Show that the figure-eight knot cannot arise as the link obtained by intersecting an analytic curve in \(\mathbb C^2\) through the origin with a sufficiently small \(3\)-sphere centered at the origin.
Solution
1. A knot arising from a singular analytic plane-curve germ has Alexander polynomial whose zeros are roots of unity.
The claim follows from the Milnor fibration and the monodromy theorem for isolated plane-curve singularities.
After replacing a defining equation by its reduced equation, a singular plane-curve germ has an isolated singularity at the origin. For a sufficiently small sphere, its link complement is fibered over \(S^1\) by the Milnor fibration. If \(h\) denotes the monodromy of the fiber, then, up to multiplication by a unit \(\pm t^m\), \[ \Delta_K(t)=\det\bigl(tI-h_*\mid H_1(F;\mathbb Z)\bigr). \] The monodromy theorem states that every eigenvalue of \(h_*\) is a root of unity. Hence every zero of \(\Delta_K\) is a root of unity.
If the analytic curve is smooth at the origin instead, the holomorphic implicit-function theorem makes it locally equivalent to a coordinate axis, whose link is the unknot. Thus every nontrivial knot arising from an analytic curve is the link of a singular germ.
2. The Alexander polynomial of the figure-eight knot is\[ \Delta_{4_1}(t)=t^2-3t+1 \]up to multiplication by a unit \(\pm t^m\).
A standard genus-one Seifert surface for the figure-eight knot has Seifert matrix \[ V=\begin{pmatrix}1&0\\-1&-1\end{pmatrix}. \] Therefore \[ \begin{aligned} \det(V-tV^T) &= \det\begin{pmatrix} 1-t&t\\ -1&t-1 \end{pmatrix}\\ &=-(t^2-3t+1). \end{aligned} \] Since the Alexander polynomial is defined only up to \(\pm t^m\), this gives the stated normalization.
3. The zeros of \(\Delta_{4_1}\) are not roots of unity.
The quadratic formula gives \[ \lambda_\pm=\frac{3\pm\sqrt5}{2}. \] In particular, \[ \lambda_+>1. \] Every root of unity has absolute value \(1\), so \(\lambda_+\) is not a root of unity.
4. Hence the figure-eight knot is not the link of an analytic plane curve singularity.
If the figure-eight knot arose as such a link, 1 would force every zero of its Alexander polynomial to be a root of unity. This contradicts <1>2–<1>3. The smooth case was already excluded in 1 because it gives the unknot. Therefore no analytic curve germ in \(\mathbb C^2\) has the figure-eight knot as its sufficiently small link.