Recognitions:

## Conformal group, Conformal algebra and Conformal invariance in field theory

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

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

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

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

invariant up to a scale:

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

I.e.,

$$d \bar{s}^{2} \left ( = \eta_{ab} d\bar{x}^{a}d\bar{x}^{b} \right ) = S(x) ds^2 \ \ (1.1')$$

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;

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

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

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

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

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

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

or, after integration,

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

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

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

where $$a^{a}$$ 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 $$\partial_{a}f_{b} + \partial_{b}f_{a} = 0$$ .

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

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

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

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

now, use the conformal Killing equation to find

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

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

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

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

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

When n > 2, this implies

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

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

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

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

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

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

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....
 PhysOrg.com science news on PhysOrg.com >> New language discovery reveals linguistic insights>> US official: Solar plane to help ground energy use (Update)>> Four microphones, computer algorithm enough to produce 3-D model of simple, convex room
 Recognitions: Science Advisor 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; $$[ . , . ] : \cal{L} \times \cal{L} \rightarrow \cal{L}$$ 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 $$\cal{L}$$ . 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; $$g = e^{X} = e^{a_{i}X^{i}} \ \ (2.1)$$ Where $$\{X^{i}\}$$ is a basis for $$\cal{L}$$ .The components $$a_{i}$$ 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 $$e^{X} e^{Y} = e^{X + Y + \frac{1}{12}[X,[X,Y]] + ...} \ \ (2.2)$$ i.e., $$g(a).g(b) = g(f(a,b))$$ more to come, please dont reply to my posts just yet....
 Recognitions: Science Advisor 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; $$f^{a} = \lambda_{A} \delta_{( \lambda )}^{A} x^{a} \ \ (3.1)$$ where $$\lambda_{A} \equiv ( a^{a}, \frac{1}{2}\omega^{ab}, \alpha , c^{a})$$ and $$\delta^{A}$$ are the generators of the following infinitesimal coordinate transformations; Translation; $$\delta^{a}_{(T)} x^{c} = \eta^{ab} \ \ (3.2a)$$ Lorentz; $$\delta^{ab}_{(L)} x^{c} = \eta^{bc}x^{a} - \eta^{ac}x^{b} \ \ (3.2b)$$ Dilation(scale); $$\delta_{(D)}x^{c} = -x^{c} \ \ (3.2c)$$ Conformal; $$\delta^{a}_{(C)}x^{c} = 2 x^{a}x^{c} - \eta^{ac}x^{2} \ \ (3.2d)$$ Ex(3.1) Show that; $$\left[\delta^{a}_{(T)},\delta^{b}_{(T)}\right] = 0 \ \ (3.3a)$$ $$\left[\delta^{ab}_{(L)},\delta^{c}_{(T)}\right] = \eta^{bc}\delta^{a}_{(T)} - \eta^{ac}\delta^{b}_{(T)} \ \ (3.3b)$$ $$\left[\delta^{ab}_{(L)},\delta^{cd}_{(L)}\right] = \eta^{ac}\delta^{bd}_{(L)} - \eta^{bc}\delta^{ad}_{(L)} + \eta^{ad}\delta^{cb}_{(L)} - \eta^{bd}\delta^{ca}_{(L)} \ \ (3.3c)$$ $$\left[\delta_{(D)},\delta_{(D)}\right] = \left[\delta^{ab}_{(L)},\delta_{(D)}\right] = 0 \ \ (3.3d)$$ $$\left[\delta_{(D)},\delta^{a}_{(T)}\right] = \delta^{a}_{(T)} \ \ (3.3e)$$ $$\left[\delta_{(D)},\delta^{a}_{(C)}\right] = -\delta^{a}_{(C)} \ \ (3.3f)$$ $$\left[\delta^{a}_{(C)},\delta^{b}_{(C)}\right] = 0 \ \ (3.3g)$$ $$\left[\delta^{ab}_{(L)},\delta^{c}_{(C)}\right] = \eta^{bc}\delta^{a}_{(C)} - \eta^{ac}\delta^{b}_{(C)} \ \ (3.3h)$$ $$\left[\delta^{a}_{(C)},\delta^{b}_{(T)}\right] = 2 \eta^{ab}\delta_{(D)} - 2 \delta^{ab}_{(L)} \ \ (3.3i)$$ Thus the operators $$\delta^{A}_{(\lambda )}$$ 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...

Recognitions:

## Conformal group, Conformal algebra and Conformal invariance in field theory

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

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

with

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

gives the vector field $$f_{(3)}$$ 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

