Mixing
For which $P,Q in text{SO}$ $T_Ptext{SO}$ and $T_Qtext{SO}$ are parallel?
$begingroup$
I am curious: For which $P,Q in text{SO}_n$ does $T_Qtext{SO}_n=T_Ptext{SO}_n$ hold?
This reduces to the question at the identity,i.e. for which $Q in text{SO}_n$, $T_Qtext{SO}_n=T_{Id}text{SO}_n=text{skew}$. I will now prove that $Q^2=Id$ is a necessary condition. Is it sufficient?
Since $T_Qtext{SO}_n=QT_{Id}text{SO}_n=Qtext{skew}$, this happens if and only if
$Qtext{skew}=text{skew}$, i.e. $-AQ^T=(QA)^T=-QA$ for every $A in text{skew}$, or $AQ^T=QA$. Taking traces we get
$$ langle Q,Arangle=text{tr}(Q^TA)=text{tr}(AQ^T)=text{tr}(QA)=text{tr}(AQ)=-text{tr}(A^TQ)= langle A,Qrangle,$$
so $langle Q,Arangle=0$ for every $A in text{skew}$, i.e. $Q in text{skew}^{perp}=text{sym}$, so $Q^T=Q$, or $Q^2=Id$.
Note that at even dimensions $Q=-Id$ is always a solution. For dimension $n=2$, this is indeed the only non-trivial solution (since $text{SO}_2$ is the circle). In that case $Q^2=Id$, and $Q=pm Id$ are equivalent.
differential-geometry lie-groups riemannian-geometry orthogonal-matrices tangent-spaces
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
add a comment |
$begingroup$
I am curious: For which $P,Q in text{SO}_n$ does $T_Qtext{SO}_n=T_Ptext{SO}_n$ hold?
This reduces to the question at the identity,i.e. for which $Q in text{SO}_n$, $T_Qtext{SO}_n=T_{Id}text{SO}_n=text{skew}$. I will now prove that $Q^2=Id$ is a necessary condition. Is it sufficient?
Since $T_Qtext{SO}_n=QT_{Id}text{SO}_n=Qtext{skew}$, this happens if and only if
$Qtext{skew}=text{skew}$, i.e. $-AQ^T=(QA)^T=-QA$ for every $A in text{skew}$, or $AQ^T=QA$. Taking traces we get
$$ langle Q,Arangle=text{tr}(Q^TA)=text{tr}(AQ^T)=text{tr}(QA)=text{tr}(AQ)=-text{tr}(A^TQ)= langle A,Qrangle,$$
so $langle Q,Arangle=0$ for every $A in text{skew}$, i.e. $Q in text{skew}^{perp}=text{sym}$, so $Q^T=Q$, or $Q^2=Id$.
Note that at even dimensions $Q=-Id$ is always a solution. For dimension $n=2$, this is indeed the only non-trivial solution (since $text{SO}_2$ is the circle). In that case $Q^2=Id$, and $Q=pm Id$ are equivalent.
differential-geometry lie-groups riemannian-geometry orthogonal-matrices tangent-spaces
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
add a comment |
$begingroup$
I am curious: For which $P,Q in text{SO}_n$ does $T_Qtext{SO}_n=T_Ptext{SO}_n$ hold?
This reduces to the question at the identity,i.e. for which $Q in text{SO}_n$, $T_Qtext{SO}_n=T_{Id}text{SO}_n=text{skew}$. I will now prove that $Q^2=Id$ is a necessary condition. Is it sufficient?
Since $T_Qtext{SO}_n=QT_{Id}text{SO}_n=Qtext{skew}$, this happens if and only if
$Qtext{skew}=text{skew}$, i.e. $-AQ^T=(QA)^T=-QA$ for every $A in text{skew}$, or $AQ^T=QA$. Taking traces we get
$$ langle Q,Arangle=text{tr}(Q^TA)=text{tr}(AQ^T)=text{tr}(QA)=text{tr}(AQ)=-text{tr}(A^TQ)= langle A,Qrangle,$$
so $langle Q,Arangle=0$ for every $A in text{skew}$, i.e. $Q in text{skew}^{perp}=text{sym}$, so $Q^T=Q$, or $Q^2=Id$.
