Unfolding Conformal Geometry

Let us begin with setting the convention for the conformal algebra: the Lie algebra $\mathfrak{so}(2,d)$ is generated by anti-Hermitian generators $\hat{M}_{AB}$ with the Lie bracket,

$$[\hat{M}_{AB},\hat{M}_{CD}]=\eta_{AD}\hat{M}_{BC}+\eta_{BC}\hat{M}_{AD}-\eta_{AC}\hat{M}_{BD}-\eta_{BD}\hat{M}_{AC}\,,$$

(2.1)

where $\eta_{AB}$ is the flat metric with signature $(2,d)$ . Taking the basis with indices $A=+,-,a$ and $a=0,1,\ldots,d-1$ where $\eta_{+-}=1$ and $\eta^{ab}$ is the $d$-dimensional flat metric with signature $(1,d-1)$, the Lie bracket of $\hat{M}_{ab}=\hat{J}_{ab}$, $\hat{M}_{a+}=\hat{P}_{a}$, $\hat{M}_{a-}=\hat{K}_{a}$ and $\hat{M}_{+-}=\hat{D}$ read

		$\displaystyle[\hat{J}_{ab},\hat{J}_{cd}]=\eta_{ad}\hat{J}_{bc}+\eta_{bc}\hat{J}_{ad}-\eta_{ac}\hat{J}_{bd}-\eta_{bd}\hat{J}_{ac}\,,\qquad$	$\displaystyle[\hat{J}_{ab},\hat{D}]=0\,,$		(2.2)
		$\displaystyle[\hat{J}_{ab},\hat{P}_{c}]=\eta_{bc}\hat{P}_{a}-\eta_{ac}\hat{P}_{b}\,,\qquad$	$\displaystyle[\hat{J}_{ab},\hat{K}_{c},]=\eta_{bc}\hat{K}_{a}-\eta_{ac}\hat{K}_{b}\,,$
		$\displaystyle[\hat{D},\hat{P}_{a}]=\hat{P}_{a}\,,\qquad[\hat{D},\hat{K}_{a}]=-\hat{K}_{a}\,,\qquad$	$\displaystyle[\hat{K}_{a},\hat{P}_{b}]=\eta_{ab}\,\hat{D}-\hat{J}_{ab}\,.$

From now on, all the Latin indices $a,b,c,d,\ldots$ are lowered and raised by $\eta_{ab}$ and $\eta^{ab}$.

Let us review now the gauge formulation of conformal geometry. We consider the gauge one-form taking value in $\mathfrak{so}(2,d)$ algebra,

$$\hat{A}=e^{a}\,\hat{P}_{a}+\frac{1}{2}\,\omega^{ab}\,\hat{J}_{ab}+f^{a}\,\hat{K}_{a}+b\,\hat{D}\,,$$

(2.3)

where $e^{a},\omega^{ab},f^{a},b$ are one-form fields which are all independent at this stage. For geometric interpretation, we assume the one-form $e^{a}$ has components $e^{a}_{\mu}$ which are invertible, and the inverse is denoted by $E_{a}^{\mu}$, which define vector fields $E_{a}=E_{a}^{\mu}\,\partial_{\mu}$ . The curvature two form,

$$\hat{F}={\rm d}\hat{A}+\hat{A}\wedge\hat{A}=F_{\hat{P}}^{a}\,\hat{P}_{a}+\frac{1}{2}\,F_{\hat{J}}^{ab}\,\hat{J}_{ab}+F_{\hat{K}}^{a}\,\hat{K}_{a}+F_{\hat{D}}\,\hat{D}\,.$$

(2.4)

has the components,

		$\displaystyle F_{\hat{P}}^{a}=({{\rm D}^{L}}+b)\,e^{a}\,,\qquad$	$\displaystyle F_{\hat{J}}^{ab}=R^{ab}-2\,f^{[a}\wedge e^{b]}\,,$		(2.5)
		$\displaystyle F_{\hat{K}}^{a}=({{\rm D}^{L}}-b)\,f^{a}\,,\qquad$	$\displaystyle F_{\hat{D}}={\rm d}b+f^{a}\wedge e_{a}\,.$

