Which matrices commute with $operatorname{SO}_n$?
$begingroup$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
Let $n>2$, and Let $A in GLp$ be an invertible real $n times n$ matrix, which commutes with $SO$.
Is it true that $A= lambda Id$ for some $lambda in mathbb{R}$ ?
An equivalent requirement is that $A$ commutes with every skew-symmetric matrix.
One direction is obtained by differentiating a path of orthogonal matrices starting at the identity. The converse implication comes from the fact that every element of $SO$ equals to $exp(M)$ for some skew-symmetric $M$.
Note that if we assume that $A in SO$, then the answer is positive: we must have $A=pm Id$ .
group-theory lie-groups symmetry orthogonal-matrices
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
$endgroup$
|
show 3 more comments
$begingroup$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
Let $n>2$, and Let $A in GLp$ be an invertible real $n times n$ matrix, which commutes with $SO$.
Is it true that $A= lambda Id$ for some $lambda in mathbb{R}$ ?
An equivalent requirement is that $A$ commutes with every skew-symmetric matrix.
One direction is obtained by differentiating a path of orthogonal matrices starting at the identity. The converse implication comes from the fact that every element of $SO$ equals to $exp(M)$ for some skew-symmetric $M$.
Note that if we assume that $A in SO$, then the answer is positive: we must have $A=pm Id$ .
group-theory lie-groups symmetry orthogonal-matrices
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
$endgroup$
$begingroup$
If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = pm Id$, we must have $A^{-1}B = pm id implies B = pm A$. Oh I think this is the answer
$endgroup$
– TrostAft
Dec 20 '18 at 20:39
2
$begingroup$
You could have $B=kA$ for any nonzero $k$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:42
3
$begingroup$
When $n=2$, $SO_n$ consists of the $cos t I+sin t J$ for $J=pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $Bbb RI +Bbb RJ$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:46
1
$begingroup$
As every element of $SO_n$ is $exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:55
1
$begingroup$
The proof sketch in the linked question applies to this case as well
$endgroup$
– Dap
Dec 21 '18 at 14:30
|
show 3 more comments
$begingroup$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
Let $n>2$, and Let $A in GLp$ be an invertible real $n times n$ matrix, which commutes with $SO$.
Is it true that $A= lambda Id$ for some $lambda in mathbb{R}$ ?
An equivalent requirement is that $A$ commutes with every skew-symmetric matrix.
One direction is obtained by differentiating a path of orthogonal matrices starting at the identity. The converse implication comes from the fact that every element of $SO$ equals to $exp(M)$ for some skew-symmetric $M$.
Note that if we assume that $A in SO$, then the answer is positive: we must have $A=pm Id$ .
group-theory lie-groups symmetry orthogonal-matrices
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
$endgroup$
$newcommand{GLp}{operatorname{GL}_n^+}$
$newcommand{SO}{operatorname{SO}_n}$
Let $n>2$, and Let $A in GLp$ be an invertible real $n times n$ matrix, which commutes with $SO$.
Is it true that $A= lambda Id$ for some $lambda in mathbb{R}$ ?
An equivalent requirement is that $A$ commutes with every skew-symmetric matrix.
One direction is obtained by differentiating a path of orthogonal matrices starting at the identity. The converse implication comes from the fact that every element of $SO$ equals to $exp(M)$ for some skew-symmetric $M$.
Note that if we assume that $A in SO$, then the answer is positive: we must have $A=pm Id$ .
