Conformal group, Conformal algebra and Conformal invariance in field theory

**samalkhaiat** · Jun1-07, 03:56 PM

I have noticed that questions about this subject get either ignored or receive some confusing answers. So I decided to write a "brief" but self-contained introduction to the subject. I'm sure you will find it useful.
It is going to take about 13 or 14 post to complete the work. Be patient with me as my time allow me to post only 2 or 3 posts a day.
Equations and exercises will be numbered by the post number; for example Eq(1.6) means equation 6 in post#1 and Ex(5.7) stands for exercise 7 in post#5.

SO PLEASE DO NOT POST YOUR COMMENTS, QUESTIONS etc.,IN BETWEEN MY POSTS, AS THIS WOULD MESS UP THE NUMBERING.

CONFORMAL TRANSFORMATIONS

Consider a flat n-dimensional Minkowski spacetime $LaTeX Code: (M^{n}, \\eta)$ . The conformal group C(1,n-1) can be formally realized as a group of (nonlinear) coordinate transformations:

$LaTeX Code: f: x \\rightarrow \\bar{x} = \\bar{x}(x)$

which leaves the metric $LaTeX Code: (f* \\bar{g})_{ab}$ , where

$LaTeX Code: \\bar{g} = \\eta_{ab} d \\bar{x}^{a} \\otimes d\\bar{x}^{b},$

invariant up to a scale:

$LaTeX Code: \\bar{g}_{ab}(x) \\left ( = \\partial_{a} \\bar{x}^{c} \\partial_{b} \\bar{x}^{d} \\eta_{cd} \\right ) = S(x) \\eta_{ab} \\ \\ (1.1)$

I.e.,

$LaTeX Code: d \\bar{s}^{2} \\left ( = \\eta_{ab} d\\bar{x}^{a}d\\bar{x}^{b} \\right ) = S(x) ds^2 \\ \\ (1.1single-quote)$

and we say that the conformal group preserves the light-cone structure. This excludes the conformal group as a symmetry of massive particle theories. If massive particles are included, the condition S=1 must be imposed which restricts the symmetry to the Poicare subgroup.
For spacetime with n>2, the conformal group is finite-dimensional. To see this, let us solve (1.1) for general infinitesimal coordinate transformation;

$LaTeX Code: \\bar{x}^{a} = x^{a} + f^{a}(x)$

which leads to

$LaTeX Code: \\partial_{a} f_{b} + \\partial_{b} f_{a} = \\eta_{ab} (S-1),$

or, taking the trace to obtain $LaTeX Code: (S-1) = 2/n \\partial .f \\equiv F$ ,

$LaTeX Code: \\partial_{a} f_{b} + \\partial_{b} f_{a} = \\eta_{ab} F(x) \\ \\ (1.2)$

By applying an extra derivative $LaTeX Code: \\partial_{c}$ on this (conformal Killing) equation, permuting the indices and taking a linear combination, we get

$LaTeX Code: \\partial_{c} ( \\partial_{a} f_{b} - \\partial_{b} f_{a}) = \\eta_{cb} \\partial_{a} F - \\eta_{ac} \\partial_{b} F \\ \\ (1.3)$

or, after integration,

$LaTeX Code: \\partial_{a} f_{b} - \\partial_{b} f_{a} = \\int ( \\partial_{a} F dx_{b} - \\partial_{b} F dx_{a} ) + 2 \\omega_{ab} \\ \\ (1.4)$

for some constant antisymmetric tensor $LaTeX Code: \\omega_{ab}$ .
Adding (1.2) to (1.4) and integrating again, we find

$LaTeX Code: f^{a} = a^{a} + \\omega^{ba}x_{b} + \\frac{1}{2} \\int dx^{a} F + \\frac{1}{2} \\int dx^{b} \\int \\left ( \\partial_{b} F dx^{a} - \\partial^{a} F dx_{b} \\right ) \\ \\ (1.5)$

where $LaTeX Code: a^{a}$ is a constant n-vector. Notice that the first two terms represent Poincare transformation. This is expected because F = 0 corresponds to a coordinate transformations which do not change the form of the metric, i.e., a general solution to the homogeneous differential equation $LaTeX Code: \\partial_{a}f_{b} + \\partial_{b}f_{a} = 0$ .

The integral equation (1.5) determines the conformal Killing vector f once the function F(x) is found. So let us find it; By contracting the indices (c,a) in eq.(1.3), we get

$LaTeX Code: \\partial^{2} f_{b} = (1 - n/2) \\partial_{b} F \\ \\ \\ (1.6)$

Operate by $LaTeX Code: \\partial_{a}$ and form the symmetric combination;

$LaTeX Code: 2(1 - n/2) \\partial_{a} \\partial_{b} F = \\partial^{2} ( \\partial_{a}f_{b} + \\partial_{b}f_{a} )$

now, use the conformal Killing equation to find

$LaTeX Code: (2 - n) \\partial_{a} \\partial_{b} F = \\eta_{ab} \\partial^{2} F \\ \\ (1.7)$

Finally, contracting with $LaTeX Code: \\eta^{ab}$ , we end up with

$LaTeX Code: (n -1) \\partial^{2} F = 0 \\ \\ (1.8)$

Therefore $LaTeX Code: \\partial^{2} F = 0$ except for the trivial case n = 1.
Thus for n > 1, Eq(1.7) becomes

$LaTeX Code: (2 - n) \\partial_{a} \\partial_{b} F = 0 \\ \\ (1.9)$

When n > 2, this implies

$LaTeX Code: \\partial_{a} \\partial_{b} F(x) = 0 \\ \\ (1.10)$

I.e., F is at most linear in the coordinates:

$LaTeX Code: F(x) = -2 \\alpha + 4 c_{a}x^{a} \\ \\ (1.11)$

with $LaTeX Code: (\\alpha , c_{a})$ are constants.

Inserting (1.11) in (1.5), we find our conformal Killing vector

$LaTeX Code: f^{a} = a^{a} + \\omega^{ba}x_{b} - \\alpha x^{a} + c_{b} (2x^{a}x^{b} - \\eta^{ab} x^{2}) \\ \\ (1.12)$

This depends on (n + 1)(n + 2)/2 parameters: n translations, n(n-1)/2 Lorentz transformations, one dilation and n special conformal transformations.

In n = 2, Eq(1.9) does not imply Eq(1.10), which was crucial for the finiteness of the group C(1,n-1) in the n>2 case, in this case (n=2) every harmonic function F determines a solution, i.e., the group C(1,1) becomes infinite-dimensional. This C(1.1), interesting for string theories, will not be discussed in this introduction.

more to come, please be patient....

**samalkhaiat** · Jun1-07, 04:50 PM

The local structure of any Lie group contains the most essential feature of the group, namely its continuity, and it is studied by considering the group for values of the parameters in the neighbourhood of the identity (infinitesimal transformations). The remarkable property of Lie groups (discovered by S Lie himself) is that in spite of the continuity of the parameters, almost all information about the local structure can be obtained from finite discrete system, namely Lie algebras.
Lie algebras may be regarded as (finite-dimensional) vector spaces equipped with a multiplication law;
$LaTeX Code: [ . , . ] : \\cal{L} \\times \\cal{L} \\rightarrow \\cal{L}$
which satisfies the conditions

i) [X,Y] = - [Y,X]

ii) [X,[Y,Z]] + [Z,[X,Y]] + [Y,[Z,X]] = 0

for all (X,Y,Z) in $LaTeX Code: \\cal{L}$ .

Please note that the product [X,Y] (Lie bracket) need not necessarily be of the form (XY - YX), Poisson's bracket is another example of Lie bracket.

The 1-to-1 correspodence between Lie algebras and local Lie groups can be expressed explicitly by means of the exponential mapping, i.e.,(almost all) elements of a connected Lie group can be represented as exponentials of Lie algebra elements;

$LaTeX Code: g = e^{X} = e^{a_{i}X^{i}} \\ \\ (2.1)$

Where $LaTeX Code: \\{X^{i}\\}$ is a basis for $LaTeX Code: \\cal{L}$ .The components $LaTeX Code: a_{i}$ of Lie algebra elements (the parameters) play the role of local coordinates (in the neighbourhood of the identity) of the group manifold. The fact that the union of elements of the form (2.1) presents a Lie group is a consequence of the Baker-Campbell-Hausdroff formula

$LaTeX Code: e^{X} e^{Y} = e^{X + Y + \\frac{1}{12}[X,[X,Y]] + ...} \\ \\ (2.2)$

i.e.,

more to come, please dont reply to my posts just yet....

**samalkhaiat** · Jun1-07, 06:07 PM

From infinitesimal transformations to Lie algebra

By considering the combined action of various infinitesimal transformations in different orders, the Lie algebra of the conformal group can be abstructed.
Let us write the conformal Killing vector (1.12) in the form;

$LaTeX Code: f^{a} = \\lambda_{A} \\delta_{( \\lambda )}^{A} x^{a} \\ \\ (3.1)$

where

$LaTeX Code: \\lambda_{A} \\equiv ( a^{a}, \\frac{1}{2}\\omega^{ab}, \\alpha , c^{a})$

and $LaTeX Code: \\delta^{A}$ are the generators of the following infinitesimal coordinate transformations;

Translation;

$LaTeX Code: \\delta^{a}_{(T)} x^{c} = \\eta^{ab} \\ \\ (3.2a)$

Lorentz;

$LaTeX Code: \\delta^{ab}_{(L)} x^{c} = \\eta^{bc}x^{a} - \\eta^{ac}x^{b} \\ \\ (3.2b)$

Dilation(scale);

$LaTeX Code: \\delta_{(D)}x^{c} = -x^{c} \\ \\ (3.2c)$

Conformal;

$LaTeX Code: \\delta^{a}_{(C)}x^{c} = 2 x^{a}x^{c} - \\eta^{ac}x^{2} \\ \\ (3.2d)$

Ex(3.1) Show that;

$LaTeX Code: \\left[\\delta^{a}_{(T)},\\delta^{b}_{(T)}\\right] = 0 \\ \\ (3.3a)$

$LaTeX Code: \\left[\\delta^{ab}_{(L)},\\delta^{c}_{(T)}\\right] = \\eta^{bc}\\delta^{a}_{(T)} - \\eta^{ac}\\delta^{b}_{(T)} \\ \\ (3.3b)$

$LaTeX Code: \\left[\\delta^{ab}_{(L)},\\delta^{cd}_{(L)}\\right] = \\eta^{ac}\\delta^{bd}_{(L)} - \\eta^{bc}\\delta^{ad}_{(L)} + \\eta^{ad}\\delta^{cb}_{(L)} - \\eta^{bd}\\delta^{ca}_{(L)} \\ \\ (3.3c)$

