Fundamental groups of 3-dimensional spherical space forms

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1 Introduction

The purpose of this article is to describe fundamental groups of $3$${{Authors|Wpolitarczyk}}{{Stub}} == Introduction == <wikitex>; The purpose of this article is to describe fundamental groups of $-dimensional spherical space forms. For the historical context refer to [[Space forms: a history | this article]]. Today we know the whole list of groups which arise as fundamental groups of $-dimensional spherical space forms. These are exactly groups which admit a fixed-point free representation in $SO(4)$. In 1950's Milnor in {{cite|Milnor1957}} provided a list of all finite groups which could possibly act freely but not necessarilly linearly on $S^3$. All groups mentioned earlier belong to this list however there was also included a family of finite groups denoted by $Q(8n,k,l)$. Question whether these groups act on $S^3$ remained unsolved until the proof of the Geometrization Conjecture was finished by Perelman. The exposition of the first part is based on {{cite|Wolf2011}}. </wikitex> == Finite subgroups of SO(4) == <wikitex>; To determine finite subgroups of $SO(4)$ it is necessary to proceed in three steps: * determine finite subgroups of $SO(3)$, * use the covering map $S^3 \to SO(3)$ to determine finite subgroups of $S^3$, * use the fact that $SO(4)$ is doubly covered by $S^3 \times S^3$ to determine its finite groups. </wikitex> === Finite subgroups of SO(3) === <wikitex>; To classify finite subgroups of $SO(3)$ one has to analyse action of these groups on $S^2$. From [[Wikipedia:Riemann–Hurwitz_formula | Riemann-Hurwitz formula]] we obtain the following equation $ \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$$ where $N$ denotes the order of the group, $q$ denotes number of orbits with non-trivial isotropy and $n_i$ denotes the order of the respective isotropy subgroup. Solutions to this equation yield the desired list. {{beginthm|Theorem|{{cite|Wolf2011|thm 2.6.5.}}}} Every finite subgroup of $SO(3)$ is either * [[Wikipedia:Cyclic_group|cyclic]], * [[Wikipedia:Dihedral_group|dihedral]], * [[Wikipedia:Tetrahedral_group|tetrahedral]], * [[Wikipedia:Octahedral_group|octahedral]] or * [[Wikipedia:Icosahedral_group|icosahedral]]. These groups are called polyhedral groups. {{endthm}} </wikitex> === Finite subgroups of S^3 === <wikitex>; Let $\mathbb{H}$ denote the algebra of quaternions and treat $S^3$ as a subset of $\mathbb{H}$ of quaternions of norm 3$-dimensional spherical space forms. For the historical context refer to this article. Today we know the whole list of groups which arise as fundamental groups of $3$$3$-dimensional spherical space forms. These are exactly groups which admit a fixed-point free representation in $SO(4)$$SO(4)$. In 1950's Milnor in [Milnor1957] provided a list of all finite groups which could possibly act freely but not necessarilly linearly on $S^3$$S^3$. All groups mentioned earlier belong to this list however there was also included a family of finite groups denoted by $Q(8n,k,l)$$Q(8n,k,l)$. Question whether these groups act on $S^3$$S^3$ remained unsolved until the proof of the Geometrization Conjecture was finished by Perelman. The exposition of the first part is based on [Wolf2011].

2 Finite subgroups of SO(4)

To determine finite subgroups of $SO(4)$$SO(4)$ it is necessary to proceed in three steps:

• determine finite subgroups of $SO(3)$$SO(3)$,
• use the covering map $S^3 \to SO(3)$$S^3 \to SO(3)$ to determine finite subgroups of $S^3$$S^3$,
• use the fact that $SO(4)$$SO(4)$ is doubly covered by $S^3 \times S^3$$S^3 \times S^3$ to determine its finite groups.

2.1 Finite subgroups of SO(3)

To classify finite subgroups of $SO(3)$$SO(3)$ one has to analyse action of these groups on $S^2$$S^2$. From Riemann-Hurwitz formula we obtain the following equation

$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$ $$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$$

where $N$$N$ denotes the order of the group, $q$$q$ denotes number of orbits with non-trivial isotropy and $n_i$$n_i$ denotes the order of the respective isotropy subgroup. Solutions to this equation yield the desired list.

Theorem 2.1 [Wolf2011, thm 2.6.5.]. Every finite subgroup of $SO(3)$$SO(3)$ is either

• cyclic,
• dihedral,
• tetrahedral,
• octahedral or
• icosahedral.

These groups are called polyhedral groups.

