1
$\begingroup$

I am looking for all 'small' deformations of the three 3-dimensional Riemannian spaces with maximal symmetry, the pseudo-sphere, Eucidean space and the sphere. By a theorem by Fubini (1903) the deformed metric can at most have 4 Killing vectors. Let us say that a deformation is small if the deformed metric has 4 Killing vectors and if it is `infinitesimally close' to the undeformed metric with 6 Killing vectors. Let me try and explain 'infinitesimally close' with Euclidean space.

Add a 4th coordinate, 'time' $t$, and define the 'line element of the axial Bianchi I universe' $$ d \tau^2 = dt^2 - a(t)^2 (dx^2 + dy^2) - c(t)^2 dz^2 $$ (do not mind the minus signs) with two positive, differentiable functions $a(t)$ and $c(t)$, 'the scale factors'. If $c \not= a$, we have 4 Killing vectors, $\partial_x$, $\partial_y$, $\partial_z$ and $x\partial_y - y\partial_x$. If $c = a$, we have two more rotations and maximal symmetry. In a uniform limit $c \rightarrow a$ the deformed metric is as close to the undeformed one as you want.

Bianchi's classification includes two more examples: axial Bianchi V (negative curvature) and axial Bianchi IX (positive curvature, also known as Berger sphere). Note that axial Bianchi VII0 is identical to axial Bianchi I and axial Bianchi VII$h$, $h>0$, is identical to axial Bianchi V.

Are there other small deformations of maximally symmetric 3-spaces, i.e. not contained in Bianchi's classification?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

No there are no others.

Bianchi shows in ``Sugli spazi a tre dimensioni che ammettono un gruppo continuo di movimenti,'' Memorie di Matematica e di Fisica della Societa Italiana delle Scienze 11 (1898) 267352 that any 4-dimensional Lie algebra represented by Killing vectors on a 3-space contains a 3-dimensional Lie subalgebra.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ I wasn't very familiar with the setting of your question, but purely in terms of Lie algebras, indeed this is true, and better: every 4-dimensional Lie algebra over a field of characteristic zero admits a homomorphism onto the 1-dimensional abelian Lie algebra (so the kernel is a 3-dimensional ideal). $\endgroup$
    – YCor
    Commented Nov 19, 2020 at 16:52
  • $\begingroup$ Thank you, Yves, for this useful fact. Would you have a reference for it, please. $\endgroup$
    – Thomas Schucker
    Commented Dec 2, 2020 at 17:35
  • 1
    $\begingroup$ It's certainly somewhere in Jacobson's book "Lie algebras" (to prove such a fact, it's enough to prove it in the algebraically closed field, and the every non-solvable 4-dimensional Lie algebra over an algebraically closed field $K$ of char 0 is isomorphic to $\mathfrak{sl}_2(K)\times K$). $\endgroup$
    – YCor
    Commented Dec 2, 2020 at 17:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .