Dual Conformal Invariants | M.E. Irizarry-Gelpí

In this post I will discuss some aspects of dual conformal symmetry. I will assume that each external energy-momentum vector $p_{I}$ satisfies an on-shell constraint of the form

$$p_{I}^{2} = - m_{I}^{2}$$

and that the mass $m_{I}$ is positive and real. During some of the remarks I will take massless limits to illustrate some results.

Traditional Conformal Symmetry

Conformal symmetry is a large set of symmetries of spacetime. It includes translations, Lorentz transformations, dilatations, and conformal boosts. You might be more familiar with how conformal symmetry transformations act on the spacetime position vector.

Translations and Lorentz transformations are familiar; together they make Poincaré symmetry. Under a finite translation, the spacetime position vector of an event transforms as

$$x \longrightarrow x + a.$$

In order to obtain a translation-invariant quantity, you must use the spacetime position of two events and construct an interval vector. Under the translation $x \longrightarrow x + a$ and $y \longrightarrow y + a$ you can see that $x - y \longrightarrow x - y$. Thus, the simplest Poincaré-invariant is given by $(x - y)^{2}$.

Under a dilatation the spacetime position transforms as

$$x \longrightarrow \lambda x \qquad \lambda > 0$$

so an easy dilatation-invariant and Poincaré-invariant quantity requires three spacetime position vectors:

$$\frac{(x - y)^{2}}{(x - z)^{2}}.$$

It appears that the more symmetry you have, the more information you need to construct an invariant. However, the expression above is not invariant under the full conformal symmetry transformations.

In order to introduce the remaining conformal symmetry transformation, the conformal boost, it will be useful to introduce a discrete transformation that is not part of the conformal symmetry transformations: the inversion. Under an inversion, the spacetime position vector transforms as

$$x \longrightarrow \frac{1}{x^{2}} x.$$

It follows that

$$(x - y)^{2} \longrightarrow \frac{(x - y)^{2}}{x^{2} y^{2}}.$$

If you want to construct a quantity that is invariant under translations, Lorentz transformations, dilatations, and inversions, it appears that you need the spacetime position of four events in order to construct the ratio

$$\frac{(x - y)^{2} (z - w)^{2}}{(x - z)^{2} (y - w)^{2}}.$$

This quantity is called a conformal ratio. With the same four spacetime position vectors you can construct two other conformal ratios:

$$\frac{(x - y)^{2} (z - w)^{2}}{(x - w)^{2} (y - z)^{2}} \qquad \frac{(x - w)^{2} (y - z)^{2}}{(x - z)^{2} (y - w)^{2}}.$$

It is clear that the product of these two conformal ratios gives the first conformal ratio that I introduced earlier, so given four spacetime position vectors you can only construct two independent conformal ratios.

Without providing any further details, a conformal boost transformations can be done as a sequence of an inversion, a translation, and another inversion:

$$x \longrightarrow \frac{1}{x^{2}} x \longrightarrow \frac{1}{(x + b)^{2}} (x + b) \longrightarrow \frac{1}{1 + 2 (x \cdot b) + x^{2} b^{2}} (x + x^{2} b).$$

Since the conformal ratio is invariant under translations and inversions, it is actually invariant under all of the conformal symmetry transformations. Thus, the simplest conformal-invariant quantity requires four pieces of information. If you have a collection of $N$ points in spacetime, you have to count the number of ways of choosing four of them to make two conformal ratios. Thus, the number of independent conformal ratios is

$$\mathfrak{R}_{N} \equiv 2 \times {N \choose 4} = \frac{N (N - 1) (N - 2) (N - 3)}{12}, \qquad N \geq 4.$$

Note that $\mathfrak{R}_{4} = 2$ and $\mathfrak{R}_{6} = 30$.

Dual Conformal Symmetry

The origins of dual conformal symmetry are still murky. I am going to present it in a way that will make it sound as less of what it is. Dual conformal symmetry is related to even more symmetry: a Yangian symmetry, which is effectively an infinite amount of symmetry. Since I do not understand that stuff well, I am not going to say anything about it.

