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Are the 1-parameters subgroups of $SO(3)$ closed?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)$1$-parameter subgroups in $GL_n(mathbbC)$When does the $mathfrak g$-invariance of the symplectic form imply $G$-invariance?Non-solvable, closed subgroups of $mathrmPSL(2,mathbbR)$homomorphism between smooth algebraic groups of the same dimensionIs the subgroup of homotopically trivial isometries a closed subgroup of the isometry group?One-dimensional closed subgroups of $SO(3)$Are degree > 1 maps surjective on codimension 0 submanifoldsHomomorphisms of matrix Lie groupsAbstract Proof that Exponential Map is Surjective onto $mathrmGL_n(mathbbC)$Are matrix exponentials the only continuous homomorphisms $mathbb R_+tomathcal M_n(mathbb R)$?
$begingroup$
I'm trying to solve the following question
Question: Prove that all $1$-parameters subgroup of $SO(3)$ are closed. Does this statement holds for $SO(n),$ $n>3$?
Some comments
The $1$-parameters subgroups of $SO(n)$ are the groups
$$P_A := e^tA; A in M_n(mathbbR) textand A+A^T = 0. $$
Since $SO(n)$ is compact the exponential map is surjective, and the map $mathbbRto P_A (tmapsto e^tA)$ is an isomorphism of groups (it is easy to see that $0$ is the unique element of the kernel, sinse $e^B = textId$ $Leftrightarrow B =0$) homomorphism of groups.
Being honest I can't believe that $P_A$ is closed when $n=3$, the fact of $P_A cong mathbbR$ as a group makes me have no idea of what is going on, once $P_A$ closed would imply $P_A$ compact.
EDIT: After Reuns' help I realised that I wroted thing that does not make any sense.
Can anyone help me?
manifolds differential-topology lie-groups matrix-exponential
edited Apr 2 at 5:36
Travis
64.6k769152
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to solve the following question
Question: Prove that all $1$-parameters subgroup of $SO(3)$ are closed. Does this statement holds for $SO(n),$ $n>3$?
Some comments
The $1$-parameters subgroups of $SO(n)$ are the groups
$$P_A := e^tA; A in M_n(mathbbR) textand A+A^T = 0. $$
Since $SO(n)$ is compact the exponential map is surjective, and the map $mathbbRto P_A (tmapsto e^tA)$ is an isomorphism of groups (it is easy to see that $0$ is the unique element of the kernel, sinse $e^B = textId$ $Leftrightarrow B =0$) homomorphism of groups.
Being honest I can't believe that $P_A$ is closed when $n=3$, the fact of $P_A cong mathbbR$ as a group makes me have no idea of what is going on, once $P_A$ closed would imply $P_A$ compact.
EDIT: After Reuns' help I realised that I wroted thing that does not make any sense.
Can anyone help me?
manifolds differential-topology lie-groups matrix-exponential
edited Apr 2 at 5:36
Travis
64.6k769152
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
$endgroup$
$begingroup$
Are you sure that $mathbb R rightarrow P_A$ is an isomorphism? What would make more sense to me is to have $P_A cong mathbb R/mathbb Z$, since closed subsets of compacts are compact as you said.
$endgroup$
– D_S
Apr 2 at 1:50
$begingroup$
@D_S I do think so, because $e^tA=textId$, implies $tA=0$, have I made a mistake?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:01
$begingroup$
I'd say $P_A cong lambda m + BbbZ^n, m in BbbZ subset BbbR^n/ Z^n$ where $2ipi lambda_j$ are the eigenvalues of $A$, and $P_A cong BbbR/Z$ iff the ratios between those eigenvalues are rational. Otherwise $P_A cong BbbR$ and it isn't closed. $e^tA=textId$ implies $QfractA2ipiQ^-1$ is diagonal with integer entries for some $Q$.
$endgroup$
– reuns
Apr 2 at 2:14
$begingroup$
@reuns But if all entries of $A$ and $Q$ are real numbers, then $fract2pi i Q A Q^-1$ is an imaginary matrix. If $A$ is a real matrix then $e^A = I Rightarrow A=0$, isn't it?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:28
$begingroup$
You should diagonalize $pmatrix1/sqrt2 & 1/sqrt2 \ 1/sqrt2&-1/sqrt2in SO(2)$ and compute the one parameter subgroup it belongs to
$endgroup$
– reuns
Apr 2 at 2:51
|
show 1 more comment
$begingroup$
I'm trying to solve the following question
Question: Prove that all $1$-parameters subgroup of $SO(3)$ are closed. Does this statement holds for $SO(n),$ $n>3$?