$LaTeX Code: \\left[\\delta_{(D)},\\delta_{(D)}\\right] = \\left[\\delta^{ab}_{(L)},\\delta_{(D)}\\right] = 0 \\ \\ (3.3d)$

$LaTeX Code: \\left[\\delta_{(D)},\\delta^{a}_{(T)}\\right] = \\delta^{a}_{(T)} \\ \\ (3.3e)$

$LaTeX Code: \\left[\\delta_{(D)},\\delta^{a}_{(C)}\\right] = -\\delta^{a}_{(C)} \\ \\ (3.3f)$

$LaTeX Code: \\left[\\delta^{a}_{(C)},\\delta^{b}_{(C)}\\right] = 0 \\ \\ (3.3g)$

$LaTeX Code: \\left[\\delta^{ab}_{(L)},\\delta^{c}_{(C)}\\right] = \\eta^{bc}\\delta^{a}_{(C)} - \\eta^{ac}\\delta^{b}_{(C)} \\ \\ (3.3h)$

$LaTeX Code: \\left[\\delta^{a}_{(C)},\\delta^{b}_{(T)}\\right] = 2 \\eta^{ab}\\delta_{(D)} - 2 \\delta^{ab}_{(L)} \\ \\ (3.3i)$

Thus the operators $LaTeX Code: \\delta^{A}_{(\\lambda )}$ span a vector space with multiplication law as in (3.3). It is the Lie algebra of the conformal group C(1,n-1).

more to come, please dot reply just yet...

**samalkhaiat** · Jun1-07, 07:20 PM

Let $LaTeX Code: f_{(1)} = f^{a}_{(1)}\\partial_{a}$ and $LaTeX Code: f_{(2)} = f^{a}_{(2)}\\partial_{a}$ be two conformal Killing fields in Minkowsiki spacetime. Their Lie bracket

$LaTeX Code: [f_{(1)},f_{(2)}] = f_{(3)} = f^{a}_{(3)}\\partial_{a} \\ \\ (4.1)$

with

$LaTeX Code: f^{a}_{(3)} = f^{b}_{(1)}\\partial_{b}f^{a}_{(2)} - f^{b}_{(2)}\\partial_{b}f^{a}_{(1)} \\ \\ (4.2)$

gives the vector field $LaTeX Code: f_{(3)}$ which also satisfies the conformal Killing Eq(1.2). thus, the set of all conformal Killing vectors form a Lie algebra.

Now, we introduce a basis

$LaTeX Code: G^{A}_{(\\lambda )} \\equiv \\left( p^{a}, j^{ab}, d, k^{a}\\right) \\ \\ (4.3)$

for the algebra by the rule

$LaTeX Code: f = \\lambda_{A}\\left( \\delta^{A}_{(\\lambda )} x^{a} \\right) \\partial_{a} = i \\lambda_{A}G^{A}_{(\\lambda )} \\ \\ (4.4)$

From (3.2) we find

$LaTeX Code: ip^{a} = \\partial^{a} \\ \\ (4.5a)$

$LaTeX Code: ij^{ab} = x^{a}\\partial^{b} - x^{b}\\partial^{a} \\ \\ (4.5b)$

$LaTeX Code: id = -x^{a}\\partial_{a} \\ \\ (4.5c)$

$LaTeX Code: ik^{a} = 2x^{a} x^{c}\\partial_{c} - x^{2} \\partial^{a} \\ \\ (4.5d)$

And, by explicit calculations we find the algebra

$LaTeX Code: [ip^{a} , p^{b}] = [id , d] = [ij^{ab}, d] = [ik^{a},k^{b}] = 0 \\ \\ (4.6a)$

$LaTeX Code: [id , p] = p \\ \\ (4.6b)$

$LaTeX Code: [id, k] = -k \\ \\ (4.6c)$

$LaTeX Code: \\left[ ik^{a},p^{b}\\right] = 2 \\eta^{ab} d - 2 j^{ab} \\ \\ (4.6d)$

$LaTeX Code: \\left[ ij^{ab}, k^{c}\\right] = \\eta^{ac} k^{b} - \\eta^{bc} k^{a} \\ \\ (4.6e)$

$LaTeX Code: \\left[ ij^{ab}, p^{c}\\right] = \\eta^{ac} p^{b} - \\eta^{bc} p^{a} \\ \\ (4.6f)$

$LaTeX Code: \\left[ ij^{ab}, j^{cd}\\right] = \\eta^{bc}j^{ad} - \\eta^{ac}j^{bd} + \\eta^{ad}j^{bc} - \\eta^{bd}j^{ac} \\ \\ (4.6g)$

Since every Lie algebra has a faithful matrix representation, we can forget about the explicit (differential) realization (4.5) and postulate the conformal algebra as an abstract real algebra subject to 2 requirements;
1) it has a basis

$LaTeX Code: G^{A} = \\left( P^{a}, J^{ab}=-J^{ba}, D, K^{a}\\right) \\ \\ (4.7)$
with multiplication law as in (4.6);
2) its general element is of the form

$LaTeX Code: X = i \\lambda_{A}G^{A} \\ \\ (4.8)$

The conformal group C(1,n-1) is formally obtained by exponentiation of the conformal algebra

$LaTeX Code: g(\\lambda ) = \\exp (i \\lambda_{A} G^{A} ) \\ \\ (4.9)$

more to come, please dont reply just yet,.....

**samalkhaiat** · Jun2-07, 04:51 PM

This time, we cosider various subgroups of the conformal group, find their multiplication laws and derive the corresponding Lie subalgebras using a unitary representations of the laws.

Scale plus Translation

Consider the transformation;

$LaTeX Code: \\bar{x} = T(\\alpha , a) x = e^{\\alpha} x + a \\ \\ (5.1)$

It is easy to see that two such transformations is also a transformation of the same kind;

$LaTeX Code: T(\\alpha, a) T(\\beta , b) = T\\left( \\alpha + \\beta , a + be^{\\alpha} <BR>\\right) \\ \\ (5.2)$

Ex(5.1): Prove the above "easy to see statement".

From (5.2), we find;

1)

$LaTeX Code: \\{T(\\alpha , a) T(\\beta , b) \\} T(\\gamma , c) = T\\left(\\alpha + \\beta +\\gamma , a + be^{\\alpha} + ce^{\\alpha + \\beta}\\right) = T(\\alpha , a) \\{ T(\\beta , b) T(\\gamma , c) \\} \\ \\ (5.3a)$

i.e., the multiplication law (5.2) is associative.

Ex(5.2): Prove Eq(5.3a).

2)

$LaTeX Code: T(\\alpha , a) T(0,0) = T(0,0) T(\\alpha , a) = T(\\alpha , a) \\ \\ (5.3b)$

i.e., an identity element exists;

3)

$LaTeX Code: T\\left(\\alpha , a \\right) T\\left(-\\alpha , -ae^{-\\alpha} \\right) = T \\left(-\\alpha , -ae^{-\\alpha} \\right) T(\\alpha , a) = T(0,0) \\ \\ (5.3c)$

i.e., an inverse element exsists

$LaTeX Code: T^{-1}(\\alpha , a) = T\\left( -\\alpha , -ae^{-\\alpha} \\right) \\ \\ (5.4)$

Thus, the set $LaTeX Code: \\{T(\\alpha , a)\\}$ with the multiplication law (5.2) forms a group.

We know from QM that coordinate transformations, such as our T, induce a unitary linear transformation on vectors in the physical Hilbert space;

$LaTeX Code: | \\Psi_{T}\\rangle = U(T) | \\Psi \\rangle \\ \\ (5.5)$

i.e., the unitary operator U forms a representation of the group (multiplication law) in question;

$LaTeX Code: U( \\alpha , a) U(\\beta , b) = U( \\alpha + \\beta , a + be^{\\alpha} ) \\ \\ (5.6)$

Since U(T(0,0)) = U(E) carries any vector into itself, it must be proportional to the unit operator. So, in at least a finite neighbourhood of the identity, U can be represented by a power series. For infinitesimal coordinate transformations, U(T) must then differ from 1 by terms linear in the parameters $LaTeX Code: (\\alpha , a)$ , so we may write

$LaTeX Code: U(\\alpha , a) \\approx 1 + i \\alpha D + i a^{a}P_{a} \\ \\ (5.7)$

where

are Hermitian operators independent of the parameters $LaTeX Code: (\\alpha , a)$ .
In order to find the transformations laws of D and P, you need to work out the following;

Ex(5.3): Show that

$LaTeX Code: U(\\beta , b)U(\\alpha , a)U^{-1}(\\beta , b) = U\\left(\\alpha , ae^{\\beta} - \\alpha b \\right) \\ \\ (5.8)$

where $LaTeX Code: (\\beta ,b)$ are the parameters of new transformation unrelated to $LaTeX Code: (\\alpha ,a)$

Next, write (5.8) to the 1st order in $LaTeX Code: (\\alpha ,a)$ ,equate coefficients of $LaTeX Code: \\alpha$ and

on both sides and show that

$LaTeX Code: \\bar{P}_{a} = U^{-1}(\\beta , b) P_{a} U(\\beta , b) = e^{-\\beta} P_{a} \\ \\ (5.9a)$
$LaTeX Code: \\bar{D} = U^{-1} D U = D + b_{a}P^{a} e^{-\\beta} \\ \\ (5.9b)$

Now, let $LaTeX Code: (\\beta , b)$ themselves be infinitesimals and find the commutation relations;

$LaTeX Code: [iD , D] = 0 \\ \\ (5.10a)$
$LaTeX Code: [iP^{a} , P^{b}] = 0 \\ \\ (5.10b)$
$LaTeX Code: [iD , P^{a}] = P^{a} \\ \\ (5.10c)$

This is the Lie algebra of the (semi-direct product) group of scale & translation.