One way to understand dual conformal symmetry is to consider a scattering process with $N$ external quanta. The conservation constraint is

$$p_{1} + p_{2} + \ldots + p_{N} = 0.$$

This constraint can be satisfied explicitly by writting each energy-momentum vector as the difference of dual position vectors:

$$p_{1} = q_{2} - q_{1} \qquad p_{2} = q_{3} - q_{2} \qquad \cdots \qquad p_{N} = q_{1} - q_{N}.$$

After this "change of variables" from $p_{I}$ to $q_{I}$ you can consider conformal symmetry transformations acting on the dual position vectors. This conformal symmetry acts on momentum space (e.g. dilatation of the dual position vector); it is different from the traditional conformal symmetry (e.g. dilatation of the spacetime position vector). The dual conformal ratios constructed with the dual position vectors are invariant under dual conformal symmetry transformations.

However, the way that I introduced the dual position vectors above is not unique. If you think of the dual position vectors as describing the positions of the vertices of a polygon, then the edges of this polygon correspond to the energy-momentum vectors. Drawing the vertices in ascending order counter-clockwise, you can see that the energy-momentum vectors are arranged in a particular order (in the case above the order is ascending, i.e. $1 2 \ldots N$). There is nothing special about the particular order that I chose above, but choosing another order will change the explicit values of the dual conformal ratios. In other words, just like there are different scattering channels, there are also have different dual conformal channels. The number of dual conformal channels is the same as the number of cyclically-inequivalent permutations of $N$ integers (permutations that are not related by a cyclic rotation). This number is given by

$$\mathfrak{C}_{N} \equiv (N - 1)! \qquad N \geq 1.$$

Note that $\mathfrak{C}_{4} = 6$ and $\mathfrak{C}_{6} = 120$. As you will see below, only half of these dual conformal channels are interesting because the other half gives the same information due to reflection invariance.

Finding Cyclically-Inequivalent Permutations

An easy way of finding cyclically-inequivalent permutations is to construct them from smaller ones. You start with one element. There is only $1! = 1$ permutation. Then you add another element. With two elements you have $2! = 2$ possible permutations. These are $12$ and $21$. However, $21$ follows from $12$ by a cyclic rotation, so only one of these permutations is independent (these two permutations are also related by a reflection). For convenience I will pick $12$ as the representative of the cyclic class. One way to construct the permutation $12$ is to start with the one-element "subpermutation" $1$ and append the second element.

Then you add a third element. With three elements you have $3! = 6$ possible permutations. But it turns out that only two of these permutations are cyclically-inequivalent (you expect this, since $\mathfrak{C}_{3} = 2$). To construct these inequivalent permutations, you take the subpermutation $12$ and its cyclic rotation $21$, and append the third element yielding $123$ and $213$. Performing cyclic rotations on $123$ yields $231$ and $312$. Similarly, performing cyclic rotations on $213$ yields $132$ and $321$. Note that the two cyclic classes are related by a reflection.

Then you add a fourth element. With four elements you have $4! = 24$ possible permutations. You expect $\mathfrak{C}_{4} = 6$ cyclically-inequivalent permutations. To construct these, you append the fourth element to each of the permutations in the $123$ and $213$ cyclic classes:

$$1234 \qquad 2314 \qquad 3124,$$

and

$$2134 \qquad 1324 \qquad 3214.$$

Note that these six permutations can be divided into three pairs that are related by reflections (for example, $1234$ and $3214$).

This process can be repeated in order to obtain larger permutations.

Four External Quanta

In order to be concrete, I will now do some examples with explicit details. The simplest case is with four external quanta. Let me first consider the permutation $1234$. With this order of the external states, the dual position variables are introduced as

$$p_{1} = q_{2} - q_{1} \qquad p_{2} = q_{3} - q_{2} \qquad p_{3} = q_{4} - q_{3} \qquad p_{4} = q_{1} - q_{4}.$$

Other possible intervals are $q_{2} - q_{4} = p_{1} + p_{4}$ and $q_{3} - q_{1} = p_{1} + p_{2}$. Using the notation $q_{IJ} \equiv q_{I} - q_{J}$, the dual conformal ratios are given by

$$r_{1} = \frac{q_{21}^{2} q_{43}^{2}}{q_{24}^{2} q_{31}^{2}} = \frac{p_{1}^{2} p_{3}^{2}}{(p_{1} + p_{4})^{2} (p_{1} + p_{2})^{2}} \qquad r_{2} = \frac{q_{14}^{2} q_{32}^{2}}{q_{21}^{2} q_{43}^{2}} = \frac{p_{4}^{2} p_{2}^{2}}{p_{1}^{2} p_{3}^{2}} \qquad r_{3} = r_{1} r_{2} = \frac{p_{2}^{2} p_{4}^{2}}{(p_{1} + p_{4})^{2} (p_{1} + p_{2})^{2}}.$$

