# Integrating the generator of the infinitesimal special conformal transformation

(c.f Di Francesco, Conformal Field Theory chapters 2 and 4).

The expression for the full generator, $G_a$, of a transformation is $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta \omega_{a}} \partial_{\mu} \Phi - \frac{\delta F}{\delta \omega_a}$$ For an infinitesimal special conformal transformation (SCT), the coordinates transform like $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$$

If we now suppose the field transforms trivially under a SCT across the entire space, then $\delta F/\delta \omega_a = 0$.

Geometrically, a SCT comprises of a inversion, translation and then a further inversion. An inversion of a point in space just looks like a translation of the point. So the constant vector $b^{\mu}$ parametrises the SCT. Then $$\frac{\delta x^{\mu}}{\delta b^{\nu}} = \frac{\delta x^{\mu}}{\delta (x^{\rho}b_{\rho})} \frac{\delta (x^{\gamma}b_{\gamma})}{\delta b^{\nu}} = 2 x^{\mu}x_{\nu} - x^2 \delta_{\nu}^{\mu}.$$ Now moving on to my question: Di Francesco makes a point of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ I was wondering if somebody could point me to a link or explain the derivation. Is the reason for its non appearance due to complication or by being tedious?

I am also wondering how, from either of the infinitesimal or finite forms, we may express the SCT as $$\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu},$$ which is to say the SCT is an inversion $(1/x^2)$ a translation $-b^{\mu}$ and then a further inversion $(1/x'^2)$ which then gives $x'^{\mu}$, i.e the transformed coordinate.

• The expressions $\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu}$ and $x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$ are equivalent (just express $x'^2$ as a function of $x$), and the infinitesimal form is clearly $x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$ (because only quadratic terms in $x$ are allowed in infinitesimal conformal transformations) – Trimok Jun 27 '14 at 9:16

## 3 Answers

In order to determine the finite SCT from its infinitesimal version, we need to solve for the integral curves of the special conformal killing field $X$ defined by \begin{align} X(x) = 2(b\cdot x) x - x^2 b. \end{align} I explain why this is equivalent to "integrating" the infinitesimal transformation below. This means we need to solve the differential equation $X(x(t)) = \dot x(t)$ for the function $x$. Explicitly, this differential equation is \begin{align} \dot x = 2(b\cdot x) x - x^2 b. \end{align} This can be done with a trick, namely a certain change of variables. Define \begin{align} y = \frac{x}{x^2}, \end{align} then the resulting differential equation satisfied by $y$ becomes simple; \begin{align} \dot y = -b. \end{align} I urge you to perform the algebra yourself to confirm this. It's kind of magic that it works if you ask me, and the change of variables is precisely an inversion, so I think there's something deeper going on here, but I'm not sure what it is. The solution to this equation is simply $y = y_0 -tb$, so we find that the original function $x$ satisifes \begin{align} \frac{x}{x^2} = \frac{x_0}{x_0^2} - tb. \end{align} In other words, we've turned a monstrous nonlinear system of ODEs into a simple algebraic equation. In fact, one can show that the algebraic eqution $x/x^2 = A$ has the solution $x = A/A^2$, from which it follows that the solution to our original problem is \begin{align} x(t) = \frac{x_0 - x_0^2(tb)}{1-2x_0\cdot(tb) + x_0^2(tb)^2}, \end{align} as desired, since this is precisely the form of the "finite" SCT. Note that these only are local integral curves; the solution hits a singularity when $t$ is such that the denominator vanishes.

Why solve for integral curves?

If you're wondering what your original question has anything to do with solving the integral curves of the special conformal vector field I wrote down, then read on.

It helps to start with the concept of a flow.

Transforming points via flows.

Let a point $x\in\mathbb R^d$ be given, then for each $b\in\mathbb R^d$, we assume, at least in a neighborhood of that point, that there is a $\epsilon$-parameter family of transformations $\Phi_b(\epsilon):\mathbb R^d \to \mathbb R^d$ with $\epsilon\in [0,\bar\epsilon)$ such that $\Phi_b(\epsilon)(x)$ tells you what an SCT corresponding to the vector $b$ does to the point $x$ is you flow" in $\epsilon$. At $\epsilon = 0$, this flow just maps the point to itself; \begin{align} \Phi_b(0)(x) = x, \end{align} namely it starts at the identity. For $\epsilon >0$, the flow translates the point along a curve in $\mathbb R^d$. If you change $b$, then this corresponds to moving way from $x$ in a different initial direction under the flow.

