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Are the 1-parameters subgroups of $SO(3)$ closed?

2

$begingroup$

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

share|cite|improve this question

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

2

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Apr 2 at 5:36

Travis

64.6k769152

asked Apr 2 at 1:35

Matheus ManzattoMatheus Manzatto

1,3091626

$endgroup$

share|cite|improve this question

edited Apr 2 at 5:36

Travis

64.6k769152

asked Apr 2 at 1:35

Matheus ManzattoMatheus Manzatto

1,3091626

share|cite|improve this question

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

asked Apr 2 at 1:35

Matheus ManzattoMatheus Manzatto

1,3091626

asked Apr 2 at 1:35

Matheus ManzattoMatheus Manzatto

1,3091626

1 Answer
1

active

oldest

votes

2

$begingroup$

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 $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$

share|cite|improve this answer

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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2

$begingroup$

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 $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$

share|cite|improve this answer

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 | 

2

$begingroup$

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 $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$

share|cite|improve this answer

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 $$exp tA = exp t (P D P^-1) = P (exp t D) P^-1 .$$

share|cite|improve this answer

answered Apr 2 at 5:36

TravisTravis

64.6k769152

$endgroup$

share|cite|improve this answer

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