Unitary representation of GL(4)

[Jim Kata]

Let me preface this by saying, I am not a professional mathematician, and I do not know that much about Lie Algebras.

 

In this question I will be using the strong Einstein summation convention. The summation over the greek indices will be over numbers 0,1,2,3.

 

My question involves comparing and contrasting the Lie Algebras of the Poincare group versus that of the group [math]

GL(4,\mathbb{R})

[/math] with translations [math]

\mathbb{R}^4

[/math].

In special relativity coordinate transformations are given by [math]

{\mathbf{\bar x}} = \Lambda {\mathbf{x}} + {\mathbf{a}}

[/math] where [math]

\eta = \Lambda ^T \eta \Lambda

[/math] and

[math]

\eta _{\alpha \beta } = \left\{ {\begin{array}{*{20}c}

{0 if \alpha \ne \beta } \\

{-1 if \alpha ,\beta = 0} \\

{1 otherwise} \\

 

\end{array} } \right.

[/math].

A unitary representation of this transformation may be written as [math]

U\left( {\Lambda ,{\mathbf{a}}} \right)

[/math]. Infinitesimally, [math]

U\left( {1 + \omega ,\varepsilon } \right) = 1 + i\frac{1}

{2}\omega _{\alpha \beta } J^{\alpha \beta } + i\varepsilon _\rho P^\rho

[/math] where [math]

\omega _{\beta \alpha } = - \omega _{\alpha \beta }

[/math] and [math]

J^{\beta \alpha } = - J^{\alpha \beta }

[/math]. It can be worked out that [math]

U(\Lambda ,{\mathbf{a}})U(1 + \omega ,\varepsilon )U^{ - 1} (\Lambda ,{\mathbf{a}})

[/math] will give two equations:

 

[math]

U(\Lambda ,{\mathbf{a}})J^{\alpha \beta } U^{ - 1} (\Lambda ,{\mathbf{a}}) = \Lambda _\mu ^\alpha \Lambda _\tau ^\beta (J^{\mu \tau } - a^\mu P^\tau + a^\tau P^\mu )

[/math] (1)

and [math]

U(\Lambda ,{\mathbf{a}})P^\rho U^{ - 1} (\Lambda ,{\mathbf{a}}) = \Lambda _\mu ^\rho P^\mu

[/math] (2)

Where [math]

\Lambda _\mu ^\tau = \eta _{\mu \sigma } \Lambda ^\sigma _\gamma \eta ^{\gamma \tau }

[/math]

Taking [math]

U\left( {\Lambda ,{\mathbf{a}}} \right)

[/math] and [math]

U^{ - 1} \left( {\Lambda ,{\mathbf{a}}} \right)

[/math] as infinitesimals from equation (1) you obtain two Lie Algebras Namely:

 

[math]

i\left[ {J^{\alpha \beta } ,J^{\mu \tau } } \right] = \eta ^{\beta \mu } J^{\alpha \tau } - \eta ^{\alpha \mu } J^{\beta \tau } - \eta ^{\tau \alpha } J^{\mu \beta } + \eta ^{\tau \beta } J^{\mu \alpha }

[/math] (3)

and [math]

i\left[ {P^\rho ,J^{\mu \tau } } \right] = \eta ^{\rho \mu } P^\tau - \eta ^{\rho \tau } P^\mu

[/math] (4)

Doing a similar procedure with equation (2) you obtain two more Lie Algebras Namely:

[math]

i\left[ {J^{\mu \tau } ,P^\rho } \right] = - \eta ^{\rho \mu } P^\tau + \eta ^{\rho \tau } P^\mu

[/math] (5) Which is consistent with equation (4) and

[math]

\left[ {P^\alpha ,P^\beta } \right] = 0

[/math] (6)

 

Now recreating this entire process except this time using [math]

GL(4,\mathbb{R})

[/math] instead of [math] O(3,1) [/math].

 

A general coordinate change is given by [math]

{\mathbf{\bar x}} = {\mathbf{Ax}} + {\mathbf{a}}

[/math] where [math]

{\mathbf{A}}\varepsilon GL(4,\mathbb{R})

[/math] and [math]

{\mathbf{a}}\varepsilon \mathbb{R}^4

[/math]. It has the property [math]

{\mathbf{\bar g}} = {\mathbf{A}}^T {\mathbf{gA}}

[/math] where [math]

{\mathbf{g}}[/math] is the symmetric bilinear form on the space.

