isomorphism between subset of SU(2) and SO(3)
I know that there is a surjective map $Phi : SU(2)to SO(3) $.
My question is if there is a subgroup $A subset SU(2)$ such that $Phi_{|A}:Ato SO(3)$ can be a (group) isomorphism.
What would $A$ be?
Thank you
group-theory lie-groups quaternions
edited Nov 25 '18 at 23:12
Bernard
118k639112
asked Nov 25 '18 at 23:03
kot
808
add a comment |
I know that there is a surjective map $Phi : SU(2)to SO(3) $.
My question is if there is a subgroup $A subset SU(2)$ such that $Phi_{|A}:Ato SO(3)$ can be a (group) isomorphism.
What would $A$ be?
Thank you
group-theory lie-groups quaternions
edited Nov 25 '18 at 23:12
Bernard
118k639112
asked Nov 25 '18 at 23:03
kot
808
add a comment |
I know that there is a surjective map $Phi : SU(2)to SO(3) $.
My question is if there is a subgroup $A subset SU(2)$ such that $Phi_{|A}:Ato SO(3)$ can be a (group) isomorphism.
What would $A$ be?
Thank you
group-theory lie-groups quaternions
edited Nov 25 '18 at 23:12
Bernard
118k639112
asked Nov 25 '18 at 23:03
kot
808
I know that there is a surjective map $Phi : SU(2)to SO(3) $.
My question is if there is a subgroup $A subset SU(2)$ such that $Phi_{|A}:Ato SO(3)$ can be a (group) isomorphism.
What would $A$ be?
Thank you
group-theory lie-groups quaternions
group-theory lie-groups quaternions
edited Nov 25 '18 at 23:12
Bernard
118k639112
asked Nov 25 '18 at 23:03
kot
808
edited Nov 25 '18 at 23:12
Bernard
118k639112
asked Nov 25 '18 at 23:03
kot
808
edited Nov 25 '18 at 23:12
Bernard
118k639112
edited Nov 25 '18 at 23:12
Bernard
118k639112
edited Nov 25 '18 at 23:12
Bernard
118k639112
118k639112
asked Nov 25 '18 at 23:03
kot
808
asked Nov 25 '18 at 23:03
kot
808
asked Nov 25 '18 at 23:03
kot
808
808
add a comment |
add a comment |
2 Answers
2
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oldest
votes
Such a subgroup $Asubsetoperatorname{SU}(2)$ fails to exist on topological grounds; I'll assume it to be known that $operatorname{SU}(2)cong S_3$ and $operatorname{SO}(3)congBbb{RP}^3$. The usual surjective map
$$Phi: operatorname{SU}(2) longrightarrow operatorname{SO}(3)$$
is a homomorphism of topological groups, hence its restriction to $A$ is a homeomorphism. This implies that $S_3$ contains a subspace homeomorphic to $Bbb{RP}^3$, a contradiction.
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
add a comment |
Hint If there were such a subgroup $A$, then for any $g in SU(2) - A$, $A cup gA$ would be a separation of $SU(2)$---but $SU(2)$ is connected.
answered Nov 26 '18 at 1:14
Travis
59.7k767146
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
add a comment |
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2 Answers
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2 Answers
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Such a subgroup $Asubsetoperatorname{SU}(2)$ fails to exist on topological grounds; I'll assume it to be known that $operatorname{SU}(2)cong S_3$ and $operatorname{SO}(3)congBbb{RP}^3$. The usual surjective map
$$Phi: operatorname{SU}(2) longrightarrow operatorname{SO}(3)$$
is a homomorphism of topological groups, hence its restriction to $A$ is a homeomorphism. This implies that $S_3$ contains a subspace homeomorphic to $Bbb{RP}^3$, a contradiction.
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
add a comment |
Such a subgroup $Asubsetoperatorname{SU}(2)$ fails to exist on topological grounds; I'll assume it to be known that $operatorname{SU}(2)cong S_3$ and $operatorname{SO}(3)congBbb{RP}^3$. The usual surjective map
$$Phi: operatorname{SU}(2) longrightarrow operatorname{SO}(3)$$
is a homomorphism of topological groups, hence its restriction to $A$ is a homeomorphism. This implies that $S_3$ contains a subspace homeomorphic to $Bbb{RP}^3$, a contradiction.
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
add a comment |
Such a subgroup $Asubsetoperatorname{SU}(2)$ fails to exist on topological grounds; I'll assume it to be known that $operatorname{SU}(2)cong S_3$ and $operatorname{SO}(3)congBbb{RP}^3$. The usual surjective map
$$Phi: operatorname{SU}(2) longrightarrow operatorname{SO}(3)$$
is a homomorphism of topological groups, hence its restriction to $A$ is a homeomorphism. This implies that $S_3$ contains a subspace homeomorphic to $Bbb{RP}^3$, a contradiction.
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
Such a subgroup $Asubsetoperatorname{SU}(2)$ fails to exist on topological grounds; I'll assume it to be known that $operatorname{SU}(2)cong S_3$ and $operatorname{SO}(3)congBbb{RP}^3$. The usual surjective map
$$Phi: operatorname{SU}(2) longrightarrow operatorname{SO}(3)$$
is a homomorphism of topological groups, hence its restriction to $A$ is a homeomorphism. This implies that $S_3$ contains a subspace homeomorphic to $Bbb{RP}^3$, a contradiction.
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
answered Nov 26 '18 at 0:05
Servaes
22.4k33793
22.4k33793
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
add a comment |
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
Thank you very much @Servaes
– kot
Nov 26 '18 at 0:26
add a comment |
Hint If there were such a subgroup $A$, then for any $g in SU(2) - A$, $A cup gA$ would be a separation of $SU(2)$---but $SU(2)$ is connected.
answered Nov 26 '18 at 1:14
Travis
59.7k767146
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
add a comment |
Hint If there were such a subgroup $A$, then for any $g in SU(2) - A$, $A cup gA$ would be a separation of $SU(2)$---but $SU(2)$ is connected.
answered Nov 26 '18 at 1:14
Travis
59.7k767146
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
add a comment |
Hint If there were such a subgroup $A$, then for any $g in SU(2) - A$, $A cup gA$ would be a separation of $SU(2)$---but $SU(2)$ is connected.
answered Nov 26 '18 at 1:14
Travis
59.7k767146
Hint If there were such a subgroup $A$, then for any $g in SU(2) - A$, $A cup gA$ would be a separation of $SU(2)$---but $SU(2)$ is connected.
answered Nov 26 '18 at 1:14
Travis
59.7k767146
answered Nov 26 '18 at 1:14
Travis
59.7k767146
answered Nov 26 '18 at 1:14
Travis
59.7k767146
answered Nov 26 '18 at 1:14
Travis
59.7k767146
59.7k767146
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
add a comment |
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
+1 I guess this implicitly uses the fact that $operatorname{SO}(3)$ is also connected?
– Servaes
Nov 26 '18 at 8:10
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
Unless I'm missing something, the proof goes through without needing that fact, I think. In any case, that fact follows immediately from the connectedness of $SU(2)$ and the continuity of $Phi : SU(2) to SO(3)$.
– Travis
Nov 26 '18 at 9:39
add a comment |
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