Time-Orientability of Lorentzian Metrics on the Mobius Band

Given a smooth manifold M, we can endow it with a Lorentzian structure by a smooth assignment of a Lorentzian inner product (bilinear non-degenerate form of signature (-, +, …, +)) to the tangent space $T_p M$ at each point $p \in M$ , i.e. $p \mapsto \{ g_p : T_p M \times T_p M \rightarrow \mathbb{R} \}$ .

Definition of Time-Orientation:

Let (M, g) be a Lorentzian manifold. For any $p \in M$ , the time-cone in $T_p$ given by $I_p = \{ v \in T_p M : g_p(v, v) < 0 \}$ has two components (i.e. 2 cones sharing the same vertex). We can fix a $T \in I_p$ and define the positive cone to be $I_p^{+} := \{ v \in I_p : g_p(v, T) < 0 \}$ (it is not difficult to show that this is indeed well-defined).

A time-orientation on M is a smooth assignment of a positive time-cone ${I_p}^{+}$ to each tangent space $T_p M$ , i.e.

for any $p \in M$ , there exists an open neighborhood $U$ containing $p$ and a smooth vector field $\bf{V}$ defined on $U$ such that $\bf{V}_q \in {I_q}^{+}$ for any $q \in U$ .

We say $(M, g)$ is time-orientable if it admits a time-orientation.

Lemma (to be proven in a following post): $(M, g)$ is time-orientable if and only if M admits a globally time-like vector field.

An example of a time-orientable manifold is the Minkowski space; we can take the time-like vector field to be $\bf{X} = \frac{\partial}{\partial x_1}$ .

Now, we can ask: does time-orientability have anything to do with topological-orientability?

Answer: these two concepts are independent!

We will show this independence by constructing two examples: the Mobius band with two different Lorenztian metrics — one making it time-orientable and the other, not time-orientable.

Constructing the Mobius band as the quotient of a group action on $\mathbb{R}^2$ .

Define the action of $\mathbb{Z}$ on $\mathbb{R}^2$ to be:

$n \cdot (x, y) := (x+n, (-1)^n y)$ .

This group of actions is generated by $1 \cdot (x, y) = (x+1, -y)$ .

The quotient space, $\mathbb{R}^2/\mathbb{Z}$ , is the Mobius band.

Let $\pi : \mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}$ . It is not so difficult to see that $\pi$ is a local diffeomorphism.

Notice also that for any $(x, y) \in [(x_0, y_0)] = \{ (x_0 + n, (-1)^n y_0) : n \in \mathbb{Z} \}$ , $T_{(x, y)} \mathbb{R}^2$ are isomorphic to each other.

Let the Lorentzian metric on the tangent spaces of $\mathbb{R}^2$ be given by $g_1 = dx^2 - dy^2$ .

Now, how do we extend this Lorentzian metric to one on $\mathbb{R}^2 / \mathbb{Z}$ ?

Let $u, v \in T_{[(x_0, y_0)]}(\mathbb{R}^2/\mathbb{Z})$ . Since $\pi$ is a local diffeomorphism, for any $(x, y) \in [(x_0, y_0)]$ , there exists $\tilde{u}, \tilde{v} \in T_{(x, y)} \mathbb{R}^2$ such that $d\pi(\tilde{u}) = u$ and $d \pi(\tilde{v}) = v$ .

For each $p := (x, y) \in \mathbb{R}^2$ , we either have:

$d\pi_p = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$d\pi_p = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ .

We define the corresponding Lorentzian metric on $\mathbb{R}^2/\mathbb{Z}$ to be the following:

for $u, v \in T_{[(x_0, y_0)]} \mathbb{R}^2/\mathbb{Z}$ ,

$(\tilde{g}_1)_{[(x_0, y_0)]} T_{[(x_0, y_0)]} \mathbb{R}^2 / \mathbb{Z} = (g_1)_{(x_0, y_0)}(\tilde{u}, \tilde{v})$

where $d\pi_{p}(\tilde{u}) = u$ and $d\pi_p(\tilde{v}) = v$ .

Let’s check if this definition makes sense!

First, since $T_p \mathbb{R}^2$ is isomorphic to each other for any $p \in \mathbb{R}^2$ , it does not matter which representative we choose for $[(x_0, y_0)]$ .

But what about the corresponding $d\pi_(x_0, y_0)$ for the representative?

Suppose $u = (dx_u, dy_u), v = (dx_v, dy_v)$ .

If $d\pi_(x_0, y_0) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ ,

${g_1}_{(x_0, y_0)} (d\pi_(x_0, y_0)^{-1} u, d\pi_(x_0, y_0)^{-1} v) = -dx_u \cdot dx_v + dy_u \cdot dy_v$

We can similarly that this also follows for $d\pi_(x_0, y_0) = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ .

Hence, the Lorentzian metric on $\mathbb{R}^2/\mathbb{Z}$ is $\tilde{g}_1 = -dx^2 + dy^2$ as well.

Choose $X = \frac{\partial}{\partial x}$ on $\mathbb{R}^2/\mathbb{Z}$ . It is easy to check that $X$ is a time-like vector field. Hence, $(\mathbb{R}^2/\mathbb{Z}, \tilde{g}_1)$ is time-orientable.

Suppose we had chosen the metric $g_2 = dx^2 - dy^2$ on $\mathbb{R}^2$ instead.

Using the same method as above, we can see that the corresponding Lorentzian metric on $\mathbb{R}^2/\mathbb{Z}$ is given by:

$\tilde{g}_2 = dx^2 - dy^2$ .

Take $X = \frac{\partial}{\partial x}$ . It is easy to check that it is a space-like vector field.

Suppose that there exists a smooth time-like vector field $Y$ on the tangent bundle of $(\mathbb{R}^2/\mathbb{Z}, \tilde{g}_2)$ .

Then at each $[p] \in \mathbb{R}^2/\mathbb{Z}$ , $Y_[p]$ cannot be a linear multiple of $X_[p] = \frac{\partial}{\partial x}$ .

Since $T_[p](\mathbb{R}^2/\mathbb{Z})$ is a 2-dimensional vector space, it follows that $\text{span}\{Y_[p], X_[p] \} =T_[p](\mathbb{R}^2/\mathbb{Z})$ . Hence, we have an oriented global frame for the tangent bundle, i.e. we can continuously extend an orthogonal frame in the tangent bundle of $\mathbb{R}^2/\mathbb{Z}$ . This contradicts with the fact that $\mathbb{R}^2/\mathbb{Z}$ is topologically not orientable. Hence, such smooth time-like vector field does not exist. This shows that $(\mathbb{R}^2/\mathbb{Z}, \tilde{g}_2)$ is not time-orientable.

Time-Orientability of Lorentzian Metrics on the Mobius Band

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