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

for the algebra by the rule

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

From (3.2) we find

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

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

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

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

And, by explicit calculations we find the algebra

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

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

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

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

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

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

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

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

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

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

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

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

 Recognitions: Science Advisor 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; $$\bar{x} = T(\alpha , a) x = e^{\alpha} x + a \ \ (5.1)$$ It is easy to see that two such transformations is also a transformation of the same kind; $$T(\alpha, a) T(\beta , b) = T\left( \alpha + \beta , a + be^{\alpha} \right) \ \ (5.2)$$ Ex(5.1): Prove the above "easy to see statement". From (5.2), we find; 1) $$\{T(\alpha , a) T(\beta , b) \} T(\gamma , c) = T\left(\alpha + \beta +\gamma , a + be^{\alpha} + ce^{\alpha + \beta}\right) = T(\alpha , a) \{ T(\beta , b) T(\gamma , c) \} \ \ (5.3a)$$ i.e., the multiplication law (5.2) is associative. Ex(5.2): Prove Eq(5.3a). 2) $$T(\alpha , a) T(0,0) = T(0,0) T(\alpha , a) = T(\alpha , a) \ \ (5.3b)$$ i.e., an identity element exists; $$E = T(0,0)$$ 3) $$T\left(\alpha , a \right) T\left(-\alpha , -ae^{-\alpha} \right) = T \left(-\alpha , -ae^{-\alpha} \right) T(\alpha , a) = T(0,0) \ \ (5.3c)$$ i.e., an inverse element exsists $$T^{-1}(\alpha , a) = T\left( -\alpha , -ae^{-\alpha} \right) \ \ (5.4)$$ Thus, the set $$\{T(\alpha , a)\}$$ 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; $$| \Psi_{T}\rangle = U(T) | \Psi \rangle \ \ (5.5)$$ i.e., the unitary operator U forms a representation of the group (multiplication law) in question; $$U( \alpha , a) U(\beta , b) = U( \alpha + \beta , a + be^{\alpha} ) \ \ (5.6)$$ 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 $$(\alpha , a)$$ , so we may write $$U(\alpha , a) \approx 1 + i \alpha D + i a^{a}P_{a} \ \ (5.7)$$ where $$(D,P)$$ are Hermitian operators independent of the parameters $$(\alpha , a)$$ . In order to find the transformations laws of D and P, you need to work out the following; Ex(5.3): Show that $$U(\beta , b)U(\alpha , a)U^{-1}(\beta , b) = U\left(\alpha , ae^{\beta} - \alpha b \right) \ \ (5.8)$$ where $$(\beta ,b)$$ are the parameters of new transformation unrelated to $$(\alpha ,a)$$ Next, write (5.8) to the 1st order in $$(\alpha ,a)$$ ,equate coefficients of $$\alpha$$ and $$a$$ on both sides and show that $$\bar{P}_{a} = U^{-1}(\beta , b) P_{a} U(\beta , b) = e^{-\beta} P_{a} \ \ (5.9a)$$ $$\bar{D} = U^{-1} D U = D + b_{a}P^{a} e^{-\beta} \ \ (5.9b)$$ Now, let $$(\beta , b)$$ themselves be infinitesimals and find the commutation relations; $$[iD , D] = 0 \ \ (5.10a)$$ $$[iP^{a} , P^{b}] = 0 \ \ (5.10b)$$ $$[iD , P^{a}] = P^{a} \ \ (5.10c)$$ 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 $$\bar{x} = T(\Lambda , a)x = \Lambda x + a \ \ (5.11)$$ which have the multiplication law $$T_{1}T_{2} = T(\Lambda_{1} \Lambda_{2} , \Lambda_{1} a_{2} + a_{1}) \ \ (5.12)$$ For infinitesimal transformation, $$T(1 + \omega , \epsilon )$$ , we may write the corresponding unitary operator as $$U(1+\omega ,\epsilon ) = 1 + i \epsilon . P - (i/2) \omega . J \ \ ((5.13)$$ where $$\omega . J \equiv \omega_{ab}J^{ab}$$ Ex(5.4): Use Eq(5.12) to show that $$U(\Lambda ,a)U(1+\omega ,\epsilon )U^{-1}(\Lambda , a) = U(\Lambda \omega \Lambda^{-1} , \Lambda \epsilon - \Lambda \omega \Lambda^{-1} a) \ \ (5.14)$$ write this to 1st order in $$(\omega , \epsilon)$$ , then show that J is a Lorentz tensor; $$\bar{J}^{ab} = U^{-1}J^{ab}U = \Lambda_{c}{}^{a}\Lambda_{d}{}^{b}( J^{cd} + a^{c}P^{d} - a^{d}P^{c}) \ \ (5.15a)$$ and P is a vector $$\bar{P}^{a} = U^{-1}P^{a}U = \Lambda_{b}{}^{a}P^{b} \ \ (5.15b)$$ From the infinitesimal versions of Eq(5.15) deduce the Lie algebra of Poincare' group; $$[iP_{a},P_{b}] = 0 \ \ \ \ (5.16a)$$ $$[iJ^{ab},P^{c}] = \eta^{ac}P^{b} - \eta^{bc}P^{a} \ \ (5.16b)$$ $$[iJ^{ab},J^{cd}] = \eta^{bc}J^{ad} - \eta^{ac}J^{bd} + \eta^{ad}J^{bc} - \eta^{bd}J^{ac} \ \ (5.16c)$$ **** Scale & Lorentz group $$\bar{x} = T(\alpha ,\Lambda ) x = e^{\alpha} \Lambda x \ \ \(5.17)$$ For this group, we have $$u(\alpha_{1} , \Lambda_{1})u(\alpha_{2} , \Lambda_{2}) = u(\alpha_{1} + \alpha_{2} , \Lambda_{1} \Lambda_{2} ) \ \ (5.18)$$ From this, it is seen; $$u(\alpha , 1)u(0 , \Lambda )u(-\alpha , 1) = u(0 , \Lambda) \ \ (5.19)$$ By writting $$u(\alpha , 1) = 1 + i \alpha D \ \ (5.20a)$$ $$u(0 , \Lambda) = 1 + (i/2) \omega . J \ \ (5.20b)$$ we find $$[iJ^{ab}, D] = 0 \ \ (5.21)$$ Next, we postulate that the algebra of C(1,n-1) closes on the (n+1)(n+2)/2 generators (P,J,D,K) and that K is a Lorentz vecto $$[iJ^{ab}, K^{c}] = \eta^{ac}K^{b} - \eta^{bc}K^{a} \ \ (5.22)$$ Surprisingly, this postulate together with what we already know about the subalgebra (P,J,D), is sufficient to determine the remaining commutators [P,K], [D,K] and [K,K] from the Jacobi identities alone. Ex(5.5): On grounds of Lorentz invariance, i.e., the Jacobi identities found by taking the above commutators with J, conclude that $$[iD, K^{a}] = aK^{a} + bP^{a} \ \ (5.23a)$$ $$[iK^{a},K^{b}] = cJ^{ab} \ \ (5.23b)$$ $$[iK^{a},P^{b}] = d\eta^{ab} D - dJ^{ab} \ \ (5.23c)$$ where a,b,c and d are constants. Now, use the J. identities $$\left[iD ,[iK^{a},P^{b}] \right] + \left[iP^{b}, [iD,K^{a}] \right] + \left[iK^{a}, [iP^{b},D] \right] = 0 \ \ (5.24a)$$ $$\left[iD, [iK^{a},K^{b}] \right] + .... + ... = 0 \ \ (5.24b)$$ $$\left[iP^{a}, [iK^{b},K^{c}] \right] + ... + ... = 0 \ \ (5.24c)$$ to find that a = -1 and c = bd = 0 . Notice that a vanishing [P,K] would also be compatible with the Jacobi identities, in that case there would be an arbitrary real factor in [D,K]. However the choice b = 0 and d = 2 is compatible with our previous work $$[iD,K^{a}] = - K^{a} \ \ (5.25a)$$ $$[iK^{a},K^{b}] = 0 \ \ (5.25b)$$ $$[iK^{a},P^{b}] = 2\eta^{ab} D - 2J^{ab} \ \ (5.25c)$$ more to come.....
 Recognitions: Science Advisor Here we will show that the algebra of C(1,n-1) is isomorphic to that of the "Lorentz" group SO(2,n). The latter may be cosidered as a set of pseudoorthognal transformations in a (n+2)-dimensional flat spacetime $$M^{n+2}$$ with the metric; $$\eta^{\mu \nu} = (-1,1,\eta^{ab}) \ \ (6.1)$$ where $$(\mu , \nu) = -2, -1, 0, 1, ...,n-1$$ $$(a , b) = 0, 1, ..., n-1$$ the algebra of SO(2,n) is given by $$[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)$$ Ex(6.1) Define the following generators $$D = M^{-2,-1} \ \ (6.3a)$$ $$J^{ab} = M^{ab} \ \ (6.3b)$$ $$\frac{1}{2}(P^{a} - K^{a}) = M^{-2,a} \ \ (6.3c)$$ $$\frac{1}{2}(P^{a} + K^{a}) = M^{-1,a} \ \ (6.3d)$$ then show that Eq(6.2) gives the conformal algebra. The origin of the nonlinearity of conformal transformations can now be understood by considering the relation between C(1,n-1) and SO(2,n). In $$M^{n+2}$$ ,the SO(2,n) coordinate transformations are linear. However, by projecting these transformations on $$M^{n}$$ ,we get a nonlinear realization of the group SO(2,n), that coincides with the action of C(1,n-1) on $$M^{n}$$ . more to come...