Note that at even dimensions $Q=-Id$ is always a solution. For dimension $n=2$, this is indeed the only non-trivial solution (since $text{SO}_2$ is the circle). In that case $Q^2=Id$, and $Q=pm Id$ are equivalent.
differential-geometry lie-groups riemannian-geometry orthogonal-matrices tangent-spaces
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
I am curious: For which $P,Q in text{SO}_n$ does $T_Qtext{SO}_n=T_Ptext{SO}_n$ hold?
This reduces to the question at the identity,i.e. for which $Q in text{SO}_n$, $T_Qtext{SO}_n=T_{Id}text{SO}_n=text{skew}$. I will now prove that $Q^2=Id$ is a necessary condition. Is it sufficient?
Since $T_Qtext{SO}_n=QT_{Id}text{SO}_n=Qtext{skew}$, this happens if and only if
$Qtext{skew}=text{skew}$, i.e. $-AQ^T=(QA)^T=-QA$ for every $A in text{skew}$, or $AQ^T=QA$. Taking traces we get
$$ langle Q,Arangle=text{tr}(Q^TA)=text{tr}(AQ^T)=text{tr}(QA)=text{tr}(AQ)=-text{tr}(A^TQ)= langle A,Qrangle,$$
so $langle Q,Arangle=0$ for every $A in text{skew}$, i.e. $Q in text{skew}^{perp}=text{sym}$, so $Q^T=Q$, or $Q^2=Id$.
Note that at even dimensions $Q=-Id$ is always a solution. For dimension $n=2$, this is indeed the only non-trivial solution (since $text{SO}_2$ is the circle). In that case $Q^2=Id$, and $Q=pm Id$ are equivalent.
differential-geometry lie-groups riemannian-geometry orthogonal-matrices tangent-spaces
differential-geometry lie-groups riemannian-geometry orthogonal-matrices tangent-spaces
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
asked Jan 6 at 14:55
Asaf ShacharAsaf Shachar
5,3843941
5,3843941
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think this works: the equation $AQ^T=QA$, along with symmetry of $Q$, means that $Q$ commutes with every skew-symmetric matrix. We can write any matrix $M$ as a sum $M = S + R$ of a symmetric and a skew-symmetric matrix. Looking at the commutator $[Q,M]$:
begin{equation}
begin{split}
Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\
&=QS-SQ\
&=QS - (QS)^T
end{split}
end{equation}
Where we used commutativity of $Q$ with skew-symmetric matrices, and then symmetry of $Q$ and $S$. The result in the last line is clearly antisymmetric, so $Q$ commutes with that as well:
begin{equation}
begin{split}
0 &= Q(QS-SQ) - (QS-SQ)Q\
&=QQS-QSQ-QSQ+SQQ\
&=2S-2QSQ
end{split}
end{equation}
So that $S = QSQ$, or $QS=SQ$, since $Q^2=Id$. But then $Q$ commutes with the symmetric part of the arbitrary matrix $M$ as well, so in fact $Q$ commutes with every matrix! This implies that $Q$ is a multiple of the identity (this should be a common result but see e.g. here). Of course then $Q=lambda I$ with $lambda^2 = 1$, so $Q=pm I$.
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
$endgroup$
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
add a comment |
$begingroup$
Here is a different approach to proceed: We shall in fact prove the following more general claim:
Suppose that $AQ^T=QA$ for every $A in text{skew}$. Then $Q=lambda Id$. (Here we do not assume $Q$ is orthogonal).
Proof:
The argument in the question shows $Q^T=Q$, so $AQ=QA$, i.e. $Q$ commutes with every $A in text{skew}$. By exponentiating, we get that $Q$ commutes with $e^A$ for every $A in text{skew}$, i.e. $Q$ commutes with all rotations. So, if $n>2$, then $Q=lambda Id$. If $n=2$, then $Q$ must be a scaled rotation; since it is also symmetric, this forces $Q=lambda Id$.