Here, ${{\rm D}^{L}}$ is the Lorentz covariant differential,

$${{\rm D}^{L}}\,V^{a}={\rm d}\,V^{a}+\omega^{a}{}_{b}\,V^{b}\,,$$

(2.6)

and $R^{ab}$ is given by

$$R^{ab}={\rm d}\,\omega^{ab}+\omega^{ac}\wedge\omega_{c}{}^{b}\,.$$

(2.7)

This system has $\mathfrak{so}(2,d)$ gauge symmetry,

$$\delta\hat{A}={\rm d}\hat{\Lambda}+[\hat{A},\hat{\Lambda}]\,,\qquad\delta\hat{F}=[\hat{F},\hat{\Lambda}]\,.$$

(2.8)

Labeling the components of the gauge parameter $\hat{\Lambda}$ as

$$\hat{\Lambda}=\epsilon^{a}\,\hat{P}_{a}+\frac{1}{2}\,\lambda^{ab}\,\hat{J}_{ab}+\kappa^{a}\,\hat{K}_{a}+\sigma\,\hat{D}\,,$$

(2.9)

the one-forms transform as

	$\displaystyle\delta e^{a}=({{\rm D}^{L}}+b)\,\epsilon^{a}-\lambda^{a}{}_{b}\,e^{b}-\sigma\,e^{a}\,,$		(2.10)
	$\displaystyle\delta\omega^{ab}={{\rm D}^{L}}\lambda^{ab}+2\,e^{[a}\,\kappa^{b]}+2\,f^{[a}\,\epsilon^{b]}\,,$		(2.11)
	$\displaystyle\delta f^{a}=({{\rm D}^{L}}-b)\,\kappa^{a}-\lambda^{a}{}_{b}\,f^{b}+\sigma\,f^{a}\,,$		(2.12)
	$\displaystyle\delta b={\rm d}\sigma-e^{a}\,\kappa_{a}+\epsilon^{a}\,f_{a}\,.$		(2.13)

We would need a partial gauge fixing as well as imposing constraints to recover the usual geometry based on metric tensor.

Let us review the set of constraints which bring the $\mathfrak{so}(2,d)$-gauge theory to conformal geometry. First, we impose the torsionless constraint,

$${\rm C}_{\hat{P}}\,:\qquad F_{\hat{P}}^{a}=({{\rm D}^{L}}+b)\,e^{a}\overset{!}{=}0\,,$$

(2.14)

which is modified by the presence of $b$. Here, we use the notation $\overset{!}{=}$ to emphasize that it is a constraint that we decided to impose. We also impose the curvature for dilation $\hat{D}$ to vanish

$${\rm C}_{\hat{D}}\,:\qquad F_{\hat{D}}={\rm d}\,b+f^{a}\wedge e_{a}\overset{!}{=}0\,.$$

(2.15)

From the Bianchi identity ${\rm d}\hat{F}+[\hat{A},\hat{F}]=0$, we find

		$\displaystyle{\rm C}_{\hat{P}}\ +\ {\rm BI}_{\hat{P}}\,:\qquad$	$\displaystyle F^{ab}_{\hat{J}}\wedge e_{b}=0\,,$		(2.16)
		$\displaystyle{\rm C}_{\hat{D}}\ +\ {\rm BI}_{\hat{D}}\,:\qquad$	$\displaystyle F^{a}_{\hat{K}}\wedge e_{a}=0\,,$		(2.17)
		$\displaystyle{\rm BI}_{\hat{J}}\,:\qquad$	$\displaystyle{{\rm D}^{L}}\,F^{ab}_{\hat{J}}-2\,e^{[a}\wedge F_{\hat{K}}^{b]}=0\,,$		(2.18)
		$\displaystyle{\rm BI}_{\hat{K}}\,:\qquad$	$\displaystyle({{\rm D}^{L}}-b)\,F_{\hat{K}}^{a}-f_{b}\wedge F_{\hat{J}}^{ab}=0\,,$		(2.19)