group-theory lie-groups symmetry orthogonal-matrices
group-theory lie-groups symmetry orthogonal-matrices
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
edited Dec 20 '18 at 21:02
Asaf Shachar
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
asked Dec 20 '18 at 20:18
Asaf ShacharAsaf Shachar
5,46531141
5,46531141
$begingroup$
If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = pm Id$, we must have $A^{-1}B = pm id implies B = pm A$. Oh I think this is the answer
$endgroup$
– TrostAft
Dec 20 '18 at 20:39
2
$begingroup$
You could have $B=kA$ for any nonzero $k$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:42
3
$begingroup$
When $n=2$, $SO_n$ consists of the $cos t I+sin t J$ for $J=pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $Bbb RI +Bbb RJ$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:46
1
$begingroup$
As every element of $SO_n$ is $exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:55
1
$begingroup$
The proof sketch in the linked question applies to this case as well
$endgroup$
– Dap
Dec 21 '18 at 14:30
|
show 3 more comments
$begingroup$
If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = pm Id$, we must have $A^{-1}B = pm id implies B = pm A$. Oh I think this is the answer
$endgroup$
– TrostAft
Dec 20 '18 at 20:39
2
$begingroup$
You could have $B=kA$ for any nonzero $k$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:42
3
$begingroup$
When $n=2$, $SO_n$ consists of the $cos t I+sin t J$ for $J=pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $Bbb RI +Bbb RJ$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:46
1
$begingroup$
As every element of $SO_n$ is $exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:55
1
$begingroup$
The proof sketch in the linked question applies to this case as well
$endgroup$
– Dap
Dec 21 '18 at 14:30
$begingroup$
If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = pm Id$, we must have $A^{-1}B = pm id implies B = pm A$. Oh I think this is the answer
$endgroup$
– TrostAft
Dec 20 '18 at 20:39
$begingroup$
If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = pm Id$, we must have $A^{-1}B = pm id implies B = pm A$. Oh I think this is the answer
$endgroup$
– TrostAft
Dec 20 '18 at 20:39
2
2
$begingroup$
You could have $B=kA$ for any nonzero $k$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:42
$begingroup$
You could have $B=kA$ for any nonzero $k$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:42
3
3
$begingroup$
When $n=2$, $SO_n$ consists of the $cos t I+sin t J$ for $J=pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $Bbb RI +Bbb RJ$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:46
$begingroup$
When $n=2$, $SO_n$ consists of the $cos t I+sin t J$ for $J=pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $Bbb RI +Bbb RJ$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:46
1
1
$begingroup$
As every element of $SO_n$ is $exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:55
$begingroup$
As every element of $SO_n$ is $exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 20:55
1
1
$begingroup$
The proof sketch in the linked question applies to this case as well
$endgroup$
– Dap
Dec 21 '18 at 14:30
$begingroup$
The proof sketch in the linked question applies to this case as well
$endgroup$
– Dap
Dec 21 '18 at 14:30
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is a representation theory question: slightly generalized (there's no need to restrict our attention to $GL_n^{+}$), you're asking what the endomorphisms of $mathbb{R}^n$ as a representation of the Lie group $SO(n)$ (or, equivalently, the Lie algebra $mathfrak{so}(n)$) are.
This representation is always irreducible, so by Schur's lemma the endomorphisms form a division algebra over $mathbb{R}$, which by the Frobenius theorem must be $mathbb{R}, mathbb{C}$, or $mathbb{H}$. The latter two cases can't happen if $n$ is odd (because $mathbb{C}$ and $mathbb{H}$ only act on $mathbb{R}^n$ when $n$ is divisible by $2$ or $4$ respectively).
If $n = 2k ge 4$ is even we can argue as follows: if the endomorphism ring contains $mathbb{C}$, then $SO(2k)$ must embed into $GL_k(mathbb{C})$ and hence into the unitary group $U(k)$, by compactness, and similarly on the level of Lie algebras. But this is impossible by a dimension count: $SO(2k)$ has dimension $k(2k-1)$, but $U(k)$ has dimension $k^2$, and for $k ge 2$ we have $2k-1 > k$. (For $k = 1$ they are equal, reflecting the coincidence $SO(2) = U(1)$.) So the endomorphism ring must be $mathbb{R}$. Probably a simpler argument is possible here.