Infinitesimal generator of a flow.

When we talk of an infinitesimal generator of such a transformation, we are talking about the term that generates the linear approximation for the flow in the parameter $\epsilon$. In other words, we expand \begin{align} \Phi_b(\epsilon)(x) = x + \epsilon G_b(x) + O(\epsilon^2), \end{align} and the vector field $G_b$ is called the infinitesimal generator of the flow. What you have pointed out in your question is that we know this infinitesimal generator; \begin{align} G_b(x) = 2(b\cdot x)x - x^2 b, \end{align} and we now want to reconstruct the whole flow simply by knowing this information corresponding to its linear approximation at every point. This is equivalent to solving some first order ordinary differential equations, which is why people often say we want to "integrate" the infinitesimal transformation to determine the finite one; integration is a perhaps somewhat archaic way of solving the corresponding differential equation.

Finding the whole flow given its generator.

Ok, so what differential equation do we solve? Well, note that the vector field $G_b$ is tangent to the curves generated by the flow by its very construction (we took a derivative with respect to $\epsilon$ with is the "velocity" of the curve generated by the flow), so the differential equation we want to solve is \begin{align} \dot x(\epsilon) = G_b(x(\epsilon)), \end{align} and we want to solve for $x(\epsilon)$. The solutions to this differential equation are referred to as integral curves of the vector field $G_b$.

Acknowledgements.

I didn't figure out the first part of this answer completely on my own. The idea for making the substitution $y=x/x^2$, which is really the crux of everything, came from here http://www.physicsforums.com/showthread.php?t=518316, namely from user Bill_K.

The idea for how to solve the algebraic equation $x/x^2 = A$ came from math.SE user @HansLundmark after I posted essentially your question in mathy language on math.SE here.

I should another math.SE user @Kirill solved for the integral curves in a totally different way in his answer to the question I posted.

Addendum.

How does one get $\dot y = -b$ from the change of variable $y=x/x^2$ as claimed in the first section? Well, let's calculate: \begin{align} \dot y &= \frac{x^2\dot x - x(2x\cdot \dot x)}{(x^2)^2} \\ &= \frac{x^2(2(b\cdot x) x - x^2 b) - x(2x\cdot (2(b\cdot x) x - x^2 b))}{(x^2)^2} \\ &= \frac{2x^2(b\cdot x) x - (x^2)^2b - 4x^2(b\cdot x) x + 2x^2(b\cdot x)x}{(x^2)^2} \\ &= -\frac{(x^2 )^2b}{(x^2)^2} \\ &= -b \end{align} Magic!

• Many thanks! I think I followed most of what you wrote - the addendums at the end about flow really clarified the argument for me. Can I check a couple of things? Does $\epsilon$ denote the parameter of the transformation? (I.e in this case, it would correspond to the vector $b^{\mu}$?) Also, the equation can be rewritten like $\dot{x}/x^2 = 2(b \cdot x)x/x^2 - b$ which means, to make contact with your result, $\dot{y} = \dot{x}/x^2 - 2(b \cdot x)y$. I see how the first term is recovered but not the second. Here is my working: (See next post) – CAF Jun 30 '14 at 9:59
• $$\frac{d}{dt} y^{\mu} = \left(\frac{d}{dt}x^{\mu} \right)\frac{1}{x^2} + x^{\mu} \frac{d}{dt}\left(\frac{1}{x^2}\right) = \frac{\dot{x}^{\mu}}{x^2} + x^{\mu}\frac{d}{dt} \left(\frac{1}{x^2}\right)$$ What did you do to the second term to get the $2(b\cdot x)y$ term? – CAF Jun 30 '14 at 10:00
• @CAF $\epsilon\geq 0$ is a real number. It is the flow parameter, but in my notation, it doesn't exactly correspond to $b$. $b$ tells you "in what direction" you're performing the transformation, the value of $\epsilon$ tells you how "large" the transformation is, namely how far along in the flow you are. If you want to get $\dot y = -b$, continue further with the explicit calculation you wrote down, then anywhere you see an $\dot x$, replace that with what it is equal to according to the original equation. A lot of cancellation will happen, and in the end only $-b$ will remain. – joshphysics Jun 30 '14 at 10:07