****

Poincare' group

The same can be applied to the Poincare transformations

$LaTeX Code: \\bar{x} = T(\\Lambda , a)x = \\Lambda x + a \\ \\ (5.11)$

which have the multiplication law

$LaTeX Code: T_{1}T_{2} = T(\\Lambda_{1} \\Lambda_{2} , \\Lambda_{1} a_{2} + a_{1}) \\ \\ (5.12)$

For infinitesimal transformation, $LaTeX Code: T(1 + \\omega , \\epsilon )$ , we may write the corresponding unitary operator as

$LaTeX Code: U(1+\\omega ,\\epsilon ) = 1 + i \\epsilon . P - (i/2) \\omega . J \\ \\ ((5.13)$

where $LaTeX Code: \\omega . J \\equiv \\omega_{ab}J^{ab}$

Ex(5.4): Use Eq(5.12) to show that

$LaTeX Code: U(\\Lambda ,a)U(1+\\omega ,\\epsilon )U^{-1}(\\Lambda , a) = U(\\Lambda \\omega \\Lambda^{-1} , \\Lambda \\epsilon - \\Lambda \\omega \\Lambda^{-1} a) \\ \\ (5.14)$

write this to 1st order in $LaTeX Code: (\\omega , \\epsilon)$ , then show that J is a Lorentz tensor;

$LaTeX Code: \\bar{J}^{ab} = U^{-1}J^{ab}U = \\Lambda_{c}{}^{a}\\Lambda_{d}{}^{b}( J^{cd} + a^{c}P^{d} - a^{d}P^{c}) \\ \\ (5.15a)$

and P is a vector

$LaTeX Code: \\bar{P}^{a} = U^{-1}P^{a}U = \\Lambda_{b}{}^{a}P^{b} \\ \\ (5.15b)$

From the infinitesimal versions of Eq(5.15) deduce the Lie algebra of Poincare' group;

$LaTeX Code: [iP_{a},P_{b}] = 0 \\ \\ \\ \\ (5.16a)$
$LaTeX Code: [iJ^{ab},P^{c}] = \\eta^{ac}P^{b} - \\eta^{bc}P^{a} \\ \\ (5.16b)$
$LaTeX Code: [iJ^{ab},J^{cd}] = \\eta^{bc}J^{ad} - \\eta^{ac}J^{bd} + \\eta^{ad}J^{bc} - \\eta^{bd}J^{ac} \\ \\ (5.16c)$

****

Scale & Lorentz group

$LaTeX Code: \\bar{x} = T(\\alpha ,\\Lambda ) x = e^{\\alpha} \\Lambda x \\ \\ \\(5.17)$

For this group, we have

$LaTeX Code: u(\\alpha_{1} , \\Lambda_{1})u(\\alpha_{2} , \\Lambda_{2}) = u(\\alpha_{1} + \\alpha_{2} , \\Lambda_{1} \\Lambda_{2} ) \\ \\ (5.18)$

From this, it is seen;

$LaTeX Code: u(\\alpha , 1)u(0 , \\Lambda )u(-\\alpha , 1) = u(0 , \\Lambda) \\ \\ (5.19)$

By writting

$LaTeX Code: u(\\alpha , 1) = 1 + i \\alpha D \\ \\ (5.20a)$
$LaTeX Code: u(0 , \\Lambda) = 1 + (i/2) \\omega . J \\ \\ (5.20b)$

we find

$LaTeX Code: [iJ^{ab}, D] = 0 \\ \\ (5.21)$

Next, we postulate that the algebra of C(1,n-1) closes on the (n+1)(n+2)/2 generators (P,J,D,K) and that K is a Lorentz vecto

$LaTeX Code: [iJ^{ab}, K^{c}] = \\eta^{ac}K^{b} - \\eta^{bc}K^{a} \\ \\ (5.22)$

Surprisingly, this postulate together with what we already know about the subalgebra (P,J,D), is sufficient to determine the remaining commutators
[P,K], [D,K] and [K,K] from the Jacobi identities alone.

Ex(5.5): On grounds of Lorentz invariance, i.e., the Jacobi identities found by taking the above commutators with J, conclude that

$LaTeX Code: [iD, K^{a}] = aK^{a} + bP^{a} \\ \\ (5.23a)$
$LaTeX Code: [iK^{a},K^{b}] = cJ^{ab} \\ \\ (5.23b)$
$LaTeX Code: [iK^{a},P^{b}] = d\\eta^{ab} D - dJ^{ab} \\ \\ (5.23c)$

where a,b,c and d are constants.

Now, use the J. identities

$LaTeX Code: \\left[iD ,[iK^{a},P^{b}] \\right] + \\left[iP^{b}, [iD,K^{a}] \\right] + \\left[iK^{a}, [iP^{b},D] \\right] = 0 \\ \\ (5.24a)$
$LaTeX Code: \\left[iD, [iK^{a},K^{b}] \\right] + .... + ... = 0 \\ \\ (5.24b)$
$LaTeX Code: \\left[iP^{a}, [iK^{b},K^{c}] \\right] + ... + ... = 0 \\ \\ (5.24c)$

to find that a = -1 and c = bd = 0 .

Notice that a vanishing [P,K] would also be compatible with the Jacobi identities, in that case there would be an arbitrary real factor in [D,K]. However the choice b = 0 and d = 2 is compatible with our previous work

$LaTeX Code: [iD,K^{a}] = - K^{a} \\ \\ (5.25a)$
$LaTeX Code: [iK^{a},K^{b}] = 0 \\ \\ (5.25b)$
$LaTeX Code: [iK^{a},P^{b}] = 2\\eta^{ab} D - 2J^{ab} \\ \\ (5.25c)$

more to come.....

**samalkhaiat** · Jun2-07, 07:46 PM

Here we will show that the algebra of C(1,n-1) is isomorphic to that of the "Lorentz" group SO(2,n). The latter may be cosidered as a set of pseudoorthognal transformations in a (n+2)-dimensional flat spacetime $LaTeX Code: M^{n+2}$ with the metric;

$LaTeX Code: \\eta^{\\mu \\nu} = (-1,1,\\eta^{ab}) \\ \\ (6.1)$

where

$LaTeX Code: (\\mu , \\nu) = -2, -1, 0, 1, ...,n-1$

the algebra of SO(2,n) is given by

$LaTeX Code: [iM^{\\mu \\nu},M^{\\rho \\sigma}] = \\eta^{\\nu \\rho}M^{\\mu \\sigma} - \\eta^{\\mu \\rho}M^{\\nu \\sigma} + \\eta^{\\mu \\sigma}M^{\\nu \\rho} - \\eta^{\\nu \\sigma}M^{\\mu \\rho} \\ \\ (6.2)$

Ex(6.1) Define the following generators

$LaTeX Code: D = M^{-2,-1} \\ \\ (6.3a)$
$LaTeX Code: J^{ab} = M^{ab} \\ \\ (6.3b)$
$LaTeX Code: \\frac{1}{2}(P^{a} - K^{a}) = M^{-2,a} \\ \\ (6.3c)$
$LaTeX Code: \\frac{1}{2}(P^{a} + K^{a}) = M^{-1,a} \\ \\ (6.3d)$

then show that Eq(6.2) gives the conformal algebra.

The origin of the nonlinearity of conformal transformations can now be understood by considering the relation between C(1,n-1) and SO(2,n). In $LaTeX Code: M^{n+2}$ ,the SO(2,n) coordinate transformations are linear. However, by projecting these transformations on $LaTeX Code: M^{n}$ ,we get a nonlinear realization of the group SO(2,n), that coincides with the action of C(1,n-1) on $LaTeX Code: M^{n}$ .

more to come...

**samalkhaiat** · Jun2-07, 09:00 PM

Can we arrive at the conformal coordinate transformations from the conformal algebra? Yes, we can. This is because the Minkowski space $LaTeX Code: M^{n}$ can be identified with the coset manifold C(1,n-1)/H, with H the subgroup generated by $LaTeX Code: (J^{ab},D,K^{a})$ ,i.e., it can be parametrized by;

$LaTeX Code: T(x) = \\exp(iP.x) \\ \\ (7.1)$

In general, a set of group elements T(x), labelled by as many parameters as necessary, parametrizes the manifold if each coset contains one of the T's. Once a parametrization T(x) has been chosen, each group element g can be uniquely decomposed into a product

$LaTeX Code: g = T(x) . L \\ \\ (7.2)$

where T is the representative member of the coset to which g belongs and L connect T to g within the coset. A product of g with an arbitrary element, and in particular with some T(x) will therefore give another T and an L according to

$LaTeX Code: g . T(x) = T(\\bar{x}) . L \\ \\ (7.3)$

where
$LaTeX Code: \\bar{x} = \\bar{x}(x,g) \\ \\ (7.4a)$
$LaTeX Code: L = L(x,g) \\ \\ (7.4b)$

We will see that the conformal algebra is sufficient to determine $LaTeX Code: \\bar{x}$ and L explicitly from

$LaTeX Code: g . e^{iP.x} = e^{iP. \\bar{x}} . L \\ \\ (7.3single-quote)$

Translations:

$LaTeX Code: g = T(a) = \\exp(ia.P)$

gives
$LaTeX Code: \\bar{x} = x + a$

Lorentz trans

$LaTeX Code: g = \\exp(\\frac{i}{2} \\omega . J)$

For infinitesimal $LaTeX Code: \\omega$ , we can write

$LaTeX Code: (1 + \\frac{i}{2} \\omega . J) e^{ix.P} = e^{ix.P}\\left[ e^{-ix.P}(1 + i/2 \\omega . J) e^{ix.P}\\right] \\ \\ (7.5)$

The expression in [ ] can be expanded by use of the Hausdorff formula

$LaTeX Code: e^{-A} B e^{A} = B + [B,A] + 1/2 [[B,A],A] + ... \\ \\ (7.6)$

this gives

$LaTeX Code: (1 + \\frac{i}{2}\\omega .J) e^{ix.P} = e^{ix.P}\\left(1 + \\frac{i}{2}\\omega .J + \\frac{i}{2}\\omega_{ab}x_{c}[iJ^{ab},P^{c}] \\right) \\ \\ (7.7)$

upon inserting the algebra (5.16b), we find

$LaTeX Code: (1 + i/2 \\omega .J) e^{ix.P} = e^{i(x^{a} - \\omega^{ab}x_{b})P_{a}} (1 + i/2 \\omega .J) + o(\\omega^{2}) \\ \\ (7.8)$

For finite $LaTeX Code: \\omega$ we find the expected Lorentz transformations

$LaTeX Code: \\bar{x}^{a} = \\Lambda^{a}{}_{b}x^{b}$

and
$LaTeX Code: L = g =\\exp(i/2 \\omega .J) \\ \\ (7.9)$

Ex(7.1): For infinitesimal D-transformations (dilations) and K-transformations (special conformal trans), show that

$LaTeX Code: (1 - i \\alpha D) e^{ix.P} = e^{i(1 - \\alpha )x.P} (1 - i \\alpha D) + o(\\alpha^{2}) \\ \\ (7.10)$

$LaTeX Code: (1 + i c.K) e^{ix.P} = e^{iP_{a}(x^{a}+2c.x x^{a}-c^{a}x^{2})}\\left(1 + ic.K + 2ic.xD - 2ic_{b}x_{a}J^{ab}\\right) + o(c^{2}) \\ \\ (7.11)$

