Are the Euclidean plane and the Minkowski plane conformally equivalent?

Question

Are the Euclidean plane $R^{2,0}$ and the Minkowski plane $R^{1,1}$ conformally equivalent?

Answer 1

This is just linear algebra.

In a coordinate-free form, a conformal linear map between two $n$-dimensional vector spaces $V_1, V_2$ equipped with nondegenerate symmetric bilinear forms $Q_1, Q_2$ is a linear isomorphism $$ A: V_1\to V_2 $$ such that there exists a positive (this is negotiable) real number $\lambda$ such that for every pair of vectors $u, v\in V_1$, we have $$ Q_2(Au, Av)= \lambda Q_1(u, v). $$ Therefore, replacing $A$ with the linear map $$ B= \sqrt{\lambda}^{-1} A, $$ we have $$ Q_2(Bu, Bv)= Q_1(u, v). $$ Thus, the existence of a conformal linear map implies existence of an isometric linear map. Recall that the complete isometry invariant of a nondegenerate bilinear form $Q$ is its rank $n$ (the dimension of the vector space) and signature i.e. the pair $(p,q)$ satisfying $p+q=n$, such that in a suitable orthogonal basis $p$ of the basis vectors $v_i$ satisfy $Q(v_i,v_i)=1$, and $q$ of the basis vector satisfy $Q(v_i,v_j)=-1$. This is discussed in any advanced linear algebra book (say, S.Lang, "Linear Algebra"). You can use the ones listed by wikipedia as alternatives.

Now, by definition, $R^{p,q}$ is the real vector space of dimension $p+q$ equipped with a nondegenerate bilinear form of signature $(p,q)$. Therefore, $R^{2,0}$ is not linearly conformally isomorphic to $R^{1,1}$. (For this matter, $R^{p,q}$ is not linearly conformally isomorphic to $R^{p',q'}$ unless $p=p', q=q'$.)

A smooth conformal isomorphism between two pseudo-Riemannian manifolds $M_1, M_2$ is a diffeomorphism $f: M_1\to M_2$ such that at each point $x\in M$ the differential $$ df_x: T_xM_1\to T_{f(x)}M_2 $$ is a conformal linear isomorphism. Thus, unless $M_1, M_2$ have the same signature, they cannot be conformally diffeomorphic.

An aside: There is an interesting question of conformal classification of simply-connected open subsets of $R^{1,1}$. In the signature $(2,0)$ the answer is given by the Riemann mapping theorem which states that there are precisely two conformal types. But in the Lorentzian signature case, there is a continuum of conformally inequivalent simply-connected domains.