 Recognitions: Science Advisor Can we arrive at the conformal coordinate transformations from the conformal algebra? Yes, we can. This is because the Minkowski space $$M^{n}$$ can be identified with the coset manifold C(1,n-1)/H, with H the subgroup generated by $$(J^{ab},D,K^{a})$$ ,i.e., it can be parametrized by; $$T(x) = \exp(iP.x) \ \ (7.1)$$ In general, a set of group elements T(x), labelled by as many parameters as necessary, parametrizes the manifold if each coset contains one of the T's. Once a parametrization T(x) has been chosen, each group element g can be uniquely decomposed into a product $$g = T(x) . L \ \ (7.2)$$ where T is the representative member of the coset to which g belongs and L connect T to g within the coset. A product of g with an arbitrary element, and in particular with some T(x) will therefore give another T and an L according to $$g . T(x) = T(\bar{x}) . L \ \ (7.3)$$ where $$\bar{x} = \bar{x}(x,g) \ \ (7.4a)$$ $$L = L(x,g) \ \ (7.4b)$$ We will see that the conformal algebra is sufficient to determine $$\bar{x}$$ and L explicitly from $$g . e^{iP.x} = e^{iP. \bar{x}} . L \ \ (7.3')$$ Translations: $$g = T(a) = \exp(ia.P)$$ gives $$\bar{x} = x + a$$ $$L = 1$$ Lorentz trans $$g = \exp(\frac{i}{2} \omega . J)$$ For infinitesimal $$\omega$$ , we can write $$(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)$$ The expression in [ ] can be expanded by use of the Hausdorff formula $$e^{-A} B e^{A} = B + [B,A] + 1/2 [[B,A],A] + ... \ \ (7.6)$$ this gives $$(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)$$ upon inserting the algebra (5.16b), we find $$(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)$$ For finite $$\omega$$ we find the expected Lorentz transformations $$\bar{x}^{a} = \Lambda^{a}{}_{b}x^{b}$$ and $$L = g =\exp(i/2 \omega .J) \ \ (7.9)$$ Ex(7.1): For infinitesimal D-transformations (dilations) and K-transformations (special conformal trans), show that $$(1 - i \alpha D) e^{ix.P} = e^{i(1 - \alpha )x.P} (1 - i \alpha D) + o(\alpha^{2}) \ \ (7.10)$$ $$(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)$$ Next, we will talk physics
 Recognitions: Science Advisor [You don't have to read this post if you are not familiar with QFT] Whereas conformal transformations are well understood in mathematics, this is not the case in physics! The physical meaning of the conformal transformations in n>2 spacetime dimensions is far from clear. Many believe that conformal invariance has no physical meaning and consequently should not play "any" role in physics!! So, what is the problem? Well, apart from the Poincare' algebra, we can not say that the conformal algebra (invariance) is realized in nature. For if $$[iD,P^{a}] = P^{a} \ \ (8.1)$$ holds in nature, then it is also true that; $$e^{-i \alpha D}P^{2}e^{i \alpha D} = e^{-2 \alpha}P^{2} \ \ (8.2)$$ So, if |P> is some 1-particle state of given positive mass m ; $$P^{2}|P \rangle = m^{2}|P \rangle \ \ (8.3)$$ then the state $$|\bar{P} \rangle = e^{-i \alpha D} |P \rangle \ \ (8.4)$$ carries the mass $$\bar{m}^{2} = e^{2\alpha} m^{2} \ \ (8.5)$$ i.e., $$P^{2}|\bar{P}\rangle = e^{2\alpha}m^{2}|\bar{P}\rangle \ \ (8.6)$$ If we assume that scale invariance is not spontaneously broken, i.e., $$e^{-i\alpha D}|0 \rangle = |0 \rangle \ \ (8.7)$$ then we conclude that $$|\bar{P}\rangle = e^{-i\alpha D}a^{\dagger}(p) |0 \rangle \propto a^{\dagger}(e^{\alpha}p) |0 \rangle \ \ (8.8)$$ This means that the state $$|\bar{P}\rangle$$ is a quantum of the same field as the state $$|P\rangle$$ but with a rescaled momentum, i.e., by virtue of conformal invariance, both states must belong to the same Hilbert space. Therefore, by Eq(8.5), conformal invariance implies that the mass spectrum is either continuous or all masses vanish! In order to avoid this physically absurd conclusion, conformal symmetry must be explicitly or spontaneously broken (mathematically this problem is equivalent to the fact that $$P^{2}$$ is not a Casmir's operator). Notice though, if the vacuum was not unique then the state $$|\bar{P}>$$ would belong to a different Hilbert space than the state |P> and our conclusion would be avoided. Another problem is that conformal invariance at the quantum level does not follow from conformal invariance at the classical level. Operationally this means that the renormalized energy-momentum tensor possesses a non-vanishing trace and there are anomalies in the dilation and conformal currents. We will see that scale invariance requires that there be no dimensionfull parameters in the Lagrangian. But QFT does not make sense without a regularization prescription that introduces a scale in the theory. This scale, at which the theory is renormalized, breaks conformal invariance. Ex(8.1): Derive (8.2) and (8.6). more to come...
 Recognitions: Science Advisor Before considering conformal field theory and the conditions under which an action integral is conformally invariant, we must decide how the fields transform under the conformal transformations. If the configuration variables are transformed according to $$x \rightarrow \bar{x} = g x \ \ (9.1)$$ with g belongs to some Lie group, state vectors and ,therefore, local fields will be in general subject to a unitary transformations; $$|\Phi \rangle \rightarrow |\Phi_{g} \rangle = u(g) |\Phi \rangle \ \ (9.2)$$ $$\bar{\phi}_{i}(x) = u^{-1}(g) \phi_{i} u(g) = \mathcal{D}_{i}{}^{j}\phi_{j}(g^{-1}x) \ \ (9.3)$$ where $$\mathcal{D}(g)$$ is a finite-dimensional (matrix) representation of the group, i.e., set of matrices satisfying the Lie algebra of the group. They need not necessarily be unitary. Their effect is a reshuffling of field components. Eq(9.3) can be understood as follow; Since the expectation values; $$C_{i}(x) = \langle \Phi | \phi_{i}(x) | \Phi \rangle \ \ (9.4)$$ are C-numbers, we expect them to transform like classical fields (which belong to some representation D): $$\bar{C}_{i}(\bar{x}) = \mathcal{D}_{i}{}^{j}C_{j}(x) \ \ (9.5)$$ and, by Eq(9.2), the expectation value of the transformed field operator $$\bar{\phi}(\bar{x})$$ in a state described by $$|\Phi \rangle$$ is the same as the expectation value calculated using the untransformed operator $$\phi(\bar{x})$$ and the transformed state $$|\Phi_{g}\rangle$$ : $$\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)$$ Since this is true for any state vector, we conclude that; $$\bar{\phi}(\bar{x}) = u^{-1}\phi (\bar{x}) u = \mathcal{D} \phi(x) \ \ (9.7)$$ or, by (9.1), $$\bar{\phi}(\bar{x}) = u^{-1}\phi(\bar{x}) u = \mathcal{D} \phi (g^{-1}\bar{x}) \ \ (9.8)$$ This leads to Eq(9.3) after renaming the coordinates $$\bar{x}=x$$ . Now I want you to work out the action of infinitesimal generators on local fi eld. Ex(9.1): Let $$g^{-1}x^{a} = x^{a} + \lambda_{A}f^{Aa}(x) \ \ (9.9)$$ $$u(g) = \exp(- i \lambda_{A}G^{A}) \ \ (9.10)$$ and $$\mathcal{D}^{i}_{j} = \exp(\lambda_{A}D^{Ai}{}_{j}) \ \ (9.11)$$ where $$\lambda_{A}$$ [A = 1,2,....,dim(g)] are the infinitesimal parameters and $$D^{A}$$ forms a matrix representation of the generators $$G^{A}$$ . Write Eq(9.3) to 1st order in lambda and show that; $$\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)$$ where $$\delta \phi_{i} = \lambda_{A}\delta^{A}_{(\lambda)}\phi_{i} = \bar{\phi}_{i}(x) - \phi_{i}(x) \ \ (9.13)$$ is the infinitesimal variation (change) in the form of the field function. more to come soon...