2.2 Finite subgroups of S^3

Let $\mathbb{H}$$\mathbb{H}$ denote the algebra of quaternions and treat $S^3$$S^3$ as a subset of $\mathbb{H}$$\mathbb{H}$ of quaternions of norm $1$$1$. Consider an action of $S^3$$S^3$ on $\mathbb{H}$$\mathbb{H}$ by conjugation

$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$ $$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$$

This action preserves $1$$1$, so it induces an action on the set of imaginary quaternions which preserves the norm. Therefore this action yields a representation $\pi \colon S^3 \to SO(3)$$\pi \colon S^3 \to SO(3)$ with kernel equal to $\{\pm 1\}$$\{\pm 1\}$.

If $G$$G$ is a finite subgroup of $S^3$$S^3$, then let $F = \pi(G)$$F = \pi(G)$. If $F = G$$F = G$, then, since $-1$$-1$ is the only element of $S^3$$S^3$ of order $2$$2$, $F$$F$ and $G$$G$ must be both of odd order. Therefore comparing this with the list of finite subgroups of $SO(3)$$SO(3)$ yields that $F$$F$ and $G$$G$ are both cyclic of odd order. On the other hand, if $F \neq G$$F \neq G$, then $G$$G$ is an extension of the form

$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$ $$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$$

These considerations yields the following theorem.

Theorem 2.2 [Wolf2011]. Every finite subgroup of $S^3$$S^3$ is either

• cyclic group,
• binary dihedral group,
• binary tetrahedral group,
• binary octahedral group or
• binary icosahedralhedral group.

These groups are called binary polyhedral groups.

2.3 Finite subgroups of SO(4)

To perform the final step, consider a homomophism

$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$ $$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$$

Its kernel is equal to $\{(1,1), (-1,-1)\}$$\{(1,1), (-1,-1)\}$.

Finite subgroups of $S^3 \times S^3$$S^3 \times S^3$ can be determined by Goursat's lemma. This lemma says, that every finite subgroup of $S^3 \times S^3$$S^3 \times S^3$ is isomorphic to the fibre product $G \times_{Q} H$$G \times_{Q} H$, where $G$$G$ and $H$$H$ are finite subgroups of $S^3$$S^3$ and $Q$$Q$ is a common quotient of $G$$G$ and $H$$H$.

2.4 Finite fixed-point free subgroups of SO(4)

Not every finite subgroup of $SO(4)$$SO(4)$ act freely on $S^3$$S^3$. Following lemma gives necessary and sufficient condition for $F(q_1,q_2)$$F(q_1,q_2)$ to be fixed point free for $q_1, q_2 \in S^3$$q_1, q_2 \in S^3$.

Lemma 2.3 [Wolf2011]. Let $q_1, q_2$$q_1, q_2$ be unit quaternions, then $F(q_1,q_2)$$F(q_1,q_2)$ has a fixed point on $S^3$$S^3$ if, and only if, $q_1$$q_1$ is conjgate to $q_2$$q_2$ in $S^3$$S^3$.

Proof. This is a simple observation

$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$ $$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$$

$\square$$\square$

Theorem 2.4 [Wolf2011]. Finite fixed-point free subgroup of $SO(4)$$SO(4)$ belongs to the following list

• cyclic group,
• generalised quaternion group $Q_{8k}$$Q_{8k}$,
• binary tetrahedral group $T^{\ast}$$T^{\ast}$,
• binary icosahedral group $I^{\ast}$$I^{\ast}$,
• groups $D_{2^k(2n+1)}$$D_{2^k(2n+1)}$, for $k \geq 2$$k \geq 2$ and $n \geq 1$$n \geq 1$, with presentation

$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$ $$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$$

• groups $P_{8 \cdot 3^{k}}'$$P_{8 \cdot 3^{k}}'$ defined by the following presentation

$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$ $$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$$

• direct product of any of the above group with a cyclic group of relatively prime order.

3 Milnor's contribution

In [Milnor1957] author proves the following theorem.

Theorem 3.1. If $G$$G$ is a finite group which admits a fixed-point free action on a sphere $S^{2n+1}$$S^{2n+1}$, then for every prime $p$$p$ every subgroup of $G$$G$ of order $2p$$2p$ is cyclic.

This theorem allowed him to choose all possible candidates, from the list compiled by Zassenhaus and Suzuki of groups with periodic cohomology, which could possibly act on $S^3$$S^3$. Apart from groups which admit a fixed point free representation in $SO(4)$$SO(4)$ he obtained the following family of groups.

$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$ $$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$$

where $8n,k,l$$8n,k,l$ are relatively prime integers and

$\displaystyle r \equiv -1 \pmod{k},$ $$\displaystyle r \equiv -1 \pmod{k},$$

$\displaystyle r \equiv 1 \pmod{l}.$ $$\displaystyle r \equiv 1 \pmod{l}.$$

Groups $Q(8n,k,l)$$Q(8n,k,l)$ were excluded from the list of fundamental groups of $3$$3$-manifolds only after resolution of the Geometrization Conjecture.