where we implemented the constraints C${}_{\hat{P}}$ (2.14) and C${}_{\hat{D}}$ (2.15). Let us express

$$F_{\hat{J}\,ab}=\frac{1}{2}\,C_{ab,cd}\,e^{c}\wedge e^{d}\,,\qquad F_{\hat{K}\,a}=\frac{1}{2}\,C_{a,bc}\,e^{b}\wedge e^{c}\,.$$

(2.20)

Then $C_{ab,cd}$ and $C_{a,bc}$ are given in terms of $R_{cd,ab}=i_{a}\,i_{b}\,R_{cd}$ and $f_{b,a}=i_{a}\,f_{b}$ with $i_{a}:=i_{E_{a}}$ (or $R_{ab}=\frac{1}{2}\,R_{ab,cd}\,e^{c}\wedge e^{d}$ and $f_{a}=f_{a,b}\,e^{b}$) as

	$\displaystyle C_{ab,cd}$	$\displaystyle=$	$\displaystyle R_{ab,cd}-\eta_{ad}\,f_{b,c}+\eta_{bd}\,f_{a,c}+\eta_{ac}\,f_{b,d}-\eta_{cb}\,f_{a,d}\,,$		(2.21)
	$\displaystyle C_{a,bc}$	$\displaystyle=$	$\displaystyle 2\left({{\rm D}^{L}}_{[b|}f_{a,|c]}-2\,b_{[b|}\,f_{a,|c]}\right),$		(2.22)

where ${{\rm D}^{L}}=e^{a}\,{{\rm D}^{L}}_{a}$ . The first two identities, namely, the algebraic Bianchi identities are equivalent to

$${\rm C}_{\hat{P}}\ +\ {\rm BI}_{\hat{P}}\qquad\Longrightarrow\qquad C_{a[b,cd]}=0\,,$$

(2.23)

$${\rm C}_{\hat{D}}\ +\ {\rm BI}_{\hat{D}}\qquad\Longrightarrow\qquad C_{[ab,c]}=0\,,$$

(2.24)

so they are irreducible $GL_{d}$ tensors:

$$C_{ab,cd}\sim{\small\young(ac,bd)}\,,\qquad C_{c,ab}\sim{\small\young(ac,b)}\,.$$

(2.25)

Moreover if we impose the trace-free constraint on $C_{ab,cd}$ ,

$${\rm C}_{\hat{J}}\,:\qquad\eta^{ac}\,C_{ab,cd}\overset{!}{=}0\quad\Longleftrightarrow\quad i_{a}\,F^{ab}_{\hat{J}}\overset{!}{=}0\,,$$

(2.26)

the trace of the differential Bianchi identity (2.18) requires $C_{a,bc}$ to be trace-free as well:

$${\rm C}_{\hat{J}}\ +\ {\rm BI}_{\hat{J}}\qquad\Longrightarrow\qquad\eta^{ab}\,C_{a,bc}=0\,.$$

(2.27)

In fact, the constraints ${\rm C}_{\hat{D}}$ and ${\rm C}_{\hat{J}}$ are not independent, and the former is a consequence of the latter together with other constraints. To recapitulate, we impose the following set of constraints,¹¹1In parabolic geometry [19], these constraints are what define “normal connection”. See Section 7 for further discussions.

$$F^{a}_{\hat{P}}\overset{!}{=}0\,,\qquad i_{a}\,F^{ab}_{\hat{J}}\overset{!}{=}0\,,\qquad(F_{\hat{D}}\overset{!}{=}0),$$

(2.28)

and the resulting algebraic Bianchi identities are

$$C_{a[b,cd]}=0\,,\qquad C_{[a,bc]}=0\,,\qquad\eta^{ab}\,C_{a,bc}=0\,.$$

(2.29)