 Recognitions: Science Advisor Here, we will use Eq(9.3) or its infinitesimal version (9.12) to determine the action of the generators (P,J,D,K) on local field: For translation; $$g^{-1}x = x - a \ \ (10.1)$$ $$u(a) = \exp(i a.P) \ \ (10.2)$$ $$\mathcal{D}^{i}_{j} = \delta^{i}_{j} \ \ (10.3)$$ we find $$\bar{\phi}_{i}(x) = \phi_{i}(x) - a_{a}[i P^{a},\phi_{i}] = \phi_{i}(x) - a_{a}\partial^{a}\phi_{i}$$ or $$\delta^{a}_{(T)}\phi = [i P^{a},\phi (x)] = \partial^{a}\phi \ \ (10.4)$$ For Lorentz transformations; $$g^{-1}x^{a} = x^{a} - \omega^{ab}x_{b} \ \ (10.5)$$ $$u(\omega) = \exp(-\frac{i}{2}\omega . J) \ \ (10.6)$$ $$\mathcal{D}^{i}_{j} = \exp(\frac{1}{2}\omega . \Sigma^{i}_{j}) \ \ (10.7)$$ where $$\Sigma$$ is the appropriate spin matrix for the field, we get; $$\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)$$ Ex(10.1): Check Eq(10.8). For Scale transformations (Dilations); $$g^{-1}x = x + \alpha x \ \ (10.9)$$ $$u(\alpha) = \exp(-i \alpha D) \ \ (10.10)$$ $$\mathcal{D} = \exp(d \alpha ) \ \ (10.11)$$ where d is (real number) called the canonical dimension of the field. We will see how to find its values for fermions and bosons. Inserting (10.9), (10.10) and (10.11) in Eq(9.3), we find $$\bar{\phi}_{i} = \phi_{i} + \alpha [i D, \phi_{i}] = \phi_{i} + \alpha (x.\partial + d ) \phi_{i}$$ or $$\delta_{(D)}\phi = [i D, \phi ] = ( d + x . \partial ) \phi \ \ (10.12)$$ For free fields, we have the canonical equal-time (anti)commutation relations: $$\{ \psi (x,t) , \psi^{\dagger}(y,t) \} = \delta^{3}(x-y) \ \ (10.13a)$$ $$[\phi(x,t) , \dot{\phi}(y,t) ] = i \delta^{3}(x-y) \ \ (10.13b)$$ The scale dimension d is defined so that (10.13) remain invariant under scale transformation. Transforming the fields in (10.13a) according to Eq(9.7) and (10.11), i.e., $$\psi(x,t) = e^{- d \alpha} \bar{\psi}(\bar{x},\bar{t}) \ \ (10.14)$$ one finds $$\{\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)$$ Thus the invariance of (10.13a) implies that the canonical scale dimension for fermion field is $$d_{\psi} = \frac{3}{2} \ \ (10.16)$$ Ex(10.2): show that for boson field, the invariqance of (10.13b) under scale transformations, implies $$d_{\phi} = 1 \ \ (10.17)$$ These values for d correspond to the natural dimention of the fields in units of mass. Special conformal transformations; In Eq(9.12), if we put A = a (a spacetime index), $$f^{ab} = 2 x^{a}x^{b} - \eta^{ab} x^{2} \ \ (10.18)$$ $$D^{aj}_{i} = -2 d x^{a}\delta^{j}_{i} + 2 x_{b}\Sigma^{baj}_{i} \ \ (10.19)$$ and $$G^{a} = K^{a} \ \ (10.20)$$ we find $$\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)$$ In offering Eq(10.19) we assumed that the field carries irreducible representation of Lorentz group. This will become clear in the next post when we rederive the equations (10.8),(10.12) and (10.21), using the theory of induced representations. At this moment in time, Eq(5.25c) and (7.11) are the only hints we have for (10.19). more to come soon..
 Recognitions: Science Advisor Let $$\mathcal{G}$$ be a Lie group such that the Minkowski space can be identified with the coset space $$M^{n} = \mathcal{G} / H \equiv \frac{(G^{A},P^{a})}{(G^{A})} \ \ (11.1)$$ where the subgroup H (generated by $$G^{A}$$ ) leaves the origin x = 0 invariant, and P generates the usual abelian group of translation. Now, given any representation of the "little" group H on $$\phi (0)$$ , we can induce it to the whole group, i.e., to a representation of $$\mathcal{G}$$ on $$\phi (x)$$ . This is done using the definition $$\phi (x) = e^{i P.x}\phi (0) e^{-i P.x} \ \ (11.2a)$$ i.e., $$\partial \phi (x) = [iP_{a}, \phi (x)] \ \ (11.2b)$$ and the algebra of $$\mathcal{G}$$ , i.e., the commutators [P,G], [P,P] and [G,G]. We take the action of H to be $$[iG^{A},\phi_{i}(0)] = D^{Aj}_{i}\phi_{j}(0) \ \ (11.3)$$ where D's are matrix representation of the generators G's of the little group H. Ex(11.1): Show that the D's in (11.3) form a matrix representation of the algebra $$[iG^{A},G^{B}] = f^{AB}{}_{C}G^{C} \ \ (11.4)$$ Now, Eq(11.2a) and (11.3) can be used to write $$[i (e^{iP.x} G^{A} e^{-iP.x}) , \phi_{i}(x)] = D^{Aj}_{i}\phi_{j}(x) \ \ (11.5)$$ Applying Hausdroff formula Eq(7.6) to the lefthand-side, gives $$[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)$$ Notice that the RHS can be evaluated using only the algebra of the group $$\mathcal{G}$$ . Let us apply this method to the conformal group C(1,n-1). Here the subgroup (H) that leaves the point x = 0 invariant is generated by; $$G^{A} = (J^{ab}, D, K^{a}) \ \ (11.7)$$ If we remove the translation generator (P) from the conformal algebra, we get something identical to Poicare' algebra augmented by dilations (D), because of the similar role played by P and K. Next we introduce a set of matrices; $$D^{A} = (\Sigma^{ab}, \Delta , \mathcal{K}^{a}) \ \ (11.8)$$ to define the action of this little group on $$\phi (0)$$ ; $$[iJ^{ab},\phi_{i}(0)] = \Sigma^{abj}_{i}\phi_{j}(0) \ \ (11.9a)$$ $$[iD,\phi_{i}(0)] = \Delta^{j}_{i}\phi_{j}(0) \ \ (11.9b)$$ $$[iK^{a},\phi_{i}(0) = \mathcal{K}^{aj}_{i}\phi_{j}(0) \ \ (11.9c)$$ these matrices must form arepresentation of the reduced algebra $$[\Sigma,\Delta ] = [\Delta ,\Delta ] = [\mathcal{K},\mathcal{K}] = 0 \ \ (10.10a)$$ $$[\Delta , \mathcal{K}^{a}] = \mathcal{K}^{a} \ \ (11.10b)$$ $$[\Sigma^{ab},\mathcal{K}^{c}] = \eta^{bc}\mathcal{K}^{a} - \eta^{ac}\mathcal{K}^{b} \ \ (11.10c)$$ $$[\Sigma^{ab},\Sigma^{cd}] = \eta^{ac}\Sigma^{bd} - ... \ \ (11.10d)$$ Ex(11.2) use the algebra of (J, D, K) and Eq(11.9) to derive Eq(11.10). If the field belongs to an irreducible representation of Lorentz group, then by Schur's lemma, any matrix that commutes with all the (irreducible) grnerators $$\Sigma$$ must be a multiple of the unit matrix. Consequently, the matrix $$\Delta$$ is $$\Delta^{j}_{i} = d \delta^{j}_{i} \ \ (11.11)$$ where d is a real number, and the algebra (11.10b) forces all matrices $$\mathcal{K}$$ to vanish. If we now put $$D^{Aj}_{i} = \Sigma^{abj} , d\delta^{j}_{i} , 0$$ and $$G^{A} = J^{ab} , D, K^{a}$$ in Eq(11.6), we arrive (after using the conformal algebra) at $$[iJ^{ab},\phi (x)] = (x^{a}\partial^{b} - x^{b}\partial^{a} + \Sigma ) \phi (x) \ \ (11.12a)$$ $$[iD,\phi (x)] = (d + x. \partial ) \phi (x) \ \ (11.12b)$$ and $$[iK^{a},\phi] = (2x^{a}x^{b} - \eta^{ab}x^{2})\partial_{b}\phi + (2dx^{a} + 2x_{b}\Sigma^{ab})\phi \ \ (11.12c)$$ These equations together with $$[iP^{a},\phi ] = \partial^{a}\phi \ \ (11.12d)$$ determine the action of the conformal group on local fields. In the next post we will see that, when the conditions for conformal invariance are met, Noether theorem gives us a time independent and Lorentz covariant objects satisfying the conformal algebra and effect the proper transformations on the fields, i.e., satisfing (11.12). Therefore one can identify Noether charges with the conformal generators. However, we will see that the noether charges (P,J,D,K) satisfy (11.12) even when the (K,D)-symmetry is broken! more to come tomorow...
