Lee deformation and Rasmussen’s s-invariant — Examples and Counterexamples

Under [A1], the crossingless unknot complex has one cohomological degree and basis $1,X$ in quantum degrees $1,-1$, so its two filtration levels are $1,-1$. The definition $s=s_{\max }-1=s_{\min }+1$ gives $s(U)=0$.

The standard positive trefoil diagram has $n=3$ crossings and $k=2$ circles in its oriented resolution. Applying [F2] gives $s(T_{2,3})=3-2+1=2$. By [F3], its reverse mirror has value $-2$.

The trefoil Seifert construction from the given diagram starts with two disks and attaches three bands, so its connected orientable surface has Euler characteristic $2-3=-1$ and one boundary component. The identity $\chi =1-2g$ gives genus $1$. Push its interior into a collar of the boundary three-sphere in the four-ball, using a height function that vanishes only on the boundary knot; its graph is still an embedded orientable surface with the same boundary and genus. Thus [F5] gives $g_4^{\mathrm{s}\mathrm{m}}(T_{2,3})\le 1$. The bound in [F6] gives $2=\mathrm{\mid }s(T_{2,3})\mathrm{\mid }\le 2g_4^{\mathrm{s}\mathrm{m}}(T_{2,3})$, so its smooth four-ball genus is $1$.

Apply the joint value lemma [F4] to the actual positive-clasped double with Hedden–Ording twist parameter $t=2$. Its diagram and zero-longitude convention are exactly those stipulated here, so it gives $s(D_{+}(T_{2,3},2))=2$ and $\tau (D_{+}(T_{2,3},2))=0$. Together with the unknot and trefoil calculations and the smooth genus computation above, these are every value asserted in the Example.

For this actual knot, [F1] and [L1] give $s(D_{+}(T_{2,3},2))=2$ and $\tau (D_{+}(T_{2,3},2))=0$. Hence twice its normalized tau value is $2\cdot 0=0$.

If the asserted identity held for every oriented knot, substituting this knot would give $2=0$. Since these integers are unequal, the displayed knot is an actual counterexample to the universal identity.