The differential Bianchi identities (2.18) and (2.19) read

	$\displaystyle({{\rm D}^{L}}-2\,b)_{[k}\,C^{ab,}{}_{cd]}-2\,\delta_{[k}^{[a}\,C^{b],}{}^{\phantom{]}}_{cd]}=0\,,$
	$\displaystyle({{\rm D}^{L}}-3\,b)_{[k}C^{a,}{}_{cd]}-\,f_{b,[k}\,C^{ab,}{}_{cd]}=0\,.$		(2.30)

The above is the starting point of the unfolding machinery. Before moving to that, let us review how the usual conformal geometry can be recovered from this setting.

All the constraints can be solved algebraically:

• •

  The constraint C${}_{\hat{P}}$ determines $\omega_{ab,c}=i_{c}\,\omega_{ab}$ (or $\omega_{ab}=\omega_{ab,c}\,e^{c}$) in terms of $e^{a}$ and $b$ as

  $$\omega_{ab,c}=E^{\mu}_{[b}\,E_{c]}^{\nu}\,\partial_{\mu}\,e_{a\nu}+E^{\mu}_{[c}\,E_{a]}^{\nu}\,\partial_{\mu}\,e_{b\nu}+E^{\mu}_{[b}\,E_{a]}^{\nu}\,\partial_{\mu}\,e_{c\nu}+2\,b_{[a}\,\eta_{b]c}\,.$$

  (2.31)

• •

  The constraint C${}_{\hat{D}}$ determines $f_{[a,b]}$ in terms of $b$ :

  $$f_{[a,b]}=\partial_{[a}\,b_{b]}\,,$$

  (2.32)

  where $\partial_{a}=E_{a}^{\mu}\,\partial_{\mu}=E_{a}$ .

• •

  The constraint C${}_{\hat{J}}$ determines $f_{(a,b)}$ in terms of $R_{ab}=R_{a}{}^{c}{}_{,bc}$ : from (2.21), we find

  $$f_{(a,b)}=\frac{1}{d-2}\left(R_{ab}-\frac{\eta_{ab}\,R}{2\,(d-1)}\right).$$

  (2.33)

After solving all the constraints, the $\mathfrak{so}(2,d)$-gauge symmetry reduces to²²2In fact, the constraints are not invariant under the gauge transformation (2.13) with the parameter $\epsilon^{a}$: $$\delta F^{a}_{\hat{P}}=F^{ab}_{\hat{J}}\,\epsilon_{b}\,,\qquad\delta(i_{a}\,F^{ab}_{\hat{J}})=i_{a}\,F^{[a}_{\hat{K}}\,\epsilon^{b]}\,,\qquad\delta F_{\hat{D}}=-F^{a}_{\hat{K}}\,\epsilon_{a}\,.$$ (2.34) However, the gauge symmetries can be properly modified by a “non-geometrical” curvature term [23] so as to leave all the constraints invariant. In fact, this modification naturally arises in the unfolded formulation. See Section 6.1 for the details.

	$\displaystyle\delta e^{a}=({{\rm D}^{L}}+b)\,\epsilon^{a}-\lambda^{a}{}_{b}\,e^{b}-\sigma\,e^{a}\,,$		(2.35)
	$\displaystyle\delta b={\rm d}\sigma-e^{a}\,\kappa_{a}+\epsilon^{a}\,f_{a}\,.$		(2.36)

We can fix the gauge symmetries associated with $\hat{K}_{a}$ and $\hat{J}_{ab}$ as follows.

