Knots in Manifolds: Beyond Euclidean Perception — Epoche B2
Knots in Manifolds: Beyond Euclidean Perception A knot looks like a property of a piece of string [1] . Tie a trefoil, and no amount of pulling will undo it; the knottedness seems to be stored in the string itself. It is not. Every statement of knot theory is a statement about a loop and the space the loop sits in, and the standard tools — the polynomials, the colouring counts, the very word unknotted — quietly assume that this space is $\mathbb{R}^3$. This essay makes the assumption visible. It first computes an invariant from scratch to show that the trefoil is genuinely knotted in $\mathbb{R}^3$, then exhibits an invariant that exists only when the ambient space has a hole, then settles what changing the ambient space can and cannot do — because the two directions are not symmetric, and one of them is routinely stated backwards. Knots, and what sameness means A knot in a 3-manifold $M$ — a space in which every point has a neighbourhood looking like an open ball of $\mathbb{R}^3$ — is a smooth injective map $K$ of the circle $S^1$ into $M$ whose derivative never vanishes. Two knots $K_0,K_1\subset M$ are equivalent when there is an ambient isotopy carrying one to the other [2] : a smooth family of diffeomorphisms $h_t\colon M\to M$, $t\in[0,1]$, with $$ h_0=\mathrm{id}_M,\qquad h_1(K_0)=K_1 . $$ The whole space is deformed, not just the loop. Deforming the loop alone would be useless, since any loop can be shrunk to a point if the rest of space is allowed to stand still, and every knot would be equivalent to every other. A knot is trivial , or an unknot , when it bounds an embedded disc in $M$ — a disc whose boundary circle is exactly $K$ and whose interior meets $K$ nowhere. One simplification is free of charge. The 3-sphere $S^3$ is $\mathbb{R}^3$ with a single point added at infinity, and any knot can be pushed off that point, so knot theory in $\mathbb{R}^3$ and in $S^3$ are the same theory. Below, $S^3$ is used whenever a compact ambient space is convenient. Counting colourings: a proof that the trefoil is knotted Nothing said so far shows that any knot is non-trivial. Here is a complete argument, using only arithmetic modulo $3$. Draw a knot as a diagram in the plane, with the strand passing underneath broken at each crossing. The diagram falls into arcs : the unbroken pieces running from one under-crossing to the next. Assign to each arc a colour from $\{0,1,2\}$, and demand at every crossing that the over-arc's colour $x$ and the two under-arc colours $y,z$ satisfy $$ 2x\equiv y+z \pmod 3 . $$ Call an assignment satisfying this at every crossing a 3-colouring , and write $\mathrm{Col}_3(K)$ for the number of them. Reidemeister showed that any two diagrams of equivalent knots differ by a finite sequence of three local moves [3] ; checking the rule above against each of the three moves shows the count is unchanged by all of them. So $\mathrm{Col}_3$ is a knot invariant: equivalent knots have equal counts, and unequal counts prove inequivalence. Compute both sides. The unknot has a diagram with no crossings and therefore one arc, freely coloured: $\mathrm{Col}_3=3$. The standard trefoil diagram has three crossings and three arcs $a,b,c$, each arc passing over exactly one crossing, giving three conditions. Since $2\equiv-1\pmod 3$, the first reads $-a\equiv b+c$, that is $a+b+c\equiv 0$; the other two collapse to the same condition. The solutions are the triples with $a+b+c\equiv0\pmod 3$: choose $a$ and $b$ freely and $c$ is forced, so there are $3\times 3=9$ of them. Hence $$ \mathrm{Col}_3(\text{trefoil})=9\;\neq\;3=\mathrm{Col}_3(\text{unknot}) . $$ The trefoil is not the unknot. The same conclusion follows from the Alexander polynomial $\Delta_K(t)$, defined up to multiplication by $\pm t^{k}$, which takes the values $$ \Delta_{\text{unknot}}(t)=1,\qquad \Delta_{\text{trefoil}}(t)=t-1+t^{-1} . $$ Both computations are performed on a planar diagram, and a planar diagram is a record of how the loop sits in $\mathbb{R}^3$. That is the first sign that the ambient space is doing work. An invariant that $\mathbb{R}^3$ cannot support The dependence becomes explicit once the ambient space is allowed a hole. Take the solid torus $V=S^1\times D^2$, the product of a circle with a disc: a doughnut including its filling. Its first homology group, which records how many independent loops fail to bound a surface, is $$ H_1(V;\mathbb{Z})\cong\mathbb{Z},\qquad\text{generated by the core }S^1\times\{0\} . $$ Any knot $K\subset V$ therefore carries an integer: its winding number $w(K)$, the algebraic count of the intersections of $K$ with a meridian disc $\{p\}\times D^2$ [4] , signed by direction of passage. Ambient isotopies of $V$ move $K$ and the disc together and cannot change the count, so $w$ is an invariant of knots in $V$. It settles a question immediately. If $K$ bounds an embedded disc in $V$, then $K$ is null-homologous, so $w(K)=0$. The core circle has $w=1$. The core is therefore a non-trivial knot in the solid torus, while a small round circle contained in one ball inside $V$ has $w=0$ and is trivial. Yet in $\mathbb{R}^3$ both are ordinary unknotted circles and are equivalent to each other. Triviality is not a property of the loop. The contrast is structural rather than accidental. For every knot in the 3-sphere, Alexander duality — the general theorem, proved in Hatcher's Algebraic Topology , and not anything about knots in particular — forces $$ H_1(S^3\setminus K;\mathbb{Z})\cong\mathbb{Z}, $$ independently of which knot $K$ is — every knot bounds a surface, and the winding number of the previous paragraph has no analogue. That uniform answer is also what makes the Alexander polynomial available: it is built from the infinite cyclic cover of the complement — the construction set out in Rolfsen's Knots and Links — and that cover exists precisely because the first homology of the complement is $\mathbb{Z}$. In a general 3-manifold $H_1(M\setminus K)$ can be anything, the canonic