Some comments
The $1$-parameters subgroups of $SO(n)$ are the groups
$$P_A := e^tA; A in M_n(mathbbR) textand A+A^T = 0. $$
Since $SO(n)$ is compact the exponential map is surjective, and the map $mathbbRto P_A (tmapsto e^tA)$ is an isomorphism of groups (it is easy to see that $0$ is the unique element of the kernel, sinse $e^B = textId$ $Leftrightarrow B =0$) homomorphism of groups.
Being honest I can't believe that $P_A$ is closed when $n=3$, the fact of $P_A cong mathbbR$ as a group makes me have no idea of what is going on, once $P_A$ closed would imply $P_A$ compact.
EDIT: After Reuns' help I realised that I wroted thing that does not make any sense.
Can anyone help me?
manifolds differential-topology lie-groups matrix-exponential
edited Apr 2 at 5:36
Travis
64.6k769152
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
$endgroup$
I'm trying to solve the following question
Question: Prove that all $1$-parameters subgroup of $SO(3)$ are closed. Does this statement holds for $SO(n),$ $n>3$?
Some comments
The $1$-parameters subgroups of $SO(n)$ are the groups
$$P_A := e^tA; A in M_n(mathbbR) textand A+A^T = 0. $$
Since $SO(n)$ is compact the exponential map is surjective, and the map $mathbbRto P_A (tmapsto e^tA)$ is an isomorphism of groups (it is easy to see that $0$ is the unique element of the kernel, sinse $e^B = textId$ $Leftrightarrow B =0$) homomorphism of groups.
Being honest I can't believe that $P_A$ is closed when $n=3$, the fact of $P_A cong mathbbR$ as a group makes me have no idea of what is going on, once $P_A$ closed would imply $P_A$ compact.
EDIT: After Reuns' help I realised that I wroted thing that does not make any sense.
Can anyone help me?
manifolds differential-topology lie-groups matrix-exponential
manifolds differential-topology lie-groups matrix-exponential
edited Apr 2 at 5:36
Travis
64.6k769152
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
edited Apr 2 at 5:36
Travis
64.6k769152
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
edited Apr 2 at 5:36
Travis
64.6k769152
edited Apr 2 at 5:36
Travis
64.6k769152
edited Apr 2 at 5:36
Travis
64.6k769152
64.6k769152
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
asked Apr 2 at 1:35
Matheus ManzattoMatheus Manzatto
1,3091626
1,3091626
$begingroup$
Are you sure that $mathbb R rightarrow P_A$ is an isomorphism? What would make more sense to me is to have $P_A cong mathbb R/mathbb Z$, since closed subsets of compacts are compact as you said.
$endgroup$
– D_S
Apr 2 at 1:50
$begingroup$
@D_S I do think so, because $e^tA=textId$, implies $tA=0$, have I made a mistake?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:01
$begingroup$
I'd say $P_A cong lambda m + BbbZ^n, m in BbbZ subset BbbR^n/ Z^n$ where $2ipi lambda_j$ are the eigenvalues of $A$, and $P_A cong BbbR/Z$ iff the ratios between those eigenvalues are rational. Otherwise $P_A cong BbbR$ and it isn't closed. $e^tA=textId$ implies $QfractA2ipiQ^-1$ is diagonal with integer entries for some $Q$.
$endgroup$
– reuns
Apr 2 at 2:14
$begingroup$
@reuns But if all entries of $A$ and $Q$ are real numbers, then $fract2pi i Q A Q^-1$ is an imaginary matrix. If $A$ is a real matrix then $e^A = I Rightarrow A=0$, isn't it?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:28
$begingroup$
You should diagonalize $pmatrix1/sqrt2 & 1/sqrt2 \ 1/sqrt2&-1/sqrt2in SO(2)$ and compute the one parameter subgroup it belongs to
$endgroup$
– reuns
Apr 2 at 2:51
|
show 1 more comment
$begingroup$
Are you sure that $mathbb R rightarrow P_A$ is an isomorphism? What would make more sense to me is to have $P_A cong mathbb R/mathbb Z$, since closed subsets of compacts are compact as you said.