4 References

• [Milnor1957] J. Milnor, Groups which act on $S^n$$S^n$ without fixed points, Amer. J. Math. 79 (1957), 623–630. MR0090056 (19,761d)
• [Wolf2011] J. A. Wolf, Spaces of constant curvature, AMS Chelsea Publishing, Providence, RI, 2011. MR2742530 (2011j:53001) Zbl 05830219

$. Consider an action of $S^3$ on $\mathbb{H}$ by conjugation $$q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$$ This action preserves 3-dimensional spherical space forms. For the historical context refer to this article. Today we know the whole list of groups which arise as fundamental groups of $3$$3$-dimensional spherical space forms. These are exactly groups which admit a fixed-point free representation in $SO(4)$$SO(4)$. In 1950's Milnor in [Milnor1957] provided a list of all finite groups which could possibly act freely but not necessarilly linearly on $S^3$$S^3$. All groups mentioned earlier belong to this list however there was also included a family of finite groups denoted by $Q(8n,k,l)$$Q(8n,k,l)$. Question whether these groups act on $S^3$$S^3$ remained unsolved until the proof of the Geometrization Conjecture was finished by Perelman. The exposition of the first part is based on [Wolf2011].

2 Finite subgroups of SO(4)

To determine finite subgroups of $SO(4)$$SO(4)$ it is necessary to proceed in three steps:

• determine finite subgroups of $SO(3)$$SO(3)$,
• use the covering map $S^3 \to SO(3)$$S^3 \to SO(3)$ to determine finite subgroups of $S^3$$S^3$,
• use the fact that $SO(4)$$SO(4)$ is doubly covered by $S^3 \times S^3$$S^3 \times S^3$ to determine its finite groups.

2.1 Finite subgroups of SO(3)

To classify finite subgroups of $SO(3)$$SO(3)$ one has to analyse action of these groups on $S^2$$S^2$. From Riemann-Hurwitz formula we obtain the following equation

$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$ $$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$$

where $N$$N$ denotes the order of the group, $q$$q$ denotes number of orbits with non-trivial isotropy and $n_i$$n_i$ denotes the order of the respective isotropy subgroup. Solutions to this equation yield the desired list.

Theorem 2.1 [Wolf2011, thm 2.6.5.]. Every finite subgroup of $SO(3)$$SO(3)$ is either

• cyclic,
• dihedral,
• tetrahedral,
• octahedral or
• icosahedral.

These groups are called polyhedral groups.

2.2 Finite subgroups of S^3

Let $\mathbb{H}$$\mathbb{H}$ denote the algebra of quaternions and treat $S^3$$S^3$ as a subset of $\mathbb{H}$$\mathbb{H}$ of quaternions of norm $1$$1$. Consider an action of $S^3$$S^3$ on $\mathbb{H}$$\mathbb{H}$ by conjugation

$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$ $$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$$

This action preserves $1$$1$, so it induces an action on the set of imaginary quaternions which preserves the norm. Therefore this action yields a representation $\pi \colon S^3 \to SO(3)$$\pi \colon S^3 \to SO(3)$ with kernel equal to $\{\pm 1\}$$\{\pm 1\}$.

If $G$$G$ is a finite subgroup of $S^3$$S^3$, then let $F = \pi(G)$$F = \pi(G)$. If $F = G$$F = G$, then, since $-1$$-1$ is the only element of $S^3$$S^3$ of order $2$$2$, $F$$F$ and $G$$G$ must be both of odd order. Therefore comparing this with the list of finite subgroups of $SO(3)$$SO(3)$ yields that $F$$F$ and $G$$G$ are both cyclic of odd order. On the other hand, if $F \neq G$$F \neq G$, then $G$$G$ is an extension of the form

$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$ $$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$$

These considerations yields the following theorem.

Theorem 2.2 [Wolf2011]. Every finite subgroup of $S^3$$S^3$ is either

• cyclic group,
• binary dihedral group,
• binary tetrahedral group,
• binary octahedral group or
• binary icosahedralhedral group.

These groups are called binary polyhedral groups.

2.3 Finite subgroups of SO(4)

To perform the final step, consider a homomophism

$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$ $$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$$

Its kernel is equal to $\{(1,1), (-1,-1)\}$$\{(1,1), (-1,-1)\}$.

Finite subgroups of $S^3 \times S^3$$S^3 \times S^3$ can be determined by Goursat's lemma. This lemma says, that every finite subgroup of $S^3 \times S^3$$S^3 \times S^3$ is isomorphic to the fibre product $G \times_{Q} H$$G \times_{Q} H$, where $G$$G$ and $H$$H$ are finite subgroups of $S^3$$S^3$ and $Q$$Q$ is a common quotient of $G$$G$ and $H$$H$.