• •

  The gauge symmetry of $\hat{K}_{a}$ allows us to set $b_{a}$ to zero:

  $$\delta_{\kappa}\,b_{a}=\kappa_{a}\qquad\Longrightarrow\qquad b_{a}=0\,.$$

  (2.37)

  Note that in this gauge, the $\hat{K}_{a}$ symmetry must transformation under the $\hat{D}$ symmetry with the parameter,

  $$\kappa_{a}=\partial_{a}\sigma\,.$$

  (2.38)

• •

  The gauge symmetry of $\hat{J}_{ab}$ is

  $$\delta_{\lambda}e^{a}_{\mu}=\lambda^{ab}\,e_{b\mu}\,,$$

  (2.39)

  which allows us to fix the degrees of freedom of $e^{a}_{\mu}$ besides those in

  $$g_{\mu\nu}=\eta_{ab}\,e^{a}_{\mu}\,e^{b}_{\nu}\,.$$

  (2.40)

The residual gauge symmetries are those of $\hat{D}$ and $\hat{P}$,

• •

  The gauge symmetry of $\hat{D}$ gives the Weyl rescaling,

  $$\delta_{\sigma}\,g_{\mu\nu}=2\,\sigma\,g_{\mu\nu}\,.$$

  (2.41)

• •

  The gauge symmetry of $\hat{P}$ gives the diffeomorphism,

  $$\delta_{\epsilon}\,g_{\mu\nu}=\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}\,,\qquad\epsilon_{\mu}=e^{a}_{\mu}\,\epsilon_{a}\,.$$

  (2.42)

After the reduction, we find that $R_{ab,cd}$ and $C_{ab,cd}$ coincide with the usual Riemann and Weyl tensor, and $P_{ab}=f_{(a,b)}$ and $C_{a,bc}=\nabla_{b}\,P_{ac}-\nabla_{c}\,P_{ab}$ with the Schouten and Cotton tensors.

Let us now consider the unfolding of conformal geometry. Remind that we have used the equations,

	$\displaystyle{\rm D}^{K}e^{a}=0\,,$
	$\displaystyle{\rm D}^{K}\omega^{ab}-2\,e^{[a}\wedge f^{b]}=\frac{1}{2}\,e_{c}\wedge e_{d}\,C^{ab,cd}\,,$
	$\displaystyle{\rm D}^{K}b+e^{a}\wedge f_{a}=0\,,$
	$\displaystyle{\rm D}^{K}f^{a}=\frac{1}{2}\,e_{b}\wedge e_{c}\,C^{a,bc}\,,$		(2.43)

with

$$C_{a[b,cd]}=0\,,\qquad C_{[a,bc]}=0\,,\qquad\eta^{ab}\,C_{ab,cd}=0\,,\qquad\eta^{ab}\,C_{a,bc}=0\,.$$

(2.44)

In the former set of equations, we slightly simplified the expressions by introducing $K=SO(1,1)\times SO(1,d-1)$ or $\mathfrak{k}=\mathfrak{so}(1,1)\oplus\mathfrak{so}(1,d-1)$³³3The maximal compact subalgebra of $\mathfrak{so}(2,d)$ is not $\mathfrak{so}(1,1)\oplus\mathfrak{so}(1,d-1)$ but $\mathfrak{so}(2)\oplus\mathfrak{so}(d)$. However, the two subalgebras are intimately related as we shall comment later in Section 4.1. covariant derivative ${\rm D}^{K}$, which acts on a $\mathfrak{so}(1,d-1)$-tensor with conformal dimension $\Delta$ as

$${\rm D}^{K}W^{[\Delta]ab\cdots}={\rm D}^{L}W^{[\Delta]ab\cdots}-\Delta\,b\,W^{[\Delta]ab\cdots}\,,$$

(2.45)

and assigning the conformal dimensions $\Delta=-1,0,0,1$ to $e^{a},\omega^{ab},b,f^{a}$, respectively.⁴⁴4Assigning the conformal dimension $-1$ to ${\rm d}x^{\mu}$, the fields $e^{a}_{\mu},\omega^{ab}_{\mu},b_{\mu},f^{a}_{\mu}$ have conformal dimensions $0,1,1,2$ identical to the numbers of derivatives of the corresponding fields in the metric formulation. Note here that the eigenvalue of the dilation operator $\hat{D}$ is $-\Delta$. This reflects that the fields $W^{[\Delta]ab\cdots}$ carry in fact a dual (or contragredient) representation which is obtained by the anti-involution $(\hat{P}^{a},\hat{J}^{ab},\hat{D},\hat{K}^{a})\to(\hat{P}^{a},-\hat{J}^{ab},-\hat{D},\hat{K}^{a})$ . See Section 4.2 for more discussions on this point.