Next, we will talk physics

**samalkhaiat** · Jun2-07, 10:16 PM

[You don't have to read this post if you are not familiar with QFT]

Whereas conformal transformations are well understood in mathematics, this is not the case in physics! The physical meaning of the conformal transformations in n>2 spacetime dimensions is far from clear. Many believe that conformal invariance has no physical meaning and consequently should not play "any" role in physics!!
So, what is the problem? Well, apart from the Poincare' algebra, we can not say that the conformal algebra (invariance) is realized in nature. For if

$LaTeX Code: [iD,P^{a}] = P^{a} \\ \\ (8.1)$

holds in nature, then it is also true that;

$LaTeX Code: e^{-i \\alpha D}P^{2}e^{i \\alpha D} = e^{-2 \\alpha}P^{2} \\ \\ (8.2)$

So, if |P> is some 1-particle state of given positive mass m ;

$LaTeX Code: P^{2}|P \\rangle = m^{2}|P \\rangle \\ \\ (8.3)$

then the state

$LaTeX Code: |\\bar{P} \\rangle = e^{-i \\alpha D} |P \\rangle \\ \\ (8.4)$

carries the mass
$LaTeX Code: \\bar{m}^{2} = e^{2\\alpha} m^{2} \\ \\ (8.5)$

i.e.,

$LaTeX Code: P^{2}|\\bar{P}\\rangle = e^{2\\alpha}m^{2}|\\bar{P}\\rangle \\ \\ (8.6)$

If we assume that scale invariance is not spontaneously broken, i.e.,

$LaTeX Code: e^{-i\\alpha D}|0 \\rangle = |0 \\rangle \\ \\ (8.7)$

then we conclude that

$LaTeX Code: |\\bar{P}\\rangle = e^{-i\\alpha D}a^{\\dagger}(p) |0 \\rangle \\propto a^{\\dagger}(e^{\\alpha}p) |0 \\rangle \\ \\ (8.8)$

This means that the state $LaTeX Code: |\\bar{P}\\rangle$ is a quantum of the same field as the state $LaTeX Code: |P\\rangle$ but with a rescaled momentum, i.e., by virtue of conformal invariance, both states must belong to the same Hilbert space. Therefore, by Eq(8.5), conformal invariance implies that the mass spectrum is either continuous or all masses vanish! In order to avoid this physically absurd conclusion, conformal symmetry must be explicitly or spontaneously broken (mathematically this problem is equivalent to the fact that $LaTeX Code: P^{2}$ is not a Casmir's operator).

Notice though, if the vacuum was not unique then the state $LaTeX Code: |\\bar{P}>$ would belong to a different Hilbert space than the state |P> and our conclusion would be avoided.

Another problem is that conformal invariance at the quantum level does not follow from conformal invariance at the classical level. Operationally this means that the renormalized energy-momentum tensor possesses a non-vanishing trace and there are anomalies in the dilation and conformal currents.
We will see that scale invariance requires that there be no dimensionfull parameters in the Lagrangian. But QFT does not make sense without a regularization prescription that introduces a scale in the theory. This scale, at which the theory is renormalized, breaks conformal invariance.

Ex(8.1): Derive (8.2) and (8.6).

more to come...

**samalkhaiat** · Jun3-07, 12:38 PM

Before considering conformal field theory and the conditions under which an action integral is conformally invariant, we must decide how the fields transform under the conformal transformations.
If the configuration variables are transformed according to

$LaTeX Code: x \\rightarrow \\bar{x} = g x \\ \\ (9.1)$

with g belongs to some Lie group, state vectors and ,therefore, local fields will be in general subject to a unitary transformations;

$LaTeX Code: |\\Phi \\rangle \\rightarrow |\\Phi_{g} \\rangle = u(g) |\\Phi \\rangle \\ \\ (9.2)$

$LaTeX Code: \\bar{\\phi}_{i}(x) = u^{-1}(g) \\phi_{i} u(g) = \\mathcal{D}_{i}{}^{j}\\phi_{j}(g^{-1}x) \\ \\ (9.3)$

where $LaTeX Code: \\mathcal{D}(g)$ is a finite-dimensional (matrix) representation of the group, i.e., set of matrices satisfying the Lie algebra of the group. They need not necessarily be unitary. Their effect is a reshuffling of field components.

Eq(9.3) can be understood as follow;

Since the expectation values;

$LaTeX Code: C_{i}(x) = \\langle \\Phi | \\phi_{i}(x) | \\Phi \\rangle \\ \\ (9.4)$

are C-numbers, we expect them to transform like classical fields (which belong to some representation D):

$LaTeX Code: \\bar{C}_{i}(\\bar{x}) = \\mathcal{D}_{i}{}^{j}C_{j}(x) \\ \\ (9.5)$
and, by Eq(9.2), the expectation value of the transformed field operator $LaTeX Code: \\bar{\\phi}(\\bar{x})$ in a state described by $LaTeX Code: |\\Phi \\rangle$ is the same as the expectation value calculated using the untransformed operator $LaTeX Code: \\phi(\\bar{x})$ and the transformed state $LaTeX Code: |\\Phi_{g}\\rangle$ :

$LaTeX Code: \\langle \\Phi |\\bar{\\phi}(\\bar{x}) |\\Phi \\rangle \\equiv \\langle \\Phi |u^{-1}\\phi(\\bar{x}) u |\\Phi \\rangle = \\mathcal{D} \\langle \\Phi |\\phi(x) |\\Phi \\rangle \\ \\ (9.6)$

Since this is true for any state vector, we conclude that;

$LaTeX Code: \\bar{\\phi}(\\bar{x}) = u^{-1}\\phi (\\bar{x}) u = \\mathcal{D} \\phi(x) \\ \\ (9.7)$

or, by (9.1),

$LaTeX Code: \\bar{\\phi}(\\bar{x}) = u^{-1}\\phi(\\bar{x}) u = \\mathcal{D} \\phi (g^{-1}\\bar{x}) \\ \\ (9.8)$

This leads to Eq(9.3) after renaming the coordinates $LaTeX Code: \\bar{x}=x$ .
Now I want you to work out the action of infinitesimal generators on local fi eld.
Ex(9.1): Let

$LaTeX Code: g^{-1}x^{a} = x^{a} + \\lambda_{A}f^{Aa}(x) \\ \\ (9.9)$
$LaTeX Code: u(g) = \\exp(- i \\lambda_{A}G^{A}) \\ \\ (9.10)$

and

$LaTeX Code: \\mathcal{D}^{i}_{j} = \\exp(\\lambda_{A}D^{Ai}{}_{j}) \\ \\ (9.11)$

where $LaTeX Code: \\lambda_{A}$ [A = 1,2,....,dim(g)] are the infinitesimal

parameters and $LaTeX Code: D^{A}$ forms a matrix representation of the

generators $LaTeX Code: G^{A}$ .
Write Eq(9.3) to 1st order in lambda and show that;

$LaTeX Code: \\delta^{A}_{(\\lambda)}\\phi_{i}(x) = [iG^{A} , \\phi_{i}(x)] = f^{Aa}\\partial_{a}\\phi_{i} + D_{i}^{Aj}\\phi_{j}(x) \\ \\ (9.12)$

where

$LaTeX Code: \\delta \\phi_{i} = \\lambda_{A}\\delta^{A}_{(\\lambda)}\\phi_{i} = \\bar{\\phi}_{i}(x) - \\phi_{i}(x) \\ \\ (9.13)$

is the infinitesimal variation (change) in the form of the field function.

more to come soon...

**samalkhaiat** · Jun4-07, 04:31 PM

Here, we will use Eq(9.3) or its infinitesimal version (9.12) to determine the action of the generators (P,J,D,K) on local field:

For translation;

$LaTeX Code: g^{-1}x = x - a \\ \\ (10.1)$
$LaTeX Code: u(a) = \\exp(i a.P) \\ \\ (10.2)$
$LaTeX Code: \\mathcal{D}^{i}_{j} = \\delta^{i}_{j} \\ \\ (10.3)$

we find

$LaTeX Code: \\bar{\\phi}_{i}(x) = \\phi_{i}(x) - a_{a}[i P^{a},\\phi_{i}] = \\phi_{i}(x) - a_{a}\\partial^{a}\\phi_{i}$
or
$LaTeX Code: \\delta^{a}_{(T)}\\phi = [i P^{a},\\phi (x)] = \\partial^{a}\\phi \\ \\ (10.4)$

For Lorentz transformations;

$LaTeX Code: g^{-1}x^{a} = x^{a} - \\omega^{ab}x_{b} \\ \\ (10.5)$
$LaTeX Code: u(\\omega) = \\exp(-\\frac{i}{2}\\omega . J) \\ \\ (10.6)$
$LaTeX Code: \\mathcal{D}^{i}_{j} = \\exp(\\frac{1}{2}\\omega . \\Sigma^{i}_{j}) \\ \\ (10.7)$

where $LaTeX Code: \\Sigma$ is the appropriate spin matrix for the field,

we get;

$LaTeX Code: \\delta^{ab}_{(L)}\\phi_{i} = [i J^{ab},\\phi_{i}] = (x^{a}\\partial^{b} - x^{b}\\partial^{a})\\phi_{i} + \\Sigma^{abj}_{i}\\phi_{j} \\ \\ (10.8)$

Ex(10.1): Check Eq(10.8).

For Scale transformations (Dilations);

$LaTeX Code: g^{-1}x = x + \\alpha x \\ \\ (10.9)$
$LaTeX Code: u(\\alpha) = \\exp(-i \\alpha D) \\ \\ (10.10)$
$LaTeX Code: \\mathcal{D} = \\exp(d \\alpha ) \\ \\ (10.11)$

where d is (real number) called the canonical dimension of the field. We will see how to find its values for fermions and bosons.