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
$endgroup$
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
This is a representation theory question: slightly generalized (there's no need to restrict our attention to $GL_n^{+}$), you're asking what the endomorphisms of $mathbb{R}^n$ as a representation of the Lie group $SO(n)$ (or, equivalently, the Lie algebra $mathfrak{so}(n)$) are.
This representation is always irreducible, so by Schur's lemma the endomorphisms form a division algebra over $mathbb{R}$, which by the Frobenius theorem must be $mathbb{R}, mathbb{C}$, or $mathbb{H}$. The latter two cases can't happen if $n$ is odd (because $mathbb{C}$ and $mathbb{H}$ only act on $mathbb{R}^n$ when $n$ is divisible by $2$ or $4$ respectively).
If $n = 2k ge 4$ is even we can argue as follows: if the endomorphism ring contains $mathbb{C}$, then $SO(2k)$ must embed into $GL_k(mathbb{C})$ and hence into the unitary group $U(k)$, by compactness, and similarly on the level of Lie algebras. But this is impossible by a dimension count: $SO(2k)$ has dimension $k(2k-1)$, but $U(k)$ has dimension $k^2$, and for $k ge 2$ we have $2k-1 > k$. (For $k = 1$ they are equal, reflecting the coincidence $SO(2) = U(1)$.) So the endomorphism ring must be $mathbb{R}$. Probably a simpler argument is possible here.
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
$endgroup$
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
add a comment |
$begingroup$
This is a representation theory question: slightly generalized (there's no need to restrict our attention to $GL_n^{+}$), you're asking what the endomorphisms of $mathbb{R}^n$ as a representation of the Lie group $SO(n)$ (or, equivalently, the Lie algebra $mathfrak{so}(n)$) are.
This representation is always irreducible, so by Schur's lemma the endomorphisms form a division algebra over $mathbb{R}$, which by the Frobenius theorem must be $mathbb{R}, mathbb{C}$, or $mathbb{H}$. The latter two cases can't happen if $n$ is odd (because $mathbb{C}$ and $mathbb{H}$ only act on $mathbb{R}^n$ when $n$ is divisible by $2$ or $4$ respectively).
If $n = 2k ge 4$ is even we can argue as follows: if the endomorphism ring contains $mathbb{C}$, then $SO(2k)$ must embed into $GL_k(mathbb{C})$ and hence into the unitary group $U(k)$, by compactness, and similarly on the level of Lie algebras. But this is impossible by a dimension count: $SO(2k)$ has dimension $k(2k-1)$, but $U(k)$ has dimension $k^2$, and for $k ge 2$ we have $2k-1 > k$. (For $k = 1$ they are equal, reflecting the coincidence $SO(2) = U(1)$.) So the endomorphism ring must be $mathbb{R}$. Probably a simpler argument is possible here.
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
$endgroup$
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
add a comment |
$begingroup$
This is a representation theory question: slightly generalized (there's no need to restrict our attention to $GL_n^{+}$), you're asking what the endomorphisms of $mathbb{R}^n$ as a representation of the Lie group $SO(n)$ (or, equivalently, the Lie algebra $mathfrak{so}(n)$) are.
This representation is always irreducible, so by Schur's lemma the endomorphisms form a division algebra over $mathbb{R}$, which by the Frobenius theorem must be $mathbb{R}, mathbb{C}$, or $mathbb{H}$. The latter two cases can't happen if $n$ is odd (because $mathbb{C}$ and $mathbb{H}$ only act on $mathbb{R}^n$ when $n$ is divisible by $2$ or $4$ respectively).
If $n = 2k ge 4$ is even we can argue as follows: if the endomorphism ring contains $mathbb{C}$, then $SO(2k)$ must embed into $GL_k(mathbb{C})$ and hence into the unitary group $U(k)$, by compactness, and similarly on the level of Lie algebras. But this is impossible by a dimension count: $SO(2k)$ has dimension $k(2k-1)$, but $U(k)$ has dimension $k^2$, and for $k ge 2$ we have $2k-1 > k$. (For $k = 1$ they are equal, reflecting the coincidence $SO(2) = U(1)$.) So the endomorphism ring must be $mathbb{R}$. Probably a simpler argument is possible here.
