mathcal{H}\cong E_{48}\ast_{Q_{16}}Q_{4n}$ and $\operatorname{{\rm Ad}}^{-1}(\overline{\operatorname{S}\!\mathcal{G}}_{n})=\operatorname{S}\!\mathcal{G}_{n}$. Since $[\overline{\mathcal{G}}_{n}:\overline{\operatorname{S}\!\mathcal{G}}_{n}]=2$, it follows that $[\mathcal{H}:\operatorname{S}\!\mathcal{G}_{n}]=2$.

∎

6. Euler-Poincaré characteristics

In this section we determine the Euler-Poincaré characteristics of unitary groups over cyclotomic rings and Clifford-cyclotomic groups. These results will then be used in the proof of Theorem 1.3. General references for Euler-Poincaré characteristics are [b, Chapter 9] and [s1].

Definition 6.1** ([b, Section IX.6]).**

A group $\Gamma$ is of finite homological type if $\Gamma$ has finite virtual cohomological dimension and, for every $\Gamma$-module $M$ that is finitely generated as an abelian group and every natural number $i$, the homology group $H_{i}(\Gamma,M)$ is finitely generated.

Proposition 6.2**.**

1. (a)

Suppose $1\rightarrow\Gamma^{\prime}\rightarrow\Gamma\rightarrow\Gamma^{\prime\prime}\rightarrow 1$ is a short exact sequence of groups with $\Gamma^{\prime}$, $\Gamma^{\prime\prime}$ of finite homological type. If $\Gamma$ is virtually torsion-free, then $\Gamma$ is of finite homological type and

[TABLE] 2. (b)

Suppose $\Gamma^{\prime}$ is a subgroup of $\Gamma$ of finite index and $\chi(\Gamma)$ is defined. Then $\chi(\Gamma^{\prime})$ is defined and $\chi(\Gamma^{\prime})=[\Gamma:\Gamma^{\prime}]\chi(\Gamma)$. 3. (c)

Suppose $\Gamma^{\prime}\leq\Gamma$ with $\chi(\Gamma)$ and $\chi(\Gamma^{\prime})$ both defined. If $|\chi(\Gamma^{\prime})|/|\chi(\Gamma)|$ is not a positive integer, then $\Gamma^{\prime}$ has infinite index in $\Gamma$. In particular this holds if $|\chi(\Gamma^{\prime})|<|\chi(\Gamma)|$. 4. (d)

Suppose $\Gamma^{\prime}$ and $\Gamma^{\prime\prime}$ are finite groups with $A\leq\Gamma^{\prime}$ and $A\leq\Gamma^{\prime\prime}$. Let $\Gamma=\Gamma^{\prime}\ast_{A}\Gamma^{\prime\prime}$. Then

[TABLE]

Proof.

(a), (b) are parts (d) and (c) of [b, Proposition 7.3]. (c) is an immediate consequence of (a), while (d) is [s1, Corollaire 1, p. 104]. ∎

Theorem 6.3**.**

Assume $4|n$. Then $\chi(\operatorname{S}\!\mathcal{G}_{n})=\chi(G(4,n))=\chi(\operatorname{P\!}\mathcal{G}_{n})=-\frac{1}{12}+\frac{1}{2n}$.

Proof.

Let $\mathcal{H}$ be as in Corollary 5.2. By (16) and Proposition 6.2(d) we have

[TABLE]

But $\operatorname{S}\!\mathcal{G}_{n}$ is an index-$2$ subgroup of $\mathcal{H}$ from Corollary 5.2, so by Proposition 6.2(b)

[TABLE]

We have $\chi(G(4,n))=\chi(\operatorname{P\!}\mathcal{G}_{n})=-1/12+1/2n$ from (17) and Proposition 6.2(d). ∎

We will need the following in the proof of Theorem 6.6 below.

Remark 6.4**.**

Recall that a connected linear algebraic group $G$ over a perfect field is reductive if it admits a representation with finite kernel that is a direct sum of irreducible representations. An alternative definition sufficient for this paper is that $G$ over an algebraically closed field is reductive if and only if every smooth connected unipotent normal subgroup of $G$ is trivial, and if $k$ is perfect then $G$ is reductive over $k$ if and only if it is over $\bar{k}$.