2.4 Finite fixed-point free subgroups of SO(4)

Not every finite subgroup of $SO(4)$$SO(4)$ act freely on $S^3$$S^3$. Following lemma gives necessary and sufficient condition for $F(q_1,q_2)$$F(q_1,q_2)$ to be fixed point free for $q_1, q_2 \in S^3$$q_1, q_2 \in S^3$.

Lemma 2.3 [Wolf2011]. Let $q_1, q_2$$q_1, q_2$ be unit quaternions, then $F(q_1,q_2)$$F(q_1,q_2)$ has a fixed point on $S^3$$S^3$ if, and only if, $q_1$$q_1$ is conjgate to $q_2$$q_2$ in $S^3$$S^3$.

Proof. This is a simple observation

$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$ $$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$$

$\square$$\square$

Theorem 2.4 [Wolf2011]. Finite fixed-point free subgroup of $SO(4)$$SO(4)$ belongs to the following list

• cyclic group,
• generalised quaternion group $Q_{8k}$$Q_{8k}$,
• binary tetrahedral group $T^{\ast}$$T^{\ast}$,
• binary icosahedral group $I^{\ast}$$I^{\ast}$,
• groups $D_{2^k(2n+1)}$$D_{2^k(2n+1)}$, for $k \geq 2$$k \geq 2$ and $n \geq 1$$n \geq 1$, with presentation

$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$ $$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$$

• groups $P_{8 \cdot 3^{k}}'$$P_{8 \cdot 3^{k}}'$ defined by the following presentation

$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$ $$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$$

• direct product of any of the above group with a cyclic group of relatively prime order.

3 Milnor's contribution

In [Milnor1957] author proves the following theorem.

Theorem 3.1. If $G$$G$ is a finite group which admits a fixed-point free action on a sphere $S^{2n+1}$$S^{2n+1}$, then for every prime $p$$p$ every subgroup of $G$$G$ of order $2p$$2p$ is cyclic.

This theorem allowed him to choose all possible candidates, from the list compiled by Zassenhaus and Suzuki of groups with periodic cohomology, which could possibly act on $S^3$$S^3$. Apart from groups which admit a fixed point free representation in $SO(4)$$SO(4)$ he obtained the following family of groups.

$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$ $$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$$

where $8n,k,l$$8n,k,l$ are relatively prime integers and

$\displaystyle r \equiv -1 \pmod{k},$ $$\displaystyle r \equiv -1 \pmod{k},$$

$\displaystyle r \equiv 1 \pmod{l}.$ $$\displaystyle r \equiv 1 \pmod{l}.$$

Groups $Q(8n,k,l)$$Q(8n,k,l)$ were excluded from the list of fundamental groups of $3$$3$-manifolds only after resolution of the Geometrization Conjecture.

4 References

• [Milnor1957] J. Milnor, Groups which act on $S^n$$S^n$ without fixed points, Amer. J. Math. 79 (1957), 623–630. MR0090056 (19,761d)
• [Wolf2011] J. A. Wolf, Spaces of constant curvature, AMS Chelsea Publishing, Providence, RI, 2011. MR2742530 (2011j:53001) Zbl 05830219

$, so it induces an action on the set of imaginary quaternions which preserves the norm. Therefore this action yields a representation $\pi \colon S^3 \to SO(3)$ with kernel equal to $\{\pm 1\}$. If $G$ is a finite subgroup of $S^3$, then let $F = \pi(G)$. If $F = G$, then, since $-1$ is the only element of $S^3$ of order $, $F$ and $G$ must be both of odd order. Therefore comparing this with the list of finite subgroups of $SO(3)$ yields that $F$ and $G$ are both cyclic of odd order. On the other hand, if $F \neq G$, then $G$ is an extension of the form $3-dimensional spherical space forms. For the historical context refer to this article. Today we know the whole list of groups which arise as fundamental groups of $3$$3$-dimensional spherical space forms. These are exactly groups which admit a fixed-point free representation in $SO(4)$$SO(4)$. In 1950's Milnor in [Milnor1957] provided a list of all finite groups which could possibly act freely but not necessarilly linearly on $S^3$$S^3$. All groups mentioned earlier belong to this list however there was also included a family of finite groups denoted by $Q(8n,k,l)$$Q(8n,k,l)$. Question whether these groups act on $S^3$$S^3$ remained unsolved until the proof of the Geometrization Conjecture was finished by Perelman. The exposition of the first part is based on [Wolf2011].

