1
$\begingroup$

I'm wondering if it is possible for the isometry group of a Lorentzian metric to preserve a Riemannian metric. So if $(M,g)$ is a Lorentz manifold and $Iso(M,g)$ its isometry group, I ask myself if there are situations where $Iso(M,g)$ acts by isometries for a Riemannian metric $h$ on $M$.

If we assume $M$ to be compact, then it may happen that for a Lorentz metric $g$ on $M$ the isometry group $Iso(M,g)$ is noncompact. So in this case I think it is not possible for $Iso(M,g)$ to preserve a Riemannian metric on $M$, because the isometry group of a compact Riemannian manifold is compact. But what about Lorentz metrics on $M$ which have compact isometry groups. Can they always preserve a Riemannian metric on $M$? Or is this a very special (and restrictive) thing?

$\endgroup$
4
  • $\begingroup$ Did you look at examples? My Lorentzian geometry is very rusty but I think you can construct a Lorentzian metric on the $3$-sphere through the Hopf fibration. Wouldn't the isometry group of that Lorentz metric just be a proper subgroup of the isometry group of the standard Riemannian metric on $S^3$? Edit: a flat torus might be an easier example to study. $\endgroup$
    – quarague
    Feb 8, 2019 at 9:28
  • $\begingroup$ Iam acutally looking for examples, because I only know of examples where $Iso(M,g)$ can not preserve a Riemannian metric. I already thought about a flat torus. If you take a quotient of the flat minkowski space $\mathbb{R}^{1,n-1}$ by the standard lattice, then you end up with a Lorentz n-torus which has noncompact isometry group, if $n>2$. And if $n=2$ the isometry group will be $O(1,1)_{\mathbb{Z}}\ltimes T^2$, but I don't see how this can preserve a Riemannian metric (of course there are subgroups which can preserve a Riemannian metric). $\endgroup$
    – user450093
    Feb 8, 2019 at 9:35
  • $\begingroup$ Maybe I'm missing something here, but isn't $O(1,1)_{\mathbb{Z}}$ just the trivial $1$-element group whereas for the $n=2$ torus with the flat Riemannian metric you instead get $O(2)_{\mathbb{Z}}$ which is the two element group. Essentially in the Riemannian setting you can switch the two axis of the torus but in the Lorentzian you cannot. $\endgroup$
    – quarague
    Feb 8, 2019 at 10:41
  • 2
    $\begingroup$ A compact Lie group acting smoothly on a manifold always preserves some Riemannian metric. Hence, what you are effectively asking is if there are examples of semi-Riemannian manifolds with compact isometry groups. The thing is that a "generic" metric (of any signature) has no isometries whatsoever. I am sure there are more interesting examples as well. $\endgroup$ Feb 8, 2019 at 18:07

1 Answer 1

2
$\begingroup$

I do not have a complete answer to this question, but I believe that I have an example which shows that the Isometry group of a Lorentzian manifold $(M, g_{L})$ can act by isometries on a corresponding Riemannian structure $(M, g_{R})$ and vice-a-versa. (For simplicity I am only considering the connected component of the isometry group which contains the identity.)

Consider $\mathbb{R}^{3}$ with coordinates $(x, y, z)$ and consider the frame field \begin{align*} e_{1} &= \frac{\partial}{\partial x} - \frac{y}{2} \frac{\partial}{\partial z},\\ e_{2} &= \frac{\partial}{\partial y} + \frac{x}{2} \frac{\partial}{\partial z}, \textrm{ and }\\ e_{3} &= \frac{\partial}{\partial z}. \end{align*}

The corresponding dual co-frame field is \begin{align*} \omega^{1} &= \mathrm{d}x,\\ \omega^{2} &= \mathrm{d}y,\\ \omega^{3} &= \mathrm{d}z + \frac{1}{2}\left(y \mathrm{dx} - x \mathrm{d}y\right). \end{align*}

Now consider the Lorentzian metric defined by $$ g_{L} = \omega^{1} \otimes \omega^{1} + \omega^{2} \otimes \omega^{2} - \omega^{3} \otimes \omega^{3}$$ and the Riemannian metric defined by $$g_{R} = \omega^{1} \otimes \omega^{1} + \omega^{2} \otimes \omega^{2} + \omega^{3} \otimes \omega^{3}.$$

The connected component of the isometry group containing the identity for the metrics $g_{L}$ and $g_{R}$ is $$ Isom(M, g) = \left\{ A \in GL_{4}\left(\mathbb{R}\right) \Big \vert \begin{pmatrix} \cos \theta & -\sin \theta & 0 & a\\ \sin \theta & \cos \theta & 0 & b\\ \frac{1}{2}\left(a \sin \theta - b \cos \theta\right) & \frac{1}{2}\left( a \cos \theta + b\sin \theta \right) & 1 & c\\ 0 & 0&0 &1\\ \end{pmatrix}, a, b, c, \theta \in \mathbb{R} \right \}. $$

Identifying a point $\mathbf{x} = (x, y, z)$ in $\mathbb{R}^{3}$ with $\mathbf{x} =\left( x, y, z, 1\right)$ in $\mathbb{R}^{4}$, then the action of $Isom(M, g)$ on $\mathbb{R}^{3}$ is given by $A\cdot\mathbf{x}^{t} = \overline{\mathbf{x}}^{t} = \left( \overline{x}, \overline{y}, \overline{z}, 1 \right)$, where \begin{align} \overline{x} &= x\cos \theta - y \sin \theta + a\\ \overline{y} &= x \sin \theta + y \cos \theta + b\\ \overline{z} &= z + c + \frac{1}{2}\left(a \sin \theta - b \cos \theta\right)x + \frac{1}{2}\left( a \cos \theta + b \sin \theta\right) y. \end{align}

Ignoring the $z$-component you have your standard Euclidean isometries on the $xy$-plane. The key to the example is that the one-form $\omega^{3}$ is invariant under the indicated action. This allows one to toggle the sign on $\omega^{3} \otimes \omega^{3}$ to obtain either the Lorentzian metric $g_{L}$ or the Riemannian metric $g_{R}$.

(As an aside: The indicated example comes from three-dimensional Heisenberg group $H^3$ with group structure $(a, b, c)\star(x, y, z) = \left(a + x, b + y, c + z +\frac{1}{2}(ay - bx)\right)$. The frame field $e_{1}, e_{2}$, and $e_{3}$ is left invariant, as are the corresponding metrics. The connected-component of the identity of isometry group has the structure of a semi-direct product $H^3 \ltimes SO(2)$. For details on the Riemannian case see Differential geometry of curves and surfaces in 3-dimensional homogeneous spaces Part I by Inoguchi et al.)

$\endgroup$

You must log in to answer this question.

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