Inserting (10.9), (10.10) and (10.11) in Eq(9.3), we find

$LaTeX Code: \\bar{\\phi}_{i} = \\phi_{i} + \\alpha [i D, \\phi_{i}] = \\phi_{i} + \\alpha (x.\\partial + d ) \\phi_{i}$

or

$LaTeX Code: \\delta_{(D)}\\phi = [i D, \\phi ] = ( d + x . \\partial ) \\phi \\ \\ (10.12)$

For free fields, we have the canonical equal-time (anti)commutation relations:

$LaTeX Code: \\{ \\psi (x,t) , \\psi^{\\dagger}(y,t) \\} = \\delta^{3}(x-y) \\ \\ (10.13a)$
$LaTeX Code: [\\phi(x,t) , \\dot{\\phi}(y,t) ] = i \\delta^{3}(x-y) \\ \\ (10.13b)$

The scale dimension d is defined so that (10.13) remain invariant under scale transformation. Transforming the fields in (10.13a) according to Eq(9.7) and (10.11), i.e.,

$LaTeX Code: \\psi(x,t) = e^{- d \\alpha} \\bar{\\psi}(\\bar{x},\\bar{t}) \\ \\ (10.14)$

one finds

$LaTeX Code: \\{\\psi(x,t) , \\psi^{\\dagger}(y,t) \\} = e^{-2d \\alpha} \\delta^{3}(\\bar{x} - \\bar{y}) = e^{\\alpha (3 - 2d)}\\delta^{3}(x-y) \\ \\ (10.15)$

Thus the invariance of (10.13a) implies that the canonical scale dimension for fermion field is

$LaTeX Code: d_{\\psi} = \\frac{3}{2} \\ \\ (10.16)$

Ex(10.2): show that for boson field, the invariqance of (10.13b) under scale transformations, implies

$LaTeX Code: d_{\\phi} = 1 \\ \\ (10.17)$

These values for d correspond to the natural dimention of the fields in units of mass.

Special conformal transformations;

In Eq(9.12), if we put A = a (a spacetime index),

$LaTeX Code: f^{ab} = 2 x^{a}x^{b} - \\eta^{ab} x^{2} \\ \\ (10.18)$
$LaTeX Code: D^{aj}_{i} = -2 d x^{a}\\delta^{j}_{i} + 2 x_{b}\\Sigma^{baj}_{i} \\ \\ (10.19)$

and
$LaTeX Code: G^{a} = K^{a} \\ \\ (10.20)$

we find

$LaTeX Code: \\delta^{a}_{(C)}\\phi_{i} = [i K^{a}, \\phi_{i}] = (2x^{a}x^{b} - \\eta^{ab}x^{2}) \\partial_{b}\\phi_{i} + 2d x^{a}\\phi_{i} - 2 x_{b}\\Sigma^{baj}_{i}\\phi_{j} \\ \\ (10.21)$

In offering Eq(10.19) we assumed that the field carries irreducible representation of Lorentz group. This will become clear in the next post when we rederive the equations (10.8),(10.12) and (10.21), using the theory of induced representations. At this moment in time, Eq(5.25c) and (7.11) are the only hints we have for (10.19).

more to come soon..

**samalkhaiat** · Jun4-07, 06:15 PM

Let $LaTeX Code: \\mathcal{G}$ be a Lie group such that the Minkowski space can be identified with the coset space

$LaTeX Code: M^{n} = \\mathcal{G} / H \\equiv \\frac{(G^{A},P^{a})}{(G^{A})} \\ \\ (11.1)$

where the subgroup H (generated by $LaTeX Code: G^{A}$ ) leaves the origin x = 0 invariant, and P generates the usual abelian group of translation.

Now, given any representation of the "little" group H on $LaTeX Code: \\phi (0)$ , we can induce it to the whole group, i.e., to a representation of $LaTeX Code: \\mathcal{G}$ on $LaTeX Code: \\phi (x)$ . This is done using the definition

$LaTeX Code: \\phi (x) = e^{i P.x}\\phi (0) e^{-i P.x} \\ \\ (11.2a)$

i.e.,

$LaTeX Code: \\partial \\phi (x) = [iP_{a}, \\phi (x)] \\ \\ (11.2b)$

and the algebra of $LaTeX Code: \\mathcal{G}$ , i.e., the commutators [P,G], [P,P] and [G,G].

We take the action of H to be

$LaTeX Code: [iG^{A},\\phi_{i}(0)] = D^{Aj}_{i}\\phi_{j}(0) \\ \\ (11.3)$

where D's are matrix representation of the generators G's of the little group H.

Ex(11.1): Show that the D's in (11.3) form a matrix representation of the algebra

$LaTeX Code: [iG^{A},G^{B}] = f^{AB}{}_{C}G^{C} \\ \\ (11.4)$

Now, Eq(11.2a) and (11.3) can be used to write

$LaTeX Code: [i (e^{iP.x} G^{A} e^{-iP.x}) , \\phi_{i}(x)] = D^{Aj}_{i}\\phi_{j}(x) \\ \\ (11.5)$

Applying Hausdroff formula Eq(7.6) to the lefthand-side, gives

$LaTeX Code: [iG^{A},\\phi (x)] = D^{Aj}_{i}\\phi_{j}(x) + \\left[i \\left( [iG^{A},x.P] -\\frac{1}{2}\\left[i[G^{A},x.P],x.P\\right] + ...\\right) , \\phi_{i}(x) \\right] \\ \\ (11.6)$

Notice that the RHS can be evaluated using only the algebra of the group $LaTeX Code: \\mathcal{G}$ .

Let us apply this method to the conformal group C(1,n-1).
Here the subgroup (H) that leaves the point x = 0 invariant is generated by;

$LaTeX Code: G^{A} = (J^{ab}, D, K^{a}) \\ \\ (11.7)$

If we remove the translation generator (P) from the conformal algebra, we get something identical to Poicare' algebra augmented by dilations (D), because of the similar role played by P and K.
Next we introduce a set of matrices;

$LaTeX Code: D^{A} = (\\Sigma^{ab}, \\Delta , \\mathcal{K}^{a}) \\ \\ (11.8)$

to define the action of this little group on $LaTeX Code: \\phi (0)$ ;

$LaTeX Code: [iJ^{ab},\\phi_{i}(0)] = \\Sigma^{abj}_{i}\\phi_{j}(0) \\ \\ (11.9a)$

$LaTeX Code: [iD,\\phi_{i}(0)] = \\Delta^{j}_{i}\\phi_{j}(0) \\ \\ (11.9b)$

$LaTeX Code: [iK^{a},\\phi_{i}(0) = \\mathcal{K}^{aj}_{i}\\phi_{j}(0) \\ \\ (11.9c)$

these matrices must form arepresentation of the reduced algebra

$LaTeX Code: [\\Sigma,\\Delta ] = [\\Delta ,\\Delta ] = [\\mathcal{K},\\mathcal{K}] = 0 \\ \\ (10.10a)$

$LaTeX Code: [\\Delta , \\mathcal{K}^{a}] = \\mathcal{K}^{a} \\ \\ (11.10b)$

$LaTeX Code: [\\Sigma^{ab},\\mathcal{K}^{c}] = \\eta^{bc}\\mathcal{K}^{a} - \\eta^{ac}\\mathcal{K}^{b} \\ \\ (11.10c)$

$LaTeX Code: [\\Sigma^{ab},\\Sigma^{cd}] = \\eta^{ac}\\Sigma^{bd} - ... \\ \\ (11.10d)$

Ex(11.2) use the algebra of (J, D, K) and Eq(11.9) to derive Eq(11.10).

If the field belongs to an irreducible representation of Lorentz group, then by Schur's lemma, any matrix that commutes with all the (irreducible) grnerators $LaTeX Code: \\Sigma$ must be a multiple of the unit matrix. Consequently, the matrix $LaTeX Code: \\Delta$ is

$LaTeX Code: \\Delta^{j}_{i} = d \\delta^{j}_{i} \\ \\ (11.11)$

where d is a real number, and the algebra (11.10b) forces all matrices $LaTeX Code: \\mathcal{K}$ to vanish.

If we now put

$LaTeX Code: D^{Aj}_{i} = \\Sigma^{abj} , d\\delta^{j}_{i} , 0$

and

$LaTeX Code: G^{A} = J^{ab} , D, K^{a}$

in Eq(11.6), we arrive (after using the conformal algebra) at

$LaTeX Code: [iJ^{ab},\\phi (x)] = (x^{a}\\partial^{b} - x^{b}\\partial^{a} + \\Sigma ) \\phi (x) \\ \\ (11.12a)$

$LaTeX Code: [iD,\\phi (x)] = (d + x. \\partial ) \\phi (x) \\ \\ (11.12b)$

and

$LaTeX Code: [iK^{a},\\phi] = (2x^{a}x^{b} - \\eta^{ab}x^{2})\\partial_{b}\\phi + (2dx^{a} + 2x_{b}\\Sigma^{ab})\\phi \\ \\ (11.12c)$

These equations together with

$LaTeX Code: [iP^{a},\\phi ] = \\partial^{a}\\phi \\ \\ (11.12d)$

determine the action of the conformal group on local fields. In the next post we will see that, when the conditions for conformal invariance are met, Noether theorem gives us a time independent and Lorentz covariant objects satisfying the conformal algebra and effect the proper transformations on the fields, i.e., satisfing (11.12). Therefore one can identify Noether charges with the conformal generators. However, we will see that the noether charges (P,J,D,K) satisfy (11.12) even when the (K,D)-symmetry is broken!

more to come tomorow...

**samalkhaiat** · Jun5-07, 05:41 PM

An arbitrary field theory is described by a Lagrangian density which we take to be a real function of the field variables $LaTeX Code: \\phi(x)$ and of their first derivative $LaTeX Code: \\partial_{a}\\phi$ . If $LaTeX Code: \\mathcal{L}$ depends only on the state of the fields in an infinitely small neighbourhood of the point x ,i.e., on the values oh $LaTeX Code: \\phi$ and of $LaTeX Code: \\partial \\phi$ evaluated at the same point x , then it is called a local Lagrangian, and the corresponding theory is said to be a local theory.
So, we may write;

$LaTeX Code: \\mathcal{L}(x) = \\mathcal{L}\\left(\\phi (x), \\partial_{a}\\phi (x)\\right) \\ \\ (12.1)$

It is assumed that $LaTeX Code: \\phi (x) \\rightarrow 0$ sufficiently fast as $LaTeX Code: |\\vec{x}| \\rightarrow \\infty$ .
The integral of the Lagrangian over bounded, arbitrary contractible, region in spacetime;
$LaTeX Code: S = \\int_{D}d^{4}x \\mathcal{L}(x) \\ \\ (12.2)$
where
$LaTeX Code: D \\equiv ( V \\cup \\partial V ) \\subset \\mathbb{R}^{4} \\ \\ (12.3)$

is called the action.