2 Finite subgroups of SO(4)

To determine finite subgroups of $SO(4)$$SO(4)$ it is necessary to proceed in three steps:

• determine finite subgroups of $SO(3)$$SO(3)$,
• use the covering map $S^3 \to SO(3)$$S^3 \to SO(3)$ to determine finite subgroups of $S^3$$S^3$,
• use the fact that $SO(4)$$SO(4)$ is doubly covered by $S^3 \times S^3$$S^3 \times S^3$ to determine its finite groups.

2.1 Finite subgroups of SO(3)

To classify finite subgroups of $SO(3)$$SO(3)$ one has to analyse action of these groups on $S^2$$S^2$. From Riemann-Hurwitz formula we obtain the following equation

$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$ $$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$$

where $N$$N$ denotes the order of the group, $q$$q$ denotes number of orbits with non-trivial isotropy and $n_i$$n_i$ denotes the order of the respective isotropy subgroup. Solutions to this equation yield the desired list.

Theorem 2.1 [Wolf2011, thm 2.6.5.]. Every finite subgroup of $SO(3)$$SO(3)$ is either

• cyclic,
• dihedral,
• tetrahedral,
• octahedral or
• icosahedral.

These groups are called polyhedral groups.

2.2 Finite subgroups of S^3

Let $\mathbb{H}$$\mathbb{H}$ denote the algebra of quaternions and treat $S^3$$S^3$ as a subset of $\mathbb{H}$$\mathbb{H}$ of quaternions of norm $1$$1$. Consider an action of $S^3$$S^3$ on $\mathbb{H}$$\mathbb{H}$ by conjugation

$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$ $$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$$

This action preserves $1$$1$, so it induces an action on the set of imaginary quaternions which preserves the norm. Therefore this action yields a representation $\pi \colon S^3 \to SO(3)$$\pi \colon S^3 \to SO(3)$ with kernel equal to $\{\pm 1\}$$\{\pm 1\}$.

If $G$$G$ is a finite subgroup of $S^3$$S^3$, then let $F = \pi(G)$$F = \pi(G)$. If $F = G$$F = G$, then, since $-1$$-1$ is the only element of $S^3$$S^3$ of order $2$$2$, $F$$F$ and $G$$G$ must be both of odd order. Therefore comparing this with the list of finite subgroups of $SO(3)$$SO(3)$ yields that $F$$F$ and $G$$G$ are both cyclic of odd order. On the other hand, if $F \neq G$$F \neq G$, then $G$$G$ is an extension of the form

$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$ $$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$$

These considerations yields the following theorem.

Theorem 2.2 [Wolf2011]. Every finite subgroup of $S^3$$S^3$ is either

• cyclic group,
• binary dihedral group,
• binary tetrahedral group,
• binary octahedral group or
• binary icosahedralhedral group.

These groups are called binary polyhedral groups.

2.3 Finite subgroups of SO(4)

To perform the final step, consider a homomophism

$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$ $$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$$

Its kernel is equal to $\{(1,1), (-1,-1)\}$$\{(1,1), (-1,-1)\}$.

Finite subgroups of $S^3 \times S^3$$S^3 \times S^3$ can be determined by Goursat's lemma. This lemma says, that every finite subgroup of $S^3 \times S^3$$S^3 \times S^3$ is isomorphic to the fibre product $G \times_{Q} H$$G \times_{Q} H$, where $G$$G$ and $H$$H$ are finite subgroups of $S^3$$S^3$ and $Q$$Q$ is a common quotient of $G$$G$ and $H$$H$.

2.4 Finite fixed-point free subgroups of SO(4)

Not every finite subgroup of $SO(4)$$SO(4)$ act freely on $S^3$$S^3$. Following lemma gives necessary and sufficient condition for $F(q_1,q_2)$$F(q_1,q_2)$ to be fixed point free for $q_1, q_2 \in S^3$$q_1, q_2 \in S^3$.

Lemma 2.3 [Wolf2011]. Let $q_1, q_2$$q_1, q_2$ be unit quaternions, then $F(q_1,q_2)$$F(q_1,q_2)$ has a fixed point on $S^3$$S^3$ if, and only if, $q_1$$q_1$ is conjgate to $q_2$$q_2$ in $S^3$$S^3$.

Proof. This is a simple observation

$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$ $$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$$

$\square$$\square$

Theorem 2.4 [Wolf2011]. Finite fixed-point free subgroup of $SO(4)$$SO(4)$ belongs to the following list

• cyclic group,
• generalised quaternion group $Q_{8k}$$Q_{8k}$,
• binary tetrahedral group $T^{\ast}$$T^{\ast}$,
• binary icosahedral group $I^{\ast}$$I^{\ast}$,
• groups $D_{2^k(2n+1)}$$D_{2^k(2n+1)}$, for $k \geq 2$$k \geq 2$ and $n \geq 1$$n \geq 1$, with presentation

$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$ $$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$$

• groups $P_{8 \cdot 3^{k}}'$$P_{8 \cdot 3^{k}}'$ defined by the following presentation

$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$ $$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$$

• direct product of any of the above group with a cyclic group of relatively prime order.