It's unitary representation is [math]

U\left( {{\mathbf{A}},{\mathbf{a}}} \right)

[/math]. Infinitesimally, [math]

U\left( {1 + \xi ,\varepsilon } \right) = 1 + i\xi _{\alpha \beta } A^{\alpha \beta } + i\varepsilon _\rho P^\rho

[/math] where [math]

\xi _{\alpha \beta } = g_{\alpha \mu } \xi ^\mu ,_\beta

[/math] and [math]

\varepsilon _\rho = g_{\rho \mu } \varepsilon ^\mu

[/math]. Doing a coordinate change [math]

\xi _{\alpha \beta } = \xi _{[\alpha ,\beta ]} + \xi _{\{ \alpha ,\beta \} }

[/math] and [math]

A^{\alpha \beta } = J^{\alpha \beta } + D^{\alpha \beta }

[/math] where [math]

\xi _{[\alpha ,\beta ]} = \frac{1}

{2}(\xi _{\alpha \beta } - \xi _{\beta \alpha } )

[/math],[math]

\xi _{\{ \alpha ,\beta \} } = \frac{1}

{2}(\xi _{\alpha \beta } + \xi _{\beta \alpha } )

[/math], [math]

J^{\alpha \beta } = A^{\alpha \beta } - A^{\beta \alpha }

[/math], and [math]

D^{\alpha \beta } = A^{\alpha \beta } + A^{\beta \alpha }

[/math] you obtain [math]

U(1 + \xi ,\varepsilon ) = 1 + \frac{1}

{2}i\xi _{[\alpha ,\beta ]} J^{\alpha \beta } + \frac{1}

{2}i\xi _{\{ \alpha ,\beta \} } D^{\alpha \beta } + i\varepsilon _\rho P^\rho

[/math] (7)

Now doing the same procedure as done for the Poincare group three equations are obtained for [math]

U({\mathbf{A}},{\mathbf{a}})U(1 + \xi ,\varepsilon )U^{ - 1} ({\mathbf{A}},{\mathbf{a}})

[/math] namely:

[math]

U(A,a)J^{\alpha \beta } U^{ - 1} (A,a) = A^{ - 1\alpha } _\mu A^{ - 1\beta } _\tau (J^{\mu \tau } + a^\mu P^\tau - a^\tau P^\mu )

[/math] (8)

[math]

U(A,a)D^{\alpha \beta } U^{ - 1} (A,a) = A^{ - 1\alpha } _\mu A^{ - 1\beta } _\tau (D^{\mu \tau } - a^\mu P^\tau - a^\tau P^\mu )

[/math] (9)

[math]

U(A,a)P^\rho U^{ - 1} (A,a) = A^{ - 1\rho } _\mu P^\mu

[/math] (10)

Taking [math]

U({\mathbf{A}},{\mathbf{a}})

[/math] and [math]

U^{ - 1} ({\mathbf{A}},{\mathbf{a}})

[/math] as infinitesimals from equation (8) you obtain three Lie Algebras namely:

[math]

i\left[ {J^{\alpha \beta } ,J^{\mu \tau } } \right] = g ^{\beta \mu } J^{\alpha \tau } - g ^{\alpha \mu } J^{\beta \tau } - g ^{\tau \alpha } J^{\mu \beta } + g ^{\tau \beta } J^{\mu \alpha }

[/math] (11) which is the equivalent of equation (3) for a general metric

[math]

i\left[ {P^\rho ,J^{\mu \tau } } \right] = g ^{\rho \mu } P^\tau - g ^{\rho \tau } P^\mu

[/math] (12) which is the equivalent of equation (4) for a general metric

[math]

i[J^{\alpha \beta } ,D^{\mu \tau } ] = g^{\beta \mu } J^{\alpha \tau } + g^{\beta \tau } J^{\alpha \mu } - g^{\alpha \tau } J^{\beta \mu } - g^{\alpha \mu } J^{\beta \tau }

[/math] (13)

Doing the same procedure with equation (9), you obtain:

