How do i show the lie algebra of the group

[latentcorpse]

how do i show the lie algebra of the group SO(p,q) is

\mathfrak{so}(p,q) = \Biggl\{ \left( \begin{array}{cc} X & Z \\ Z^t & Y \end{array} \right) \vline X \in M_p ( \mathbb{R} ), Y \in M_q ( \mathbb{R} ), Z \in M_{p \times q} ( \mathbb{R} ), X^t=-X, Y^t=-Y \Biggr\}

i'm not really getting anywhere so any help is appreciated. thanks.

 

[latentcorpse]

nobody? aargh.

 

[Dick]

A matrix g is in the Lie group if g^(T)*eta*g=eta, where eta is a unit diagonal matrix with signature (p,q), right? So if G is in the Lie algebra then exp(t*G^T)*eta*exp(t*G)=eta. Do the usual thing and differentiate and set t=0 to get the condition on G. Now split G up into block form and see what that condition is on each block. There's no special tricks here.

 

[latentcorpse]

A matrix g is in the Lie group if g^(T)*eta*g=eta, where eta is a unit diagonal matrix with signature (p,q), right? So if G is in the Lie algebra then exp(t*G^T)*eta*exp(t*G)=eta. Do the usual thing and differentiate and set t=0 to get the condition on G. Now split G up into block diagonal form and see what that condition is on each block. There's no special tricks here.

yeah. this is pretty much what I've been trying.
(disclaimer: my notation differs a bit from yours but it should be understandable!)

using the above defn, if we write A=exp(tau X) then
A^t \eta A = \eta \Rightarrow X=-\eta^{-1} X^t \eta=-\eta X^t \eta since \eta^{-1}=\eta (since it's a diagonal matrix)

thus if we write X as a block matrix X= \left( \begin{array}{cc} B & C \\ D & E \end{array} \right) and if we write \eta = \left( \begin{array}{cc} \mathbb{I}_{p} & 0 \\ 0 & \mathbb{I}_{q} \end{array} \right)

then if we follow through the couple of lines of matrix multiplication, our condition on X simplifies to

\left( \begin{array}{cc} B & C \\ D & E \end{array} \right) = \left( \begin{array}{cc} -B & 0 \\ 0 & - E \end{array} \right)

this doesn't look to promising at this point. also i haven't even made use of the determinant condition yet and I've already established that the upper right and lower left blocks are zero, which according to the answer, they shouldn't be.

 

[Dick]

For one thing eta=diag(-I_p,I_q) or diag(I_p,-I_q) depending on where you like to put the sign, right? For another X^(T)=[[B^T,D^T],[C^T,E^T]]. I think you are forgetting to transpose the block matrices.

 

[latentcorpse]

For one thing eta=diag(-I_p,I_q) or diag(I_p,-I_q) depending on where you like to put the sign, right? For another X^(T)=[[B^T,D^T],[C^T,E^T]]. I think you are forgetting to transpose the block matrices.

ok. I've managed to get to the answer without using the condition of unit determinant though. surely I've done something wrong?
or perhaps this condition is used on the next part of the question, which is to establish that the lie algebra has dimesnion \frac{n(n-1)}{2} where n=p+q.

do you have any advice on how to go about provingthe dimension? thanks.
so far i' ve said that because of the condition on B it will have dimension p/2 and similarly E has dim q/2 but don't know how to deal with Z?

p.s. i also have to do the same for the group SU(p,q). i don't actually know the definition of SU(p,q) though? i can define SU(n)= \{ a \in M_n ( \mathbb{C} ) | \bar{a}^t a = \mathbb{I} , \text{det } a = 1 \} alright but the p and q bit messes it up. unlike the SO case where we had a eta matrix to give the p,q signature to.

 

[Dick]

Just add up the number of free variables in your matrix blocks. The unit determinant choice only selects components of the Lie group, it doesn't affect the dimension of the Lie algebra.

 

[latentcorpse]

Just add up the number of free variables in your matrix blocks. The unit determinant choice only selects components of the Lie group, it doesn't affect the dimension of the Lie algebra.

(i)so the unit determinant must affect the lie algebra in some way otherwise O(p,q) and SO(p,q) would have identical lie algebras, no?

(ii)okay. as for the dimension:
there are p free variables in B and q free variables in E. C has (pq)^2 free variables. i think I've gone wrong since this is looking nothing like what i want!