Definition 6.5**.**

Set

[TABLE]

Theorem 6.6**.**

Suppose $n\geq 8$ and $4|n$ with $r(n)$, $r_{+}(n)$ as in Definition 4.10(b).

1. (a)

$\chi(\operatorname{SU_{2}}(R_{n}))=-M_{n}/2$. 2. (b)

$\chi(\operatorname{PSU_{2}}(R_{n}))=2\chi(\operatorname{SU_{2}}(R_{n}))=-M_{n}$. 3. (c)

$\chi(\operatorname{PU_{2}^{\zeta}}(R_{n}))=\chi(\operatorname{SU_{2}}(R_{n}))=-M_{n}/2$. 4. (d)

[TABLE] 5. (e)

Put $c_{n}=c(R_{n})$ and $\overline{c}_{n}=\overline{c}(R_{n})$ as in Definition 4.10. Then

[TABLE]

Proof.

(a) follows from a result of Harder [s1, Section 3.7, (*)].

The claims (b), (c), (d), (e) are obtained by combining this with Proposition 6.2, once we verify that all groups involved are of finite homological type. We start from the fact, due to Borel and Serre [b, page 218], that a torsion-free reductive $S$-arithmetic group $\Gamma$ is of type FL (i.e., that the ${\mathbb{Z}}\Gamma$-module ${\mathbb{Z}}$ has a finite free resolution). This is stated for arithmetic groups over ${\mathbb{Q}}$; however, the result follows more generally by restriction of scalars. The only problem is to verify that restriction of scalars preserves reductivity. As noted in Remark 6.4, in characteristic [math] an algebraic group is reductive over $k$ if and only if it is reductive over $\bar{k}$, and for a finite extension $K/F$ we have $\operatorname{Res}_{K/F}G\otimes_{F}\bar{F}\equiv(G\otimes_{K}\bar{F})^{[K:F]}$. A finite direct product of reductive groups is clearly reductive, so it follows that if $G$ is reductive, then so is $\operatorname{Res}_{K/F}G$. As $\operatorname{SU_{2}}$ is a simple group, it is certainly reductive. This shows that our groups are all VFL (as usual, the subgroup of matrices congruent to $1$ modulo a large prime is torsion-free).

Since free modules are projective, VFL implies VFP, and groups of type VFP are of finite homological type. This is enough to apply [b, Proposition 7.3].

(b): Apply Proposition 6.2(a) to

[TABLE]

using the fact that $\operatorname{SU_{2}}(R_{n})$ is virtually torsion-free because it is arithmetic (so a sufficiently small congruence subgroup is torsion-free).

(c): Apply Proposition 6.2(a) to

[TABLE]

from Proposition 4.11(c). To show that $\operatorname{PU_{2}^{\zeta}}(R_{n})$ is virtually torsion-free, it suffices to show that the finite-index subgroup $\operatorname{PSU_{2}}(R_{n})$ is virtually torsion-free. But $\operatorname{PSU_{2}}(R_{n})$ is virtually torsion-free since it is a finite quotient $\operatorname{SU_{2}}(R_{n})$, which is virtually torsion-free from (b).

(d): Apply Proposition 6.2(a) to

[TABLE]

from Proposition 4.11(a). The group $\operatorname{PU_{2}}(R_{n})$ is virtually torsion-free because its finite-index subgroup $\operatorname{PSU_{2}}(R_{n})$ is virtually torsion-free from (c).

(e): Apply Proposition 4.11a to

[TABLE]

as in Definition 4.10; $\#\overline{C}(R_{n})=\overline{c}(R_{n})=\overline{c}_{n}$ and $\#C(R_{n})=c(R_{n})=c_{n}$. The group $\operatorname{SO_{3}}(R_{n}^{+})$ is virtually torsion-free since it is arithmetic.

∎

Remark 6.7**.**

Suppose $n=2^{s}\geq 8$. Then $r(n)=r_{+}(n)=1$. Serre [s2, p. 48] uses Tamagawa numbers to show in this case that $\chi(\operatorname{SO_{3}}(R_{n}^{+}))=-M_{n}/2$ as in Theorem 1.2(b). Theorem 6.6(e) then shows that $c(R_{n})=2$ and $\overline{c}(R_{n})=1$, giving an independent proof of Theorem 4.13(a).

