The fundamental groups of 3-dimensional spherical space forms The 2019 Stack Overflow...
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The fundamental groups of 3-dimensional spherical space forms
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$begingroup$
Let $S^3/Gamma_i,(i=1,2)$ be a $3$-dimensional spherical space form, where $Gamma_i subset SO(4)$ is a finite subgroup acting freely on $S^3$. If $S^3/Gamma_1$ is homotopy equivalent to $S^3/Gamma_2$, can we find a $g in SO(4)$ such that
$$
gGamma_1 g^{-1}=Gamma_2?
$$
group-theory geometric-topology orthogonal-matrices low-dimensional-topology
asked Mar 21 at 14:38
TotoroTotoro
28915
$endgroup$
add a comment |
$begingroup$
Let $S^3/Gamma_i,(i=1,2)$ be a $3$-dimensional spherical space form, where $Gamma_i subset SO(4)$ is a finite subgroup acting freely on $S^3$. If $S^3/Gamma_1$ is homotopy equivalent to $S^3/Gamma_2$, can we find a $g in SO(4)$ such that
$$
gGamma_1 g^{-1}=Gamma_2?
$$
group-theory geometric-topology orthogonal-matrices low-dimensional-topology
asked Mar 21 at 14:38
TotoroTotoro
28915
$endgroup$
$begingroup$
On the other hand, see en.wikipedia.org/wiki/Mostow_rigidity_theorem.
$endgroup$
– Qiaochu Yuan
Mar 23 at 19:36
add a comment |
$begingroup$
Let $S^3/Gamma_i,(i=1,2)$ be a $3$-dimensional spherical space form, where $Gamma_i subset SO(4)$ is a finite subgroup acting freely on $S^3$. If $S^3/Gamma_1$ is homotopy equivalent to $S^3/Gamma_2$, can we find a $g in SO(4)$ such that
$$
gGamma_1 g^{-1}=Gamma_2?
$$
group-theory geometric-topology orthogonal-matrices low-dimensional-topology
asked Mar 21 at 14:38
TotoroTotoro
28915
$endgroup$
Let $S^3/Gamma_i,(i=1,2)$ be a $3$-dimensional spherical space form, where $Gamma_i subset SO(4)$ is a finite subgroup acting freely on $S^3$. If $S^3/Gamma_1$ is homotopy equivalent to $S^3/Gamma_2$, can we find a $g in SO(4)$ such that
$$
gGamma_1 g^{-1}=Gamma_2?
$$
group-theory geometric-topology orthogonal-matrices low-dimensional-topology
group-theory geometric-topology orthogonal-matrices low-dimensional-topology
asked Mar 21 at 14:38
TotoroTotoro
28915
asked Mar 21 at 14:38
TotoroTotoro
28915
asked Mar 21 at 14:38
TotoroTotoro
28915
asked Mar 21 at 14:38
TotoroTotoro
28915
asked Mar 21 at 14:38
TotoroTotoro
28915
28915
$begingroup$
On the other hand, see en.wikipedia.org/wiki/Mostow_rigidity_theorem.
$endgroup$
– Qiaochu Yuan
Mar 23 at 19:36
add a comment |
$begingroup$
On the other hand, see en.wikipedia.org/wiki/Mostow_rigidity_theorem.
$endgroup$
– Qiaochu Yuan
Mar 23 at 19:36
$begingroup$
On the other hand, see en.wikipedia.org/wiki/Mostow_rigidity_theorem.
$endgroup$
– Qiaochu Yuan
Mar 23 at 19:36
$begingroup$
On the other hand, see en.wikipedia.org/wiki/Mostow_rigidity_theorem.
$endgroup$
– Qiaochu Yuan
Mar 23 at 19:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is not true in general, in fact it can fail for the lens spaces which are the quotients
$$L(p;q) = S^3 / Gamma(p;q)
$$
where $Gamma(p;q)$ is the cyclic subgroup of order $p$ in $SO(4)$ that is generated by the element described, in complex coordinates, by the formula
$$(z_1,z_2) mapsto (e^{2 pi i/p} cdot z_1, e^{2 pi i q/p} cdot z_2)
$$
The link given explains how to determine when $L(p;q_1)$ and $L(p;q_2)$ are homotopy equivalent and when they are homeomorphic, and these do not match up. And if they are not homeomorphic then there cannot be any $g in SO(4)$ for which $g Gamma(p;q_1) g^{-1} = Gamma(p;q_2)$, because the map of $SO(4)$ determined by $g$ would descend to a homeomorphism between $L(p;q_1)$ and $L(p;q_2)$.
