Usbrth

Can I visit the Trinity College (Cambridge) library and see some of their rare books

What is the role of 'For' here?

Simulating Exploding Dice

Huge performance difference of the command find with and without using %M option to show permissions

Am I ethically obligated to go into work on an off day if the reason is sudden?

"is" operation returns false with ndarray.data attribute, even though two array objects have same id

Windows 10: How to Lock (not sleep) laptop on lid close?

Sort list of array linked objects by keys and values

How to read αἱμύλιος or when to aspirate

Presidential Pardon

Loose spokes after only a few rides

should truth entail possible truth

Do ℕ, mathbbN, BbbN, symbbN effectively differ, and is there a "canonical" specification of the naturals?

Why are PDP-7-style microprogrammed instructions out of vogue?

One-dimensional Japanese puzzle

What do I do when my TA workload is more than expected?

Variable with quotation marks "$()"

What is the padding with red substance inside of steak packaging?

Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?

Can each chord in a progression create its own key?

Could an empire control the whole planet with today's comunication methods?

Did the new image of black hole confirm the general theory of relativity?

Can the DM override racial traits?

Single author papers against my advisor's will?

dimension of $U_p,q$ the pseudo unitary group

0

$begingroup$

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

share|cite|improve this question

edited Apr 8 at 7:17

PerelMan

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 | 

0

$begingroup$

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

share|cite|improve this question

edited Apr 8 at 7:17

PerelMan

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

share|cite|improve this question

edited Apr 8 at 7:17

PerelMan

asked Apr 8 at 6:39

PerelManPerelMan

751414

$endgroup$

share|cite|improve this question

edited Apr 8 at 7:17

PerelMan

asked Apr 8 at 6:39

PerelManPerelMan

751414

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

1

$begingroup$

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.

share|cite|improve this answer

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 | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3179261%2fdimension-of-u-p-q-the-pseudo-unitary-group%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

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.

share|cite|improve this answer

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 | 

1

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Apr 8 at 11:26

random123random123

1,2951720

$endgroup$

share|cite|improve this answer

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