7. Proof of Theorem 1.3

We first prove Theorem 1.3(b). It is already known that $\operatorname{S}\!\mathcal{G}_{n}=\operatorname{SU_{2}}(R_{n})$ for $n=8,12,16,24$ (Theorem 1.1(a)). We will prove that $\operatorname{S}\!\mathcal{G}_{n}$ is not a finite-index subgroup of $\operatorname{SU_{2}}(R_{n})$ otherwise. By Proposition 6.2(b), to do this it suffices to show $\lvert\chi(\operatorname{S}\!\mathcal{G}_{n})\rvert<\lvert\chi(\operatorname{SU_{2}}(R_{n}))\rvert$ for $n\notin\{8,12,16,24\}$

Let $S$ be the places of $F_{n}=K^{+}_{n}$ above $2\infty$ and denote by $\zeta_{F_{n},S}(s)$ the Dedekind zeta function of $F_{n}$ with the Euler factors at finite places in $S$ omitted. Then the Euler-Poincaré characteristic of $\Gamma_{n}=\operatorname{SU_{2}}(R_{n})$ is given in [s1, Section 3.7]:

[TABLE]

By the functional equation for $\zeta_{F_{n}}$,

[TABLE]

and by [w, Proposition 2.7]

[TABLE]

As for $F_{n}$, let $f=\sqrt{\lvert N_{F_{n}/{\mathbb{Q}}}\operatorname{Disc}(K_{n}/F_{n})\rvert}$. Then $f=1$ unless $n$ is a power of $2$, in which case $f=2$. Now we have

[TABLE]

using standard properties of the discriminant in towers [neu, Corollary 2.10, p. 202].

Hence,

[TABLE]

which is greater than $\frac{1}{2\sqrt{2}}$ and hence greater than $\frac{1}{12}-\frac{1}{2n}$ as long as $n>134.5>(2\pi)^{8/3}$. If $n\notin\{8,12,16,24\}$ , $4|n$, and $8<n\leq 132$, then it can be manually checked from (19) that we still have $|\chi(\Gamma_{n})|>\frac{1}{12}-\frac{1}{2n}$. Reassuringly $\chi(\Gamma_{n})=1/12-1/(2n)$ for $n=8,12,16,24$. Hence $[\operatorname{SU_{2}}(R_{n}):\operatorname{S}\!\mathcal{G}_{n}]=\infty$ for $4|n$, $n\geq 8$, and $n\notin\{8,12,16,24\}$ by Proposition 6.2(b), proving Theorem 1.3(b).

To prove Theorem 1.3(a), note that $n=[\mathcal{G}_{n}:\operatorname{S}\!\mathcal{G}_{n}]=[\operatorname{U_{2}^{\zeta}}(R_{n}):\operatorname{SU_{2}}(R_{n})]$ by Proposition 2.1(a) and (7). Hence $[\operatorname{SU_{2}}(R_{n}):\operatorname{S}\!\mathcal{G}_{n}]=[\operatorname{U_{2}^{\zeta}}(R_{n}):\mathcal{G}_{n}]$, and so Theorem 1.3(b) together with Theorem 1.1(a) implies Theorem 1.3(a).

The proof of Theorem 1.3(c) is similar: both surjections $\mathcal{G}_{n}\twoheadrightarrow\operatorname{P\!}\mathcal{G}_{n}\cong\overline{\mathcal{G}}_{n}$ and $\operatorname{U_{2}^{\zeta}}(R_{n})\twoheadrightarrow\operatorname{PU_{2}^{\zeta}}(R_{n})$ have kernel of order $n$. Hence $[\operatorname{U_{2}^{\zeta}}(R_{n}):\mathcal{G}_{n}]=[\operatorname{PU_{2}^{\zeta}}(R_{n}):\operatorname{P\!}\mathcal{G}_{n}]$. But then we have

[TABLE]

Hence $[\operatorname{U_{2}^{\zeta}}(R_{n}):\mathcal{G}_{n}]=\infty$ implies $[\operatorname{SO_{3}}(R_{n}^{+}):G(4,n)]=\infty$. Theorem 1.3(c) then follows from Theorem 1.3 (a) and Theorem 1.1(a), concluding the proof of Theorem 1.3.