dimension of $U_p,q$ the pseudo unitary group The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Pseudo inverse interpretationThe dimension of $SU(n)$Calculating Moore-Penrose pseudo inverseUnitary matrix for matrix representationOver the reals the intersection of the orthogonal and symplectic groups in even dimension is isomorphic to the unitary group in the half dimension.Form for pseudo-unitary matrices of particular dimensionDimension of centralizer of unitary matrixDimension of Lorentz groupThe special unitary group is a Lie Groupunitary matrix with an unusual constraint
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dimension of $U_p,q$ the pseudo unitary group
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Pseudo inverse interpretationThe dimension of $SU(n)$Calculating Moore-Penrose pseudo inverseUnitary matrix for matrix representationOver the reals the intersection of the orthogonal and symplectic groups in even dimension is isomorphic to the unitary group in the half dimension.Form for pseudo-unitary matrices of particular dimensionDimension of centralizer of unitary matrixDimension of Lorentz groupThe special unitary group is a Lie Groupunitary matrix with an unusual constraint
$begingroup$
Dimension of the pseudo unitary group is $(p+q)^2$ where $U_p,q = M ∈ M_p+q (Bbb C) : ^tMJ_p,qM = J_p,q.$ where $J_p,q=Diag(1...1,-1...-1)$ with $1 p$ times and $-1 q$ times.
How can we see that the dimension of this group is $(p+q)^2$? I can't see any reference or proof of this?
What about the dimension of $SU_p,q$?
Thank you for your help
linear-algebra lie-groups
asked Apr 8 at 6:39
PerelManPerelMan
751414
$endgroup$
add a comment |
$begingroup$
Dimension of the pseudo unitary group is $(p+q)^2$ where $U_p,q = M ∈ M_p+q (Bbb C) : ^tMJ_p,qM = J_p,q.$ where $J_p,q=Diag(1...1,-1...-1)$ with $1 p$ times and $-1 q$ times.
How can we see that the dimension of this group is $(p+q)^2$? I can't see any reference or proof of this?
What about the dimension of $SU_p,q$?
Thank you for your help
linear-algebra lie-groups
asked Apr 8 at 6:39
PerelManPerelMan
751414
$endgroup$
2
$begingroup$
One can compute the Lie algebra and from there it is easier to see the dimension. Also the groups $SU(p,q)$ are inner forms of $SL(p+q)$ and hence are all of the same dimension as $SL(p+q)$. Either of these ways should work.
$endgroup$
– random123
Apr 8 at 7:26
add a comment |
$begingroup$
Dimension of the pseudo unitary group is $(p+q)^2$ where $U_p,q = M ∈ M_p+q (Bbb C) : ^tMJ_p,qM = J_p,q.$ where $J_p,q=Diag(1...1,-1...-1)$ with $1 p$ times and $-1 q$ times.
How can we see that the dimension of this group is $(p+q)^2$? I can't see any reference or proof of this?
What about the dimension of $SU_p,q$?
Thank you for your help
linear-algebra lie-groups
asked Apr 8 at 6:39
PerelManPerelMan
751414
$endgroup$
Dimension of the pseudo unitary group is $(p+q)^2$ where $U_p,q = M ∈ M_p+q (Bbb C) : ^tMJ_p,qM = J_p,q.$ where $J_p,q=Diag(1...1,-1...-1)$ with $1 p$ times and $-1 q$ times.
How can we see that the dimension of this group is $(p+q)^2$? I can't see any reference or proof of this?
What about the dimension of $SU_p,q$?
Thank you for your help
linear-algebra lie-groups
linear-algebra lie-groups
asked Apr 8 at 6:39
PerelManPerelMan
751414
asked Apr 8 at 6:39
PerelManPerelMan
751414
edited Apr 8 at 7:17
PerelMan
asked Apr 8 at 6:39
PerelManPerelMan
751414
asked Apr 8 at 6:39
PerelManPerelMan
751414
asked Apr 8 at 6:39
PerelManPerelMan
751414
751414
2
$begingroup$
One can compute the Lie algebra and from there it is easier to see the dimension. Also the groups $SU(p,q)$ are inner forms of $SL(p+q)$ and hence are all of the same dimension as $SL(p+q)$. Either of these ways should work.
