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I think that every lightlike curve in $\mathbb{S}_1^2 \subseteq \mathbb{L}^3$ must be a line. But I'm having trouble concluding it.

Let $\alpha\colon I \subseteq \Bbb R \to \Bbb S^2_1 \subseteq \Bbb L^3$ be a lightlike curve. Then: $$\langle \alpha, \alpha \rangle = 1, \quad \langle \alpha', \alpha' \rangle = 0.$$From here, we obtain: $$\langle \alpha, \alpha' \rangle = 0, \quad \langle \alpha', \alpha '' \rangle = 0.$$So far all of this is trivial. Differentiating $\langle \alpha, \alpha'\rangle = 0$ we get: $$0 = \langle \alpha',\alpha'\rangle + \langle \alpha, \alpha''\rangle \implies \langle \alpha, \alpha''\rangle = 0.$$

Intuition says that $\alpha''$ is in the plane spanned by $\alpha$ and $\alpha'$ (since $\alpha''$ would point to the origin), so the above calculation would give that $\langle \alpha'',\alpha''\rangle = 0$, but I'm not sure of this. And even if this is true, it doesn't mean that $\alpha'' = 0$.

Can someone help me or give a counter-example? This is a self-posed question, made after noticing that every line contained in $\mathbb{S}^2_1$ must be lightlike. I just wondered about the reciprocal.


The Lorentz product is given by $\langle {\bf x},{\bf y}\rangle = x_1y_1+x_2y_2-z_1z_2$, and $\mathbb{S}^2_1$ is the set of all vectors $\bf v$ such that $\langle {\bf v},{\bf v}\rangle = 1$, in case I didn't make it clear before. Use whatever signature for $\langle \cdot,\cdot\rangle$ you want, though, it probably won't appear in the computations. The result seems to be true for $\Bbb L^n$ in general, if that's really the case.

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In order to fix the ideas, let the Lorentzian metric on the Minkowski space $\mathbb{L}^3$ have signature $-++$. The Lorentz group $O(1,2)$ of $\mathbb{L}^3$, on the one hand, acts transitively on the subset $\mathbb{S}^2_1 \subset \mathbb{L}^3$ and, on the other hand, is a subgroup of the linear group $\mathrm{Gl}(3, \mathbb{R})$.

The first fact implies that null-curves of $\mathbb{S}^2_1$ are sent to null-curves by the action of $O(1,2)$. Furthermore, since $\mathbb{S}^2_1$ is a 2-dimensional manifold, there are at any point only two null-directions, so that there are (up to parametrization) only two null-curves going through any fix point. This observation allows one to show that $O(1,2)$ acts transitively on the set of null-curves of $\mathbb{S}^2_1$.

The second fact implies that affine lines in $\mathbb{R}^3$ are sent to affine lines under the action of $O(1,2)$.

Combining these two consequences, in order to conclude that any null-curve is a straight line, we only have to show that a null straight line passes through the point $p(t,x,y) = (0,0,1)$. At $T_p\mathbb{S}^2_1$, the vector $v = \partial_t + \partial_x = (1,1,0)$ is null, so we would like to show that the line $\gamma(t) = p + tv$ (i) belongs to $\mathbb{S}^2_1$ and (ii) is everywhere lightlike. Fix any $t \in \mathbb{R}$.

(i) $\langle \gamma(t), \gamma(t) \rangle = \langle p, p \rangle + 2 t \langle p, v \rangle = \langle p, p \rangle = 1 \, .$

(ii) It is clear that $v = \dot{\gamma}(t) \in T_{\gamma(t)} \mathbb{L}^3$ is null. According to (i), $\dot{\gamma}(t) \in T_{\gamma(t)} \mathbb{S}^2_1$, so it also has to be null inside this subplane.


This argument uses in an essential way the homogeneity of the de Sitter space. However, in order to deduce that the space of null-curves is also an $O(1,2)$-homogeneous space, it uses the fact that $\mathbb{S}^2_1$ is 2-dimensional. Indeed, in dimension three or more, the lightcone without the origin is a dimension two, etc. manifold with two connected components (in dimension two, it is a 1-dimensional manifold with four components), so that there is a lot of room for non-straight null-curve.

In order to get a similar result for the various $\mathbb{S}^n_1$, one would need to put some further constraints on the null-curves, like being geodesics. Indeed, $O(1,n)$ also acts transitively on $\mathbb{S}^n_1$ ; the stabilizer of the point $p(t, x_1, \dots, x_n) = (0, 0, \dots, 0,1)$ is easily seen to be isomorphic to $O(1,n-1)$. Now, one shows that the stabilizer acts transitively on the lightcone inside $T_p \mathbb{S}^n_1$ i.e. $O(1,n-1)$ acts transitively on the lightcone inside $\mathbb{L}^{n-1}$. Hence, $O(1,n)$ acts transitively on the set of null-vectors in $T\mathbb{S}^n_1$. Moreover, since $O(1,n)$ acts through isometries, it sends (null-)geodesic to (null-)geodesic. Since (null-)geodesic going through some given point are in bijection with the set of (null-)vectors at that point, we deduce that $O(1,n)$ acts transitively on the set of (null-)geodesics. Since, as argued above, there exists a straight null-curve and as it is certainly a geodesic*, we deduce that all null-geodesics of $\mathbb{S}^n_1$ are straight lines.

$^*$ Geodesics are curves of vanishing geodesic curvature, which is a measure of the amount of the ambient curvature that is 'tangent' to the hypersurface. A straight line having no ambient curvature, it doesn't have geodesic curvature either.

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  • $\begingroup$ Wow, great answer. Very well explained, thanks a lot. $\endgroup$ – Ivo Terek Jul 31 '15 at 16:25

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