From the variation principle of stationary action;

$LaTeX Code: \\delta \\int_{D} d^{4}x \\mathcal{L}(x) = 0 \\ \\ (12.4)$

together with the assumption that the variation of the fields $LaTeX Code: \\delta \\phi$ vanish on $LaTeX Code: \\partial V$ (the surface of the 4-volume over which the integral is taken) but arbitrary in

, we obtain the field equations in V:

$LaTeX Code: \\frac{\\delta \\mathcal{L}}{\\delta \\phi} = 0 \\ \\ (12.5a)$

where

$LaTeX Code: \\frac{\\delta \\mathcal{L}}{\\delta \\phi} \\equiv \\frac{\\partial \\mathcal{L}}{\\partial \\phi} - \\partial_{a}\\left(\\frac{\\partial \\mathcal{L}}{\\partial \\partial_{a}\\phi}\\right) \\ \\ (12.5b)$

is the Euler derivative.
In the canonical formalism, the field and its conjugate momentum

$LaTeX Code: \\pi = \\frac{\\partial \\mathcal{L}}{\\partial \\partial_{0}\\phi} \\ \\ (12.6)$

satisfy the equal-time commutation relations

$LaTeX Code: [\\phi(t,x) , \\pi (t,y)] = i\\delta^{3}(x-y) \\ \\ (12.7a)$
$LaTeX Code: [\\phi , \\phi ] = [\\pi ,\\pi ] = 0 \\ \\ (12.7b)$

The connection between continuous symmetries and conserved quantities is highlighted by Noether's theorem which states: to every Lie group of transformations which leaves the action unchanged, there corresponds a definite combination of the Lagrangian derivatives that determines the field invariants when the fields satisfy the E-L equation, i.e., when the field equations are satisfied, a systematic procedure for obtaining conservation laws can be developed from a direct study of the invariance properties of the action integral.
Therefore, it is interesting to ask what conditions on $LaTeX Code: \\mathcal{L}$ insure the invariance of the action? In other words, if G is a given Lie group, then the question will be; how to formulate a G-invariant theory?

To answer this question, let us consider an infinitesimal coordinate transformations
$LaTeX Code: x^{a} \\rightarrow \\bar{x}^{a} = x^{a} + \\lambda_{A}f^{Aa}(x) \\ \\ (12.8)$
such that the fields transform according to some known representation of G :

$LaTeX Code: \\phi (x) \\rightarrow \\bar{\\phi}(\\bar{x}) = \\phi (x) + \\lambda_{A}\\bar{\\delta}^{A}\\phi (x) \\ \\ (12.9)$

where $LaTeX Code: \\lambda_{A}$ are the infinitesimal parameters, A=1,2,..,dim(G),
and $LaTeX Code: \\bar{\\delta}^{A}\\phi$ is the local variation in the field due to the change in the form of the field function and its argument:

$LaTeX Code: \\bar{\\delta}^{A}\\phi = \\delta^{A}\\phi + f^{Aa}\\partial_{a}\\phi \\ \\ (12.10)$

where the variation in the form of the field is defined by

$LaTeX Code: \\delta \\phi \\equiv \\lambda_{A}\\delta^{A}\\phi = \\bar{\\phi}(x) - \\phi (x) \\ \\ (12.11)$

Please note that

$LaTeX Code: \\partial_{a}\\delta^{A}\\phi = \\delta^{A}\\partial_{a}\\phi \\ \\ (12.12a)$

but

$LaTeX Code: \\partial_{a}\\bar{\\delta}^{A}\\phi = \\bar{\\delta}^{A}\\partial_{a}\\phi + \\partial_{a}f^{Ab}\\partial_{b}\\phi \\ \\ (12.12b)$

Ex(12.1): Derive the above two equations.

Under these transformations (the action of the group G), the form of a local Lagrangian would change according to

$LaTeX Code: \\delta^{A}\\mathcal{L} = \\frac{\\partial \\mathcal{L}}{\\partial \\phi}\\delta^{A}\\phi + \\pi^{a}\\partial_{a}\\delta^{A}\\phi \\ \\ (12.13)$

where

$LaTeX Code: \\pi^{a} = \\frac{\\partial \\mathcal{L}}{\\partial \\partial_{a}\\phi} \\ \\ (12.14a)$

$LaTeX Code: \\pi^{0}\\equiv \\pi \\ \\ (12.14b)$

Since the explicit form for $LaTeX Code: \\bar{\\delta}^{A}\\phi$ is assumed known (matrix representation of G), we can use (12.10) to put (12.13) in a form like;

$LaTeX Code: \\delta^{A}\\mathcal{L} + \\partial_{a}(f^{Aa}\\mathcal{L}) = C^{A}\\left(\\mathcal{L},\\frac{\\partial \\mathcal{L}}{\\partial \\phi},\\pi^{a};D \\right) \\ \\ (12.15)$

with $LaTeX Code: C^{A}$ has no exiplicit x-dependence and

is the matrix representation of G.

If we can construct a Lagrangian (i.e.,formulate a theory) in such a way that

$LaTeX Code: C^{A} = 0 \\ \\ (12.16)$

then our theory will be invariant under the given Lie group G. Mathematically speaking, the solution $LaTeX Code: (\\mathcal{L})$ of the system of 1st order PDE (12.16), describes the most general G-invariant theory. Also, whether or not a given theory is G-invariant can be decided by examining the Lagrangian against (12.16).

To show that (12.16) is invariance condition, look at the variation of the action;

$LaTeX Code: \\bar{\\delta}\\int_{D}\\mathcal{L} d^{4}x = \\int_{\\bar{D}} \\bar{\\mathcal{L}}(\\bar{x}) d^{4}\\bar{x} - \\int_{D}\\mathcal{L}(x) d^{4}x \\ \\ (12.17)$

it is made of the sum of the local variation of $LaTeX Code: \\mathcal{L}$ and of the variation in the region of integration:

$LaTeX Code: \\bar{\\delta}^{A}\\int_{D}\\mathcal{L} d^{4}x = \\int_{D}\\left(\\bar{\\delta}^{A}\\mathcal{L}\\right) d^{4}x + \\int_{D} \\mathcal{L}(x) \\bar{\\delta}^{A}(d^{4}x) \\ \\ (12.18)$

Symbolically, this can be understood in terms of the algebric property of the

local variation symbol $LaTeX Code: \\bar{\\delta}$ .
Indeed, to the 1st order, it is a derivation;

$LaTeX Code: \\bar{\\delta}(fg) = (\\bar{\\delta}f)g + f(\\bar{\\delta}g)$

Now, the local variation in $LaTeX Code: \\mathcal{L}$ can be expressed as
a variation in the form of the lagrangian plus "drag";

$LaTeX Code: \\bar{\\delta}^{A}\\mathcal{L} = \\delta^{A}\\mathcal{L} + f^{Aa}\\partial_{a}\\mathcal{L} \\ \\ (12.19)$

[see Eq(12.10)], and

$LaTeX Code: \\bar{\\delta}(d^{4}x) \\equiv d^{4}\\bar{x} - d^{4}x = J(\\frac{\\bar{x}}{x}) d^{4}x - d^{4}x \\ \\ (12.20)$

J-is the Jacobian of the transformations (12.8). To the 1st order we can write;

$LaTeX Code: \\bar{\\delta}^{A}(d^{4}x) = \\partial_{a}f^{Aa} d^{4}x \\ \\ (12.21)$

Putting (12.19) and (12.21) in Eq(12.18) leads to;

$LaTeX Code: \\bar{\\delta}^{A}\\int_{D}\\mathcal{L} d^{4}x = \\int_{D} C^{A} d^{4}x \\ \\ (12.22)$

Since $LaTeX Code: D \\subset \\mathbb{R}^{4}$ is an arbitrary contractible domain, we conclude that the action is invariant if and only if $LaTeX Code: C^{A}= 0$ .

So what is the link between the invariance condition (12.16) and Noether theorem?

In terms of the Euler derivative (12.5b), we can rewrite (12.13) as;

$LaTeX Code: \\delta^{A}\\mathcal{L} = \\partial_{a}(\\pi^{a}\\delta^{A}\\phi ) + \\frac{\\delta \\mathcal{L}}{\\delta \\phi}\\delta^{A}\\phi \\ \\ (12.23)$

Putting this in (12.15) we find;

$LaTeX Code: C^{A} = \\frac{\\delta \\mathcal{L}}{\\delta \\phi}\\delta^{A}\\phi + \\partial_{a}(\\pi^{a}\\delta^{A}\\phi + f^{Aa}\\mathcal{L}) \\ \\ (12.24)$

Thus, our invariance condition $LaTeX Code: C^{A} = 0$ is equivalent to the Noether identity;

$LaTeX Code: \\frac{\\delta \\mathcal{L}}{\\delta \\phi}\\delta^{A} = \\partial_{a}\\mathcal{J}^{aA} \\ \\ (12.25)$

where

$LaTeX Code: \\mathcal{J}^{aA} = -\\pi^{a}\\delta^{A}\\phi - f^{Aa}\\mathcal{L} \\ \\ (12.26)$

is the canonical Noether current. the conservation of this current;

$LaTeX Code: \\partial_{a}\\mathcal{J}^{aA} = 0 \\ \\ (12.27)$

follows from (12.25), when the field satisfies E-L equations.

Thus, in terms of the invariance condition (12.16), the statement of Noether theorem becomes

$LaTeX Code: \\left( C^{A}= 0, \\frac{\\delta \\mathcal{L}}{\\delta \\phi}= 0 \\right) \\Rightarrow \\left(\\partial_{a}\\mathcal{J}^{aA}= 0 \\right) \\ \\ (12.28)$

When G is a symmetry group (C = 0), we can use (12.27) to show that the Noether charges

$LaTeX Code: Q^{A}= \\int d^{3}x \\mathcal{J}^{A0} \\ \\ (12.29)$

1) are time-independent.
2) are G-covariant.
3) generate the correct transformation on the field;

$LaTeX Code: \\delta^{A}\\phi = [i Q^{A}, \\phi ] \\ \\ (12.30)$

4) satisfy the Lie algebra of G:

$LaTeX Code: [iQ^{A},Q^{B}] = f^{AB}{}_{C}Q^{C} \\ \\ (12.31)$

Because of (3) and (4), Q is called the generator of the symmetry transformations.

Ex(12.2) Derive Eq(12.30).

It is clear that when G is not a symmetry operation, i.e., $LaTeX Code: C^{A}\\ne 0$ , it is still possible to define the current (12.26) which is not conserved, and the charge (12.29). In this case, two situations are distinguished. If G is an internal Lie group, i.e., $LaTeX Code: f^{Aa}= 0$ , the Noether charges would still have the properties (3) & (4) but not (1) or (2). For Spacetime groups (1),(2) & (4) are fulse but (3) is still true.