$endgroup$
– D_S
Apr 2 at 1:50
$begingroup$
@D_S I do think so, because $e^tA=textId$, implies $tA=0$, have I made a mistake?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:01
$begingroup$
I'd say $P_A cong lambda m + BbbZ^n, m in BbbZ subset BbbR^n/ Z^n$ where $2ipi lambda_j$ are the eigenvalues of $A$, and $P_A cong BbbR/Z$ iff the ratios between those eigenvalues are rational. Otherwise $P_A cong BbbR$ and it isn't closed. $e^tA=textId$ implies $QfractA2ipiQ^-1$ is diagonal with integer entries for some $Q$.
$endgroup$
– reuns
Apr 2 at 2:14
$begingroup$
@reuns But if all entries of $A$ and $Q$ are real numbers, then $fract2pi i Q A Q^-1$ is an imaginary matrix. If $A$ is a real matrix then $e^A = I Rightarrow A=0$, isn't it?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:28
$begingroup$
You should diagonalize $pmatrix1/sqrt2 & 1/sqrt2 \ 1/sqrt2&-1/sqrt2in SO(2)$ and compute the one parameter subgroup it belongs to
$endgroup$
– reuns
Apr 2 at 2:51
$begingroup$
Are you sure that $mathbb R rightarrow P_A$ is an isomorphism? What would make more sense to me is to have $P_A cong mathbb R/mathbb Z$, since closed subsets of compacts are compact as you said.
$endgroup$
– D_S
Apr 2 at 1:50
$begingroup$
Are you sure that $mathbb R rightarrow P_A$ is an isomorphism? What would make more sense to me is to have $P_A cong mathbb R/mathbb Z$, since closed subsets of compacts are compact as you said.
$endgroup$
– D_S
Apr 2 at 1:50
$begingroup$
@D_S I do think so, because $e^tA=textId$, implies $tA=0$, have I made a mistake?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:01
$begingroup$
@D_S I do think so, because $e^tA=textId$, implies $tA=0$, have I made a mistake?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:01
$begingroup$
I'd say $P_A cong lambda m + BbbZ^n, m in BbbZ subset BbbR^n/ Z^n$ where $2ipi lambda_j$ are the eigenvalues of $A$, and $P_A cong BbbR/Z$ iff the ratios between those eigenvalues are rational. Otherwise $P_A cong BbbR$ and it isn't closed. $e^tA=textId$ implies $QfractA2ipiQ^-1$ is diagonal with integer entries for some $Q$.
$endgroup$
– reuns
Apr 2 at 2:14
$begingroup$
I'd say $P_A cong lambda m + BbbZ^n, m in BbbZ subset BbbR^n/ Z^n$ where $2ipi lambda_j$ are the eigenvalues of $A$, and $P_A cong BbbR/Z$ iff the ratios between those eigenvalues are rational. Otherwise $P_A cong BbbR$ and it isn't closed. $e^tA=textId$ implies $QfractA2ipiQ^-1$ is diagonal with integer entries for some $Q$.
$endgroup$
– reuns
Apr 2 at 2:14
$begingroup$
@reuns But if all entries of $A$ and $Q$ are real numbers, then $fract2pi i Q A Q^-1$ is an imaginary matrix. If $A$ is a real matrix then $e^A = I Rightarrow A=0$, isn't it?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:28
$begingroup$
@reuns But if all entries of $A$ and $Q$ are real numbers, then $fract2pi i Q A Q^-1$ is an imaginary matrix. If $A$ is a real matrix then $e^A = I Rightarrow A=0$, isn't it?
$endgroup$
– Matheus Manzatto
Apr 2 at 2:28
$begingroup$
You should diagonalize $pmatrix1/sqrt2 & 1/sqrt2 \ 1/sqrt2&-1/sqrt2in SO(2)$ and compute the one parameter subgroup it belongs to
$endgroup$
– reuns
Apr 2 at 2:51
$begingroup$
You should diagonalize $pmatrix1/sqrt2 & 1/sqrt2 \ 1/sqrt2&-1/sqrt2in SO(2)$ and compute the one parameter subgroup it belongs to
$endgroup$
– reuns
Apr 2 at 2:51
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Hint In this case one can compute the $1$-parameter subgroups rather explicitly.