In the following, we sketch the key reasoning of the unfolding scheme.

• •

  The system (2.43) can be regarded as a set of equations for one-forms $e^{a},\omega^{ab},b$ and $f^{a}$ as well as zero-forms $C^{ab,cd}$ and $C^{a,bc}$. Note that $C^{ab,cd}$ and $C^{a,bc}$ have conformal dimensions $\Delta=2$ and $3$, respectively. The zero-forms are completely, namely algebraically, determined by the equations, and hence no new degrees of freedom are introduced by employing them. About the one-forms, basically the equations tell how the (covariant) derivatives of the one-forms are determined by the other fields without any derivatives. Consequently, the one-forms are subject to certain conditions which are necessary for the system to be equivalent to conformal geometry.

• •

  Viewing (2.43) as a dynamical system for the associated fields, that is, the one-forms $e^{a},\omega^{ab},b$ and $f^{a}$ and the zero-forms $C^{ab,cd}$ and $C^{a,bc}$, it is more natural to introduce a new set of equations which determine the evolution—that is, the (covariant) derivatives—of $C^{ab,cd}$ and $C^{a,bc}$:

  	$\displaystyle{\rm D}^{K}C^{ab,cd}=({\rm D}^{L}-2\,b)C^{ab,cd}=e_{f}\,C^{ab,cd,e}+(\textrm{pre-existing fields})\,,$
  	$\displaystyle{\rm D}^{K}C^{a,bc}=({\rm D}^{L}-3\,b)C^{a,bc}=e_{d}\,C^{a,bc,d}+(\textrm{pre-existing fields})\,.$		(2.46)

  On the right hand side of the equations, we have introduced a new set of zero-form fields $C^{ab,cd,e}$ and $C^{a,bc,d}$ besides what can be expressed in terms of pre-existing fields. The new fields $C^{ab,cd,e}$ and $C^{a,bc,d}$ should be subject to a proper set of conditions so that the new equations (2.46) with the new fields neither introduce any new degrees of freedom nor remove any pre-existing degrees of freedom. For this, one need to examine the compatibility of the new equations (2.46) with the Bianchi identities (2.30).

• •

  Viewing (2.43) and (2.46) as a dynamical system for the one-forms $e^{a},\omega^{ab},b$ and $f^{a}$, and the zero-forms $C^{ab,cd},C^{ab,cd,e},C^{a,bc}$ and $C^{a,bc,d}$, we can again introduce ‘evolution equations’ for $C^{ab,cd,e}$ and $C^{a,bc,d}$ in a similar manner as we did for $C^{ab,cd}$ and $C^{a,bc}$ .

• •

  This procedure can be continued iteratively, and introduces infinitely many zero-form fields with infinitely many equations in a way that such an extension of the fields and equations does not alter the content of degrees of the freedom of the system.

In order to work with an infinite number of additional zero-form fields, we need to label them efficiently, and the subalgebra $\mathfrak{k}=\mathfrak{so}(1,1)\oplus\mathfrak{so}(1,d-1)$ can provide such a good label:⁵⁵5At this stage, it is not clear whether the $K$-label would be sufficient without necessitating an additional label to distinguish two fields of the same $K$-label. As we will show below, the $K$-label is sufficient to describe all the fields in the system. In other words, in the decomposition of the zero-form module into $K$-irreps, there is no multiplicty. in the following any zero-form fields will be labeled as traceless fiberwise tensors with two groups of totally symmetric indices,

$$C^{[\Delta]a_{1}\cdots a_{m},b_{1}\cdots b_{n}},$$

(2.47)

subject to the Young projection condition,

$$C^{[\Delta](a_{1}\cdots a_{m},b_{1})b_{2}\cdots b_{n}}=0\,.$$

(2.48)

