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This is a continuation of Classification of conformal diffeomorphisms of Minkowski space

Consider $\mathbb{R}^{n+1}$ equipped with the Minkowski (sign indefinite) metric: $$g=(x^0)^2-(x^1)^2-\dots -(x^n)^2.$$

Is there a classification of diffeomorphisms $F\colon \mathbb{R}^{n+1}\tilde\to \mathbb{R}^{n+1}$ with the property $F^*g=a\cdot g$, where $a$ is a function?

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    $\begingroup$ If $n=1$, let $u=x^0+x^1$, $v=x^0-x^1$. Then the maps $(U,V)=(U(u),V(v))$, for any diffeomorphisms $U(u)$ and $V(v)$ of the real number line, are conformal, an infinite dimensional group. For $n>1$, the group is finite dimensional, as this is a Klein geometry, and consists of those conformal maps of the conformal compactification which preserve this affine open set. But I will have to think of a good reference. $\endgroup$
    – Ben McKay
    Commented Sep 8, 2020 at 18:59
  • $\begingroup$ @BenMcKay Maybe something is in the book on Cartan geometries by Sharpe? $\endgroup$
    – Vít Tuček
    Commented Sep 8, 2020 at 19:28
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    $\begingroup$ My impression is that the answer is that, for dimensions $p+q\geq 3$, strictly speaking every global conformal transformation of $\mathbb R^{p,q}$ is a similarity transformation -- you need to compactify for the "interesting" conformal transformations to be globally defined. A reference I was told has this information is Schottenloher, "A mathematical introduction to conformal field theory". Once you compactify, the global conformal group for $\mathbb R^{p,q}$ is supposed to be $SO(p+1,q+1)$. $\endgroup$
    – Tim Campion
    Commented Sep 8, 2020 at 20:57

2 Answers 2

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  1. For every $n\ge 2$, each conformal transformation of ${\mathbb R}^{1,n}$ has constant conformal factor; in other words, it is an affine transformation of the form ${\mathbf x}\mapsto aU{\mathbf x} +{\mathbf b}$, where $a>0$, $U\in O(1,n)$, ${\mathbf b}\in {\mathbb R}^{n+1}$.

  2. When $n=1$, there are more conformal transformations; after your rotate the coordinates so that the invariant bilinear form becomes $xy$, the group $Conf_+({\mathbb R}^{1,1})$ (of orientation-preserving conformal diffeomorphisms) consists exactly of diffeomorphisms of the form $(x,y)\mapsto (f(x),g(y))$, where $f, g$ are diffeomorphisms ${\mathbb R}\to {\mathbb R}$ and they either both preserve or both reverse orientation.

  3. The case $n=0$, I hope, is clear.

For details, see:

Martin Schottenloher, The conformal group, chapter 2 of "A mathematical introduction to conformal field theory," 2008.

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See my answer to the conformal group of $S^n$ where you can find explanation for the the positive definite case and a link to a paper that actually works in arbitrary signature.

Here is a short summary of the argument: Rewrite the problem using definition of the pullback as a system of PDEs for $F$. Differentiate these equations and take clever sums. Use the original system to keep only derivatives which are not determined by the original equations. Repeat. Lo and behold! You obtained no more "undetermined" derivatives! It then follows that the solution space is finite-dimensional and you can write it's basis explicitly.

It's roughly a one page calculation plus half a page of text. For details, see the article by Slovák that I link to in my previous answer.

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  • $\begingroup$ Do you have a more direct and explicit answer? Thank you. $\endgroup$
    – asv
    Commented Sep 8, 2020 at 19:11
  • $\begingroup$ @makt This argument is as direct as it gets. Expand the definitions. Calculate. Calculate some more. Write down all the solutions. The end. It's all pretty explicit in the article by Slovák. $\endgroup$
    – Vít Tuček
    Commented Sep 8, 2020 at 19:27
  • $\begingroup$ The Liouville theorem 5.4 in Slovak's article gives the conformal group by giving a list of generators. I believe makt would like a parametrization of the group, e.g. the isometry group is more precisely described as mappings "Ax+b" rather than "the group generated by rotations and translations" $\endgroup$
    – Quarto Bendir
    Commented Sep 8, 2020 at 19:46

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