 Recognitions: Science Advisor An arbitrary field theory is described by a Lagrangian density which we take to be a real function of the field variables $$\phi(x)$$ and of their first derivative $$\partial_{a}\phi$$ . If $$\mathcal{L}$$ depends only on the state of the fields in an infinitely small neighbourhood of the point x ,i.e., on the values oh $$\phi$$ and of $$\partial \phi$$ evaluated at the same point x , then it is called a local Lagrangian, and the corresponding theory is said to be a local theory. So, we may write; $$\mathcal{L}(x) = \mathcal{L}\left(\phi (x), \partial_{a}\phi (x)\right) \ \ (12.1)$$ It is assumed that $$\phi (x) \rightarrow 0$$ sufficiently fast as $$|\vec{x}| \rightarrow \infty$$ . The integral of the Lagrangian over bounded, arbitrary contractible, region in spacetime; $$S = \int_{D}d^{4}x \mathcal{L}(x) \ \ (12.2)$$ where $$D \equiv ( V \cup \partial V ) \subset \mathbb{R}^{4} \ \ (12.3)$$ is called the action. From the variation principle of stationary action; $$\delta \int_{D} d^{4}x \mathcal{L}(x) = 0 \ \ (12.4)$$ together with the assumption that the variation of the fields $$\delta \phi$$ vanish on $$\partial V$$ (the surface of the 4-volume over which the integral is taken) but arbitrary in $$V$$ , we obtain the field equations in V: $$\frac{\delta \mathcal{L}}{\delta \phi} = 0 \ \ (12.5a)$$ where $$\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)$$ is the Euler derivative. In the canonical formalism, the field and its conjugate momentum $$\pi = \frac{\partial \mathcal{L}}{\partial \partial_{0}\phi} \ \ (12.6)$$ satisfy the equal-time commutation relations $$[\phi(t,x) , \pi (t,y)] = i\delta^{3}(x-y) \ \ (12.7a)$$ $$[\phi , \phi ] = [\pi ,\pi ] = 0 \ \ (12.7b)$$ The connection between continuous symmetries and conserved quantities is highlighted by Noether's theorem which states: to every Lie group of transformations which leaves the action unchanged, there corresponds a definite combination of the Lagrangian derivatives that determines the field invariants when the fields satisfy the E-L equation, i.e., when the field equations are satisfied, a systematic procedure for obtaining conservation laws can be developed from a direct study of the invariance properties of the action integral. Therefore, it is interesting to ask what conditions on $$\mathcal{L}$$ insure the invariance of the action? In other words, if G is a given Lie group, then the question will be; how to formulate a G-invariant theory? To answer this question, let us consider an infinitesimal coordinate transformations $$x^{a} \rightarrow \bar{x}^{a} = x^{a} + \lambda_{A}f^{Aa}(x) \ \ (12.8)$$ such that the fields transform according to some known representation of G : $$\phi (x) \rightarrow \bar{\phi}(\bar{x}) = \phi (x) + \lambda_{A}\bar{\delta}^{A}\phi (x) \ \ (12.9)$$ where $$\lambda_{A}$$ are the infinitesimal parameters, A=1,2,..,dim(G), and $$\bar{\delta}^{A}\phi$$ is the local variation in the field due to the change in the form of the field function and its argument: $$\bar{\delta}^{A}\phi = \delta^{A}\phi + f^{Aa}\partial_{a}\phi \ \ (12.10)$$ where the variation in the form of the field is defined by $$\delta \phi \equiv \lambda_{A}\delta^{A}\phi = \bar{\phi}(x) - \phi (x) \ \ (12.11)$$ Please note that $$\partial_{a}\delta^{A}\phi = \delta^{A}\partial_{a}\phi \ \ (12.12a)$$ but $$\partial_{a}\bar{\delta}^{A}\phi = \bar{\delta}^{A}\partial_{a}\phi + \partial_{a}f^{Ab}\partial_{b}\phi \ \ (12.12b)$$ Ex(12.1): Derive the above two equations. Under these transformations (the action of the group G), the form of a local Lagrangian would change according to $$\delta^{A}\mathcal{L} = \frac{\partial \mathcal{L}}{\partial \phi}\delta^{A}\phi + \pi^{a}\partial_{a}\delta^{A}\phi \ \ (12.13)$$ where $$\pi^{a} = \frac{\partial \mathcal{L}}{\partial \partial_{a}\phi} \ \ (12.14a)$$ $$\pi^{0}\equiv \pi \ \ (12.14b)$$ Since the explicit form for $$\bar{\delta}^{A}\phi$$ is assumed known (matrix representation of G), we can use (12.10) to put (12.13) in a form like; $$\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)$$ with $$C^{A}$$ has no exiplicit x-dependence and $$D$$ is the matrix representation of G. If we can construct a Lagrangian (i.e.,formulate a theory) in such a way that $$C^{A} = 0 \ \ (12.16)$$ then our theory will be invariant under the given Lie group G. Mathematically speaking, the solution $$(\mathcal{L})$$ of the system of 1st order PDE (12.16), describes the most general G-invariant theory. Also, whether or not a given theory is G-invariant can be decided by examining the Lagrangian against (12.16). To show that (12.16) is invariance condition, look at the variation of the action; $$\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)$$ it is made of the sum of the local variation of $$\mathcal{L}$$ and of the variation in the region of integration: $$\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)$$ Symbolically, this can be understood in terms of the algebric property of the local variation symbol $$\bar{\delta}$$ . Indeed, to the 1st order, it is a derivation; $$\bar{\delta}(fg) = (\bar{\delta}f)g + f(\bar{\delta}g)$$ Now, the local variation in $$\mathcal{L}$$ can be expressed as a variation in the form of the lagrangian plus "drag"; $$\bar{\delta}^{A}\mathcal{L} = \delta^{A}\mathcal{L} + f^{Aa}\partial_{a}\mathcal{L} \ \ (12.19)$$ [see Eq(12.10)], and $$\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)$$ J-is the Jacobian of the transformations (12.8). To the 1st order we can write; $$\bar{\delta}^{A}(d^{4}x) = \partial_{a}f^{Aa} d^{4}x \ \ (12.21)$$ Putting (12.19) and (12.21) in Eq(12.18) leads to; $$\bar{\delta}^{A}\int_{D}\mathcal{L} d^{4}x = \int_{D} C^{A} d^{4}x \ \ (12.22)$$ Since $$D \subset \mathbb{R}^{4}$$ is an arbitrary contractible domain, we conclude that the action is invariant if and only if $$C^{A}= 0$$ . So what is the link between the invariance condition (12.16) and Noether theorem? In terms of the Euler derivative (12.5b), we can rewrite (12.13) as; $$\delta^{A}\mathcal{L} = \partial_{a}(\pi^{a}\delta^{A}\phi ) + \frac{\delta \mathcal{L}}{\delta \phi}\delta^{A}\phi \ \ (12.23)$$ Putting this in (12.15) we find; $$C^{A} = \frac{\delta \mathcal{L}}{\delta \phi}\delta^{A}\phi + \partial_{a}(\pi^{a}\delta^{A}\phi + f^{Aa}\mathcal{L}) \ \ (12.24)$$ Thus, our invariance condition $$C^{A} = 0$$ is equivalent to the Noether identity; $$\frac{\delta \mathcal{L}}{\delta \phi}\delta^{A} = \partial_{a}\mathcal{J}^{aA} \ \ (12.25)$$ where $$\mathcal{J}^{aA} = -\pi^{a}\delta^{A}\phi - f^{Aa}\mathcal{L} \ \ (12.26)$$ is the canonical Noether current. the conservation of this current; $$\partial_{a}\mathcal{J}^{aA} = 0 \ \ (12.27)$$ follows from (12.25), when the field satisfies E-L equations. Thus, in terms of the invariance condition (12.16), the statement of Noether theorem becomes $$\left( C^{A}= 0, \frac{\delta \mathcal{L}}{\delta \phi}= 0 \right) \Rightarrow \left(\partial_{a}\mathcal{J}^{aA}= 0 \right) \ \ (12.28)$$ When G is a symmetry group (C = 0), we can use (12.27) to show that the Noether charges $$Q^{A}= \int d^{3}x \mathcal{J}^{A0} \ \ (12.29)$$ 1) are time-independent. 