In this way, the fiberwise tensor carries an irrep under $\mathfrak{so}(1,d-1)$ corresponding to a two-row Young diagram. We adopt the following common short-hand notation,

$$C^{[\Delta]a(m),b(n)}=C^{[\Delta]a_{1}\cdots a_{m},b_{1}\cdots b_{n}}\,.$$

(2.49)

Sometimes, it will be more convenient to use what we will refer to as “depth” $\delta$, than the conformal dimension $\Delta$ :

$$C^{\{\delta\}a(m),b(n)}=C^{[\Delta]a(m),b(n)}\,,\qquad\delta=\frac{\Delta-m+n}{2}\,.$$

(2.50)

The zero-form fields $C^{[2]a(2),b(2)}$ and $C^{[3]a(2),b}$ should be identified with the usual Weyl and Cotton tensors:

$$C^{[2]a(2),b(2)}=C^{(a_{1}|b_{1},|a_{2})b_{2}}\,,\qquad C^{[3]a(2),b}=C^{(a_{1},a_{2})b}\,.$$

(2.51)

If we relax the above identification condition, the zero-form equations that we will elaborate below has the capacity to describe a system of any integral spin.

The infinite amount of the $K$-covariant equations for zero-forms that we need to identify will take the following form,

$${\rm D}^{K}C^{[\Delta]a(m),b(n)}=e_{c}\,{\cal E}^{[\Delta+1]a(m),b(n)|c}(C)+f_{c}\,{\cal F}^{[\Delta-1]a(m),b(n)|c}(C)\,,$$

(2.52)

where ${\cal E}^{[\Delta+1]a(m),b(n)|c}(C)$ and ${\cal F}^{[\Delta-1]a(m),b(n)|c}(C)$ are functions of the zero-forms with total conformal weight $\Delta+1$ and $\Delta-1$. Let us recall that the conformal dimensions of $e^{a},f^{a}$ and ${\rm D}^{K}$ are $-1,1$ and 0. We can consider the Taylor expansion of ${\cal E}^{[\Delta+1]a(m),b(n)|c}(C)$,

	$\displaystyle{\cal E}^{[\Delta+1]a(m),b(n)|c}(C)={\cal E}^{[\Delta+1]a(m),b(n)|c}_{d(p),e(q)}\,C^{[\Delta+1]d(p),e(q)}$
	$\displaystyle\qquad+\sum_{\underset{\Delta_{1}+\Delta_{2}=\Delta+1}{\Delta_{1},\Delta_{2}}}{\cal E}^{[\Delta_{1},\Delta_{2}]a(m),b(n)|c}_{d(p),e(q)|f(s),g(t)}\,C^{[\Delta_{1}]d(p),e(q)}\,C^{[\Delta_{2}]f(s),g(t)}+\cdots\,,$		(2.53)

where the expansion coefficients ${\cal E}^{[\Delta+1]a(m),b(n)|c}_{d(p),e(q)}$ and ${\cal E}^{[\Delta_{1},\Delta_{2}]a(m),b(n)|c}_{d(p),e(q)|f(s),g(t)}$ are made only by Kronecker delta symbols so that they only rearrange or contract indices. The function ${\cal F}^{[\Delta-1]a(m),b(n)|c}(C)$ can be expanded analogously. Identifying the general form of ${\cal E}^{[\Delta+1]a(m),b(n)|c}(C)$ and ${\cal F}^{[\Delta-1]a(m),b(n)|c}(C)$ is a highly non-trivial task, and hence we first identify the linear parts, which will determine the content of the zero-forms. The identification of non-linear terms can be worked out, in principle, order by order in $\Delta$. Due to the boundness of $\Delta\geq 2$, ${\rm D}^{K}C^{[\Delta]a(m)b(n)}$ will involve at most $[(\Delta+1)/2]$ order terms.