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
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$begingroup$
I think this works: the equation $AQ^T=QA$, along with symmetry of $Q$, means that $Q$ commutes with every skew-symmetric matrix. We can write any matrix $M$ as a sum $M = S + R$ of a symmetric and a skew-symmetric matrix. Looking at the commutator $[Q,M]$:
begin{equation}
begin{split}
Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\
&=QS-SQ\
&=QS - (QS)^T
end{split}
end{equation}
Where we used commutativity of $Q$ with skew-symmetric matrices, and then symmetry of $Q$ and $S$. The result in the last line is clearly antisymmetric, so $Q$ commutes with that as well:
begin{equation}
begin{split}
0 &= Q(QS-SQ) - (QS-SQ)Q\
&=QQS-QSQ-QSQ+SQQ\
&=2S-2QSQ
end{split}
end{equation}
So that $S = QSQ$, or $QS=SQ$, since $Q^2=Id$. But then $Q$ commutes with the symmetric part of the arbitrary matrix $M$ as well, so in fact $Q$ commutes with every matrix! This implies that $Q$ is a multiple of the identity (this should be a common result but see e.g. here). Of course then $Q=lambda I$ with $lambda^2 = 1$, so $Q=pm I$.
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
$endgroup$
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
add a comment |
$begingroup$
I think this works: the equation $AQ^T=QA$, along with symmetry of $Q$, means that $Q$ commutes with every skew-symmetric matrix. We can write any matrix $M$ as a sum $M = S + R$ of a symmetric and a skew-symmetric matrix. Looking at the commutator $[Q,M]$:
begin{equation}
begin{split}
Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\
&=QS-SQ\
&=QS - (QS)^T
end{split}
end{equation}
Where we used commutativity of $Q$ with skew-symmetric matrices, and then symmetry of $Q$ and $S$. The result in the last line is clearly antisymmetric, so $Q$ commutes with that as well:
begin{equation}
begin{split}
0 &= Q(QS-SQ) - (QS-SQ)Q\
&=QQS-QSQ-QSQ+SQQ\
&=2S-2QSQ
end{split}
end{equation}
So that $S = QSQ$, or $QS=SQ$, since $Q^2=Id$. But then $Q$ commutes with the symmetric part of the arbitrary matrix $M$ as well, so in fact $Q$ commutes with every matrix! This implies that $Q$ is a multiple of the identity (this should be a common result but see e.g. here). Of course then $Q=lambda I$ with $lambda^2 = 1$, so $Q=pm I$.
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
$endgroup$
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
add a comment |
$begingroup$
I think this works: the equation $AQ^T=QA$, along with symmetry of $Q$, means that $Q$ commutes with every skew-symmetric matrix. We can write any matrix $M$ as a sum $M = S + R$ of a symmetric and a skew-symmetric matrix. Looking at the commutator $[Q,M]$:
begin{equation}
begin{split}
Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\
&=QS-SQ\
&=QS - (QS)^T
end{split}
end{equation}
Where we used commutativity of $Q$ with skew-symmetric matrices, and then symmetry of $Q$ and $S$. The result in the last line is clearly antisymmetric, so $Q$ commutes with that as well:
begin{equation}
begin{split}
0 &= Q(QS-SQ) - (QS-SQ)Q\
&=QQS-QSQ-QSQ+SQQ\
&=2S-2QSQ
end{split}
end{equation}
So that $S = QSQ$, or $QS=SQ$, since $Q^2=Id$. But then $Q$ commutes with the symmetric part of the arbitrary matrix $M$ as well, so in fact $Q$ commutes with every matrix! This implies that $Q$ is a multiple of the identity (this should be a common result but see e.g. here). Of course then $Q=lambda I$ with $lambda^2 = 1$, so $Q=pm I$.