2) are G-covariant. 3) generate the correct transformation on the field; $$\delta^{A}\phi = [i Q^{A}, \phi ] \ \ (12.30)$$ 4) satisfy the Lie algebra of G: $$[iQ^{A},Q^{B}] = f^{AB}{}_{C}Q^{C} \ \ (12.31)$$ Because of (3) and (4), Q is called the generator of the symmetry transformations. Ex(12.2) Derive Eq(12.30). It is clear that when G is not a symmetry operation, i.e., $$C^{A}\ne 0$$ , it is still possible to define the current (12.26) which is not conserved, and the charge (12.29). In this case, two situations are distinguished. If G is an internal Lie group, i.e., $$f^{Aa}= 0$$ , the Noether charges would still have the properties (3) & (4) but not (1) or (2). For Spacetime groups (1),(2) & (4) are fulse but (3) is still true. Ex(12.3): Let G be a broken internal symmetry with $$f^{Aa}=0 \ \ (12.32a)$$ $$\delta^{A}\phi = i T^{A}\phi \ \ (12.32b)$$ and $$[iT^{A},T^{B}] = f^{AB}{}_{C}T^{C} \ \ (12.32c)$$ Define the non-conserved current $$\mathcal{J}^{Aa} = \pi^{a}\delta^{A}\phi \ \ (12.33)$$ and its (time-dependent) charge $$Q^{A}(t) = i \int d^{3}x \pi T^{A}\phi \ \ (12.34)$$ then show that Q satisfies the Lie algebra of G and generates the transformation (12.32b). In the next post (the last I hope) we will describe the conditions of conformal invariance.
 Recognitions: Science Advisor In this and next post, we will apply the general formalism of post #12 to the conformal group C(1,3). we will derive the explicit form of the invariance condition Eq(12.16) for translations, Lorentz, scale and special conformal transformations, and find the corresponding Noether currents and charges. Translations $$\bar{x}^{a}= x^{a} + \eta^{ab}a_{b} \ \ (13.1a)$$ $$\delta^{b}_{(T)}\phi = - \eta^{ba}\partial_{a}\phi \ \ (13.1b)$$ I.e., the notation of post #12 becomes $$A = b \ \ (13.2a)$$ $$f^{Aa}\rightarrow \eta^{ba} \ \ (13.2b)$$ $$\delta^{A}\rightarrow \delta^{b}_{(T)} \ \ (13.2c)$$ [(T) is not an index, it stands for translation] In this case Eq(12.13) becomes; $$\delta^{b}_{(T)}\mathcal{L} = - \partial_{a}(\eta^{ab}\mathcal{L}) + \frac{\delta \mathcal{L}}{\delta x_{b}} \ \ (13.3)$$ From Eq(12.15) & (12.16), we see; $$C^{b}_{(T)} \equiv \frac{\delta \mathcal{L}}{\delta x_{b}} = 0 \ \ (13.4)$$ Thus, as expected, the condition for translation invariance is equivalent to the statement that $$\mathcal{L}$$ has no explicit x-dependence. From Eq(12.26) we find the conserved translation current; $$T^{ab} = \pi^{a}\partial^{b}\phi - \eta^{ab}\mathcal{L} \ \ (13.5)$$ This is the canonical energy momentum tensor. The corresponding charge $$P^{a} =\int d^{3}x T^{0a} \ \ (13.6)$$ represents the energy momentum 4-vector. Ex(13.1): Given; $$\partial_{a}T^{ab}= 0 \ \ (13.7)$$ show that $$P^{a}$$ 1) is time-independent. 2) is 4-vector. 3) grnerates the transformation; $$\delta^{a}_{(T)}\phi = [i P^{a},\phi] \ \ (13.8)$$ 4) satisfies the Lie algebra of the translation group; $$[i P^{a},P^{b}] = 0 \ \ (13.9)$$ Lorentz Transformation Here, the index A of post #12 is a double spacetime index; $$A = ab$$ $$f^{(bc)a}(x) = \eta^{ab}x^{c} - \eta^{ac}x^{b} \ \ (13.10a)$$ $$\delta^{bc}_{(L)}\phi = ( x^{b}\partial^{c} - x^{c}\partial^{b} + \Sigma^{bc}) \phi \ \ (13.10b)$$ Inserting these in (12.13) and using the translation invariance Eq(13.4), leads to; $$\delta^{bc}_{(L)}\mathcal{L} = - \partial_{a}(f^{bca}\mathcal{L}) + C^{bc}_{(L)} \ \ (13.11a)$$ where $$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)$$ Hence the condition for Lorentz invariance of the theory is; $$C^{bc}_{(L)} = 0 \ \ (13.12)$$ and the canonical Noether current is $$M^{abc} = x^{b}T^{ac} - x^{c}T^{ac} + \pi^{a}\Sigma^{bc}\phi \ \ (13.13)$$ The integrated charge (the ungular momentum of the field) is $$J^{bc} = \int d^{3}x M^{0bc} = L^{bc} + S^{bc} \ \ (13.14a)$$ where $$L^{bc} = \int d^{3}x (x^{b}T^{0c} - x^{c}T^{0b}) \ \ (13.14b)$$ is the intrinsic "orbital" angular momentum of the field, and $$S^{bc} = \int d^{3}x \pi \Sigma^{bc}\phi \ \ (13.14c)$$ describes the polarization properties of the field. It corresponds to the spin of particles described by the quantized field. Warning: some textbooks call S; the spin "tensor", and L; the orbital angular momentum "tensor". This is not correct, in general, L and S are not covariant quantities, the sum of them is. Ex(13.2): Check, by explicit calculation, that $$\partial_{a}M^{abc} = 0 \ \ (13.15)$$ follows from the invariance condition (13.12) and E-L equation. Ex(13.3): Show that $$[i J^{ab},\phi] = \delta^{ab}_{(L)}\phi$$ $$[i J^{ab}, P^{c}] = \eta^{ac}P^{b} - \eta^{bc}P^{a}$$ and $$[i J^{ab},J^{cd}] = \eta^{bc}J^{ad} - \eta^{ac}J^{bd} + \eta^{ad}J^{bc} - \eta^{bd}J^{ac}$$ Ex(13.4): Show that the Belinfante tensor, defined by $$T^{ab}_{(B)} = T^{ab} + \frac{1}{2}\partial_{c}X^{cab} \ \ (13.16a)$$ where $$X^{cab} = \pi^{c}\Sigma^{ab}\phi - \pi^{a}\Sigma^{cb}\phi - \pi^{b}\Sigma^{ca}\phi \ \ (13.16b)$$ is conserved and leads to the same momentum 4-vector $$P^{b}$$ . Hence it may be used instead of the canonical tensor $$T^{ab}$$ . Use the Lorentz-invariance condition (13.12) as well as E-L equation to show that $$T^{ab}_{(B)} = T^{ba}_{(B)} \ \ (13.17)$$ Ex(13.5): Define the tensor $$M^{abc}_{(B)} = x^{b}T^{ac}_{(B)} - x^{c}T^{ab}_{(B)} \ \ (13.18)$$ then show that it is conserved when $$M^{abc}$$ is. Also show that it leads to the same angular momentum $$J^{ab}$$ as $$M^{abc}$$ does. more to come...
 Recognitions: Science Advisor Cosider an infinitesimal scale transformation; $$\bar{x}^{a} = x^{a} - \alpha x^{a} \ \ 14.1$$ In here, the parameter carries no index, so the index A of post #12 disapears: $$f^{a}(x) = - x^{a} \ \ (14.2)$$ $$\delta_{(D)}\phi = ( d + x^{c}\partial_{c})\phi \ \ (14.3)$$ Following the general formalism of post #12, it is easy to obtain $$\delta_{(D)}\mathcal{L} = \partial_{a}(x^{a}\mathcal{L}) + C_{(D)} \ \ (14.4)$$ with $$C_{(D)} \equiv d \frac{\partial \mathcal{L}}{\partial \phi}\phi + (d+1) \pi_{c}\partial^{c}\phi - 4 \mathcal{L} = 0 \ \ (14.5)$$ is the condition for dilation invariance. Since d(boson) = 1 and d(fermion) = 3/2, the kinetic term of the Lagrangian satisfies (14.5). Examples of scale invariant interactions are; $$\phi^{4}\ \mbox{and}\ \bar{\psi}\ \psi \phi .