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
$endgroup$
This is a representation theory question: slightly generalized (there's no need to restrict our attention to $GL_n^{+}$), you're asking what the endomorphisms of $mathbb{R}^n$ as a representation of the Lie group $SO(n)$ (or, equivalently, the Lie algebra $mathfrak{so}(n)$) are.
This representation is always irreducible, so by Schur's lemma the endomorphisms form a division algebra over $mathbb{R}$, which by the Frobenius theorem must be $mathbb{R}, mathbb{C}$, or $mathbb{H}$. The latter two cases can't happen if $n$ is odd (because $mathbb{C}$ and $mathbb{H}$ only act on $mathbb{R}^n$ when $n$ is divisible by $2$ or $4$ respectively).
If $n = 2k ge 4$ is even we can argue as follows: if the endomorphism ring contains $mathbb{C}$, then $SO(2k)$ must embed into $GL_k(mathbb{C})$ and hence into the unitary group $U(k)$, by compactness, and similarly on the level of Lie algebras. But this is impossible by a dimension count: $SO(2k)$ has dimension $k(2k-1)$, but $U(k)$ has dimension $k^2$, and for $k ge 2$ we have $2k-1 > k$. (For $k = 1$ they are equal, reflecting the coincidence $SO(2) = U(1)$.) So the endomorphism ring must be $mathbb{R}$. Probably a simpler argument is possible here.
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
edited Dec 21 '18 at 23:25
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
answered Dec 21 '18 at 23:19
Qiaochu YuanQiaochu Yuan
278k32585922
278k32585922
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
add a comment |
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
Hi, I know it has been some time ago, but coming to this question again, I see that I don't understand two things here: (1) Why the fact the endomorphism ring contains $mathbb C$ implies that $text{SO}(2k)$ embeds in $text{GL}_k(mathbb C)$? Can you describe the embedding more explicitly (I guess in terms of an endomorphism $J:mathbb R^n to mathbb R^n$ whose square is $-1$)? (2) Why can $mathbb H$ only act on $mathbb R^n$ when $n$ is divisible by $4$? (If I understand correctly, the reason why $mathbb C$ can only act on $mathbb R^n$ for even $n$ is that...
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
$begingroup$
if you have a $J in text{GL}(mathbb R^n)$ , $J^2=-1$ you can take determinants and see what happens. What is the argument for the quaternionic case? I appreciate your help.
$endgroup$
– Asaf Shachar
21 hours ago
add a comment |
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$begingroup$
If you write it instead as $Q = A^{-1}B Q B^{-1} A = A^{-1} B Q ( A^{-1} B)^{-1}$. Then denote $C = A^{-1} B$ and then we have the system $C Q C^{-1} = Q$. So it seems to be true that if the matrix $A^{-1}B$ is commutative things work out. So perhaps instead of saying $B = pm Id$, we must have $A^{-1}B = pm id implies B = pm A$. Oh I think this is the answer
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– TrostAft
Dec 20 '18 at 20:39
2
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You could have $B=kA$ for any nonzero $k$.
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– Lord Shark the Unknown
Dec 20 '18 at 20:42
3
$begingroup$
When $n=2$, $SO_n$ consists of the $cos t I+sin t J$ for $J=pmatrix{0&1\-1&0}$. Then the centraliser of all these matrices consists of $Bbb RI +Bbb RJ$.
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– Lord Shark the Unknown
Dec 20 '18 at 20:46
1
$begingroup$
As every element of $SO_n$ is $exp(M)$ for a skew-symmetric $M$, then $C$ commutes with all of $SO_n$ iff $C$ commutes with all skew-symmetric matrices.
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– Lord Shark the Unknown
Dec 20 '18 at 20:55
1
$begingroup$
The proof sketch in the linked question applies to this case as well
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– Dap
Dec 21 '18 at 14:30