[math]

i[D^{\alpha \beta } ,D^{\mu \tau } ] = g^{\beta \mu } D^{\alpha \tau } + g^{\beta \tau } D^{\alpha \mu } + g^{\alpha \tau } D^{\beta \mu } + g^{\alpha \mu } D^{\beta \tau }

[/math] (14)

[math]

i[P^\rho ,D^{\mu \tau } ] = - g^{\rho \mu } P^\tau - g^{\rho \tau } P^\mu

[/math] (15)

[math]

i[J^{\alpha \beta } ,D^{\mu \tau } ] = g^{\beta \mu } D^{\alpha \tau } + g^{\beta \tau } D^{\alpha \mu } - g^{\alpha \tau } D^{\beta \mu } - g^{\alpha \mu } D^{\beta \tau }

[/math] (16)

and doing the same procedure with equation (10) you obtain

[math]

i\left[ {J^{\mu \tau},P^\rho } \right] = -g ^{\rho \mu } P^\tau + g ^{\rho \tau } P^\mu

[/math] (17) which is consistent with equation (12)

[math]

i[P^\rho ,D^{\mu \tau } ] = g^{\rho \mu } P^\tau + g^{\rho \tau } P^\mu

[/math] (18)

and [math]

\left[ {P^\alpha ,P^\beta } \right] = 0

[/math]

 

Looking at equations (15) and (18) led me to conclude that

[math]

[P^\rho ,D^{\mu \tau } ] = 0[/math] looking at equation (14) I also had to conclude that [math]

[D^{\alpha \beta } ,D^{\mu \tau } ] = 0[/math] Now equations (13) and (16), do not appear to be consistent. The only way I can think of in which they would be consistent is if [math]

[J^{\alpha \beta } ,D^{\mu \tau } ] = 0[/math] is there an easy way to prove that, maybe using the Jacobi identity? Am I going about this all wrong? What is going on? Why do I get these inconsistencies in the Lie Algebras? I know the answer is probably simple.

[w=f[z]]

Hi Jim,

I see from your post history that you don't get too many responses. This is probably because you know more about this stuff than most (if not all) of us.

 

It's been a few years since I had to think about relativity in the sense of actually working through tensor equations and such. And then I would have to wonder how much of it I understood in the first place. So I am beyond rusty.

 

But...

I have a couple friends on another forum that seem to enjoy getting their hands dirty with the stuff you ask about. I'll send the word for them to come over here and specifically have a look at your posts and see if it's something they would like to get into.

 

That's the best I can do for now.

 

Nice TeXing by the way.

 

Cheers,

w=f[z]

[BenTheMan]

Jim---

 

I think the error is here:

 

It has the property [math]

{\mathbf{\bar g}} = {\mathbf{A}}^T {\mathbf{gA}}

[/math] where [math]

{\mathbf{g}}[/math] is the symmetric bilinear form on the space.

It's unitary representation is [math]

U\left( {{\mathbf{A}},{\mathbf{a}}} \right)

[/math].

 

The Lie Algebras have special properties which govern what types of representations you're allowed to use. For example, for [math]U \in SO(N)[/math], we know [math]U^TU = \mathbb I [/math], where [math]\mathbb{I}[/math] is the identity. The matrices which live in the group [math]GL(N,\mathbb{R})[/math] don't obey the same properties.

 

The group SO(3,1) consists of special (S) orthogonal (O) matrices. Special means determinant one, and orthogonal means [math]U^TU = \mathbb I [/math]. The GL consist of general(G) Linear (L) matrices. Linear in the sense that one can solve linear equations (i.e. the matrices are invertible and don't have a zero determinant) and general in the sense that anything goes.

 

So, I think that one can't find unitary representations of the GL group. I DO think that there is an embedding of SO(N) in [math]GL(N,\mathbb{R})[/math]. That is, I think you may be able to find a subset of general linear matrices which obey the SO(N) algebra.

 

Check out this: http://en.wikipedia.org/wiki/General_linear_group.

 

I am wrong a lot, though, especially when it comes to math, so if anyone else has a better answer please don't hesitate to point this out to me.

[Jim Kata]

I believe I can prove from my previous post that [math]

[J^{\alpha \beta } ,D^{\mu \tau } ] = 0

[/math] tell me if you think my argument works. First just by looking at equation (13) and (16) in my previous post, it should be obvious that they would be consistent if [math]

[J^{\alpha \beta } ,D^{\mu \tau } ] = 0

[/math]. Here's a sketch of a proof. I'm sorry if it is not very rigorous.