(iii) what is the definition of SU(p,q)?

thanks again.

 

[Dick]

O(p,q) and SO(p,q) do have the same Lie algebra don't they? SO(p,q) is the component of O(p,q) containing the identity. They have the same dimension. You'd better try counting those variables again. What are the dimensions of B, C, D and E? SU(p,q) is the similar to SO(p,q) except the matrices can be complex and transpose is replaced by conjugate transpose.

 

[latentcorpse]

O(p,q) and SO(p,q) do have the same Lie algebra don't they? SO(p,q) is the component of O(p,q) containing the identity. They have the same dimension. You'd better try counting those variables again. What are the dimensions of B, C, D and E? SU(p,q) is the similar to SO(p,q) except the matrices can be complex and transpose is replaced by conjugate transpose.

ok. i agree on the first point that O(p,q) and SO(p,q) have the same lie algebra. the det 1 condition neatly falls into place from the fact that matrices in the lie algebra have 0's all down the diagonal.

right next bit:

B has dim p^2. we only want to count the upper triangular components (not including the diagonal). this give (p^2-p)/2. and E has dimension q^2. again count just that upper triangle. this gives (q^2-q)/2.

add these together we get

(p^2+q^2 - (p+q))/2 = (n^2-n)/2 which is the answer.

the only problem i have with this is that we haven't counted any components of C, have we? i don't understand where this piece has gone? yet it seems to work. for example consider a 4x4 matrix with 2x2 blocks then a basis for B would be [0,1][-1,0] and a basis for E would be [0,1][-1,0] and for C it would be {[1,0][0,0],[0,1][0,0],[0,0][1,0],[0,0][0,1]} this is a 6 dimensional basis and (4^2-4)/2=6 as required.
why is this working when we didn't count any of the basis elements for C?


ok.

should i work with the definition that

SU(p,q) = \{ A \in M_n ( \mathbb{C} ) | \bar{A}^t \eta A = \eta \}

 

[Dick]

p^2+q^2 isn't equal to (p+q)^2. Come on, that's a rookie mistake. Your definition is U(p,q). For SU(p,q) you need to add the determinant condition. Unlike O(p,q) the determinant does affect the Lie algebra.

 

[latentcorpse]

p^2+q^2 isn't equal to (p+q)^2. Come on, that's a rookie mistake. Your definition is U(p,q). For SU(p,q) you need to add the determinant condition. Unlike O(p,q) the determinant does affect the Lie algebra.

ok. yeah i missed the det condition when i wrote it out but put it in in the calculation and got the right answer for the SU(p,q) one.

okay so back to this dimension business.
we have the p^2 +q^2 already counted. C has entirely free components. that means it contributes pq=2pq/2. and then that all combines to give us the p^2+2pq+q^2=(p+q)^2.

finally i have to work out the lie algebra of S0*(n). i have been working with the defn

SO*(n) = \{ A \in M_n ( \mathbb{H} ) | \bar{A}^t j \mathbb{I} A = j \mathbb{I} \quad \text{ & det } A = +1 \} is this correct?

 

[Dick]

ok. yeah i missed the det condition when i wrote it out but put it in in the calculation and got the right answer for the SU(p,q) one.

okay so back to this dimension business.
we have the p^2 +q^2 already counted. C has entirely free components. that means it contributes pq. i think we count another pq from D giving 2pq and then that all combines to give us the p^2+2pq+q^2=(p+q)^2.

i still have a problem with the fact that we've counted D components when we've just proved that D is essentially entirely determined by choice of C.

Stop being so sloppy in counting. You had (p^2+q^2)/2 and you added 2pq and got (p^2+2pq+q^2)/2. You tell me what went wrong.

 

[latentcorpse]

Stop being so sloppy in counting. You had (p^2+q^2)/2 and you added 2pq and got (p^2+2pq+q^2)/2. You tell me what went wrong.

sorry i corrected this almost as soon as i posted it! see my amended post.

