As is well known Minkowski spacetime (which is four dimensional vector space with scalar product $\eta _{\mu \nu}$ of signature $-+++$) is maximally symmetric, which manifests itself in presence of ten Killing vector fields. Those are generators of one parameter groups of isometries, which can be understood as translations, boosts (hyperbolic rotations) and Euclidean rotations. There are also five conformal Killing vectors fields, which are generators of one parameter groups of conformal transformations. One of those transformations is simply scaling whole space. However there are four others, with generators:

$K _ {\mu} = 2 \eta _{\mu \alpha} x^{\alpha} x^{\nu} \partial _{\nu} - x^{\alpha} x_{\alpha} \partial _{\mu}$

With standard identification of tangent vectors and partial derivatives understood in this formula. My question is what are those transformations? Can I visualize them or explicitly find those groups? What is their interpretation or significance? Any insight will be appreciated.


Let us first look at the 1-dimensional complex case. In the Riemann sphere, you can define a conjugate-addition, $| : \hat{\mathbb{C}} \times \hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$, by

$$ \frac1x + \frac1y = \frac1{x | y} $$

In explicit form,

$$ x | y = \frac{xy}{x+y} $$

and $\infty | \infty = \infty$. It's easy to prove that conjugate-addition behaves exactly like addition: it is commutative, associative, distributive with respect to multiplication, and has an identity $\frac10 = \infty$ (and zero is now the absorbing element). This operation may already be familiar to you, it is for example implicitly used in defining the harmonic mean and when summing resistors in parallel.

Geometrically, what we really are doing is interchanging the behavior of the Northern and Southern hemispheres of the Riemann sphere with respect to addition, by means of the Möbius transformation $ x \rightarrow 1/x $. So you can imagine it as "addition from the point of view of $\infty$ ", if you take ordinary addition as being from the point of view of $0$.

In higher dimensions you can use spherical inversion to do the same thing, defining:

$$ \frac{x^{\mu}}{x^2} + \frac{y^{\mu}}{y^2} = \frac{x^{\mu} | y^{\mu}}{(x|y)^2} $$

It is a bit more tricky, but you can also get the explicit form from the definition. First, take the scalar product of the defining equation with itself:

$$ \left (\frac{x^{\mu}}{x^2} + \frac{y^{\mu}}{y^2} \right)^2 = \frac1{x^2} + \frac1{y^2} + 2 \frac{x^{\mu} y_{\mu}}{x^2 y^2} = \frac1{(x|y)^2} $$

Inverting we have

$$ (x|y)^2 = \frac{x^2 y^2}{x^2 + y^2 + 2 x^{\mu} y_{\mu}} $$

And substituting this back into the defining equation, we have finally:

$$x^{\mu} | y^{\mu} = (x|y)^2 \left( \frac{x^{\mu}}{x^2} + \frac{y^{\mu}}{y^2} \right) = \frac{x^2 y^{\mu} + y^2 x^{\mu}}{x^2 + y^2 + 2 x^{\mu} y_{\mu}} $$

Then, the special conformal generators $b^{\mu} K_{\mu}$ are just the generators of conjugate-translations:

$$x^{\mu} \rightarrow x^{\mu} | \left( - \frac{b^{\mu}}{b^2} \right) = \frac{x^{\mu} - x^2 b^{\mu}}{1 + b^2 x^2 - 2 b^{\mu} x_{\mu}} $$

The interpretation is the same, we are interchanging the behavior of the two "hemispheres" in the compactified Minkowski spacetime, and performing addition "with respect to the point at infinity".

With that image in mind, we can get the explicit form of $K_{\mu}$ by expanding

$$\frac{\partial}{\partial(x^{\mu}/x^2)} $$

using the chain rule.

