In this thread Integrating the generator of the infinitesimal special conformal transformation, the generator of the 'flow' of the transformation is written as $$G_b = 2(b \cdot x)x – x^2 b,$$ where $b$ parametrises the SCT. Now, we know that under an infinitesimal SCT, the coordinates transform like so $$x'^{\mu} = x^{\mu} + 2(b \cdot x)x – x^2b.$$ So, in the same vein that, under a translation $x'^{\mu} = x^{\mu} + a^{\mu}$ we don't speak of $a^{\mu}$ being a generator but rather a parameter, what merits the calling of $2(b \cdot x)x – x^2b$ a generator?
[Physics] Generator of the Special Conformal Transformation
conformal-field-theorylie-algebra
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The special conformal transformations as well as the translations, dilations and rotations are all continuously connected to the identity. This means that they contain parameters such that at some particular value the trasformation becomes trivial. For example, for $b=0$ the special conformal trasformation you write is simply $x_i\mapsto x_i$. The inversion map $$x_i \mapsto \frac{x_i}{x^2},$$ on the other hand, is not connected to the identity.
We can also note that an inversion changes the orientation of the space, while the other conformal transformations preserve the orientation. In $D$ (Euclidean) dimensions the special conformal transformations, translations, dilations and rotations together form the Lie group $SO(D+1,1)$. This is what, at least in physics, is normally referred to as "the conformal group". If we also allow for inversions this group is extended to $O(D+1,1)$.
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!
Best Answer
General remarks on flows and their generators.
Let an $\epsilon$-parameter flow $\Phi(\epsilon):\mathbb R^d\to \mathbb R^d$ be given. Let the flow be defined on some $\epsilon$-neighborhood containing $0$. Provided the flow is sufficiently smooth, we can expand the flow in the parameter $\epsilon$; \begin{align} \Phi(\epsilon) = \Phi(0) + \epsilon\Phi^{(1)} + O(\epsilon^2). \end{align} If the flow "starts at the identity," namely if $\Phi(0) = \mathrm{id}$, where $\mathrm{id}$ is the identity on $\mathbb R^d$, then $\Phi(0) = \mathrm{id}$. In addition, in anticipation of the terminology to come, we define $G = \Phi^{(1)}$, namely $G$ is just the function that when multiplied by $\epsilon$ gives the first order change caused by the flow. So we have \begin{align} \Phi(\epsilon) = \mathrm{id} + \epsilon G + O(\epsilon^2). \end{align} At this point, one usually defines $G$ as the generator (or infinitesimal generator depending on who you talk to ) of the flow, but of course we gain no insight into why it's called that. To see why this first order coefficient is called a generator, consider some point $x_0\in \mathbb R^d$, and suppose that we apply the flow to $x_0$, then as $\epsilon$ increases from $0$ to some $\epsilon >0$, the point $x_0$ travels along a curve, $\gamma$ defined by \begin{align} \gamma(\epsilon) = \Phi(\epsilon)(x_0). \end{align} Now consider taking the derivative of $\gamma$ with respect to $\epsilon$ at $\epsilon = 0$; \begin{align} \dot\gamma(0) = \frac{d}{d\epsilon}\Phi(\epsilon)(x_0)\Big|_{\epsilon=0} = \frac{d}{d\epsilon} \big(x_0 + \epsilon G(x_0) + O(\epsilon^2)\big)\Big|_{\epsilon=0} = G(x_0) \end{align} but recall also that $\gamma(0) = x_0$, so we find that \begin{align} \dot \gamma(0) = G(\gamma(0)). \end{align} Since the derivative $\dot\gamma(0)$ is just the tangent vector to $\gamma$ at zero, this equation means that the tangent vector of $\gamma$ at zero agrees with $G$, the generator (a vector field) at $\gamma(0)$. Actually, we can say something much stronger than this. Notice that we can compute the derivative of $\gamma$ at any parameter value $\epsilon$, not just at $\epsilon = 0$, as follows: \begin{align} \dot \gamma(\epsilon) &= \frac{d}{dt} \gamma(\epsilon+t)\Big|_{t=0} \\ &= \frac{d}{dt} \Phi(\epsilon + t)(x_0) \Big|_{t=0} \\ &= \frac{d}{dt} \Phi(t)(\Phi(\epsilon)(x_0)) \Big|_{t=0} \\ &= \frac{d}{dt} \Big(\Phi(\epsilon)(x_0) + t G(\Phi(\epsilon)(x_0)) + O(t^2)\Big) \Big|_{t=0} \\ &= G(\Phi(\epsilon)(x_0)), \end{align} where we have used the property $\Phi(t+s) = \Phi(t)\circ \Phi(s)$ of flows. But $\Phi(\epsilon)(x_0)$ is precisely $\gamma(\epsilon)$! So we get \begin{align} \dot\gamma(\epsilon) = G(\gamma(\epsilon)). \tag{$\star$} \end{align} In other words
This is a very powerful fact, because it tells us that if we are given the infinitesimal generator of a flow (which is a vector field), then we can reconstruct the entire flow by solving $(\star)$, a first order system or ordinary differential equations!
