If you've read any spacetime topology, you know that spacetime. It is the amazing rotating lightcone identified after half a rotation. And outside of De Sitter space with some identifications, it is the only non-time-orientable spacetime ever described.

For instance, here it is in Wald :

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And here it is in Sanchez :

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It is however never explicitely described. Obviously of topology $\Bbb R \times S$, but beyond that it is up to anyone's guess what the metric actually is.

Sanchez actually almost gives the metric, by giving us scalar products :

$$X_1 = \cos(\pi x) \partial_x + \sin(\pi x) \partial t$$ $$X_2 = -\sin(\pi x) \partial_x + \cos(\pi x) \partial t$$ $$g(X_1, X_2) = -1$$ $$g(X_1, X_1) = g(X_2, X_2) = 0$$

Which means

$$-\sin(\pi x)\cos(\pi x)g_{xx} + \sin(\pi x)\cos(\pi x) g_{tt}+ (\cos^2(\pi x) - \sin^2(\pi x)) g_{tx} = -1$$

$$\cos^2(\pi x) g_{xx} + \sin^2(\pi x) g_{tt} + 2 \cos(\pi x)\sin(\pi x) g_{tx} = 0$$

$$\sin^2(\pi x) g_{xx} + \cos^2(\pi x) g_{tt} - 2 \cos(\pi x)\sin(\pi x) g_{tx} = 0$$

Summing the last two, this gives us

$$g_{xx} = -g_{tt}$$


$$2 \sin(\pi x)\cos(\pi x) g_{tt}+ (\cos^2(\pi x) - \sin^2(\pi x)) g_{tx} = -1$$

$$(-\cos^2(\pi x) +\sin^2(\pi x)) g_{tt} + 2 \cos(\pi x)\sin(\pi x) g_{tx} = 0$$

$$(-\sin^2(\pi x) + \cos^2(\pi x)) g_{tt} - 2 \cos(\pi x)\sin(\pi x) g_{tx} = 0$$

As the lightcone, and hence the direction of time and space, is rotating, my instinct tells me to put $g_{tt} = \cos(\pi x)$. The determinant of the metric will then be $\cos^2(\pi x) - g_{tx}^2$. To keep the signature, this means $\cos^2(\pi x) < g_{tx}^2$. If I use that ansatz,

$$g_{tx} = -2 \frac{\sin(\pi x)\cos(\pi x)^2 + 1}{(\cos^2(\pi x) - \sin^2(\pi x)) }$$

$$g_{tx} = \frac{( \cos^2(\pi x)-\sin^2(\pi x) ) }{2 \sin(\pi x)}$$

A quick plotting tells me that those are not the same functions. Any ideas what the actual metric would be?


You can get the answer pretty quickly by first solving for $\partial_x$ and $\partial_t$ in terms of $X_1$ and $X_2$. To ease notation, let $c = \cos\pi x$ and $s= \sin\pi x$. It's easy to recognize the relation between the two pairs of vectors as a rotation, so you can simply write down the inverse, $$ \begin{pmatrix} \partial_x \\ \partial_t\end{pmatrix} = \begin{pmatrix} c&-s\\s&c\end{pmatrix} \begin{pmatrix} X_1\\X_2\end{pmatrix}. $$

Now we can simply compute the components of the metric: $$ g_{xx} = g(\partial_x,\partial_x) = g(c X_1-s X_2, c X_1-s X_2) = -2 cs g(X_1,X_2) = 2cs. $$ Here we have used that $g(X_1,X_1) = g(X_2,X_2)=0$, $g(X_1,X_2) = -1$.

The other components follow in the same way $$ g_{xt} = g(cX_1-sX_2,sX_1+cX_2) = s^2-c^2 \\ g_{tt} = g(sX_1+cX_2,sX_1+cX_2) = -2sc. $$


Don't use your instincts. For a fixed $x$ you have a linear system of three equations in three unknowns. Solve.


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