Freedman’s theorem and topological sliceness — Examples and Counterexamples

The right-handed trefoil matrix is $V=\left(\begin{array}{cc}-1& 0\\ -1& -1\end{array}\right)$ by [F1] in the convention [F2]. Choose the connected-sum balls around short boundary arcs where the Seifert surfaces are standard collars, disjoint from their basis curves and small positive push-offs. The fixed gluing [F5] joins those collars. A basis cycle in one summand and its push-off lie in a ball disjoint from the other summand's curves, so the first cycle bounds a $2$-chain there away from the second; hence all cross linking entries vanish. The combined matrix is therefore $W=\mathrm{diag}(V,-V)$ by [F3].

Direct calculation gives $\det (V-t{V}^T)=\det \left(\begin{array}{cc}t-1& t\\ -1& t-1\end{array}\right)=(1-t{)}^2+t=t^2-t+1$. For the reverse-mirror block, $-V-t(-V^T)=-(V-t{V}^T)$, whose determinant equals $\det (V-t{V}^T)$ because the rank is two. Since $W-t{W}^T$ is block diagonal, its determinant is $(t^2-t+1{)}^2$. By [F2] and [F4], this is a representative of ${\mathrm{\Delta }}_S(t)$.

Expanding $(t^2-t+1{)}^2$ gives $t^4-2t^3+3t^2-2t+1$, which has nonzero coefficients in degrees $0$ and $1$ and is therefore not a monomial $\pm t^k$. Since these monomials are exactly the units of $\mathbb{Z}[t^{\pm 1}]$, the Alexander polynomial is nonunit. On the other hand, [A2] supplies AC${}_{\omega }$ from full AC, so [F7] gives a smooth ribbon disk; by [F8] this disk is locally flat. Thus $S$ is topologically slice despite its nonunit Alexander polynomial, refuting the asserted converse. For $T$, the same calculation gives ${\mathrm{\Delta }}_T=t^2-t+1\doteq ̸1$, which only means that the Alexander-one theorem does not apply.

For the unknot, the Seifert surface is a disk, so $H_1(F;\mathbb{Z})=0$ and the empty determinant is $1$; the standard flat disk in $B^4$ proves sliceness. For the trefoil, [F1] and [F2] give $\det (V-t{V}^T)=t^2-t+1$, which is not a Laurent unit.

Construct the actual surface inside the preferred-longitude tube of $T$ from [F3]. Take a narrow annulus $A$ about its core with the zero product framing. In a small ball meeting $A$ in one square $Q$, plumb $A$ to a positive Hopf annulus $H$: identify their squares so the two annular core arcs are the two crossing coordinate axes in $Q$, and place the parts outside the square on opposite sides. Smooth the four square corners. Outside that ball the boundary is the two oppositely oriented zero-framed parallels of $T$; inside the ball the positive Hopf band joins them by its positive clasp. This is precisely the positive-clasped untwisted Whitehead pattern in the specified tube. The untwisted condition is the zero framing of $A$, and the construction uses the preferred longitude rather than an arbitrary longitude. It is also the $k=0$ surface of the cited Figure 2; that figure identifies the convention but is not a substitute for the following matrix calculation.

Let $a$ be the annulus core and $b$ the Hopf core, with orientations chosen so their sole square intersection has $a{\cdot }_{F}b=1$. The plumbed surface is connected and oriented. Its Euler characteristic is $0+0-1=-1$, and tracing the alternating four attaching edges gives one boundary component. The two annuli retract to their two core loops with the square intersection identified, so $(a,b)$ is an integral handle basis and the surface has genus one. Choose the plumbing side in which the positive push of the $a$-arc is above the square. The unknotted Hopf core $b$ has a spanning disk in the local ball lying below the square except for its boundary arc; use the lower side for the remainder of $H$. Thus this disk misses $a^{+}$, including the part of $a^{+}$ outside the ball. This is a local disk for $b$, and does not assert that the companion $a=T$ bounds a disk.

The positive push of $a$ is a small parallel with the product annulus framing; its linking with $a$ is zero because that product longitude is the preferred longitude. Equivalently move the push to the annular side direction through nonzero normal directions, keeping it disjoint from $a$. The local square does not add a twist to this annular framing. Hence $V_{11}=0$. The two boundary components of the positive Hopf annulus, with induced orientations, link $+1$. Equally oriented core parallels therefore link $-1$, since one induced boundary orientation is reversed. Moving one core parallel to the positive normal push preserves this linking, so $V_{22}=-1$. The local spanning disk from step 2.1 and symmetry of linking give $V_{12}=\mathrm{lk}(a^{+},b)=0$. Finally [F5] gives $V_{21}=-1$. Thus the ACTUAL Whitehead double of $T$, not just the pattern applied to the unknot, has matrix $V_0=\left(\begin{array}{cc}0& 0\\ -1& -1\end{array}\right)$.

Direct calculation gives $\det (V_0-t{V}_0^T)=t$. This is a Laurent unit, with inverse $t^{-1}$, so [F2] gives ${\mathrm{\Delta }}_{W{h}^{+}(T)}\doteq 1$. Applying [F4] proves the stated locally flat proper slice disk in the standard $B^4$. The argument establishes the announced topological conclusion; it contains no smooth slice-disk construction and no inference that the trefoil is nonslice from its nonunit Alexander polynomial. Every claim in the example follows without a general satellite determinant theorem.