$endgroup$
– random123
Apr 8 at 7:26
add a comment |
2
$begingroup$
One can compute the Lie algebra and from there it is easier to see the dimension. Also the groups $SU(p,q)$ are inner forms of $SL(p+q)$ and hence are all of the same dimension as $SL(p+q)$. Either of these ways should work.
$endgroup$
– random123
Apr 8 at 7:26
2
2
$begingroup$
One can compute the Lie algebra and from there it is easier to see the dimension. Also the groups $SU(p,q)$ are inner forms of $SL(p+q)$ and hence are all of the same dimension as $SL(p+q)$. Either of these ways should work.
$endgroup$
– random123
Apr 8 at 7:26
$begingroup$
One can compute the Lie algebra and from there it is easier to see the dimension. Also the groups $SU(p,q)$ are inner forms of $SL(p+q)$ and hence are all of the same dimension as $SL(p+q)$. Either of these ways should work.
$endgroup$
– random123
Apr 8 at 7:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $U(p,q) = lbrace M in M_p+q(mathbbC) | ^toverlineMJM = J rbrace$. One can compute the Lie algebra to be
$mathfraku(p,q) = lbrace M in M(p,q) | ^toverlineMJ + JM = 0 rbrace$. Now we will compute the explicit form as block matrix as below.
$M = beginbmatrix A & B \ C & Dendbmatrix$.
One can compute with above equations to get that the $M in mathfraku(p,q)$ if and only if the matrix $M$ satisfies
$A + (^toverlineA) = 0 = D + (^toverlineD)$ and $B = -(^toverlineC)$. That is $A in mathfraku(p)$, $D in mathfraku(q)$ and $B$ is any $p times q$ matrix with complex entries. Now the only thing left is to add up the dimension of the spaces of matrices $A,B$ and $D$ which are $p^2, 2pq$ and $q^2$ respectively. Thus we have the answer.
One can caluculate the dimension of $SU(p,q)$ from here. Since the condition of determinant one imposes a linear equation in the Lie algebra of matrices having trace zero.
answered Apr 8 at 11:26
random123random123
1,2951720
$endgroup$
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
add a comment |
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1 Answer
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$begingroup$
Let $U(p,q) = lbrace M in M_p+q(mathbbC) | ^toverlineMJM = J rbrace$. One can compute the Lie algebra to be
$mathfraku(p,q) = lbrace M in M(p,q) | ^toverlineMJ + JM = 0 rbrace$. Now we will compute the explicit form as block matrix as below.
$M = beginbmatrix A & B \ C & Dendbmatrix$.
One can compute with above equations to get that the $M in mathfraku(p,q)$ if and only if the matrix $M$ satisfies
$A + (^toverlineA) = 0 = D + (^toverlineD)$ and $B = -(^toverlineC)$. That is $A in mathfraku(p)$, $D in mathfraku(q)$ and $B$ is any $p times q$ matrix with complex entries. Now the only thing left is to add up the dimension of the spaces of matrices $A,B$ and $D$ which are $p^2, 2pq$ and $q^2$ respectively. Thus we have the answer.
One can caluculate the dimension of $SU(p,q)$ from here. Since the condition of determinant one imposes a linear equation in the Lie algebra of matrices having trace zero.
answered Apr 8 at 11:26
random123random123
1,2951720
$endgroup$
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
add a comment |
$begingroup$
Let $U(p,q) = lbrace M in M_p+q(mathbbC) | ^toverlineMJM = J rbrace$. One can compute the Lie algebra to be
$mathfraku(p,q) = lbrace M in M(p,q) | ^toverlineMJ + JM = 0 rbrace$. Now we will compute the explicit form as block matrix as below.