edited Mar 23 at 18:51
janmarqz
6,29241630
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not true in general, in fact it can fail for the lens spaces which are the quotients
$$L(p;q) = S^3 / Gamma(p;q)
$$
where $Gamma(p;q)$ is the cyclic subgroup of order $p$ in $SO(4)$ that is generated by the element described, in complex coordinates, by the formula
$$(z_1,z_2) mapsto (e^{2 pi i/p} cdot z_1, e^{2 pi i q/p} cdot z_2)
$$
The link given explains how to determine when $L(p;q_1)$ and $L(p;q_2)$ are homotopy equivalent and when they are homeomorphic, and these do not match up. And if they are not homeomorphic then there cannot be any $g in SO(4)$ for which $g Gamma(p;q_1) g^{-1} = Gamma(p;q_2)$, because the map of $SO(4)$ determined by $g$ would descend to a homeomorphism between $L(p;q_1)$ and $L(p;q_2)$.
edited Mar 23 at 18:51
janmarqz
6,29241630
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
$endgroup$
add a comment |
$begingroup$
This is not true in general, in fact it can fail for the lens spaces which are the quotients
$$L(p;q) = S^3 / Gamma(p;q)
$$
where $Gamma(p;q)$ is the cyclic subgroup of order $p$ in $SO(4)$ that is generated by the element described, in complex coordinates, by the formula
$$(z_1,z_2) mapsto (e^{2 pi i/p} cdot z_1, e^{2 pi i q/p} cdot z_2)
$$
The link given explains how to determine when $L(p;q_1)$ and $L(p;q_2)$ are homotopy equivalent and when they are homeomorphic, and these do not match up. And if they are not homeomorphic then there cannot be any $g in SO(4)$ for which $g Gamma(p;q_1) g^{-1} = Gamma(p;q_2)$, because the map of $SO(4)$ determined by $g$ would descend to a homeomorphism between $L(p;q_1)$ and $L(p;q_2)$.
edited Mar 23 at 18:51
janmarqz
6,29241630
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
$endgroup$
add a comment |
$begingroup$
This is not true in general, in fact it can fail for the lens spaces which are the quotients
$$L(p;q) = S^3 / Gamma(p;q)
$$
where $Gamma(p;q)$ is the cyclic subgroup of order $p$ in $SO(4)$ that is generated by the element described, in complex coordinates, by the formula
$$(z_1,z_2) mapsto (e^{2 pi i/p} cdot z_1, e^{2 pi i q/p} cdot z_2)
$$
The link given explains how to determine when $L(p;q_1)$ and $L(p;q_2)$ are homotopy equivalent and when they are homeomorphic, and these do not match up. And if they are not homeomorphic then there cannot be any $g in SO(4)$ for which $g Gamma(p;q_1) g^{-1} = Gamma(p;q_2)$, because the map of $SO(4)$ determined by $g$ would descend to a homeomorphism between $L(p;q_1)$ and $L(p;q_2)$.
edited Mar 23 at 18:51
janmarqz
6,29241630
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
$endgroup$
This is not true in general, in fact it can fail for the lens spaces which are the quotients
$$L(p;q) = S^3 / Gamma(p;q)
$$
where $Gamma(p;q)$ is the cyclic subgroup of order $p$ in $SO(4)$ that is generated by the element described, in complex coordinates, by the formula
$$(z_1,z_2) mapsto (e^{2 pi i/p} cdot z_1, e^{2 pi i q/p} cdot z_2)
$$
The link given explains how to determine when $L(p;q_1)$ and $L(p;q_2)$ are homotopy equivalent and when they are homeomorphic, and these do not match up. And if they are not homeomorphic then there cannot be any $g in SO(4)$ for which $g Gamma(p;q_1) g^{-1} = Gamma(p;q_2)$, because the map of $SO(4)$ determined by $g$ would descend to a homeomorphism between $L(p;q_1)$ and $L(p;q_2)$.
edited Mar 23 at 18:51
janmarqz
6,29241630
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
edited Mar 23 at 18:51
janmarqz
6,29241630
edited Mar 23 at 18:51
janmarqz
6,29241630
edited Mar 23 at 18:51
janmarqz
6,29241630
6,29241630
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
answered Mar 21 at 14:55
Lee MosherLee Mosher
51.8k33889
51.8k33889
add a comment |
add a comment |
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$begingroup$
On the other hand, see en.wikipedia.org/wiki/Mostow_rigidity_theorem.
$endgroup$
– Qiaochu Yuan
Mar 23 at 19:36