This also explains the terminology "infinitesimal generator" when referring to $G$. It is "infinitesimal" because it tells us how the flow behaves to first order, which is a good approximation when $\epsilon$ is small, and it is a "generator" in the sense that the flow can be reconstructed from it by solving the system $(\star)$.
Example. Special conformal transformation
Recall that in the other physics.SE post you refer to in your question:
Integrating the generator of the infinitesimal special conformal transformation
we saw that the special conformal flow is given by \begin{align} \Phi_b(\epsilon)(x) = \frac{x - x^2 (\epsilon b)}{1-2x\cdot (\epsilon b) + x^2(\epsilon b)^2}. \end{align} (although there we used $t$ instead of $\epsilon$ as the flow parameter. If we Taylor expand this around $\epsilon = 0$, then we find that \begin{align} \Phi_b(\epsilon)(x) = x + \epsilon(2(x\cdot b)x-x^2b) + O(\epsilon^2), \end{align} so we can immediately identify the infinitesimal generator of this flow as \begin{align} G_b(x) = 2(x\cdot b)x-x^2b. \end{align}
How does the SCT act on fields?
Consider a scalar field $\phi:\mathbb R^d\to\mathbb R^d$. Suppose that $\phi$ is such that the action of a conformal transformation $f$ on $\phi$ is \begin{align} \phi_f(x) = \phi(f^{-1} x) \tag{$\star\star$}. \end{align} Then we can ask the following question:
In other words, we are asking what $\phi_f$ is to first order in $\epsilon$ when $f = \Phi_b(\epsilon)$. Well, let's calculate: \begin{align} \phi_{\Phi_b(\epsilon)}(x) &= \phi(\Phi_b(-\epsilon)(x)) \\ &= \phi(x - \epsilon G_b(x) + O(\epsilon^2)) \\ &= \phi(x) - \epsilon (G_b)(x)\cdot\partial_\mu\phi(x) + O(\epsilon^2) \\ &= \Big(\mathrm{id} - \epsilon G_b(x)\cdot \partial+ O(\epsilon)^2\Big)\phi(x) \end{align} which shows that for a given $b$, the differential operator \begin{align} -G_b(x) \cdot \partial \end{align} is the infinitesimal generator of the action of a special conformal transformation on such scalar fields as opposed to on points in $\mathbb R^d$. You'll see that this agrees with eq.4.18 in Di Francesco since in that book, as is often conventional, he strips off the $b$-dependence and adds a factor of $i$ when defining the infinitesimal generator, namely he defines the infinitesimal generators $K_\mu$ such that \begin{align} ib^\mu K_\mu = - G_b^\mu(x) \cdot \partial. \end{align} Ok so this object is just the infinitesimal generator for the action of SCTs on spinless fields that obey $(\star\star)$ (or as Di Francesco writes it, fields for which $\mathcal F(\Phi) = \Phi$), namely there is no-nontrivial target space transformation that happens when such a field transforms.
However, if the field has spin, then there will by definition be a non-trivial target space transformation in which case the differential operator that represents SCTs on such fields picks up the following extra terms: \begin{align} \kappa_\mu +2 x_\mu \tilde\Delta - x^\nu S_{\mu\nu}. \end{align}