3 Milnor's contribution

In [Milnor1957] author proves the following theorem.

Theorem 3.1. If $G$$G$ is a finite group which admits a fixed-point free action on a sphere $S^{2n+1}$$S^{2n+1}$, then for every prime $p$$p$ every subgroup of $G$$G$ of order $2p$$2p$ is cyclic.

This theorem allowed him to choose all possible candidates, from the list compiled by Zassenhaus and Suzuki of groups with periodic cohomology, which could possibly act on $S^3$$S^3$. Apart from groups which admit a fixed point free representation in $SO(4)$$SO(4)$ he obtained the following family of groups.

$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$ $$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$$

where $8n,k,l$$8n,k,l$ are relatively prime integers and

$\displaystyle r \equiv -1 \pmod{k},$ $$\displaystyle r \equiv -1 \pmod{k},$$

$\displaystyle r \equiv 1 \pmod{l}.$ $$\displaystyle r \equiv 1 \pmod{l}.$$

Groups $Q(8n,k,l)$$Q(8n,k,l)$ were excluded from the list of fundamental groups of $3$$3$-manifolds only after resolution of the Geometrization Conjecture.

4 References

• [Milnor1957] J. Milnor, Groups which act on $S^n$$S^n$ without fixed points, Amer. J. Math. 79 (1957), 623–630. MR0090056 (19,761d)
• [Wolf2011] J. A. Wolf, Spaces of constant curvature, AMS Chelsea Publishing, Providence, RI, 2011. MR2742530 (2011j:53001) Zbl 05830219

\to \{\pm 1\} \to G \to F \to 1.$$ These considerations yields the following theorem. {{beginthm|Theorem|{{cite|Wolf2011}}}} Every finite subgroup of $S^3$ is either * [[Wikipedia:Cyclic_group|cyclic group]], * [[Wikipedia:Dicyclic_group|binary dihedral group]], * [[Wikipedia:Binary_tetrahedral_group|binary tetrahedral group]], * [[Wikipedia:Binary_octahedral_group|binary octahedral group]] or * [[Wikipedia:Binary_icosahedral_group|binary icosahedralhedral group]]. These groups are called binary polyhedral groups. {{endthm}} === Finite subgroups of SO(4) === ; To perform the final step, consider a homomophism $$F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}. $$ Its kernel is equal to $\{(1,1), (-1,-1)\}$. Finite subgroups of $S^3 \times S^3$ can be determined by [[Wikipedia:Goursat%27s_lemma|Goursat's lemma]]. This lemma says, that every finite subgroup of $S^3 \times S^3$ is isomorphic to the fibre product $G \times_{Q} H$, where $G$ and $H$ are finite subgroups of $S^3$ and $Q$ is a common quotient of $G$ and $H$. === Finite fixed-point free subgroups of SO(4) === ; Not every finite subgroup of $SO(4)$ act freely on $S^3$. Following lemma gives necessary and sufficient condition for $F(q_1,q_2)$ to be fixed point free for $q_1, q_2 \in S^3$. {{beginthm|Lemma|{{cite|Wolf2011}}}} Let $q_1, q_2$ be unit quaternions, then $F(q_1,q_2)$ has a fixed point on $S^3$ if, and only if, $q_1$ is conjgate to $q_2$ in $S^3$. {{endthm}} {{beginproof}} This is a simple observation $$q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$$ {{endproof}} {{beginthm|Theorem|{{cite|Wolf2011}}}} Finite fixed-point free subgroup of $SO(4)$ belongs to the following list * [[Wikipedia:Cyclic_group|cyclic group]], * [[Wikipedia:Quaternion_group|generalised quaternion group]] $Q_{8k}$, * [[Wikipedia:Binary_tetrahedral_group|binary tetrahedral group]] $T^{\ast}$, * [[Wikipedia:Binary_icosahedral_group|binary icosahedral group]] $I^{\ast}$, * groups $D_{2^k(2n+1)}$, for $k \geq 2$ and $n \geq 1$, with presentation $$\langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$$ * groups $P_{8 \cdot 3^{k}}'$ defined by the following presentation $$\langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$$ * direct product of any of the above group with a cyclic group of relatively prime order. {{endthm}} == Milnor's contribution == ; In {{cite|Milnor1957}} author proves the following theorem. {{beginthm|Theorem|}} If $G$ is a finite group which admits a fixed-point free action on a sphere $S^{2n+1}$, then for every prime $p$ every subgroup of $G$ of order p$ is cyclic. {{endthm}} This theorem allowed him to choose all possible candidates, from the list compiled by Zassenhaus and Suzuki of groups with periodic cohomology, which could possibly act on $S^3$. Apart from groups which admit a fixed point free representation in $SO(4)$ he obtained the following family of groups. $$Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$$ where n,k,l$ are relatively prime integers and $$r \equiv -1 \pmod{k},$$ $$r \equiv 1 \pmod{l}.$$ Groups $Q(8n,k,l)$ were excluded from the list of fundamental groups of $-manifolds only after resolution of the Geometrization Conjecture. == References == {{#RefList:}} [[Category:Theory]]3-dimensional spherical space forms. For the historical context refer to this article. Today we know the whole list of groups which arise as fundamental groups of $3$$3$-dimensional spherical space forms. These are exactly groups which admit a fixed-point free representation in $SO(4)$$SO(4)$. In 1950's Milnor in [Milnor1957] provided a list of all finite groups which could possibly act freely but not necessarilly linearly on $S^3$$S^3$. All groups mentioned earlier belong to this list however there was also included a family of finite groups denoted by $Q(8n,k,l)$$Q(8n,k,l)$. Question whether these groups act on $S^3$$S^3$ remained unsolved until the proof of the Geometrization Conjecture was finished by Perelman. The exposition of the first part is based on [Wolf2011].