  • $\begingroup$ This looks exactly like the thing I'm looking for, but I have trouble understanding definition of spherical inversion that you give. It seems like there is both $x^{\mu} | y ^{\mu}$ and square of the vectors on RHS. How do I extract the $x|y$ vector itself from that? $\endgroup$ – Blazej Sep 10 '15 at 12:16
  • $\begingroup$ @Blazej I updated the answer with the derivation. Spherical inversion is the multidimensional analogue of inversion: the spherical inverse of a vector $x^{\mu}$ is a vector $x^{\mu} /x^2$ whose scalar product with the original vector gives 1. $\endgroup$ – pregunton Sep 10 '15 at 12:40
  • $\begingroup$ This is a great answer, thanks! It seems like this gets slightly more complicated in Minkowski spacetime since set of points that "go to infinity" under inversion is now whole light cone rather than a single point. I remember seeing something about inversions and stereographic projections in Minkowski space in Penrose's book, I will need to look into it. $\endgroup$ – Blazej Sep 10 '15 at 14:21

This is all taken from Conformal field theory by Francesco, Mathieu, and Senechal. A conformal symmetry, $\phi^*g_{\mu\nu}=\Lambda^2g_{\mu\nu}$, implies $\mathit L _{\epsilon}g_{\mu\nu}=f(x)g_{\mu\nu} $, so, for a Minkowski metric $$\partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu=f(x)\eta_{\mu\nu}.$$ We can apply another partial derivative, permute indices and obtain $$2\partial_{\mu}\partial_\nu\epsilon_\rho=\eta_{\mu\rho}\partial_\nu f+\eta_{\nu\rho}\partial_\mu f-\eta_{\mu\nu}\partial_\rho f$$ which we can contract with the inverse metric: $$2\partial^2\epsilon_\mu=(2-d)\partial_\mu f$$ so that, appying another partial derivative and using the first equation, $$(2-d)\partial_\mu\partial_\nu f = \eta_{\mu\nu}\partial^2f.$$ Contracting: $(d-1)\partial^2f=0$. Therefore, $f(x)=A+B_\mu x^\mu$, and by the second equation, $\epsilon$ is quadratic, $\epsilon_\mu = a_\mu + b_{\mu\nu} x^\nu + c_{\mu\nu\rho} x^\nu x^\rho$. The constant amounts to an infinitesimal translation, whereas substituting the linear term into the first equation gives a scale transformation and an infinitesimal rotation. Substituting the quadratic part into the second equation gives $$c_{\mu\nu\rho}=\eta_{\mu\rho}\beta_\nu+\eta_{\mu\nu}\beta_\rho-\eta_{\nu\rho}\beta_\mu, \:\beta_\mu:=\frac{1}{d}{c^{\lambda}}_{\lambda\mu}$$ which means that this corresponds to $$x^\mu\mapsto x^\mu + 2 (x\cdot\beta) x^\mu - \beta^{\mu} x^2$$

This is the infinitesimal transformation corresponding to a special conformal transformation, $$x'^\mu = \frac{x^\mu-\beta^\mu x^2}{1 - 2 x\cdot\beta - \beta^2 x^2}.$$

You can easily verify that this can be written as: $$\frac{x'^\mu}{x'^2}=\frac{x^\mu}{x^2}-\beta^\mu$$ so these transformations are an inversion ($x^\mu\mapsto x^\mu/x^2$), followed by a translation, and another inversion.

  • $\begingroup$ I don't think it's true that $\partial ^2 f = 0$ implies the function is linear, although it clearly is a solution. In fact this is just another notation for wave equation, isn't it. This is still very nice answer, I can only answer one answer though. $\endgroup$ – Blazej Sep 10 '15 at 14:25
  • $\begingroup$ The contraction gives you $\partial^2 f = 0$, and you can then sub it into $(2-d) \partial_\mu \partial_\nu f = \eta_{\mu \nu}\partial^2 f$ to get $\partial_\mu \partial_\nu f = 0$, which does imply the function is linear. (giving very delayed clarification because this thread comes up on a reasonable Google search!) $\endgroup$ – Indivicivet Feb 19 '18 at 2:44

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