Ex(12.3): Let G be a broken internal symmetry with

$LaTeX Code: f^{Aa}=0 \\ \\ (12.32a)$

$LaTeX Code: \\delta^{A}\\phi = i T^{A}\\phi \\ \\ (12.32b)$

and

$LaTeX Code: [iT^{A},T^{B}] = f^{AB}{}_{C}T^{C} \\ \\ (12.32c)$

Define the non-conserved current

$LaTeX Code: \\mathcal{J}^{Aa} = \\pi^{a}\\delta^{A}\\phi \\ \\ (12.33)$

and its (time-dependent) charge

$LaTeX Code: Q^{A}(t) = i \\int d^{3}x \\pi T^{A}\\phi \\ \\ (12.34)$

then show that Q satisfies the Lie algebra of G and generates the transformation (12.32b).

In the next post (the last I hope) we will describe the conditions of conformal invariance.

**samalkhaiat** · Jun6-07, 04:29 PM

In this and next post, we will apply the general formalism of post #12 to the conformal group C(1,3).
we will derive the explicit form of the invariance condition Eq(12.16) for translations, Lorentz, scale and special conformal transformations, and find the corresponding Noether currents and charges.

Translations

$LaTeX Code: \\bar{x}^{a}= x^{a} + \\eta^{ab}a_{b} \\ \\ (13.1a)$

$LaTeX Code: \\delta^{b}_{(T)}\\phi = - \\eta^{ba}\\partial_{a}\\phi \\ \\ (13.1b)$

I.e., the notation of post #12 becomes

$LaTeX Code: A = b \\ \\ (13.2a)$

$LaTeX Code: f^{Aa}\\rightarrow \\eta^{ba} \\ \\ (13.2b)$

$LaTeX Code: \\delta^{A}\\rightarrow \\delta^{b}_{(T)} \\ \\ (13.2c)$

[(T) is not an index, it stands for translation]

In this case Eq(12.13) becomes;

$LaTeX Code: \\delta^{b}_{(T)}\\mathcal{L} = - \\partial_{a}(\\eta^{ab}\\mathcal{L}) + \\frac{\\delta \\mathcal{L}}{\\delta x_{b}} \\ \\ (13.3)$

From Eq(12.15) & (12.16), we see;

$LaTeX Code: C^{b}_{(T)} \\equiv \\frac{\\delta \\mathcal{L}}{\\delta x_{b}} = 0 \\ \\ (13.4)$

Thus, as expected, the condition for translation invariance is equivalent to the statement that $LaTeX Code: \\mathcal{L}$ has no explicit x-dependence.
From Eq(12.26) we find the conserved translation current;

$LaTeX Code: T^{ab} = \\pi^{a}\\partial^{b}\\phi - \\eta^{ab}\\mathcal{L} \\ \\ (13.5)$

This is the canonical energy momentum tensor. The corresponding charge

$LaTeX Code: P^{a} =\\int d^{3}x T^{0a} \\ \\ (13.6)$

represents the energy momentum 4-vector.

Ex(13.1): Given;

$LaTeX Code: \\partial_{a}T^{ab}= 0 \\ \\ (13.7)$

show that $LaTeX Code: P^{a}$

1) is time-independent.
2) is 4-vector.
3) grnerates the transformation;

$LaTeX Code: \\delta^{a}_{(T)}\\phi = [i P^{a},\\phi] \\ \\ (13.8)$

4) satisfies the Lie algebra of the translation group;

$LaTeX Code: [i P^{a},P^{b}] = 0 \\ \\ (13.9)$

Lorentz Transformation

Here, the index A of post #12 is a double spacetime index;

$LaTeX Code: f^{(bc)a}(x) = \\eta^{ab}x^{c} - \\eta^{ac}x^{b} \\ \\ (13.10a)$

$LaTeX Code: \\delta^{bc}_{(L)}\\phi = ( x^{b}\\partial^{c} - x^{c}\\partial^{b} + \\Sigma^{bc}) \\phi \\ \\ (13.10b)$

Inserting these in (12.13) and using the translation invariance Eq(13.4), leads to;

$LaTeX Code: \\delta^{bc}_{(L)}\\mathcal{L} = - \\partial_{a}(f^{bca}\\mathcal{L}) + C^{bc}_{(L)} \\ \\ (13.11a)$

where

$LaTeX Code: C^{bc}_{(L)} \\equiv ( \\pi^{b}\\partial^{c} - \\pi^{c}\\partial^{b}) \\phi + \\pi^{a}\\Sigma^{bc}\\partial_{a}\\phi + \\frac{\\partial \\mathcal{L}}{\\partial \\phi}\\Sigma^{bc}\\phi \\ \\ (13.11b)$

Hence the condition for Lorentz invariance of the theory is;

$LaTeX Code: C^{bc}_{(L)} = 0 \\ \\ (13.12)$

and the canonical Noether current is

$LaTeX Code: M^{abc} = x^{b}T^{ac} - x^{c}T^{ac} + \\pi^{a}\\Sigma^{bc}\\phi \\ \\ (13.13)$

The integrated charge (the ungular momentum of the field) is

$LaTeX Code: J^{bc} = \\int d^{3}x M^{0bc} = L^{bc} + S^{bc} \\ \\ (13.14a)$

where

$LaTeX Code: L^{bc} = \\int d^{3}x (x^{b}T^{0c} - x^{c}T^{0b}) \\ \\ (13.14b)$

is the intrinsic "orbital" angular momentum of the field, and

$LaTeX Code: S^{bc} = \\int d^{3}x \\pi \\Sigma^{bc}\\phi \\ \\ (13.14c)$

describes the polarization properties of the field. It corresponds to the spin of particles described by the quantized field.

Warning: some textbooks call S; the spin "tensor", and L; the orbital angular momentum "tensor". This is not correct, in general, L and S are not covariant quantities, the sum of them is.

Ex(13.2): Check, by explicit calculation, that

$LaTeX Code: \\partial_{a}M^{abc} = 0 \\ \\ (13.15)$

follows from the invariance condition (13.12) and E-L equation.

Ex(13.3): Show that

$LaTeX Code: [i J^{ab},\\phi] = \\delta^{ab}_{(L)}\\phi$

$LaTeX Code: [i J^{ab}, P^{c}] = \\eta^{ac}P^{b} - \\eta^{bc}P^{a}$
and
$LaTeX Code: [i J^{ab},J^{cd}] = \\eta^{bc}J^{ad} - \\eta^{ac}J^{bd} + \\eta^{ad}J^{bc} - \\eta^{bd}J^{ac}$

Ex(13.4): Show that the Belinfante tensor, defined by

$LaTeX Code: T^{ab}_{(B)} = T^{ab} + \\frac{1}{2}\\partial_{c}X^{cab} \\ \\ (13.16a)$

where

$LaTeX Code: X^{cab} = \\pi^{c}\\Sigma^{ab}\\phi - \\pi^{a}\\Sigma^{cb}\\phi - \\pi^{b}\\Sigma^{ca}\\phi \\ \\ (13.16b)$

is conserved and leads to the same momentum 4-vector $LaTeX Code: P^{b}$ . Hence it may be used instead of the canonical tensor $LaTeX Code: T^{ab}$ .
Use the Lorentz-invariance condition (13.12) as well as E-L equation to show that

$LaTeX Code: T^{ab}_{(B)} = T^{ba}_{(B)} \\ \\ (13.17)$

Ex(13.5): Define the tensor

$LaTeX Code: M^{abc}_{(B)} = x^{b}T^{ac}_{(B)} - x^{c}T^{ab}_{(B)} \\ \\ (13.18)$

then show that it is conserved when $LaTeX Code: M^{abc}$ is. Also show that it leads to the same angular momentum $LaTeX Code: J^{ab}$ as $LaTeX Code: M^{abc}$ does.

more to come...

**samalkhaiat** · Jun8-07, 03:54 PM

Cosider an infinitesimal scale transformation;

$LaTeX Code: \\bar{x}^{a} = x^{a} - \\alpha x^{a} \\ \\ 14.1$

In here, the parameter carries no index, so the index A of post #12 disapears:

$LaTeX Code: f^{a}(x) = - x^{a} \\ \\ (14.2)$

$LaTeX Code: \\delta_{(D)}\\phi = ( d + x^{c}\\partial_{c})\\phi \\ \\ (14.3)$

Following the general formalism of post #12, it is easy to obtain

$LaTeX Code: \\delta_{(D)}\\mathcal{L} = \\partial_{a}(x^{a}\\mathcal{L}) + C_{(D)} \\ \\ (14.4)$

with
$LaTeX Code: C_{(D)} \\equiv d \\frac{\\partial \\mathcal{L}}{\\partial \\phi}\\phi + (d+1) \\pi_{c}\\partial^{c}\\phi - 4 \\mathcal{L} = 0 \\ \\ (14.5)$

is the condition for dilation invariance.
Since d(boson) = 1 and d(fermion) = 3/2, the kinetic term of the Lagrangian satisfies (14.5). Examples of scale invariant interactions are;

$LaTeX Code: \\phi^{4}\\ \\mbox{and}\\ \\bar{\\psi}\\ \\psi \\phi .$

It is easy to see that (14.5) requires that the scale dimension of $LaTeX Code: \\mathcal{L}$ be 4, i.e., that there be no dimensionful parameters in the lagrangian. Terms like $LaTeX Code: m^{2}\\phi^{2}$ would of course break the scale symmetry explicitly as their scale dimension is 2. Notice the difference between scale dimension and ordinary dimention which is 4 for the term $LaTeX Code: m^{2}\\phi^{2}$ . Scale transformation effects the dynamical variables but not the dimensionful parameters.