Consider the one parameter subgroup generated by the skew-symmetric (and hence diagonalizable) matrix $A$. Its eigenvalues are imaginary and (since $A$ is real) closed under conjugation, so the eigenvalues are $+ lambda i, -lambda i, 0$ for some $lambda geq 0 $. Diagonalizing $A$ then gives
$$A = P D P^-1, qquad D := pmatrixlambda i\&-lambda i\&&0 ,$$
for some matrix $P$. Now, compute $exp t A$ in terms of $D$.
Additional hint Using, e.g., the usual series formula for $exp$ gives that $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$
answered Apr 2 at 5:36
TravisTravis
64.6k769152
$endgroup$
1
$begingroup$
Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
$endgroup$
– Matheus Manzatto
Apr 2 at 17:46
1
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
$endgroup$
– Travis
Apr 2 at 18:52
1
$begingroup$
The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
1
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint In this case one can compute the $1$-parameter subgroups rather explicitly.
Consider the one parameter subgroup generated by the skew-symmetric (and hence diagonalizable) matrix $A$. Its eigenvalues are imaginary and (since $A$ is real) closed under conjugation, so the eigenvalues are $+ lambda i, -lambda i, 0$ for some $lambda geq 0 $. Diagonalizing $A$ then gives
$$A = P D P^-1, qquad D := pmatrixlambda i\&-lambda i\&&0 ,$$
for some matrix $P$. Now, compute $exp t A$ in terms of $D$.
Additional hint Using, e.g., the usual series formula for $exp$ gives that $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$
answered Apr 2 at 5:36
TravisTravis
64.6k769152
$endgroup$
1
$begingroup$
Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
$endgroup$
– Matheus Manzatto
Apr 2 at 17:46
1
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
$endgroup$
– Travis
Apr 2 at 18:52
1
$begingroup$
The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
1
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
add a comment |
$begingroup$
Hint In this case one can compute the $1$-parameter subgroups rather explicitly.
Consider the one parameter subgroup generated by the skew-symmetric (and hence diagonalizable) matrix $A$. Its eigenvalues are imaginary and (since $A$ is real) closed under conjugation, so the eigenvalues are $+ lambda i, -lambda i, 0$ for some $lambda geq 0 $. Diagonalizing $A$ then gives
$$A = P D P^-1, qquad D := pmatrixlambda i\&-lambda i\&&0 ,$$
for some matrix $P$. Now, compute $exp t A$ in terms of $D$.
Additional hint Using, e.g., the usual series formula for $exp$ gives that $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$
answered Apr 2 at 5:36
TravisTravis
64.6k769152
$endgroup$
1
$begingroup$
Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
$endgroup$
– Matheus Manzatto
Apr 2 at 17:46
1
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
$endgroup$
– Travis
Apr 2 at 18:52
1
$begingroup$
The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
1
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
add a comment |
$begingroup$
Hint In this case one can compute the $1$-parameter subgroups rather explicitly.
Consider the one parameter subgroup generated by the skew-symmetric (and hence diagonalizable) matrix $A$. Its eigenvalues are imaginary and (since $A$ is real) closed under conjugation, so the eigenvalues are $+ lambda i, -lambda i, 0$ for some $lambda geq 0 $. Diagonalizing $A$ then gives
$$A = P D P^-1, qquad D := pmatrixlambda i\&-lambda i\&&0 ,$$
for some matrix $P$. Now, compute $exp t A$ in terms of $D$.
Additional hint Using, e.g., the usual series formula for $exp$ gives that $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$
answered Apr 2 at 5:36
TravisTravis
64.6k769152
$endgroup$
Hint In this case one can compute the $1$-parameter subgroups rather explicitly.
Consider the one parameter subgroup generated by the skew-symmetric (and hence diagonalizable) matrix $A$. Its eigenvalues are imaginary and (since $A$ is real) closed under conjugation, so the eigenvalues are $+ lambda i, -lambda i, 0$ for some $lambda geq 0 $. Diagonalizing $A$ then gives
$$A = P D P^-1, qquad D := pmatrixlambda i\&-lambda i\&&0 ,$$
for some matrix $P$. Now, compute $exp t A$ in terms of $D$.