Writing the dual conformal ratios in terms of the masses and 2-Mandelstam invariants gives

$$r_{1} = \frac{m_{1}^{2} m_{3}^{2}}{s_{14} s_{12}} \qquad r_{2} = \frac{m_{2}^{2} m_{4}^{2}}{m_{1}^{2} m_{3}^{2}} \qquad r_{3} = \frac{m_{2}^{2} m_{4}^{2}}{s_{14} s_{12}}.$$

From a kinematical point of view, only one of these dual conformal ratios is relevant, since $r_{2}$ involves only masses, and $r_{3}$ involves the same 2-Mandelstam invariants as $r_{1}$. Also, note that $r_{1}$ involves the masses in the numerator, so when working with massless states the dual conformal ratio is fixed to vanish.

The dual conformal invariants for other cyclic classes can be obtained by permuting the labels. For example, in order to go from $1234$ to $2314$, you need the replacement

$$1 \longrightarrow 2 \qquad 2 \longrightarrow 3 \qquad 3 \longrightarrow 1 \qquad 4 \longrightarrow 4$$

Thus, the dual conformal ratios become

$$r_{1} \longrightarrow \frac{m_{2}^{2} m_{1}^{2}}{s_{13} s_{14}} \qquad r_{2} \longrightarrow \frac{m_{3}^{2} m_{4}^{2}}{m_{2}^{2} m_{1}^{2}} \qquad r_{3} \longrightarrow \frac{m_{3}^{2} m_{4}^{2}}{s_{13} s_{14}},$$

where I have used $s_{24} = s_{13}$ and $s_{23} = s_{14}$ which follow from the conservation constraint.

Similarly, in order to go from $1234$ to $3124$, I need the replacement

$$1 \longrightarrow 3 \qquad 2 \longrightarrow 1 \qquad 3 \longrightarrow 2 \qquad 4 \longrightarrow 4$$

Thus, the dual conformal ratios become

$$r_{1} \longrightarrow \frac{m_{3}^{2} m_{2}^{2}}{s_{12} s_{13}} \qquad r_{2} \longrightarrow \frac{m_{1}^{2} m_{4}^{2}}{m_{3}^{2} m_{2}^{2}} \qquad r_{3} \longrightarrow \frac{m_{1}^{2} m_{4}^{2}}{s_{12} s_{13}},$$

where I have used $s_{34} = s_{12}$ which follows from the conservation constraint.

Six External Quanta

As the number of external quanta increases, a new feature arises. When $N = 6$ you have $\mathfrak{R}_{6} = 30$ and $\mathfrak{C}_{6} = 120$. There is very little benefit to listing the representatives of all cyclic classes, so for concreteness I will work with the order $123456$. The dual position vectors are introduced via

$$p_{1} = q_{21} \qquad p_{2} = q_{32} \qquad p_{3} = q_{43} \qquad p_{4} = q_{54} \qquad p_{5} = q_{65} \qquad p_{6} = q_{16},$$

where $q_{IJ} \equiv q_{I} - q_{J}$. Other possible intervals are

$$p_{1} + p_{2} = q_{31} \qquad p_{5} + p_{6} = q_{15} \qquad p_{2} + p_{3} = q_{42} \qquad p_{1} + p_{6} = q_{26} \qquad p_{3} + p_{4} = q_{53} \qquad p_{4} + p_{5} = q_{64},$$

and

$$p_{1} + p_{2} + p_{3} = q_{41} \qquad p_{1} + p_{5} + p_{6} = q_{25} \qquad p_{1} + p_{2} + p_{6} = q_{36}.$$

There are fifteen ways of choosing four elements from a set of six:

$$1234 \quad 1235 \quad 1236 \quad 1245 \quad 1246 \quad 1256 \quad 1345 \quad 1346 \quad 1356 \quad 1456 \quad 2345 \quad 2346 \quad 2356 \quad 2456 \quad 3456.$$