 

It should be noted that according to the commutation relation (12) in my previous post that the commutator of [math]J^{\alpha \beta }[/math]and [math]P^\rho[/math] is a function of [math]P[/math]. Because of this fact and the fact that [math][D^{\alpha \beta } ,P^\rho ] = 0[/math], it is true that [math][D^{\alpha \beta } ,[J^{\mu \tau } ,P^\rho ]] = 0[/math]. Now using the Jacobi identity we have: [math]

[D^{\alpha \beta } ,[J^{\mu \tau } ,P^\rho ]] + [J^{\mu \tau } ,[P^\rho ,D^{\alpha \beta } ]] + [P^\rho ,[D^{\alpha \beta } ,J^{\mu \tau } ]] = 0

[/math]. The first term is obviously zero for reasons mentioned above and the second term in the Jacobi is also zero since [math][P^\rho ,D^{\alpha \beta } ] = 0[/math] as mentioned in my previous post. This leaves the identity [math][P^\rho ,[D^{\alpha \beta } ,J^{\mu \tau } ]] = 0[/math]. Now there is a theorem that says:A linear operator that commutes with each of a complete set of commuting observables is a function of those observable. Using this theorem it is implied that [math]

[D^{\alpha \beta } ,J^{\mu \tau } ] = f^{\alpha \beta \mu \tau } (P)

[/math]. Now since [math]D^{\alpha \beta }[/math] commutes with [math]P^\rho[/math], looking at the equation in the previous sentence it is true that [math][D^{\alpha \beta } ,[J^{\mu \tau } ,D^{\alpha \beta } ]] = 0[/math]. Lets consider the set of eigenvectors of [math]J^{\mu\tau}[/math]. So [math]

J^{\mu \tau } \left| {\mathop m\limits^{\mu \tau } } \right\rangle = \lambda ^{\mu \tau } (m)\left| {\mathop m\limits^{\mu \tau } } \right\rangle [/math], where [math]m[/math] are the labels of the states of the eigenvectors corresponding to any of the operators [math]J^{\mu \tau }[/math], and [math]m[/math] doesn’t have to be the same labeling for each of the operators. I am also discretely assuming that the eigenvalues of these operators are discrete and finite, which for a particular gauge can be shown. Taking the expectation value of the double commutator [math]

[D^{\alpha \beta } ,[J^{\mu \tau } ,D^{\alpha \beta } ]] = 0

[/math] you get [math]

\mathop \sum \limits_{m'} \left( {\lambda ^{\mu \tau } (m') - \lambda ^{\mu \tau } (m)} \right)\left| {d_{m'm}^{\alpha \beta (\mu \tau )} } \right|^2

[/math] (2.1), where [math]

d_{m'm}^{\alpha \beta (\mu \tau )} = \left\langle {\mathop {m'}\limits^{\mu \tau } } \right|D^{\alpha \beta } \left| {\mathop m\limits^{\mu \tau } } \right\rangle

[/math]. If there existed an [math]m[/math] for which [math]

\lambda ^{\mu \tau } (m') \ne \lambda ^{\mu \tau } (m)

[/math] with [math]d_{m'm}^{\alpha \beta (\mu \tau )}[/math] also not equal to zero, one could choose the [math]m[/math] that made [math]

\lambda ^{\mu \tau } (m)[/math] the smallest eigenvalue to satisfy the criteria above, this can be done since [math]m[/math] is a sum over a finite range of numbers. In which case the right hand side of equation (2.1) would be positive definite, contradicting equation (2.1). From this it can be concluded that [math]d_{m'm}^{\alpha \beta (\mu \tau )}[/math] must vanish for all [math]m'[/math],[math]m[/math] in which [math]\lambda ^{\mu \tau } (m') \ne \lambda ^{\mu \tau } (m)[/math]. This proves that the eigenvectors of [math]J^{\mu \tau }[/math] also diagonalize [math]D^{\alpha\beta}[/math] and hence [math][J^{\alpha \beta } ,D^{\mu \tau } ] = 0[/math] which means [math]D^{\alpha\beta}[/math] acts like an internal symmetry.