2 Finite subgroups of SO(4)

To determine finite subgroups of $SO(4)$$SO(4)$ it is necessary to proceed in three steps:

• determine finite subgroups of $SO(3)$$SO(3)$,
• use the covering map $S^3 \to SO(3)$$S^3 \to SO(3)$ to determine finite subgroups of $S^3$$S^3$,
• use the fact that $SO(4)$$SO(4)$ is doubly covered by $S^3 \times S^3$$S^3 \times S^3$ to determine its finite groups.

2.1 Finite subgroups of SO(3)

To classify finite subgroups of $SO(3)$$SO(3)$ one has to analyse action of these groups on $S^2$$S^2$. From Riemann-Hurwitz formula we obtain the following equation

$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$ $$\displaystyle 2 \left(1 - \frac{1}{N} \right) = \sum_{i = 1}^q \left(1 - \frac{1}{n_i} \right),$$

where $N$$N$ denotes the order of the group, $q$$q$ denotes number of orbits with non-trivial isotropy and $n_i$$n_i$ denotes the order of the respective isotropy subgroup. Solutions to this equation yield the desired list.

Theorem 2.1 [Wolf2011, thm 2.6.5.]. Every finite subgroup of $SO(3)$$SO(3)$ is either

• cyclic,
• dihedral,
• tetrahedral,
• octahedral or
• icosahedral.

These groups are called polyhedral groups.

2.2 Finite subgroups of S^3

Let $\mathbb{H}$$\mathbb{H}$ denote the algebra of quaternions and treat $S^3$$S^3$ as a subset of $\mathbb{H}$$\mathbb{H}$ of quaternions of norm $1$$1$. Consider an action of $S^3$$S^3$ on $\mathbb{H}$$\mathbb{H}$ by conjugation

$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$ $$\displaystyle q \mapsto (q' \mapsto q \cdot q' \cdot q^{-1}).$$

This action preserves $1$$1$, so it induces an action on the set of imaginary quaternions which preserves the norm. Therefore this action yields a representation $\pi \colon S^3 \to SO(3)$$\pi \colon S^3 \to SO(3)$ with kernel equal to $\{\pm 1\}$$\{\pm 1\}$.

If $G$$G$ is a finite subgroup of $S^3$$S^3$, then let $F = \pi(G)$$F = \pi(G)$. If $F = G$$F = G$, then, since $-1$$-1$ is the only element of $S^3$$S^3$ of order $2$$2$, $F$$F$ and $G$$G$ must be both of odd order. Therefore comparing this with the list of finite subgroups of $SO(3)$$SO(3)$ yields that $F$$F$ and $G$$G$ are both cyclic of odd order. On the other hand, if $F \neq G$$F \neq G$, then $G$$G$ is an extension of the form

$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$ $$\displaystyle 1 \to \{\pm 1\} \to G \to F \to 1.$$

These considerations yields the following theorem.

Theorem 2.2 [Wolf2011]. Every finite subgroup of $S^3$$S^3$ is either

• cyclic group,
• binary dihedral group,
• binary tetrahedral group,
• binary octahedral group or
• binary icosahedralhedral group.