The two questions posed above are exactly the same and one may prove both simply by substituting $x'_\mu$ and calculating. If of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ then $$(x')^2 = \frac{(x_\mu-x^2 b_\mu)(x^\mu-x^2 b^\mu)}{(1-2xb+b^2x^2)^2}=\frac{x^2(1-2bx+x^2b^2) }{(1-2xb+b^2x^2)^2}=\frac{x^2}{1-2xb+b^2x^2}$$ and therefore $$\frac{x'}{(x')^2} = \frac{x-x^2b}{x^2}$$ (some indices were omitted in a self-explanatory way) which is exactly the second equation. So the two equations you are asking about are exactly equivalent.

The special conformal transformations may be uniquely defined as the conformal transformations that map the infinity to a different point and that map a line of the multiples of $b^\mu$ onto itself. The inversion is really needed to get the point at infinity to a finite place, so that it may be moved elsewhere by the translation in the middle.

If you use a different definition of the special conformal transformation than the definition "inverse times translations time inversion" or the definition from the previous paragraph, you would have to specify what's your definition is and one could easily show the equivalence with the definitions above, too.

• Many thanks for your answer. I can show that the infinitesimal form of the finite SCT is $x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$ by simply expanding the finite form for $|b^{\mu}| \ll |x^{\mu}|$. However, I don't really see how to exponentiate the infinitesimal form to actually get to the finite form. Thanks. – CAF Jun 27 '14 at 10:06
• Hi, a good question. If you prove that the finite form reduces to the infinitesimal in the limit $b^\mu\to 0$; and if you can prove the right composition rule for the finite transformations, then you have checked everything you need to check, right? It is in principle possible to exponentiate the generator "explicitly" in this case but in some cases in maths, we must simply guess the right result and prove that it is a solution and has the required properties. This should be enough for you here, too! In mathematics, there is no universal rule to constructively find the solution to every task. – Luboš Motl Jun 27 '14 at 15:16
• @LubošMotl: Thanks for the answer, I had not noticed before that multiplies of $b_{\mu}$ map to itself, so a SCT is a bit like rotating your Riemann sphere along the longitude defined by $k b_{\mu}$,until a point on that longitude gets mapped to the north pole ? – user7757 Jun 29 '14 at 10:34
• Absolutely. This conformal transformation is literally rotating the n-dimensional sphere, or its Lorentzian cousin, a hyperboloid. The rotation of the 2D sphere obtained by stereographic projection - see youtube.com/watch?v=JX3VmDgiFnY - is a special example of these special conformal transformations. The right amount of rotation is given by a condition like yours assuming that you meant it right. – Luboš Motl Jun 29 '14 at 11:42

However, I don't really see how to exponentiate the infinitesimal form to actually get to the finite form

In addition to the detailed answer of @LubošMotl, you may notice that conformal transformations transform light cones in light cones.

This means the following. Starting from a light cone :

$$(x'-a')^2 = x'^2-2x'.a' + a'^2=0\tag{1}$$

you must find that it is also a light cone for $x$, that is, it exist $a$ such as :

$$(x-a)^2 = x^2-2x.a + a^2=0\tag{2}$$

To do that, replace simply $x'$ by its value in function of $x$, by :

$$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2} \tag{3}$$

and, with some algebra, you will find that it is possible to exhibit $a(a',b)$ ($a$ as a function of $a'$ and $b$) , which satisties ($2$).:

$$a = \dfrac{a' + a'^2b}{1-2a'.b+a'^2b^2}\tag{4}$$

This means that the transformation $(3)$ is a conformal transformation whose infinitesimal expression is clearly: $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2 \tag{5}$$ (because only quadratic terms in x are allowed in infinitesimal conformal transformations)

Any other expression of the global special conformal transformation $(3)$, having infinitesimal expression $(5)$, will not transform light cones in light cones.

• Good, thanks. It may be written down instead of just saying that it's not necessary. ;-) – Luboš Motl Jun 27 '14 at 15:17