Let us consider the system only up to the linear order,

$${\rm D}^{K}C^{[\Delta]a(m),b(n)}+e^{c}\,(\hat{P}_{c}\,C)^{[\Delta]a(m),b(n)}+f^{c}\,(\hat{K}_{c}\,C)^{[\Delta]a(m),b(n)}={\cal O}(C^{2})\,,$$

(3.1)

where we have used the notation,⁶⁶6Note that $({\cal O}\,T)^{a(m),b(n)}$ denotes the $a(m),b(n)$ components of the tensor ${\cal O}\,T$. Here, $T$ is not a tenor of type $(m,n)$ but ${\cal O}\,T$ is.

	$\displaystyle(\hat{P}^{c}\,C)^{[\Delta]a(m),b(n)}=-{\cal E}^{[\Delta+1]a(m),b(n)|c}_{d(p),e(q)}\,C^{[\Delta+1]d(p),e(q)}\,,$
	$\displaystyle(\hat{K}^{c}\,C)^{[\Delta]a(m),b(n)}=-{\cal F}^{[\Delta-1]a(m),b(n)|c}_{d(p),e(q)}\,C^{[\Delta-1]d(p),e(q)}\,.$		(3.2)

Remark that the linear terms of ${\cal E}^{[\Delta+1]a(m),b(n)|c}(C)$ and ${\cal F}^{[\Delta-1]a(m),b(n)|c}(C)$ are denoted by the actions of $\hat{P}^{c}$ and $\hat{K}^{c}$, respectively. It will become shortly clearly that they indeed correspond to the action of translation and special conformal transformation. Remind also that the action of $\hat{P}_{a}$ and $\hat{K}_{a}$ on the space of zero-form fields is not yet defined. The equation (3.1) suggests to combine the linear terms with the $K$-covariant derivative as

$${\rm D}^{G}C={\cal O}(C^{2})\,,$$

(3.3)

with

$${\rm D}^{G}={\rm D}^{K}+e^{a}\,\hat{P}_{a}+f^{a}\,\hat{K}_{a}={\rm d}+\omega^{ab}\,\hat{J}_{ab}+b\,\hat{D}+e^{a}\,\hat{P}_{a}+f^{a}\,\hat{K}_{a}\,.$$

(3.4)

The Bianchi identity is the consistency of the equation (3.3) associated with

	$\displaystyle({\rm D}^{G})^{2}$	$\displaystyle=$	$\displaystyle f^{a}\wedge e^{b}\,([\hat{K}_{a},\hat{P}_{b}]+\hat{J}_{ab}-\eta_{ab}\,\hat{D})$		(3.5)
			$\displaystyle+\,e^{a}\wedge e^{b}\,\hat{P}_{[a}\,\hat{P}_{b]}+f^{a}\wedge f^{b}\,\hat{K}_{[a}\,\hat{K}_{b]}+{\cal O}(C)\,.$

Since the action of $({\rm D}^{G})^{2}$ on $C$ is at least quadratic in $C$, the following should hold.

$$\begin{split}&(\hat{P}_{[c}\,\hat{P}_{d]}\,C)^{[\Delta]a(m),b(n)}=0\,,\qquad(\hat{K}_{[c}\,\hat{K}_{d]}\,C)^{[\Delta]a(m),b(n)}=0\,,\\ &\big{(}([\hat{K}_{a},\hat{P}_{b}]+\hat{J}_{ab}-\eta_{ab}\,\hat{D})\,C\big{)}^{[\Delta]a(m),b(n)}=0\,,\end{split}$$

(3.6)

and, we find that the consistency of the equation requires the operators $\hat{P}_{a}$ and $\hat{K}_{a}$ coincide with the actions of translation and special conformal transformation. Therefore, the linear part of ${\rm D}^{G}$ can be viewed as a $G=SO(2,d)$-covariant derivative.