 

[latentcorpse]

Stop being so sloppy in counting. You had (p^2+q^2)/2 and you added 2pq and got (p^2+2pq+q^2)/2. You tell me what went wrong.

also, i have in the next part that

\mathfrak{su}(p,q) = \Biggl\{ \left( \begin{array}{cc} X & Z \\ \bar{Z}^t & Y \end{array} \right) \vline X \in M_p ( \mathbb{C} ) , Y \in M_q ( \mathbb{C} ) , Z \in M_{p \times q} ( \mathbb{C} ) , \bar{X}^t = - X , \bar{Y}^t = - Y , \text{tr}X + \text{tr}Y=0 \Biggr\}

now i need to show the dimension of this is n^2 - 1.

i have that this time since we're working in C and not R, we must also count diagonal elements of X and Y. so we get a contribution of (p^2+p)/2 and (q^2+q)/2. now Z will contribute pq=2pq/2 as before.
therefore, we have at the moment [(p+q)^2 + (p+q)]/2=(n^2+n)/2

then the constraint of vanishing trace forces us to subtract 1 from this giving

(n^2+n)/2-1

i can't seem to get that first term to go to n^2 regardless of what i try. any hints? thanks.

 

[Dick]

also, i have in the next part that

\mathfrak{su}(p,q) = \Biggl\{ \left( \begin{array}{cc} X & Z \\ \bar{Z}^t & Y \end{array} \right) \vline X \in M_p ( \mathbb{C} ) , Y \in M_q ( \mathbb{C} ) , Z \in M_{p \times q} ( \mathbb{C} ) , \bar{X}^t = - X , \bar{Y}^t = - Y , \text{tr}X + \text{tr}Y=0 \Biggr\}

now i need to show the dimension of this is n^2 - 1.

i have that this time since we're working in C and not R, we must also count diagonal elements of X and Y. so we get a contribution of (p^2+p)/2 and (q^2+q)/2. now Z will contribute pq=2pq/2 as before.
therefore, we have at the moment [(p+q)^2 + (p+q)]/2=(n^2+n)/2

then the constraint of vanishing trace forces us to subtract 1 from this giving

(n^2+n)/2-1

i can't seem to get that first term to go to n^2 regardless of what i try. any hints? thanks.

Hint: you are counting the real dimension of the Lie algebra. Z doesn't contribute pq anymore, it contributes 2pq. Each entry has an independent real and imaginary part.

 

[latentcorpse]

Hint: you are counting the real dimension of the Lie algebra. Z doesn't contribute pq anymore, it contributes 2pq. Each entry has an independent real and imaginary part.

ahhh. i see. so dim X = p^2 + p, dim Y = q^2+q, dim Z=2pq

adding together we get n^2 + n

so to get the right answer we need to subtract n+1 as a result of the constraint tr X + tr Y = 0. why is this? surely it's just one constraint so we should only have to subtract 1?

 

[Dick]

ahhh. i see. so dim X = p^2 + p, dim Y = q^2+q, dim Z=2pq

adding together we get n^2 + n

so to get the right answer we need to subtract n+1 as a result of the constraint tr X + tr Y = 0. why is this? surely it's just one constraint so we should only have to subtract 1?

How did you get dim(X)=p^2+p? I think you want to count that again.

 

[latentcorpse]

How did you get dim(X)=p^2+p? I think you want to count that again.

well initially i had it as (p^2+p)/2 but this just gives us the number of independent components. surely each of these will need to be counted twice for its' real and imaginary part?

 

[Dick]

well initially i had it as (p^2+p)/2 but this just gives us the number of independent components. surely each of these will need to be counted twice for its' real and imaginary part?

Careless counting again. Take it from the beginning. Think carefully this time.

 

[latentcorpse]

Careless counting again. Take it from the beginning. Think carefully this time.

thanks but i can't see it. since we're working in C, the diagonals aren't necessarily 0.
so we count the (p^2-p)/2 that we had before and then we add to that the p from the diagonal.

(p^2-p)/2+2p/2 = (p^2+p)/2 and then double it for real and imaginary parts

i just can't see where this is wrong. i'd convinced myself that the error must be in the number we subtract for the trace condition. we definitely only subtract 1 for the trace condition, right?

 

[Dick]

thanks but i can't see it. since we're working in C, the diagonals aren't necessarily 0.
so we count the (p^2-p)/2 that we had before and then we add to that the p from the diagonal.

(p^2-p)/2+2p/2 = (p^2+p)/2 and then double it for real and imaginary parts

i just can't see where this is wrong. i'd convinced myself that the error must be in the number we subtract for the trace condition. we definitely only subtract 1 for the trace condition, right?