First, consider the quartet $1234$. The dual conformal ratios are

$$\frac{q_{21}^{2} q_{43}^{2}}{q_{31}^{2} q_{42}^{2}} = \frac{m_{1}^{2} m_{3}^{2}}{s_{12} s_{23}} \qquad \frac{q_{41}^{2} q_{32}^{2}}{q_{31}^{2} q_{42}^{2}} = \frac{s_{123} m_{2}^{2}}{s_{12} s_{23}} \qquad \frac{q_{21}^{2} q_{43}^{2}}{q_{41}^{2} q_{32}^{2}} = \frac{m_{1}^{2} m_{3}^{2}}{s_{123} m_{2}^{2}}.$$

You find the familiar dual conformal ratio with two masses and two Mandelstam invariants, along with dual conformal ratios with three masses and one Mandelstam invariant, as well as with one mass and three Mandelstam invariants. Five other quartets are found after a cyclic shift:

$$1234 \longrightarrow 2345 \longrightarrow 3456 \longrightarrow 1456 \longrightarrow 1256 \longrightarrow 1236. $$

Next, consider the quartet $1235$. The dual conformal ratios are

$$\frac{q_{21}^{2} q_{53}^{2}}{q_{31}^{2} q_{25}^{2}} = \frac{m_{1}^{2} s_{34}}{s_{12} s_{156}} \qquad \frac{q_{15}^{2} q_{32}^{2}}{q_{41}^{2} q_{25}^{2}} = \frac{s_{56} m_{2}^{2}}{s_{12} s_{156}} \qquad \frac{q_{21}^{2} q_{53}^{2}}{q_{15}^{2} q_{32}^{2}} = \frac{m_{1}^{2} s_{34}}{s_{56} m_{2}^{2}}.$$

One new feature is the appearance of dual conformal ratios with one mass in both the numerator and denominator. Five other quartets are found after a cyclic shift:

$$1235 \longrightarrow 2346 \longrightarrow 1345 \longrightarrow 2456 \longrightarrow 1356 \longrightarrow 1246. $$

Finally, consider the quartet $1245$. The dual conformal ratios are

$$\frac{q_{21}^{2} q_{54}^{2}}{q_{41}^{2} q_{25}^{2}} = \frac{m_{1}^{2} m_{4}^{2}}{s_{123} s_{156}} \qquad \frac{q_{21}^{2} q_{54}^{2}}{q_{15}^{2} q_{42}^{2}} = \frac{m_{1}^{2} m_{4}^{2}}{s_{56} s_{23}} \qquad \frac{q_{15}^{2} q_{42}^{2}}{q_{41}^{2} q_{25}^{2}} = \frac{s_{56} s_{23}}{s_{123} s_{156}}.$$

Two other quartets are found after a cyclic shift. One of them is $2356$, with dual conformal ratios

$$\frac{q_{32}^{2} q_{65}^{2}}{q_{53}^{2} q_{26}^{2}} = \frac{m_{2}^{2} m_{5}^{2}}{s_{34} s_{16}} \qquad \frac{q_{32}^{2} q_{65}^{2}}{q_{25}^{2} q_{36}^{2}} = \frac{m_{2}^{2} m_{5}^{2}}{s_{156} s_{126}} \qquad \frac{q_{53}^{2} q_{26}^{2}}{q_{25}^{2} q_{36}^{2}} = \frac{s_{34} s_{16}}{s_{156} s_{126}}.$$

The other one is $1346$, with dual conformal ratios

$$\frac{q_{16}^{2} q_{43}^{2}}{q_{31}^{2} q_{64}^{2}} = \frac{m_{6}^{2} m_{3}^{2}}{s_{12} s_{45}} \qquad \frac{q_{16}^{2} q_{43}^{2}}{q_{36}^{2} q_{41}^{2}} = \frac{m_{6}^{2} m_{3}^{2}}{s_{126} s_{123}} \qquad \frac{q_{31}^{2} q_{64}^{2}}{q_{36}^{2} q_{41}^{2}} = \frac{s_{12} s_{45}}{s_{126} s_{123}}.$$

A feature shared by these three results is the appearance of dual conformal ratios that do not involve any of the masses. Thus, in the massless limit there will be nontrivial dual conformal invariants.