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
$endgroup$
I think this works: the equation $AQ^T=QA$, along with symmetry of $Q$, means that $Q$ commutes with every skew-symmetric matrix. We can write any matrix $M$ as a sum $M = S + R$ of a symmetric and a skew-symmetric matrix. Looking at the commutator $[Q,M]$:
begin{equation}
begin{split}
Q(S+R)-(S+R)Q &= QS - SQ + QR - RQ\
&=QS-SQ\
&=QS - (QS)^T
end{split}
end{equation}
Where we used commutativity of $Q$ with skew-symmetric matrices, and then symmetry of $Q$ and $S$. The result in the last line is clearly antisymmetric, so $Q$ commutes with that as well:
begin{equation}
begin{split}
0 &= Q(QS-SQ) - (QS-SQ)Q\
&=QQS-QSQ-QSQ+SQQ\
&=2S-2QSQ
end{split}
end{equation}
So that $S = QSQ$, or $QS=SQ$, since $Q^2=Id$. But then $Q$ commutes with the symmetric part of the arbitrary matrix $M$ as well, so in fact $Q$ commutes with every matrix! This implies that $Q$ is a multiple of the identity (this should be a common result but see e.g. here). Of course then $Q=lambda I$ with $lambda^2 = 1$, so $Q=pm I$.
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
answered Jan 7 at 9:11
mcwigglermcwiggler
49915
49915
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
add a comment |
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
$begingroup$
Thanks! I was just thinking about a slightly different approach: By exponentiating skew-symmetric matrices, we get commutation with all the special orthogonal group, which also implies the claim.
$endgroup$
– Asaf Shachar
Jan 7 at 12:58
add a comment |
$begingroup$
Here is a different approach to proceed: We shall in fact prove the following more general claim:
Suppose that $AQ^T=QA$ for every $A in text{skew}$. Then $Q=lambda Id$. (Here we do not assume $Q$ is orthogonal).
Proof:
The argument in the question shows $Q^T=Q$, so $AQ=QA$, i.e. $Q$ commutes with every $A in text{skew}$. By exponentiating, we get that $Q$ commutes with $e^A$ for every $A in text{skew}$, i.e. $Q$ commutes with all rotations. So, if $n>2$, then $Q=lambda Id$. If $n=2$, then $Q$ must be a scaled rotation; since it is also symmetric, this forces $Q=lambda Id$.
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
add a comment |
$begingroup$
Here is a different approach to proceed: We shall in fact prove the following more general claim:
Suppose that $AQ^T=QA$ for every $A in text{skew}$. Then $Q=lambda Id$. (Here we do not assume $Q$ is orthogonal).
Proof:
The argument in the question shows $Q^T=Q$, so $AQ=QA$, i.e. $Q$ commutes with every $A in text{skew}$. By exponentiating, we get that $Q$ commutes with $e^A$ for every $A in text{skew}$, i.e. $Q$ commutes with all rotations. So, if $n>2$, then $Q=lambda Id$. If $n=2$, then $Q$ must be a scaled rotation; since it is also symmetric, this forces $Q=lambda Id$.
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
add a comment |
$begingroup$
Here is a different approach to proceed: We shall in fact prove the following more general claim:
Suppose that $AQ^T=QA$ for every $A in text{skew}$. Then $Q=lambda Id$. (Here we do not assume $Q$ is orthogonal).
Proof:
The argument in the question shows $Q^T=Q$, so $AQ=QA$, i.e. $Q$ commutes with every $A in text{skew}$. By exponentiating, we get that $Q$ commutes with $e^A$ for every $A in text{skew}$, i.e. $Q$ commutes with all rotations. So, if $n>2$, then $Q=lambda Id$. If $n=2$, then $Q$ must be a scaled rotation; since it is also symmetric, this forces $Q=lambda Id$.
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
$endgroup$
Here is a different approach to proceed: We shall in fact prove the following more general claim:
Suppose that $AQ^T=QA$ for every $A in text{skew}$. Then $Q=lambda Id$. (Here we do not assume $Q$ is orthogonal).
Proof:
The argument in the question shows $Q^T=Q$, so $AQ=QA$, i.e. $Q$ commutes with every $A in text{skew}$. By exponentiating, we get that $Q$ commutes with $e^A$ for every $A in text{skew}$, i.e. $Q$ commutes with all rotations. So, if $n>2$, then $Q=lambda Id$. If $n=2$, then $Q$ must be a scaled rotation; since it is also symmetric, this forces $Q=lambda Id$.
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
answered Jan 7 at 12:56
Asaf ShacharAsaf Shachar
5,3843941
5,3843941
add a comment |
add a comment |
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