$M = beginbmatrix A & B \ C & Dendbmatrix$.
One can compute with above equations to get that the $M in mathfraku(p,q)$ if and only if the matrix $M$ satisfies
$A + (^toverlineA) = 0 = D + (^toverlineD)$ and $B = -(^toverlineC)$. That is $A in mathfraku(p)$, $D in mathfraku(q)$ and $B$ is any $p times q$ matrix with complex entries. Now the only thing left is to add up the dimension of the spaces of matrices $A,B$ and $D$ which are $p^2, 2pq$ and $q^2$ respectively. Thus we have the answer.
One can caluculate the dimension of $SU(p,q)$ from here. Since the condition of determinant one imposes a linear equation in the Lie algebra of matrices having trace zero.
answered Apr 8 at 11:26
random123random123
1,2951720
$endgroup$
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
add a comment |
$begingroup$
Let $U(p,q) = lbrace M in M_p+q(mathbbC) | ^toverlineMJM = J rbrace$. One can compute the Lie algebra to be
$mathfraku(p,q) = lbrace M in M(p,q) | ^toverlineMJ + JM = 0 rbrace$. Now we will compute the explicit form as block matrix as below.
$M = beginbmatrix A & B \ C & Dendbmatrix$.
One can compute with above equations to get that the $M in mathfraku(p,q)$ if and only if the matrix $M$ satisfies
$A + (^toverlineA) = 0 = D + (^toverlineD)$ and $B = -(^toverlineC)$. That is $A in mathfraku(p)$, $D in mathfraku(q)$ and $B$ is any $p times q$ matrix with complex entries. Now the only thing left is to add up the dimension of the spaces of matrices $A,B$ and $D$ which are $p^2, 2pq$ and $q^2$ respectively. Thus we have the answer.
One can caluculate the dimension of $SU(p,q)$ from here. Since the condition of determinant one imposes a linear equation in the Lie algebra of matrices having trace zero.
answered Apr 8 at 11:26
random123random123
1,2951720
$endgroup$
Let $U(p,q) = lbrace M in M_p+q(mathbbC) | ^toverlineMJM = J rbrace$. One can compute the Lie algebra to be
$mathfraku(p,q) = lbrace M in M(p,q) | ^toverlineMJ + JM = 0 rbrace$. Now we will compute the explicit form as block matrix as below.
$M = beginbmatrix A & B \ C & Dendbmatrix$.
One can compute with above equations to get that the $M in mathfraku(p,q)$ if and only if the matrix $M$ satisfies
$A + (^toverlineA) = 0 = D + (^toverlineD)$ and $B = -(^toverlineC)$. That is $A in mathfraku(p)$, $D in mathfraku(q)$ and $B$ is any $p times q$ matrix with complex entries. Now the only thing left is to add up the dimension of the spaces of matrices $A,B$ and $D$ which are $p^2, 2pq$ and $q^2$ respectively. Thus we have the answer.
One can caluculate the dimension of $SU(p,q)$ from here. Since the condition of determinant one imposes a linear equation in the Lie algebra of matrices having trace zero.
answered Apr 8 at 11:26
random123random123
1,2951720
answered Apr 8 at 11:26
random123random123
1,2951720
answered Apr 8 at 11:26
random123random123
1,2951720
answered Apr 8 at 11:26
random123random123
1,2951720
1,2951720
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
add a comment |
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
$begingroup$
I think dimension of $SU(p,q)$ is $(p+q)^2-1$?
$endgroup$
– PerelMan
4 hours ago
add a comment |
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$begingroup$
One can compute the Lie algebra and from there it is easier to see the dimension. Also the groups $SU(p,q)$ are inner forms of $SL(p+q)$ and hence are all of the same dimension as $SL(p+q)$. Either of these ways should work.
$endgroup$
– random123
Apr 8 at 7:26