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An open trefoil is a trefoil tied in an infinitely long line. An open trefoil that is at rest in (flat) space, in space-time is a knotted hypersurface.

Different observers in space-time have different time slices. Is there an observer that observes a closed trefoil, instead of an open one?

This issue has 2 subcases:

a - in flat Minkowski space, is there an inertial observer that sees a closed trefoil?

b - is there a non-inertial observer that sees a closed trefoil?

Thank you for any help.

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A large 3-sphere would intersect your (long trefoil)$\times\Bbb R$ surface in a closed knot that is the connected sum of your trefoil with its mirror image. Did you perhaps want a time slice that contains this closed knot rather than a closed trefoil? Side remarks: calling a codimension two (real) surface a "hypersurface" might be misleading to some; also I don't see what your question may have to do with general topology. –  Sergey Melikhov May 5 '11 at 17:29
No, I am interested in a closed trefoil. –  Chris May 5 '11 at 18:31
In my comment above and in my answer below, I was thinking of a possibly curved space-time. The flat case is trivial as discussed in the followup thread: mathoverflow.net/questions/64172 –  Sergey Melikhov May 7 '11 at 21:01

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No, a closed trefoil cannot appear, because it's not algebraically slice.

In more detail, I understand that in the usual Minkowski space, a "time slice" $T$ would be homeomorphic to $\Bbb R^3$. In case there are problems with this in a more general situation, I'm reading the question so that $T$ is implicitly assumed to be a copy to $\Bbb R^3$ because by the trefoil knot one normally means a certain type of knot in $\Bbb R^3$ (or in $S^3$, but I understand that a time slice would anyway be non-compact; though my argument below would work for a copy of $S^3$ just as well).

Suppose then that this $T$ intersects the given surface $F=$(long trefoil)$\times\Bbb R$ in a closed trefoil $K$. The latter is a copy of $S^1$ embedded in $F$, so by the Jordan curve theorem (for polygonal curves) it bounds a disk $D$ in $F$. Since $T\cap F=K$, this $D$ lies entirely in one of the regions $T^+$, $T^-$ of $\Bbb R^4$ bounded by $T$ (in other words, either in the future or in the past of the observer). Topologically, the two possibilities are symmetric, so we may assume $D\subset T^+$. Since $F$ is connected, $T^+\cap F=D$.

If $T^+$ happens to be diffeomorphic to $\Bbb R^3\times [0,\infty)$ we conclude immediately that $K$ is a slice knot. But the trefoil is not slice.

The general case follows by unraveling a standard proof that the trefoil is not slice. By Mayer-Vietoris and Seifert-van Kampen $T^+$ is contractible. Also, a Seifert surface of $K$ in $T$ capped off by the disk $D$ bounds a $3$-manifold in $T^+$. [Indeed, if $G$ is an infinite Seifert surface (properly embedded in $\Bbb R^3$) for the long trefoil, some perturbation of $G\times\Bbb R$ will meet $T^+$ in a $3$-manifold $M$ with $\partial M=D\cup (M\cap T)$. Since $D$ is compact, it's easy to arrange that $M$ be compact.] It follows that $K$ is algebraically slice, and in particular its Arf invariant vanishes. But the trefoil has nonzero Arf-invariant.

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Sergey, Thank you. I am reading everything I can on slice knots. Allow me to ask one additional aspect. As the previous poster and you write, the closed trefoil is not slice, but the connected sum of a trefoil and its mirror is. Is there a simple way to see which knots can appear for a second observer if the first one sees an open trefoil? Probably this would be the "concordance class" of the open trefoil (but maybe this is the wrong term). Is there a way to get an idea for which closed and open knots are contained in it? (Say the simplest elements.) This would help a lot! –  Chris May 6 '11 at 6:38
Chris, I suspect that the metric restrictions on possible "time slices" would exclude most knots that are possible topologically. It is conceivable that only the long trefoil itself could appear, or it could be something like the connected sum of the long trefoil with any hyperbolic knot that is (algebraically?) slice. I think this is a problem of contact/symplectic geometry and Legendrian knot theory, but I'm a (firmly non-Legendrian) topologist. If you haven't looked at Vladimir Chernov's papers, that would be a good starting point math.dartmouth.edu/~chernov/MyVitae.html –  Sergey Melikhov May 6 '11 at 11:21
Really great argument! It's worth mentioning that there is a twist spun trefoil that is an unknotted sphere. A slice along a copy of $\mathbb{R}^3$ that contains the axis of rotation is the connected sum of the trefoil and its reflection, which is slice. Take stereographic projection of this knotted sphere in $S^3$ by a point on the twist spun trefoil, and that will be an unknotted proper $\mathbb{R}^3$ with a knotted slice. –  Charlie Frohman May 6 '11 at 12:56

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