Ex(14.1) Show that the scale-invariance condition (14.5) implies

$LaTeX Code: d(\\mathcal{L}) = 4 \\ \\ (14.6)$

In terms of the canonical energy-momentum tensor, the dilation Noether current is

$LaTeX Code: D^{a} = T^{ab}x_{b} + d \\pi^{a}\\phi \\ \\ (14.7)$

and the corresponding charge is

$LaTeX Code: D = \\int d^{3}x (T^{0b}x_{b} + d \\pi \\phi ) \\ \\ (14.8)$

Ex(14.2) Use E-L equation to put the invariance condition (14.5) in the form

$LaTeX Code: T^{a}_{a} + \\partial_{c}( \\pi^{c} d \\phi ) = \\partial_{c}D^{c} = 0 \\ \\ (14.9)$

then use this to prove

$LaTeX Code: [i D,P^{a}] = P^{a} \\ \\ (14.10)$
$LaTeX Code: [i J^{ab},D] = 0 \\ \\ (14.11)$

Ex(14.3) Show that the dilation charge D generates the correct transformation on the field

$LaTeX Code: \\delta_{(D)}\\phi = [i D ,\\phi ] \\ \\ (14.12)$

Ofcourse this is true even in the absence of scale symmetry, i.e., even if (14.5) is not true.
The trace of $LaTeX Code: T^{ab}$ lives in the invariance condition (14.9) for a reason. It did not appears there by accident! Let me explain this:
Under an arbitrary transformation of the coordinates

$LaTeX Code: x \\rightarrow x + f(x)$

the action changes as

$LaTeX Code: \\delta S = \\int d^{4}x \\frac{\\partial \\mathcal{L}}{\\partial\\eta_{ab}} \\delta \\eta_{ab} = (1/2) \\int d^{4}x \\theta^{ab}(\\partial_{a}f_{b} + \\partial_{b}f_{a}) \\ \\ (14.13)$

Where $LaTeX Code: \\theta_{ab}$ is the symmetric energy-momentum tensor(we will call it E-M tensor from now on). This is true even if the field does not satisfy the E-L equation.
From the conformal Killing equation (1.2), it follows that

$LaTeX Code: \\delta S = (1/2) \\int d^{4}x \\theta_{a}^{a} F(x) \\ \\ (14.14)$

and therefore, traceless E-M tensor implies that the action is invariant under the conformal group C(1,3). Of course the converse is not true because F(x) is not an arbitrary function.

more to come

I'm having problem with my PC, so I will break this post to 2 may be 3 smaller posts

**samalkhaiat** · Jun8-07, 06:06 PM

The Poicare' group together with dilations forms a subgroup of the full conformal group. This means that a theory invariant under this subgroup is not necessarily invariant under special conformal transformations. The conditions under which it should be invariant are determined by the tracelessness of the E-M tensor.
Under certain condition, the E-M tensor of a theory, with scale invariance, can be made traceless much in the same way as it can be made symmetric in a Lorentz-invariant theory. If this can be done, then it follows from Eq(14.14) that full conformal invariance is a consequence of scale invariance and poincare' invariance.

Let us do this trick. We start with scale invariant theory, i.e., with;

$LaTeX Code: \\partial_{a}D^{a} = T_{a}^{a} + \\partial_{a}( \\pi^{a} d \\phi ) = 0 \\ \\ (15.1)$

Next we define the field virial by

$LaTeX Code: V^{a} = \\pi^{a} d \\phi - \\pi_{c}\\Sigma^{ca}\\phi \\ \\ (15.2)$

and consider the following tensor

$LaTeX Code: \\theta^{ab} = T^{ab} + (1/2) \\partial_{c}X^{cab} + (1/2) \\partial_{c}\\partial_{d}Y^{cdab} \\ \\ (15.3)$

where the first 2 terms constitute the Belinfante tensor (13.16a).

1) from the conservation law

$LaTeX Code: \\partial_{a}\\theta^{ab} = 0 \\ \\ (15.4a)$

it follows that

$LaTeX Code: \\partial_{a}\\partial_{b}\\partial_{c}Y^{cdab} = 0 \\ \\ (15.4b)$

Hence, Y must not be completely symmetric in the first 3 indices.

2) from the symmetry property

$LaTeX Code: \\theta_{ab} = \\theta_{ba} \\ \\ \\ (15.5a)$

it follows that

$LaTeX Code: \\partial_{c}\\partial_{d}Y^{cdab} = \\partial_{c}\\partial_{d}Y^{cdba} \\ \\ \\ (15.5b)$

Thus, the part of Y antisymmetric in (a,b) must also be antisymmetric in (c,d),i.e.,

$LaTeX Code: Y^{cdab} - Y^{cdba} \\propto ( \\eta^{ca}\\eta^{db} - \\eta^{cb}\\eta^{ad}) \\ \\ \\ (15.5c)$

3) finally, we take the trace of Eq(15.3):

$LaTeX Code: \\theta_{a}^{a} = T^{a}_{a} + (1/2) \\partial_{c}X^{ca}{}_{a} + (1/2) \\partial_{a}\\partial_{b}Y^{abc}{}_{c} \\ \\ \\ (15.6)$

From Eq(13.16b) we see that

$LaTeX Code: \\partial_{c}X^{ca}{}_{a} = - 2 \\partial_{c}( \\pi_{a}\\Sigma^{ca}\\phi ) \\ \\ \\ (15.7)$

Using (15.2), this becomes

$LaTeX Code: \\partial_{c}X^{ca}{}_{a} = \\partial_{a}( \\pi^{a} d \\phi - V^{a}) \\ \\ \\ (15.8)$

Putting this in (15.6) and using (15.1), leads to

$LaTeX Code: \\theta_{a}^{a} = \\partial_{c}D^{c} - \\partial_{a}V^{a} + (1/2)\\partial_{a}\\partial_{b}Y^{abc}{}_{c} \\ \\ \\ (15.9)$

If the virial is a total divergence

$LaTeX Code: V^{a} = \\partial_{c}\\sigma^{ca} \\ \\ \\ (15.10)$

then Y can be constructed such that

$LaTeX Code: \\partial_{c}V^{c} = (1/2) \\partial_{a}\\partial_{b}X^{abc}{}_{c} \\ \\ \\ (15.11)$

This leads to our final result

$LaTeX Code: \\partial_{c}D^{c} = \\theta_{a}^{a} \\ \\ \\ (15.12)$

This means that scale invariance implies that the "improved" E-M tensor (15.3) is traceless, provided that the field virial satisfies (15.10).
This shows that for conformal invariance two conditions must be met:
1) the theory should be scale invariant,i.e., (14.5) must be true; and
2) Eq(15.10) must hold.

more to come

**samalkhaiat** · Jun9-07, 03:08 PM

In this (LAST) post, we will rederive the conditions for special conformal invariance using, this time, the general formalism of post #12.
From the infinitesimal conformal transformations:

$LaTeX Code: f^{ba}(x) = 2 x^{b}x^{a} - \\eta^{ba}x^{2} \\ \\ 16.1$

$LaTeX Code: \\delta^{b}_{(C)} = f^{ba}\\partial_{a}\\phi + 2 d x^{b} \\phi - 2 x_{a}\\Sigma^{ab}\\phi \\ \\ 16.2$

it is easy to see that

$LaTeX Code: \\delta^{b}_{(C)}\\mathcal{L} = \\partial(f^{ba}\\mathcal{L}) + 2 C_{(D)}x^{b} + V^{b} \\ \\ 16.3a$

where

$LaTeX Code: C_{(D)} = d \\frac{\\partial \\mathcal{L}}{\\partial \\phi} \\phi + (d+1) \\pi_{a}\\partial^{a}\\phi - 4 \\mathcal{L} \\ \\ 16.3b$

Inderiving (16.3), Lorntz and translation invariance conditions are used.

From (16.3) we see that conformal invariance requires that

$LaTeX Code: C_{(D)} = 0 \\ \\ 16.4a$

$LaTeX Code: V^{c} = \\partial_{b}\\sigma^{ba} \\ \\ 16.4b$

You might ask why not $LaTeX Code: C = 0 \\mbox{and} V^{a} = 0$ , after all this how it should be according to our general invariance condition Eq(12.16)? Well, explicit computation show that V vanishes identically for spin-(1/2) and spin-1 fields, but for spin-0 field, we have

$LaTeX Code: V^{a} = \\pi^{a}\\phi = (1/2) \\partial^{a}\\phi^{2} \\neq 0 \\ \\ 16.5$

I have not been able to understand this unique role of the scalar field!
Remarkably,Eq(16.4b) turns out to be true for all renormalizable theories, even though scale invariance is of course broken. Eq(16.4b) is also true in all field theories (involving spins; 0, 1/2 & 1) without derivative coupling. Consequently, for these theories conformal invariance is equivalent to scale invariance.
In terms of the canonical E-M tensor, the conformal Noether current is

$LaTeX Code: K^{ab} = f^{a}{}_{c}T^{bc} + 2 \\pi^{b}(d x^{a} - x_{c}\\Sigma^{ca}) \\phi - 2 \\sigma^{ab} \\ \\ 16.6$

By inserting the formula for $LaTeX Code: \\theta^{ab}$ in terms of $LaTeX Code: T^{ab}$ in Eq(16.6) and in Eq(14.7), one finds after many tedious steps:

$LaTeX Code: D^{a} = \\theta^{ab}x_{b} + \\partial_{b}Z^{ba} \\ \\ 16.7a$

$LaTeX Code: K^{ab} = f^{a}{}_{c}\\theta^{cb} + \\partial_{c}Z^{cab} \\ \\ 16.7b$

The total divergence terms in (16.7) are superpotentials; they may be droped and we are left with the final compact forms for the currents

$LaTeX Code: D^{a} = \\theta^{ab}x_{b} \\ \\ 16.8a$

$LaTeX Code: K^{ab} = (2 x^{a}x_{c} - \\eta^{a}_{c}x^{2}) \\theta^{cb} \\ \\ 16.8b$

From these it follows that

$LaTeX Code: \\partial_{a}K^{ab} = 2 x^{b}\\theta_{c}^{c} = 2 x^{b} \\partial_{c}D^{c} \\ \\ 16.9$

Thus in real life, we see that both scale and conformal symmetries are broken by the trace of the E-M tensor $LaTeX Code: \\theta$ .

Finally, we remark that it is also possible to deduce the compact forms of the Noether currents;

$LaTeX Code: (\\theta_{ab}, M_{(B)}^{ab}, D^{a}, K^{ab})$

directly from the action integral (14.13)

$LaTeX Code: \\delta S = \\int d^{4}x \\theta^{ab}\\partial_{a}f_{b} \\ \\ 16.10$
or, by factoring out the parameters,

$LaTeX Code: \\delta = \\lambda_{A}\\delta^{A}$

$LaTeX Code: f_{b} = f^{A}{}_{b}\\lambda_{A}$

we get

$LaTeX Code: \\delta^{A} S = \\int d^{4}x \\theta^{ab}\\partial_{a}f^{A}_{b} \\ \\ 16.11$

When the field equation is satisfied, the translation current is conserved

$LaTeX Code: \\partial_{a}\\theta^{ab} = 0$

so, we can write

$LaTeX Code: \\delta^{A} S = \\int d^{4}x \\partial_{a}(\\theta^{ab}f^{A}_{b}) \\ \\ 16.12$

and identify Noether current of the conformal group by

$LaTeX Code: \\mathcal{J}^{aA} = \\theta^{ab}f^{A}{}_{b} \\ \\ 16.13$

best regards

sam

I look forward for your inputs; comments questions etc.