# Isometry group of a lorentzian metric which preserves a Riemannian metric

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?

• 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. Feb 8, 2019 at 9:28
• 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). Feb 8, 2019 at 9:35
• 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. Feb 8, 2019 at 10:41
• 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. Feb 8, 2019 at 18:07

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.)