No, you still haven't got dim(X) right. Count the off-diagonal elements first. Then count the diagonal elements. What can you say about the diagonal elements since X is antihermitian? They aren't necessarily zero. But they are necessarily something.

 

[latentcorpse]

No, you still haven't got dim(X) right. Count the off-diagonal elements first. Then count the diagonal elements. What can you say about the diagonal elements since X is antihermitian? They aren't necessarily zero. But they are necessarily something.

ok. so the diagonal elements have to be purely imaginary?
so we count the off diagonals this gives (p^2-p)/2 we multiply by 2 for real and imaginary to get p^2-p
our on diagonals are necessarily imaginary so we count only once giving +p

this cancels our -p so dim(X)=p^2.
we add this to the q^2 and 2pq to get the n^2 and then take off the 1 from the trace condition. sweet.

can u check the definition of SO*(n) i gave in post please?

 

[Dick]

ok. so the diagonal elements have to be purely imaginary?
so we count the off diagonals this gives (p^2-p)/2 we multiply by 2 for real and imaginary to get p^2-p
our on diagonals are necessarily imaginary so we count only once giving +p

this cancels our -p so dim(X)=p^2.
we add this to the q^2 and 2pq to get the n^2 and then take off the 1 from the trace condition. sweet.

can u check the definition of SO*(n) i gave in post please?

Right. Ordinarily you'd want to count tr(X)+tr(Y)=0 as two real constraints, but since you know the diagonal is pure imaginary, it's only one. I can't really check the definition of SO*(n), since I don't know what it is. What are H and j?

 

[latentcorpse]

Right. Ordinarily you'd want to count tr(X)+tr(Y)=0 as two real constraints, but since you know the diagonal is pure imaginary, it's only one. I can't really check the definition of SO*(n), since I don't know what it is. What are H and j?

j is a quaternion and H is the hypercopmlex numbers

 

[latentcorpse]

Right. Ordinarily you'd want to count tr(X)+tr(Y)=0 as two real constraints, but since you know the diagonal is pure imaginary, it's only one. I can't really check the definition of SO*(n), since I don't know what it is. What are H and j?

im trying to show that

\mathfrak{so}*(n)= \Biggl\{ X \in M_n ( \mathbb{H} ) | \bar{X}^t = jXj \} = \{ \left( \begin{array}{cc} X & -\bar{Y} \\ Y & \bar{X} \end{array} \right) \vline X,Y \in M_n ( \mathbb{C} ) , X^t=-X, \bar{Y}^t=Y \Biggr\}

right so i can get the first bit of this using my definition which I am pretty sure is correct i have that

if \bar{A}^t j A=j (note I am just going to write j instead of jI here)
then
exp( \tau \bar{X}^t) j exp ( \tau X ) = j
differentiating wrt tau and setting tau = 0 gives

\bar{X}^t j + j X = 0 \Rightarrow jX = - \bar{X}^t j \Rightarrow j^2 X = - j \bar{X}^t j
which gives
-X = - j \bar{X}^t j \Rightarrow X = j \bar{X}^t j
as required. now i need to get teh matrix bit.

if i set X = \left( \begin{array}{cc} B & C \\ D & E \end{array} \right)

then our condition is that

X=\left( \begin{array}{cc} j & 0 \\ 0 & j \end{array} \right) \left( \begin{array}{cc} \bar{B}^t & \bar{D}^t \\ \bar{C}^t & \bar{E}^t \end{array} \right) \left( \begin{array}{cc} j & 0 \\ 0 & j \end{array} \right) = \left( \begin{array}{cc} j \bar{B}^t j & j \bar{D}^t j \\ j \bar{C}^t j & j \bar{E}^t j \end{array} \right) = \left( \begin{array}{cc} - B^t & -D^t \\ -C^t & -E^t \end{array} \right)
where i used the property that for a z \in \mathbb{C}, zj=j \bar{z}

but this doesn't really look like where i want to be going.
any ideas?

thanks.

 

[Dick]

You know, I'm not all that familiar with quaternions and hypercomplex numbers. In particular, I'm not really sure how you have mapped the nxn hypercomplex matrix X into the 4 nxn complex matrices. Since you do, I think you should be able to sort this out. I shouldn't really be all that different from the other ones. I just don't know your conventions. If you really can't do it, I'd suggest reposting it and spelling out your conventions.