These groups are called binary polyhedral groups.

2.3 Finite subgroups of SO(4)

To perform the final step, consider a homomophism

$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$ $$\displaystyle F \colon S^3 \times S^3 \to SO(4), \quad F(q_1, q_2) = q_1 \cdot q \cdot q_2^{-1}.$$

Its kernel is equal to $\{(1,1), (-1,-1)\}$$\{(1,1), (-1,-1)\}$.

Finite subgroups of $S^3 \times S^3$$S^3 \times S^3$ can be determined by Goursat's lemma. This lemma says, that every finite subgroup of $S^3 \times S^3$$S^3 \times S^3$ is isomorphic to the fibre product $G \times_{Q} H$$G \times_{Q} H$, where $G$$G$ and $H$$H$ are finite subgroups of $S^3$$S^3$ and $Q$$Q$ is a common quotient of $G$$G$ and $H$$H$.

2.4 Finite fixed-point free subgroups of SO(4)

Not every finite subgroup of $SO(4)$$SO(4)$ act freely on $S^3$$S^3$. Following lemma gives necessary and sufficient condition for $F(q_1,q_2)$$F(q_1,q_2)$ to be fixed point free for $q_1, q_2 \in S^3$$q_1, q_2 \in S^3$.

Lemma 2.3 [Wolf2011]. Let $q_1, q_2$$q_1, q_2$ be unit quaternions, then $F(q_1,q_2)$$F(q_1,q_2)$ has a fixed point on $S^3$$S^3$ if, and only if, $q_1$$q_1$ is conjgate to $q_2$$q_2$ in $S^3$$S^3$.

Proof. This is a simple observation

$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$ $$\displaystyle q_1 \cdot a \cdot q_2^{-1} = a \iff q_1 = a \cdot q_2 \cdot a^{-1}.$$

$\square$$\square$

Theorem 2.4 [Wolf2011]. Finite fixed-point free subgroup of $SO(4)$$SO(4)$ belongs to the following list

• cyclic group,
• generalised quaternion group $Q_{8k}$$Q_{8k}$,
• binary tetrahedral group $T^{\ast}$$T^{\ast}$,
• binary icosahedral group $I^{\ast}$$I^{\ast}$,
• groups $D_{2^k(2n+1)}$$D_{2^k(2n+1)}$, for $k \geq 2$$k \geq 2$ and $n \geq 1$$n \geq 1$, with presentation

$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$ $$\displaystyle \langle x, y \mid x^{2^k} = y^{2n+1} = 1, xyx^{-1} = y^{-1} \rangle.$$

• groups $P_{8 \cdot 3^{k}}'$$P_{8 \cdot 3^{k}}'$ defined by the following presentation

$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$ $$\displaystyle \langle x,y,z | x^2 = (xy)^2 = y^2, zxz^{-1} = y, zyz^{-1} = xy, z^{3^k} = 1 \rangle,$$

• direct product of any of the above group with a cyclic group of relatively prime order.

3 Milnor's contribution

In [Milnor1957] author proves the following theorem.

Theorem 3.1. If $G$$G$ is a finite group which admits a fixed-point free action on a sphere $S^{2n+1}$$S^{2n+1}$, then for every prime $p$$p$ every subgroup of $G$$G$ of order $2p$$2p$ is cyclic.

This theorem allowed him to choose all possible candidates, from the list compiled by Zassenhaus and Suzuki of groups with periodic cohomology, which could possibly act on $S^3$$S^3$. Apart from groups which admit a fixed point free representation in $SO(4)$$SO(4)$ he obtained the following family of groups.

$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$ $$\displaystyle Q(8n,k,l) = \langle x,y,z \mid x^2 = (xy)^2=y^{2n}, z^{kl}=1, xzx^{-1} = z^{r}, yzy^{-1} = z^{-1} \rangle,$$

where $8n,k,l$$8n,k,l$ are relatively prime integers and

$\displaystyle r \equiv -1 \pmod{k},$ $$\displaystyle r \equiv -1 \pmod{k},$$

$\displaystyle r \equiv 1 \pmod{l}.$ $$\displaystyle r \equiv 1 \pmod{l}.$$

Groups $Q(8n,k,l)$$Q(8n,k,l)$ were excluded from the list of fundamental groups of $3$$3$-manifolds only after resolution of the Geometrization Conjecture.

4 References

• [Milnor1957] J. Milnor, Groups which act on $S^n$$S^n$ without fixed points, Amer. J. Math. 79 (1957), 623–630. MR0090056 (19,761d)
• [Wolf2011] J. A. Wolf, Spaces of constant curvature, AMS Chelsea Publishing, Providence, RI, 2011. MR2742530 (2011j:53001) Zbl 05830219