$$ It is easy to see that (14.5) requires that the scale dimension of $$\mathcal{L}$$ be 4, i.e., that there be no dimensionful parameters in the lagrangian. Terms like $$m^{2}\phi^{2}$$ would of course break the scale symmetry explicitly as their scale dimension is 2. Notice the difference between scale dimension and ordinary dimention which is 4 for the term $$m^{2}\phi^{2}$$ . Scale transformation effects the dynamical variables but not the dimensionful parameters. Ex(14.1) Show that the scale-invariance condition (14.5) implies $$d(\mathcal{L}) = 4 \ \ (14.6)$$ In terms of the canonical energy-momentum tensor, the dilation Noether current is $$D^{a} = T^{ab}x_{b} + d \pi^{a}\phi \ \ (14.7)$$ and the corresponding charge is $$D = \int d^{3}x (T^{0b}x_{b} + d \pi \phi ) \ \ (14.8)$$ Ex(14.2) Use E-L equation to put the invariance condition (14.5) in the form $$T^{a}_{a} + \partial_{c}( \pi^{c} d \phi ) = \partial_{c}D^{c} = 0 \ \ (14.9)$$ then use this to prove $$[i D,P^{a}] = P^{a} \ \ (14.10)$$ $$[i J^{ab},D] = 0 \ \ (14.11)$$ Ex(14.3) Show that the dilation charge D generates the correct transformation on the field $$\delta_{(D)}\phi = [i D ,\phi ] \ \ (14.12)$$ Ofcourse this is true even in the absence of scale symmetry, i.e., even if (14.5) is not true. The trace of $$T^{ab}$$ lives in the invariance condition (14.9) for a reason. It did not appears there by accident! Let me explain this: Under an arbitrary transformation of the coordinates $$x \rightarrow x + f(x)$$ the action changes as $$\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)$$ Where $$\theta_{ab}$$ is the symmetric energy-momentum tensor(we will call it E-M tensor from now on). This is true even if the field does not satisfy the E-L equation. From the conformal Killing equation (1.2), it follows that $$\delta S = (1/2) \int d^{4}x \theta_{a}^{a} F(x) \ \ (14.14)$$ and therefore, traceless E-M tensor implies that the action is invariant under the conformal group C(1,3). Of course the converse is not true because F(x) is not an arbitrary function. more to come I'm having problem with my PC, so I will break this post to 2 may be 3 smaller posts
 Recognitions: Science Advisor The Poicare' group together with dilations forms a subgroup of the full conformal group. This means that a theory invariant under this subgroup is not necessarily invariant under special conformal transformations. The conditions under which it should be invariant are determined by the tracelessness of the E-M tensor. Under certain condition, the E-M tensor of a theory, with scale invariance, can be made traceless much in the same way as it can be made symmetric in a Lorentz-invariant theory. If this can be done, then it follows from Eq(14.14) that full conformal invariance is a consequence of scale invariance and poincare' invariance. Let us do this trick. We start with scale invariant theory, i.e., with; $$\partial_{a}D^{a} = T_{a}^{a} + \partial_{a}( \pi^{a} d \phi ) = 0 \ \ (15.1)$$ Next we define the field virial by $$V^{a} = \pi^{a} d \phi - \pi_{c}\Sigma^{ca}\phi \ \ (15.2)$$ and consider the following tensor $$\theta^{ab} = T^{ab} + (1/2) \partial_{c}X^{cab} + (1/2) \partial_{c}\partial_{d}Y^{cdab} \ \ (15.3)$$ where the first 2 terms constitute the Belinfante tensor (13.16a). 1) from the conservation law $$\partial_{a}\theta^{ab} = 0 \ \ (15.4a)$$ it follows that $$\partial_{a}\partial_{b}\partial_{c}Y^{cdab} = 0 \ \ (15.4b)$$ Hence, Y must not be completely symmetric in the first 3 indices. 2) from the symmetry property $$\theta_{ab} = \theta_{ba} \ \ \ (15.5a)$$ it follows that $$\partial_{c}\partial_{d}Y^{cdab} = \partial_{c}\partial_{d}Y^{cdba} \ \ \ (15.5b)$$ Thus, the part of Y antisymmetric in (a,b) must also be antisymmetric in (c,d),i.e., $$Y^{cdab} - Y^{cdba} \propto ( \eta^{ca}\eta^{db} - \eta^{cb}\eta^{ad}) \ \ \ (15.5c)$$ 3) finally, we take the trace of Eq(15.3): $$\theta_{a}^{a} = T^{a}_{a} + (1/2) \partial_{c}X^{ca}{}_{a} + (1/2) \partial_{a}\partial_{b}Y^{abc}{}_{c} \ \ \ (15.6)$$ From Eq(13.16b) we see that $$\partial_{c}X^{ca}{}_{a} = - 2 \partial_{c}( \pi_{a}\Sigma^{ca}\phi ) \ \ \ (15.7)$$ Using (15.2), this becomes $$\partial_{c}X^{ca}{}_{a} = \partial_{a}( \pi^{a} d \phi - V^{a}) \ \ \ (15.8)$$ Putting this in (15.6) and using (15.1), leads to $$\theta_{a}^{a} = \partial_{c}D^{c} - \partial_{a}V^{a} + (1/2)\partial_{a}\partial_{b}Y^{abc}{}_{c} \ \ \ (15.9)$$ If the virial is a total divergence $$V^{a} = \partial_{c}\sigma^{ca} \ \ \ (15.10)$$ then Y can be constructed such that $$\partial_{c}V^{c} = (1/2) \partial_{a}\partial_{b}X^{abc}{}_{c} \ \ \ (15.11)$$ This leads to our final result $$\partial_{c}D^{c} = \theta_{a}^{a} \ \ \ (15.12)$$ This means that scale invariance implies that the "improved" E-M tensor (15.3) is traceless, provided that the field virial satisfies (15.10). This shows that for conformal invariance two conditions must be met: 1) the theory should be scale invariant,i.e., (14.5) must be true; and 2) Eq(15.10) must hold. more to come
 Recognitions: Science Advisor In this (LAST) post, we will rederive the conditions for special conformal invariance using, this time, the general formalism of post #12. From the infinitesimal conformal transformations: $$f^{ba}(x) = 2 x^{b}x^{a} - \eta^{ba}x^{2} \ \ 16.1$$ $$\delta^{b}_{(C)} = f^{ba}\partial_{a}\phi + 2 d x^{b} \phi - 2 x_{a}\Sigma^{ab}\phi \ \ 16.2$$ it is easy to see that $$\delta^{b}_{(C)}\mathcal{L} = \partial(f^{ba}\mathcal{L}) + 2 C_{(D)}x^{b} + V^{b} \ \ 16.3a$$ where $$C_{(D)} = d \frac{\partial \mathcal{L}}{\partial \phi} \phi + (d+1) \pi_{a}\partial^{a}\phi - 4 \mathcal{L} \ \ 16.3b$$ Inderiving (16.3), Lorntz and translation invariance conditions are used. From (16.3) we see that conformal invariance requires that $$C_{(D)} = 0 \ \ 16.4a$$ $$V^{c} = \partial_{b}\sigma^{ba} \ \ 16.4b$$ You might ask why not $$C = 0 \mbox{and} V^{a} = 0$$ , after all this how it should be according to our general invariance condition Eq(12.16)? Well, explicit computation show that V vanishes identically for spin-(1/2) and spin-1 fields, but for spin-0 field, we have $$V^{a} = \pi^{a}\phi = (1/2) \partial^{a}\phi^{2} \neq 0 \ \ 16.5$$ I have not been able to understand this unique role of the scalar field! Remarkably,Eq(16.4b) turns out to be true for all renormalizable theories, even though scale invariance is of course broken. Eq(16.4b) is also true in all field theories (involving spins; 0, 1/2 & 1) without derivative coupling. Consequently, for these theories conformal invariance is equivalent to scale invariance. In terms of the canonical E-M tensor, the conformal Noether current is $$K^{ab} = f^{a}{}_{c}T^{bc} + 2 \pi^{b}(d x^{a} - x_{c}\Sigma^{ca}) \phi - 2 \sigma^{ab} \ \ 16.6$$ By inserting the formula for $$\theta^{ab}$$ in terms of $$T^{ab}$$ in Eq(16.6) and in Eq(14.7), one finds after many tedious steps: $$D^{a} = \theta^{ab}x_{b} + \partial_{b}Z^{ba} \ \ 16.7a$$ $$K^{ab} = f^{a}{}_{c}\theta^{cb} + \partial_{c}Z^{cab} \ \ 16.7b$$ The total divergence terms in (16.7) are superpotentials; they may be droped and we are left with the final compact forms for the currents $$D^{a} = \theta^{ab}x_{b} \ \ 16.8a$$ $$K^{ab} = (2 x^{a}x_{c} - \eta^{a}_{c}x^{2}) \theta^{cb} \ \ 16.8b$$ From these it follows that $$\partial_{a}K^{ab} = 2 x^{b}\theta_{c}^{c} = 2 x^{b} \partial_{c}D^{c} \ \ 16.9$$ Thus in real life, we see that both scale and conformal symmetries are broken by the trace of the E-M tensor $$\theta$$ . Finally, we remark that it is also possible to deduce the compact forms of the Noether currents; $$(\theta_{ab}, M_{(B)}^{ab}, D^{a}, K^{ab})$$ directly from the action integral (14.13) $$\delta S = \int d^{4}x \theta^{ab}\partial_{a}f_{b} \ \ 16.10$$ or, by factoring out the parameters, $$\delta = \lambda_{A}\delta^{A}$$ $$f_{b} = f^{A}{}_{b}\lambda_{A}$$ we get $$\delta^{A} S = \int d^{4}x \theta^{ab}\partial_{a}f^{A}_{b} \ \ 16.11$$ When the field equation is satisfied, the translation current is conserved $$\partial_{a}\theta^{ab} = 0$$ so, we can write $$\delta^{A} S = \int d^{4}x \partial_{a}(\theta^{ab}f^{A}_{b}) \ \ 16.12$$ and identify Noether current of the conformal group by $$\mathcal{J}^{aA} = \theta^{ab}f^{A}{}_{b} \ \ 16.13$$ best regards sam I look forward for your inputs; comments questions etc.