Additional hint Using, e.g., the usual series formula for $exp$ gives that $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$
answered Apr 2 at 5:36
TravisTravis
64.6k769152
answered Apr 2 at 5:36
TravisTravis
64.6k769152
answered Apr 2 at 5:36
TravisTravis
64.6k769152
answered Apr 2 at 5:36
TravisTravis
64.6k769152
64.6k769152
1
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Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
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– Matheus Manzatto
Apr 2 at 17:46
1
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
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– Travis
Apr 2 at 18:52
1
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The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
1
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
add a comment |
1
$begingroup$
Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
$endgroup$
– Matheus Manzatto
Apr 2 at 17:46
1
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
$endgroup$
– Travis
Apr 2 at 18:52
1
$begingroup$
The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
1
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
1
1
$begingroup$
Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
$endgroup$
– Matheus Manzatto
Apr 2 at 17:46
$begingroup$
Amazing idea, using it when $SO(n)$ $n>3$, it is clear that we can find a skew-symmetric matrix $A$ such that $$varphi: mathbbRto P_A$$ $$t mapsto e^tA, $$ is a isomorphism (to construct this map we just need to find some skew-symmetric with eigenvalues $i lambda_1$ and $i lambda_2$ such that $lambda_1,lambda_2$ are l.i. under $mathbbQ$ ). Is it clear that $P_A$ is not closed? because maybe $P_A$ is not embedded in $SO(n)$. So say that $mathbbRcong P_A$ as topological spaces (anlysing $P_A$ with subspace topology) seems a little complicated.
$endgroup$
– Matheus Manzatto
Apr 2 at 17:46
1
1
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
$endgroup$
– Travis
Apr 2 at 18:52
$begingroup$
Yes, that's exactly right. And yes, in that case $P_A$ is not closed: For example, for $n = 4$ one can find a sequence of points in $P_A$ that converges to $-I$, but this point itself cannot be in $P_A$, and thus $P_A$ is not closed. In particular, the subspace topology on $P_A$ is not the same as the topology induced by declaring $varphi$ to be a homeomorphism. Cf. en.wikipedia.org/wiki/…
$endgroup$
– Travis
Apr 2 at 18:52
1
1
$begingroup$
The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
$begingroup$
The idea of finding a sequence that converges to $ -I $ is completely superb
$endgroup$
– Matheus Manzatto
Apr 2 at 20:17
1
1
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
$begingroup$
Then we need to show for any admissible $D$ there exists $A in M_n(BbbR), A^top = -A$ such that $A = P D P^-1$, using $Q = pmatrix1 & i \ 1 & -i,QQ^* = 2I$ and $P = pmatrixQ & & \ & Q & \ & & ldots$ @MatheusManzatto
$endgroup$
– reuns
Apr 2 at 23:40
add a comment |
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$begingroup$
Are you sure that $mathbb R rightarrow P_A$ is an isomorphism? What would make more sense to me is to have $P_A cong mathbb R/mathbb Z$, since closed subsets of compacts are compact as you said.
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– D_S
Apr 2 at 1:50
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@D_S I do think so, because $e^tA=textId$, implies $tA=0$, have I made a mistake?
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– Matheus Manzatto
Apr 2 at 2:01
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I'd say $P_A cong lambda m + BbbZ^n, m in BbbZ subset BbbR^n/ Z^n$ where $2ipi lambda_j$ are the eigenvalues of $A$, and $P_A cong BbbR/Z$ iff the ratios between those eigenvalues are rational. Otherwise $P_A cong BbbR$ and it isn't closed. $e^tA=textId$ implies $QfractA2ipiQ^-1$ is diagonal with integer entries for some $Q$.
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– reuns
Apr 2 at 2:14
$begingroup$
@reuns But if all entries of $A$ and $Q$ are real numbers, then $fract2pi i Q A Q^-1$ is an imaginary matrix. If $A$ is a real matrix then $e^A = I Rightarrow A=0$, isn't it?
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– Matheus Manzatto
Apr 2 at 2:28
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You should diagonalize $pmatrix1/sqrt2 & 1/sqrt2 \ 1/sqrt2&-1/sqrt2in SO(2)$ and compute the one parameter subgroup it belongs to
$endgroup$
– reuns
Apr 2 at 2:51