Recognitions:
FANTASTIC series of posts! Great contribution.
I have plenty of older papers on the conformal group, and some
textbooks, but having it all together like you've done is great.
However, it will take me while to get familiar with your notation,
and some of the details. But I'll ask a couple of things immediately...

1st question:

Back in post #8 you wrote:
 Quote by samalkhaiat then we conclude that $$|\bar{P}\rangle = e^{-i\alpha D}a^{\dagger}(p) |0 \rangle \propto a^{\dagger}(e^{\alpha}p) |0 \rangle \ \ (8.8)$$ This means that the state $$|\bar{P}\rangle$$ is a quantum of the same field as the state $$|P\rangle$$ 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 $$P^{2}$$ is not a Casmir's operator). Notice though, if the vacuum was not unique then the state $$|\bar{P}>$$ would belong to a different Hilbert space than the state |P> and our conclusion would be avoided.
I'm familiar with unitarily inequivalent representations in QFT (i.e:
disjoint Fock spaces, and all that). But I didn't follow how you arrived
at the possibility that $$|\bar{P}>$$ could belong to a different
representation. Don't you have to construct the dilation operator D
explicitly in terms of creation/annihilation operators to show this?

2nd question: Do you know of anywhere that I could find all 3 conformal
Casimirs written out explicitly in terms of the generators P, J,D,K? Most
textbooks write them out in terms of operators in the SO(4,2) acting
in the linear space. But converting these expressions to explicit ones
in P,J,D,K is seriously tedious and error-prone, so I'm wondering
whether someone has already done it somewhere?

Thanks again,

- strangerep.

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